Mech Gate V1 Solutions

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Mechanical Engineering (Production, IM & OR, Fluid Mechanics and Hydraulic Machinery, Heat Transfer, Thermodynamics, Machine Design, Theory of Machines, Engineering Mechanics, Strength of Materials)

(Solutions for Volume : I Classroom Practice Questions)

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Production Technology Solutions for Vol – I _ Classroom Practice Questions Common Data for 04 & 05

Chapter‐ 1

04. Ans: (a)

Metal Casting 01. Ans: (a) Sol: Pouring time = 

Volume A C  Vmax 2  10 6 200  2  10000  175

05. Ans: (b) Sol: 3 castings of spherical, cylindrical and cubical Vsp = Vcube

4 3 R  a 3 3

= 5.34 sec

a =R

Vcyl = VSp

02. Ans: (d)

VH PAT For standard specimen H = D = 5.08 cm

Sol: Permeability number =



 D 2 H   R3 4 3



 D 3   R 3 (D  H) 4 3

P = 5 gm/cm2, V=2000 cc, T= 2 min PN =

2000  5.08  50.1 2  2 5   5.08  2 4

03. Ans: (a) Sol: Q = 1.6 10-3 m3/sec

A = 800 mm2 Q=AV 1.6  10-3 = (800 10-6) V V = 2 m/sec =

4  = 1.61 R 3

3

2gh 2

  2  = 0.203m h =   2  9.81 

1

D=

 SP  Cub

3

16 3  16  3 R    R  1.75R 3 3 2

2  M SP   D 6  D    =    a  M Cub   a 6  2

2

2

 2R   2R  =   1.54  =  a   1.61R 

 SP  M SP   cyl  M cyl

  

2

2

2 2 D   D Sp   2R    6    = 1.306 =   D   1.75 R   D  cyl    6

= 203 mm

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06. Ans: (c) Sol: Vol. of casting =

   As  A       s   0.4    V   V  C1 C2  

 2 D L 4

 =  150 2  200  3534291 mm3 4 ht = 200+ 50 = 250 mm AC = Amin = sprue base area 400  200 mm2 2 G.R.= 1:1.5:2

Volumetric ratio,(V.R) = Y1 =

 VR = 0.8 VC1 Now Y2 

=

Pouring time =  =

200  2  9810  250 17671 2  9810  250

 8 Sec

07. Ans: (a) Sol: Circular disc casting Squared disc casting C1 ; d  20cm t  10cm ;

C2 a  20cm t  10cm  As     V  C1  1.4 Freezing ratio (F.R) = X1 =  As     V R  As     As   V  C1    1 .4  V R

 As   As       V  C2  V  C2 X1    1.4  As   As       V R  V  C1 1.4 ACE Engineering Academy

0.8VC1 VR  VC 2 VC2

  0.8  20 2  10   = 0.628 4  20  20  10

Volume of Casting A C.  Vmax 3534291

VR  0 .8 VC

08. Ans: (b) Sol: VC = 40 × 30 × 0.3 = 360cc VSc = shrinkage volume =

3  360  10.8 cc 100

Volume of riser Vr = =

 2 d h 4  2  4  4  50.24 cc 4

Vr ≥ 3 Vsc  Vr  3  10.8  32.4cc Vr ≥ 3 VSc → Satisfied

r C where r = time taken for riser material to solidify C = time taken for casting to solidify Mr  Mc

V V        A s  r  A s  casting V 360  As 240  30  30  0.3  0.3  40 

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2183.6 1800  d d2 2183.6 d  1.21mm 1800

V d 4      = 0.666  As r 6 6 =

360  0.147 2442

 r > C

Common Data for Q.10 & Q. 11

Hence diameter of riser = 4 cm 09. Ans: (c) Sol: The dimension of pouring basin will not affect the pouring time Let V = maximum velocity of molten metal in the gating system, d = dmin = dia. Sprue bottom

10. Ans: (a) & 11. Ans: (a) Sol: In centrifugal casting Centrifugal force = FC = ma = m r 2 a = r2 75 g =

volume. of casting Pouring time = P. T  A c  Vmax 

V

35 3

353  2 d V 4

= 25

D (2 N)2 2

75 ×9810 = N 2 D 

4 2 2

Constant = N 2 D 

75  9810  37273 2 2

Constant = N 2 D  37273 D=

 2183.6 / d …… (1)  2 d  25 4 To ensure the laminar flow in the gating 2

N

0.5  0.52 = 0.51 m = 510 mm 2 37273  D

37273 = 8.55 RPS 510

system Re  2000 For limiting condition Re = 2000  V d Vd = R e  2000     2000 

Sol:

1 2

1 hpb =50 mm 2 hs= 200 mm

Vd 

2000 2000 0.9 1800 …… (2)   d d d From (1) and (2) V

12. Ans: (c)

3

h = height of sprue = 200mm A2 = 650 mm2 Q = flow rate = 6.5  105 mm3 / s g = 104 mm/sec2

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6.5  105  1000 mm 2 / Sec 650

V2 =

= 2gh pb  2  10 4  h pb hpb = 50 mm = height of molten metal in the pouring basin ht = total height of molten metal above the bottom of the sprue = 200 + 50mm

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15. Ans: (d) Sol: dtop = 225 mm ht = 250 + 100 = 350 mm Volume flow rate Q = 40×106 mm3/sec Vbottom =

= 2620 mm/s Q = Atop×Vtop=Abottom×Vbottom Abottom =

Q  A2 V2  A3V3  A3 2  10 4  250 = 6.5  105 mm 3 / s

2  g  ht = 2  9810  350

dbottom =

40  10 6 =15267.17 mm2 2620 4  15267.17



=139.42 mm

 A3 = 290.7 mm2 13. Ans: (c) Sol: Net buoyancy force =Weight of core – weight of the liquid which is displaced by core = V.g (  – d ) 

 2  d h  g    d  4

 2  0.12   0.18  9.81  11300  1600  4 = 193.6N 

14. Ans: (c) Sol: Gating ratio=As: Ar : Ag =1:2:1

Q =100 cc/sec, V=50 cm/sec Q = AV A=100/50=2 cm2 Sprue area =2 cm2 Runner area = 2  2 = 4cm2

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16. Ans: (b) Sol: A2V2 = A3V3

 4 

2252  2  9810 100

 4

db2 2  9810  350

db = 164.5mm So aspiration will not occur. 17. Sol: Casting – 1 (circular) Diameter = 20mm, length = 50mm Casting -2 (elliptical) Major/Minor = 2, length = 50mm, C.S. area of the casting -1 = C.S area of the casting -2  solidification time of casting  1     solidification time of casting  2  2

 V  A c2  M  =  c1  =  c1   M c2   Vc 2  A c1 

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 Vc1 =  d 2  h 4   =  20 2  50 = 15707.96 mm3 4   Ac1 = 2   d 2  dh  4

 2   4 20  2    20  50

= 3769 mm2 C.S area of cylinder = C.S area of ellipse

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 solidification time of casting  1   solidification time of casting  2    M  =  c1   M c2 

2

2

 V  A c 2  15707.96  4140  =  c1     Vc 2  A c1   15708  3769.9 

2

= 1.205

  maj.axis  min .axis  2   4  20   4  

=

  2  (min .axis) 2 4 1

 4 2 Minor axis =   20 2    2  4 Minor axis = 14.14mm Major axis = 2  minor axis = 28.3mm perimeter = 2

a 2  b2 2

where a = major axis /2 

28.3 = 14.14 mm 2

14.14 2 = 7.07 mm

b = minor axis / 2 

Perimeter = 70.24 mm Surface area of ellipse = perimeter  length + 2 C.S. area = 70.2450 + 314  2 = 4140 mm2 = AC2 Volume of the ellipse = C.S area  length = 314  50 =15708 mm3 = Vc2 ACE Engineering Academy

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Common data for 02 & 03

Chapter‐ 2

02.

Welding

Sol: H.G = I2 R 

= (10000)2×200×10-6×

01. Sol: Power = P = 4 + 0.8L – 0.1L2 For optimum power

5  2000 J 50

03. Sol: h = 2t – 2 × 0.1 t = 1.8 t =1.8 ×1.5 = 2.7 mm

dP  0  0.8 – 0.2L = 0 dL

D = 6 t  6 1.5 = 7.35 mm

0.8  4 mm 0 .2 P = 4 + 0.8L – 0.1 L2

L

= 4 + 0.8 ×4 – 0.1 × 42 = 5.6 kW

h

Energy losses = 20% ,  = 80% 0.1 t

Area of weld bead (WB) 1 = 2   AB  AC 2 = 5 tan 30 × 5 = 14.43

Vol. of nugget =

5 tan 30 A B 300 600

=

D2h

Heat required = Volume × ×heat required /g = 114.5  10 3  8  1380  1264 J

3

Volume of W.B = 14.43 × 1000 =14433 mm Weight of W.B = 14433 × 10-6 × 8 = 115.5 g Heat required for melting of W.B =115.5 ×1400 = 161. 66 kW 161.66 Time for welding =  36 Sec 0.8  5.6 1000 36  27.78 mm/sec

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4

 7.352  2.7 = 114.5 mm2 4

5

C

Welding speed =



04. Sol: Rated Power = Vr Ir = 50 ×103  Ir 

50  10 3  2000 A 25

Dr = 50% (rated duty cycle) If Id = 1500 A (desired current) Desired duty cycle, 2

Dd =

I 2r D r  2000     0.5  0.89 2 Id  1500 

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Dd =

Arc on time = 0.8930 Total welding time

= 26.7 sec Common data for 05, 06 & 07.

06.

05. Ans:(d)

Ans:(d)

x = 200mm (given) No of electrodes/pass =

30mm

(2) (1) (3) 30o 30o

4mm

l = 1m =1000mm; t = 30 mm d = 4mm Lt = 450 mm; LS = 50mm A1 = 4  30 = 120 mm2 1  30 tan 30  30 = 259.8 mm2 2 Total volume of weld bead = volume of weld bead + crowning A2 = A3 =

1000 5 200

140  28 5 Total Arc on time

No of passes =

=

07. Ans:(c) Sol:

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1000  28  280 minutes 100

Total weld time =

280  466.67 minutes 0.6

08. Sol: Given AC = 10 mm, O1A = O1C = 7 mm, O2A = O2C = 20 mm O2 D A

B

r=20

E

r=7 C

O1

= 1.1  volume of weld bead = 1.1  (A1+2A2)1000 = 703560 mm3

= O1D–

Volume /Electrode =

  D2  Le 4

   4 2  (450  50) 4 = 1600 No of electrodes required 

Total volume of weld bead volume / Electrode



703560  139.96  140 1600

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Height of Bead = BD = O1D – O1B

O1 A2  AB 2

= 20 – 20 2  5 2 = 0.64 mm Depth of Penetration = BE = O1E–O1B =  O1 E  

 O2 A   AB 

= 7  7 2  52

2

2

= 2.10 mm

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09.

Common Data Q. No 12 and 13

   Sol: RC = 0.85    nr   = Resistivity of metal

12. Ans: (c) Sol: I = 200, V = 25, speed =18 cm /min 2

V2 V (Heat generation)1 = I2R =    R = R R R C1

0.85  2  10 5 = 1.082105  25    0.02

R C2

0.85  2  10 5 = 5.41  106  50    0.02

52 (H.g)1 = = 2310546.04 1.082  10 5

(H.g)2 =

52 = 4621072.08 5.41 10 6

D = 1.2 mm, f = 4 m /min, = 65%, Heat input =

V  I  η 25  200  0.65  60  speed 18

= 10.83 kJ / cm 13. Ans: (b) Sol: Filling rate of weld bead = filled rate by electrode Area of W.B × Speed =

 4

d2  f

Area of W.B



10. Ans: (d) Sol: Heat supplied = Heat utilized

= 4

0.5 J = m (S.H. + L.H) =V ( SH+LH) = (a×h)  (Cp (Tm–Tr)+LH) -6

= 0.05 × 10 × h × 2700 [896  (933 – 303) + 398 × 103] h = 0.00385 m = 3.85 mm 11. Ans: (d) Sol: Welding time =

900 = 3 min 300

= 3  60 = 180 sec I = 150 A ;

V = 20 V

 = 0.8 R = 36 × 10-6 Ω Heat input = I2 R   = 1502 ×36 × 10-6 ×180×0.8 = 116.64 ACE Engineering Academy

 1.2 2  4000 180

 25.12 mm2

14. Ans: (c) Sol: Heat generated = Heat utilized I2R = Vol. of nugget × × H. R/g I 2  200  10 6  0.1  0.0052  1.5  103  8000  1400  103 4 I = 4060 A 

15. Ans: (a) Sol: V0 = 80 V, IS = 800 A Let for arc welding V = a+bL For power source, Vp = V0–

V0 I Is

For stable V = Vp

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 a  b L  V0 

18. Ans: (b) Sol: Heat dissipated = 360 – 344 = 16 J

V0 I Is

When L = 5, I = 500  a + b × 5 = 80 –

80  500  30 800

a + 5b = 30 when L = 7, I = 460

19. Ans: (c)  Sol: Volume to be melted = (110 2  100 2 )  2 4  3298.66 mm3

Total heat required = 3298.66 × 10-9 × 64.4 ×106 = 212.4 Joules

80  460  34 800 By solving, b = 2, a = 20 a  b  7  80 

V V 2 302    21.43 R R 42 Total heat required = heat to be generated

V = a + bL = 20 + 2L

P = VI = V 

16. Ans: (b) Sol: 3V + I = 240 I = 240–3V P = VI = V (240 – 3V) = 240 V –3V2 For optimum power

212.4 = Pt t=

212.4  10 sec 21.43

20. Sol: For power source,

dP 0 dV

240 – 6V = 0  V =

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I 60 Va = 2L + 27

Vp = 36 –

240  40V 6

Common Data Q. 17 & Q. 18

At equilibrium conditions Va = VP

17. Ans: (c)

27 + 2 L = 36–

Sol: I = 3000 A,  = 0.2, R = 200 Ω

= I2R

I  36  27  2 L  9  2 L 60 I = 60 (9 – 2L)

= 30002 ×200×10-6 ×0.2 = 360 J

If current is 360 Amps 360 = 60 (9 – 2L)

Volume of nugget = 20 mm3 Heat generation



Heat required = V c p Tm  Tr   LH



= 8000 20×10-9×500(1520 –20)+1400×103 = 344 J ACE Engineering Academy

I 60

360 6 60 2L = 9 – 6 = 3

9 – 2L =

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L =

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Welding time =

V = 27 + 2 ×1.5 = 27+ 3 = 30 V I = 60 (9 – 2 ×1.5) = 360 A P = 30 ×360 = 10800 W If L = 4 mm, V = 27 + 1.5 ×4 = 33 V I = 60 (9 – 1.5 ×4) = 180 A P = 33 × 180 = 5940 W Change in power = 10800 – 5940 = 4860 W If the maximum current capacity is 360A, the maximum arc length is 1.5mm

300  1.5 min  1.5  60 200 = 90sec

Heat input = 2 103  90 Joule HI =

HR 40  10 3  4.2   0.9333 HI 2  10 3  90 = 93.33%

21. Ans: (a) Sol: Frictional force F = Pressure × Area × = 200 

 4

 10 2  0.5  7854

3 Torque = F   Radius 4 3 Torque = 7854   5  10 3  29.45 4 2NT Power, P = 60000 

2  4000  29.45 = 12.33 kW 60000

22. (i) Ans: (a), (ii) Ans: (b) Sol: P = 2 KW = 2  103 Watt,

V = 200mm/min, L = 300mm Heat required (HR) = 40 Kcal = 401034.2 Joule

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800  33.690 1200   tan   tan 33.69  0.67

    tan 1

Chapter‐ 3

Metal Cutting

60 60 = 1200 W Length of shear plane = LS

Power = P = FC  VC  1200 

Common Data for Q. 01 & 02

01. Ans: (a) 02. Ans: (d) Sol: Vf

=

Vs 90



Vc

Vc = 40 m/min; Vf = 20 m/min  = 10o;

r

Vf  0 .5 Vc

 r cos     tan 1    1  r sin  



Vf  cos  sin 

06. Ans: (a) Sol: For theoretically minimum possible shear strain to occur 2    90 

07. Ans: (b)

Sol: VT a f b d c  K

a = 0, 3 f2 

Common Data for Q.03, 04 & 05

VC  0.5m / s, FC  1200,

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f1 , 2

c = 0, 15

d 2  2d

V1T1a f1b d1c  V2T2a f 2b d 2c

w = b = 15 mm =0

FT  800,   30

b = 0, 3,

T1  T2  60

04. Ans: (b)

05. Ans: (d) Sol: d = t1 = 2 mm,

90   90  6   48 o 2 2

08. Ans: (b)

20  cos10 = 41.5 m/min sin 28.33

03. Ans: (c)

t1 2   4mm sin  sin 30

Common Data for Q. 07 & 08

 0.5 cos10  o  tan 1    28.33  1  0.5 sin 10  Vs 

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V2  f1    V1  f 2 

b

1 =2   2 0.3

0

 d1     d2 

c

0.15

 1.11

V2 1.11 V1

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: 14 :

% change in speed =

V2  V1  11% V1

Productivity is proportional to MRR % change in productivity = =

MRR 2  MRR 1 MRR 1 f 2 d 2 V2  f 1d 1 V1 = 11 % f 1d 1 V1

Common Data for Q. 09, 10 & 11

09. Ans: (c) 10. Ans: (d)

12. Sol: T0 , V0 = original tool life and velocity If V1  1.2V0

T1  0.5T0

V2  0.9V0 ,

T2  ?

V1T1n  V0T0n n

 T1  V0     T0  V1 V  ln  0  ln  1  V 1.2  n  1   0.263  T1  ln (0.5) ln    T0  V0T0n  V2T2n

11. Ans: (c) Sol:  = 6 ,

ME – GATE_ Postal Coaching Solutions

VC  1 m / s

b = w = 3, d = t1 = 1mm t2 = 1.5 mm;

use 2      90 o

t 1 2   0.67 r 1  t2 1.5 3  0.67 cos 6    35.62 0   tan 1   1  0.67 sin 6 

For minimum energy condition use

V T2  T0  0  V2

  

1

n

 V  T0  0  0.9V0

1

 0.263  

= 1.4927T0 % change in tool life =

T2  T0 1.4927T0  T0   0.4927 T0 T0

Common Data for Q. 13, 14 & 15

13. Ans: (d)

14. Ans: (c)

0

2 +   = 90

  90    2  90  6  2  35.62 = 24.760   tan   tan 24.76  0.461 Vf  rv c  0.67 1  60  40.2 m / min Area of shear plane = As = Ls × b =

t1  b 1 3 =  5.2 mm 2 sin  sin 35.62

ACE Engineering Academy

15. Ans: (d) Sol: D0  32 mm,  = 35, K1 = 0.1 mm,

FC = 200 N, VC = 10 m/min, L2 = 60 mm, FT = 80 N t1 L 2 60 60     0.59 t 2 L1 D 0   32 t t 0.1  0.169 r  1  t2  1  t2 r 0.59

r

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: 15 :  0.59 cos 35   = 36.150   tan 1   1 0 . 59 sin 35   FT 80  FC 200

tan     

 80      tan    200  1

 35  21.8  56.8o   tan   tan 56.8  1.52 (In general <1) Hence by applying classical friction theorem 1  1  ln  ln  r 0.59           35  2 2 180

 Vf VC

0.5276  0.55 1.04

 r  V f  rVc = 0.59 ×10 = 5.9 m/min

Vs 

Vf 5.9 cos    cos 35 sin  sin 36.15

 8.42 m / min

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17. Sol:  = 10,

t1 = 0.125,

Fc = 517N;

FT = 217N

t2 = 0.43;

Cm = 2 +  – 

t 1 0.125   0.29 t2 0.43

r

 r cos     tan 1    1  r sin    0.29 cos10  o  tan 1   = 16.73  1  0.29 sin 10  F     tan 1  T  FC

  

 217  o  10o  tan 1   = 32.77  517  Cm = 2  16.73 + 32.77 – 10 = 56.23o 18. Ans: (b) Sol: Let Q = no. of parts produced T.C on E.L = T. C on T.L 30 60  Q  80  500   Q  160 60 60 40Q  500  16Q

40 Q  16Q  24Q  500

16. Sol:  =10

t1= f.sin = 0.15 sin75 = 0.144 t2 = 0.36, r 

t1  0.402 t2

 r cos     tan 1    1  r sin    0.402 cos10   = 23.18o  tan 1   1  0.402 sin 10  ACE Engineering Academy

Q=

500  20.83  21 24

19. Ans: (a) Sol: n = 0.12, C = 130 C1 = 1.1 × 130 = 143, V = V1 = 90 m /min 1

 130  0.12 VT  C  T     21.4 min  90  n

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: 16 : 1

V T

1n

 143  C T    90  1

1

1

0.12

 47.4 min

Increased tool life = 47.4 min Increase in tool life = 47.4 – 21.4 = 26 min 20. Ans: (i) 0.2 & 60.4 (ii) 13.81 m/min Sol: VT 0.2

9.8 D 0.4  s 0.5

22. Ans: (a) Sol: Tool life = T1 

500  50, 10

122  12.2, V2  80 rpm 10 The feed and depth of are same in both cases V1  50rpm , T2 

V1T1n  V2 T2n

V2 80 ln V1 50  0.47  0.333  n  50 T 1.41 ln ln 1 12.2 T2 ln

9.8  30 0.4 =  60.4 0.4 0.5 VT 0.2  60.4, n  0.2, c  60.4  n Lm  .  Vopt = C  1  n C g 

n

 0.2 60   60.4 . 60  1  0.2 400   

0.2

 13.81 m / min

Major cutting for, b = pz = Fc

 S0 .t.S  sec   Tan   1

S0 = 0.12,  S  400 t2 a2 0.22    1.83 t1 S0 0.12

=0 Pz = 0.12×2.0×400(1.83sec0–Tan0+1) = 272 N Power = p = FC  VC  p Z 

1

1

n

 50  0.333  29  50   60 

23. Ans: (d) Sol:  = 30, FT = 800N, Fc = 1200N

a 2  t 2  0.22

t = 2.0 mm,

V1T1n  V3 T3n V  T3  T1  1   V3 

21. Sol: S0 = 0.12 mm = t1,

t = 2-0,  

ME – GATE_ Postal Coaching Solutions

Vf r

 p Z  Vf    271

52.6 1.83 60

FC cos      cos   

Fs =

tan (  ) =

FT FC

 800      = tan 1   = 33.69  1200  Fs =

1200  cos(30  33.69)  639.23N cos 33.69

24. Ans: (b) 2.0  0.8  0.024 60  10 If wear land = 1.8mm2

Sol: Slope =

Tool life= T = 10 

1.8  0.8 0.024

= 51.67 min

= 436 W ACE Engineering Academy

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: 17 : 25. Sol:

TC = 3min, Tg = 3 min, Lm = Rs. 0.5/min Depreciation of tool regrind = Rs 0.5 C = 60, n = 0.2 Cg = 3  3  0.5 + 0.5 = 3.5  n Lm  VOpt = C  .  1  n C g 

Because the power consumption is taking place only in the cutting stroke, Velocity of tool in the cutting stroke = length of work or stroke length  RPM of the crank = 200  60 = 12m/min Power required = Fc  cutting velocity

n

 0.2 0.5  .   60 1  0.2 3.5 

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

1490  12  298Watts 60

0.2

= 30.8m/min Common Data for Q. 26 & 27

26. Ans: (a) 27. Ans: (b) Sol: D = 100 mm, f = 0.25 mm/sec, d = 4 mm V = 90 m/min FC FC = 1500 N FT FC = N = 1500 N FT = F

F

28. Ans: 298 Sol: The given problem is the oblique machining problem.

Hence, t1.b = f.d = 0.25  4 = 1mm2 Specific cutting energy =

Fc  Vc =1.49 t 1  b  Vc  1000

Fc =1.49  t1  b  1000 = 1490N

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: 18 :

Chapter‐ 4 Machining

ME – GATE_ Postal Coaching Solutions

p = 10 mm/tooth h = 0.075 mm/tooth V = 0.5 m/min Equation for time for broaching operation

01. Ans: (b) Sol: Time per hole = L/f.N = 25/(0.25 300) = 1/3 min = 20sec. Because dia of drill bit was not given, hence AP1 is zero. 02. Ans: (b) Sol: No. of D.S/min = 10 B = 300 min f = 0.3 mm /stroke B 1 Time/cut =  f No. of D.S =

300 1    100 min 0.3 10

03. Ans: (d) Sol: Hobbing process No. of teeth = 30 (Not required) Module = 3 mm Pressure angle = 200 (Not required) Radial depth= Addendum+1m+1.25m = 2.25 module = 2.25  3 Radial depth = 6.75 mm 04. Sol: Broaching machine P = 1.5 kW d1 = 20 mm enlarged to df = 26 mm t = 25 mm ACE Engineering Academy

=

Length of tool travel Linear velocity of tool

Length of tool travel = L = t + Le + AP + OR As (AP + OR) is not given so take it zero Le = effective length or cutting length 26  20 =3 2 n = no. of teeth = d/h = 3 / 0.075 = 40 Depth of cut d =

Le = np = 40  10 = 400mm Le = 400 mm Time for broaching =

t  Le V

25  400 = 8.05 min 0.5  100 Time for broaching = 8.05 min =

05. Ans: (b) Sol: L = 2m = 50 + 900 + 50 + 50 + 900 + 50 B = 300 + 5 + 5 = 310 f = 1 mm/stroke, VC = 1 m /sec, M=

1 2

Time per two pieces = =

B 1  1  M  f V

310 2000 1  0.5 = 930 sec  1 1000

Time/piece =

930  465 sec 2

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: 19 :

06. Ans: (c)

Time/cut =

Sol: 4.5 mm

230  1.2 min 0.1 159 12 ii) If offset = 5mm with asymmetrical milling

d S  n s  h s  0.0125  8  0.1

= 4.5 – 0.1 = 4.4 d r 4.4   44 teeth h r 0.1

Cutting length = effective length = Le = L r  LS  L f

07. Sol: Part size = 200 × 80 × 60 mm D = 100 mm, Z = 12, V = 50 m/min, 1000 V 000  50   159 rpm D  100

f t  0.1 mm , AP = OR = 5 mm i) With symmetrical milling

 1 = 100  2





100 2  80 2

Time/cut = 

= 44 × 22 + 8 × 20 + 4 × 20 = 1208mm

1 D  D2  w 2 2





1 100  100 2  90 2 2 = 28.2 mm L = 200 + 28.2 + 5 + 5 = 238.2

AP1 =

d r  d total   d f  d s 

AP1 



1 D  D 2  w i2 2 Where, wi = w+ 2(Of) = 80 + 2 × 5 = 90 AP1 =

df  0

N=

L f t NZ



d total  4.5 mm

nr 

Production Technology

  20 mm

L = l  AP1  AP  OR

L f t Nz 238.2  1.25 min 0.1 12 159

08. Ans: (a) Sol: For producing RH threads the direction of rotation of job and lead screw must be in the same direction, for this if the designed gear train is simple gear train use 1, 3, 5 odd number idle gear to get same direction of rotation, if the designed gear train is compound gear train use 0, 2, 4,.. even number of idle gears to get same direction. In the given problem the designed gear train is a compound gear train, to change the hand of the thread it requires to change the direction of rotation of job and lead screw for this use 1, 3, 5… odd number of idle gears.

= 200 + 20 + 5 + 5 = 230 ACE Engineering Academy

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: 20 :

09. Sol: Time / cut =

V

L 576   20 min fN 0.2  144

ii) Total time/hole = Tm + idle time + Tool change time 20 300  60 162  60 = 0.9812 min = 58.87 = 59 sec = 0.617 

DN  100 144   45.2 m / min 1000 1000

 75  VT 0.75  75  T    V

1 0.75

1.333

 75  =   45.2 

 1.96 min

20  1  9.2  10 No. of tool changes = 1.96 (Because 1 tool is already mounted on W.P) Total change time / piece = 20 + 10 × 3 = 50 min 10. Sol: D = 15 mm, Vc = 20 m/min, N

ME – GATE_ Postal Coaching Solutions

1000 V 1000  20 = = 425 rpm  D  15

N = 425rpm f = 0.2 mm /rev T = 100 min,

l  45 mm

L   0.5D  fN fN

15 2  0.617 min  0.2  425 = Tm = machining time 45 

i) No. of holes produced / drill = ACE Engineering Academy

Sol: Train value = Gear ratio =

=

N follower N Driver

pitch of job threads pitch of lead screw threads



3.175  40 127   not possible 6  40 240

=

127 1  20  40 6  20



127 20   possible 40 120

12. Ans: (b) 13. Ans: (b)

Time for idle time = 20s Tool change time = 300 s Time/hole =

11. Ans: (b)

Sol: Crank rotation =

=

40 No. of teeths 40 28

3  12  = 1  = 1 7  28  9 = 1   21  1 complete revolution and 9 holes in 21 hole circle.

100  162 0.617

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: 21 :

14. Ans: (a) Sol: Common alignment test for shaper and lathe are (1) straightness (2) Flatness. Runout is used in lathe. Parallelism used in shaper. 15. Ans: (b) Sol: Out of all conventional method grinding is one which required largest specific cutting energy. 1) Because of random orientation of abrasive particles, rubbing energy losses will be very high 2) Lower penetration of abrasive particle 3) Size effect of the larger contact areas between wheel and work.

18. Ans: (d) Sol: d = 70 mm , Z = 12 teeth V = 22 m/min ft = 0.05 mm/tooth fm = ftZN, N 

1000 V d

fm = 0.05  12 

1000  22 = 60 mm/min 3.14  70

19. Ans: (d) Sol: Gear Ratio = Train value =

Tdriver P  driver Tfollower Pfollower

= G.R =

P job PL.S

N follower N driver



Pspindle PLs.S



N L.S N Spindle

P = pitch

16. Ans: (c) Sol: 1. Plane turning 3. under cutting

2. Taper turning 4. Thread cutting

17. Ans: (d) Sol: Shaping operation M = 0.6 L = 500 mm Double stroke / time = 15 N = time / D.S = 1/15 Average speed, V = 

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L 1  M  V

500 1  0.6 =12000 mm / min 1    15 

= 12 m / min

N Spindle N L.S



PL.S PSpindle / job

=

6 3 = 2 2 2

20. Ans: (d) Sol: 1 one complete rev = 360 10 holes in 30 hole circle means that In 30 hole circle, 1 hole =

360 = 120 30

10 holes = 10 120 = 120  Work piece totally turns through = 120  3 = 4800 or Crank rotation = 1

10 1 4  1   360 30 3 3

= 480 ACE Engineering Academy

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: 22 :

21. Ans: (b) Sol: Given n = 6,

r=

n 1

Dmax = 25 mm Dmin = 6.25 mm V = 18 m/min

Nmin =

1000V 1000  18    25 D max N max = N min

O2 A

d

B

1000V 1000  18    6.25 D min

6 1

O1

N max N min

Nmax =

r=

ME – GATE_ Postal Coaching Solutions

5

25 6.25

l

N=

1000V 1000  120 = 254.64 rpm    150 D

Approach = AP1 + O1O2 =

= 1.3195 = 1.32 22. Ans: (b) Sol: Given Dtool = 15 cm = 150 mm Feed = 0.08 mm/rev Depthmax = 0.5 mm = d Length of workpiece, l = 200 mm Cutting Velocity, V = 120 m/min Total depth to be cut = 2 mm

t

AP1

=

d D  d 

0.5150  0.5 = 8.645 mm

Total time/machining = No. of cutsTime/cut No. cuts =

Total depth 2  =4 depth per cut 0.5

Time/cut =

L   AP = fN fN

=

200  8.645 = 10.227 min 0.08  255

Total time = 10.227  4 = 40.91 = 41 min 23. Ans: (c) Sol: Total depth to be removed = 30 –27 = 3 mm 2 = 0.67 3 feed = 0.5, depth = 2 V = 60 m/min Given, m =

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: 23 :

Production Technology

Approach  50 m   length wise Over time  50 min  Approach  5 m   width wise Over time  5 m  B L Time/cut = 1  M   f V l = 800, L = 800 + 50 + 50 = 900 B = 400 + 5 + 5 = 410 Time / cut =

900  2  410 1    60000  3  0.5

= 20.5 min 3  1.5  2cuts 2 Total tie = 20.5×2 = 41 mins.

No. of cuts =

= SNsin

f 8300 = SN 400  300 0

 = 3.966

25. Ans: (d) Sol: d = 2 mm, w = 150 mm, l = 400 mm, D = 250 mm, Z = 8 VC  1.2m / sec, ft = 0.1 mm 1000  1.2  1.53 RPS   250





1 250  250 2  150 2  25 mm 2 L =l+AP1= 400 + 25 = 425 mm

AP1 =

Time/cut =

=

Lead of the job 2  0.5 1   Lead of the lead screw 6 6

20 25 1 20 1 25       2 20 3 25 40 75

Sol: Time / cut 

Sol: Given f = Vsin  =  D N sin

N

26. Ans: (d) Sol: Gear ratio = Train value = Nfollower / Ndriver

27. Ans: (d)

24. Ans: (d)

sin =

425  347.2 sec 0.1  8  1.53 = 5.78 min 



1000 L  fN 0.1 1000V D 100 mm = 20.94 sec 1000  30 0.1   200

28. Ans: (a) Sol: The curvature given is the concave curvature hence it increases the stress concentration factor therefore it is used for supply of lubricating oil to bearing mounting 29. Ans: (d) Sol: With this any change in UV will also changes the speed of lead screw, the pitch of the threads produced depends on the speed of work and speed of lead screw. Us will not affect the speed of the work

L f t ZN

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: 24 :

Chapter‐ 5 Metal Forming Process 01. Ans: (b) Sol: do = 15 mm, df = 0.1 mm %Reduction=

dia reduced in the draw dia before draw



d 0  d1  Ist draw do



d1  d 2  2nd draw d1

a) 3 stages with 80% reduction at each stage d o  d1 do d1  0.2 d o  3mm 0 .8 

d2 = 0.2. d1 = 0.6mm d3 =0.2 .d2 = 0.12mm

02. Ans: (a) Sol: Given wire drawing process d0 = 6 m, d1 = 5.2 mm Die angle = 180, diameter land = 4 mm Coefficient of friction = 0.15 Yield dress = 260 MPa A0 =

 2  6 = 136 = 21.237 4 4

A1 =

 2  5.2 = = 21.237 4 4

Drawing stress = 2  1  B   A 1 = y   1   B   A 0

  

B

   

B = cot = (Error is 20%)

b) 4 stages with 80% reduction in 1st 3 stages followed by 20% in 4th stage d1 = 0.2. d0 = 3 d2 = 0.2. d1 = 0.6 d3 = 0.2. d2 = 0.12 (Error is 4%) d4 = 0.8. d3 = 0.096 c) 5 stages, with 80, 80, 40, 40, 20 etc d1 = 0.2. d0 = 3 d2 = 0.2. d1 = 0.6 d3 = 0.6. d2 = 0.36 d4 = 0.6. d3 = 0.0216 d5 = 0.8. d4 = 0.1728 (Error is 72%) From the given multiple choice B, the final diameter of wire close to 0.1 mm. ACE Engineering Academy

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1 1 Die angle =  18 =90 2 2

=9 B = 0.15  cot90 = 0.947 2 = 126.958MPa 0 0.947   1  0.947   21.27    = 260   1     0 . 947 28 . 270     

= 260(2.056)(0.2375) Total drawing stress 2 = y + (2y) e

2 L R1

(By considering friction) = 260 + (130260) e

20.154 2.6

total = 260  81.94 = 178.05 MPa Total drawing load = tA1 = 178.05  21.237 = 3.781 kN

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: 25 :

03. Sol: Given:

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for stationary mandrel B =

H0 = 4.5 mm H1 = 2.5 mm

B=

H = 2 Droll = 350, Rroller = 175 mm Strip wide = 450 mm = b Average coefficient of friction = 0.1 y = 180 MPa RSF = Pavg  projected area =

2  L    y 1  bL 3  4H  RH = 175  2 = 18.7

L= 4= 

H 0  H1 4.5  2.5  = 3.5 2 2  0.1  18.7   1801    450  18.7 4  3.5  3 

2

RSF = 1982.64 kN

1   2 Tan 0.12  0.12 =1.29 Tan (12)

B 1  B    H1     1   2   y   B    H 0   1.19 1  1.29    1.8   2 / y     1    1.129    2.6  

2/y = 0.64 05. Movable mandrel

B = cot=(0.12)cot(12)=0.564 0.564 1  0.564    1.8   2 / y      0.519 1    0.564    2.6  

06. Floating mandrel B=0

Common data for Q 04, 05 & 06 04. Ans : (b) , 05. Ans: (c), 06. Ans: (a) Sol: Initial inside diameter of tube H0 = 2.6 d0 = 52 mm, D1 = 50 mm H1 = 1.8,

h  2  n  0  y  h1 

 2.6  = n  = 0.367  1.8 

2d = 24=12, =0.12 2.6mm 1.8 mm 52mm

50 mm

Common data for Q 07. & 08. Sol: d0 = 6 mm, df = 1.34 mm Given ideal condition

 = 0.2

 = 60

f = 60 MPa Maximum reduction condition 2B 2  1  B   d1   =11=   1 y  B   d 0  

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: 26 :

B = cot;

 d1     d0 

2B

1

Common data for Q 09, 10 09. Ans: (c) & 10. Ans: (b) Sol: 400 y

B = 1.9

d  B  1   1  1 B  d0 

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2B

B 1 B

200

1 = 1 B

0.2

0.4 

d1 2 B 1  d0 1 B 1

 1  21.9 1    6 d1 = d 0  2 B    1  1.9   1 B  d1 = 4.53 ……… (1) stage d2 = d1 2 B 1 1 B 1

 1  2B C=   = 0.756 1 B  Dia of wire in 2nd stage = 3.424 mm d1 = d0  c d2 = d1 c = 4.530.756 = 3.424>1.34

Lo = 100m Lf =?

 y before  200 Mpa,  y after  400 Mpa, Ao Lo = Af Lf d A L f  L o  o  L o  o Af  df

  

2

2

 12.214   100     150m  10  True strain in the drawing process 2

d  A     n o   n  o   0.4 A1  d1  From the graph  y at   0.2 ,

d3 = d2  c

 y  300 MPa

= 3.424 0.756 = 2.589>1.34 d4 = d3c = 1.957>1.34 d5 = d4  c = 1.4797 > df d6 = d5c = 1.1186 < df  Hence No. of stages = 6

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do = 12.214, df = 10mm,

11. Ans: (a) Sol: do = 25,

di = 5mm

 y  315 

0.54

d  A   n o   n  o  A1  di 

2

2

 25    n    3.22  5  0.54  y  315 (3.22)  592 MPa.

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: 27 :

16. Ans: (b) Sol: Extrusion constant = K = 250 do = 100mm, df = 50mm

Common data for Q 12 & 13 12. & 13. Sol: do = 100mm, ho = 50 mm, h f  40mm ,  y  80 MPa df  do

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Extrusion Force = A o K ln

ho 50  100  111.8mm hf 40

2

  100   100 2  250 ln  = 2.72 MN. 4  50 

Fi min  A 0   y 

  1002  80  628.318 kN 4

 (111.8) 2  80 4  785.350 kN

Ff min  A f   y 

Fmin 

Fi min  Ff min  706.834 kN 2

17. Ans: (a) Sol: Ho = 4, H1 = 3mm, R = 150mm, N = 100 rpm. Velocity of strip at neutral point = Surface Velocity of rollers

DN   300  100  1000  60 1000  60 = 1.57 m/sec



W.D  Fmin  (h o  h f )  7068J  2 W  H H

7068  0.354 m 2  10  103

14. Ans: (c) Sol: (Extrusion force)min =  y  A 0

  10  10 2 = 78539.8N 4 Extrusion force 

(E.F) min 78539.8   ext 0.4

b = 100mm R = 250mm,

N  10 rpm,

 y  300 MPa

H  20  18  2mm H = 0.089 R



L  length of deformation zone  RH

H

15. Ans: (a)

20  18  19 2

Favg  R.S.F 

Sol:  y  1400 0.33

1  y  1400  3 ACE Engineering Academy

18. Ans: (a) Sol: Ho = 20mm, H1=18mm,

 250  2  22.36 mm

= 196346.5N = 196 Tons

At maximum load, true strain 

Ao Af

1 3

0.33

 971 MPa



 L   y b  L 1   3  4H 

2

 0.089  22.36   300  100  22.36 1   4  19 3 

2

= 795 kN.

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: 28 :

T  Favg  a ,

ME – GATE_ Postal Coaching Solutions

22. Ans : (b)

Where a = moment arm   L

 0.3L to 0.4  L T  Favg  0.4L  795  10 3  0.4  22.36 = 7110 kN-mm = 7.11 kN-m 2 NT 2  10  7.110  Pav  60 60  7.44 kW / roller

23. Ans : (a) Sol: Given rolling process Initial thickness H0 = 30 mm Final thickness = H1 = 14 mm Droller = 680 = R = 340 mm

y = 200 MPa Thickness at neutral Hn = 17.2 Forward slip =

Total Power = 7.44 × 2 = 14.88 kW = 19. Ans: (d) Sol: Ho = 16mm, H1 = 10mm, R = 200mm Angle of Bite    Tan 1

 Tan 1

H R

16  10  9 .9 200

20. Ans: (d)

V1 H 1= n 1 Vn H1

17.2  1 = 0.2285 = 23% 14

Backward slip = 1   1

V0 H 1  n Vn H0

17.2 = 42.6%  43% 30

24. Ans: (b) Sol: Roll separation distance

= 2  R + H1 = 2  300 + 25 = 625 mm

21. Ans: (c) Sol: Rolling D = 300 mm  = 0.1 Minimum possible thickness sheet, means that maximum reduction condition Hmax = 2R H0H1 = 0.12150 H0H1 = 1.5 H1 = H0 1.5 = 4  1.5 = 2.5 mm ACE Engineering Academy

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: 29 :

Production Technology Fb 

Chapter‐ 6 Sheet Metal Operation



= 12.7 + 2  0.04 = 12.78 02. Ans: (a) Sol: Die size = Blank size = 25.4mm

80  0.6  1.25  34.28 kN 0.6  1.25

F  FP  Fb  51.42 kN

Common data for Q. 1 to 5

01. Ans: (b) Sol: For punching operation Punch size = Hole size = 12.7 Die size = punch size + clearance

Fb max kt kt  I

06. Sol: B1  15  0.5  2  180 

 180

= 50.265 mm B 2  6  0.5  2  90 

 = 10.99 mm 180

L 0  98  204  92  B1  2B 2 = 466.245 mm

Punch size = Die size  2(radial clearance) = 25.4  2(0.04) Punch size = 25.32 mm

2mm 15 92

03. Ans: (b) Sol: Fmax = Fp max + Fb max

98

100 6 mm

8m

=   12.7  1.25  800    25.4  1.25  800 = 40 +80 = 120 kN

8

8

204 220m

04. Ans: (c) Sol: Force required is Max [Fpunch, Fblank]  force required is Max [40, 80]  force required = 80 kN



5 cm

h = 7.5cm

Fp max .Kt Kt  I 40  0.6  1.25  17.14 kN 0.6  1.25  1

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07. Ans: (c) Sol: d = 5cm = 50mm

05. Ans: (b) Sol: Fp 

Common data for Q. 07, 08 & 09

7.5 cm

D  d2  4d h 

5 2  4  5  7.5  13.2 cm

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: 30 :

08. Ans: (b)

 d1 =

11. Sol: Work done in blanking open = Fmax.K.t

Dia.before Dia.after

Sol: Draw Ratio =

= 83.61030.42103

13.22 = 7.34 > 5cm 1.8

= 66.88 J

7.34 = 4.08<5 cm 1.8 n=2

12. Sol: I = ? F = 24 kN Fmax = 83.6 kN

 d2 =

09. Ans: (a)

F(Kt + I) = Fmax  Kt

Sol: D  d  4d 1 h 1 2 1

4d 1 h 1  D  d 2

h1 

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I=

2 1

D 2  d 12 13.22 2  7.34 2  4  d1 4  7.34

Fmax  Kt Kt F

 83.6  0.4  2   0.4  2  = 1.98 mm. = 24  

= 4.11 mm Common data for Q. 13, 14 & 15 13. Ans: (b)

P1  Dt y =   132.22  1.5  315  196238 N = 196.238 kN

E = P1h1 = 196.2384.1110–3 = 806.6 kJ

Sol: y = 35MPa, d = 12mm, r = 0.5 mm

Blank diameter, D = =

Common data for Q. 10, 11 & 12

d 2  4dh as D/ r> 20

122  41216

 30.2 mm

10. Sol:

100

14. Ans: (b) 30

50mm 450 80mm

Sol: DRR 1  0.4 

D  d1 D

d 1  D1  0.4   30.2  0.6  18.12 20

20

P  100  30  20 2  80  50  288.28

d 2  d 1 1  0.25  18.120.75  13.59

d 3  d 2 1  0.25  13.590.75  10.19 d3 < 12  n = 3

Fmax  Pt u  288.28  2  145  83.6 kN ACE Engineering Academy

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: 31 :

Production Technology

D total  D  2  3  144.42 mm

15. Ans: (b) Sol: P1  Dt y    30.2  2  35  6641.3  21 



P1



 2 2 d 1  d 1  2 t  4 6,641.3  =65.5 MPa.  2 2 18.12  18.12  2  2 4





19. Ans: Sol: d2 = 40 + 10 + 10 = 60 d2 = 60mm , d1 = 40mm T.A = 4mm D=

d 22  4d 1 h

D=

60

2

10

10 40 30

40

 440 30 

Common data for 16 & 17 16.

D = 91.65mm D = 91.65 + 2(T.A)

Sol: D  d 2  4dh  30 2  4  30  150

D = 91.65 + 2(4)  D = 99.65mm

 137.47

d 1  D  0.6  137.47  0.6  82.48  30 d 2  82.48  0.8  65.984  30

d 3  65.984  0.8  52.7  30 d 4  52.7  0.8  42.2  30

20. Ans: (d) 21. Ans: (d) Sol: k = 0.5 I=t

d 5  42.2  0.8  33.7  30

F  k t  I   Fmax (kt )

d 6  33.7  0.8  27  30

F  t 1.5  Fmax  k t

n=6

F1.5  0.5Fmax 1 F  Fmax 3

17. Ans: Sol: d3 = 52.7 mm

22. Ans: (b) 18. Ans:

23. Ans: (a)

d 100 Sol:   16.66  15 to 20 6 r D  d 2  4dh 

r 2

 100 2  4  100  25  = 138.42 +23 ACE Engineering Academy

Sol: Fmax  5  dt  u  dt u 

5 

Fmax    1.5d  0.4t   u 6 2

   1.5  0.4  dt u    1.5  0.4 

2  3 KN 

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: 32 : Common Solution for Q. 24 & 25

24. Ans: (a) 25. Ans: (b) Sol: t = 5 mm, L = 200 mm, τu = 100 MPa, K = 0.2 W.D = Fmax Kt = L × t × τu × K.t = 200 × 5 × 100 ×0.2 × 5

ME – GATE_ Postal Coaching Solutions

28. Ans: (a) Sol: Die size = Blank size = 25 – 0.05 = 24.95 Punch size = Die size – clearance = 24.95 – 2  0.06 = 24.83

100  10 3  100 N  m (or ) J only 1000 Shear provided over a length of 

20  200 = 10 mm 400 Fmax Kt = F (Kt + I) 200 mm 

F

100  10 3  0.2  5  9.09  10 kN 0.2  5  10

26. Ans: (d) Sol: d = 25 mm, t = 2.5 mm → piercing  u  350 MPa Diameter clearance  C  0.0064 K t  0.0064  2.5 350 = 0.3 mm In piercing P.S = H.S = 25 mm. D.S = P.S + C = 25 + 0.3 = 25.3 Fmax  dt  u    25  2.5  350 = 68.72 kN. 27. Ans: (c) Sol: Number of earing defects produced =2n Where n is an integer So possible option is 64. ACE Engineering Academy

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: 33 :

Production Technology

(ii) Ans: (b) Allowance = difference between max. material limits = L.hole – H.shaft = 25.00 – 24.98 = 0.02 mm

Chapter‐ 7 Metrology

7.1 Limits, Fits & tolerances (iii) Ans: (b) 0.02

01. Sol:

Shaft  25 H

H. Limit

0.021 B.S

= (25.021)  (24.947) = 0.074 mm

B.S F.D=0.02mm f

0.021

, Hole = 250.00

Max clearance = different between minimum material limits = H.hole – L.shaft

L. Limit

H. Limit

0.053

Tol = 0.033mm

L. Limit

D  18  30  23.24mm

(iv) Ans: (a) Size of the GO plug gauge = max. material limit of hole = L.hole = 25 mm (v)

i  0.453 D  0.0010  1.3m FD of hole H = 0 FD Shaft = 5.5(23.24)0.41 = 20m Hole tolerance, IT7 =16i= 20.8m = 21m=0.021 mm

Ans: (b) Size of the NOGO plug gauge = min. material limit of hole = H.hole = 25.021 mm

(vi) Ans: (c) Size of the GO ring gauge = max. material limit of shaft = H.shaft = 24.98 mm

Shaft tolerance, IT 8 =25i = 32.5m = 33m = 0.033mm L - hole = basic size =25 mm H - hole = 25 + 0.021 = 25.021 mm H - shaft = 25 – 0.02 = 24.98 mm L - shaft = 24.98 – .033 = 24.947 mm (i) Ans: (a) L- hole > H- shaft  Clearance fit ACE Engineering Academy

(vii)

Ans: (d) Size of the NOGO ring gauge = min. material limit of shaft = L.shaft = 24.947 mm

(viii) Ans: (a) 02. Ans: (a) Sol: For Clearance fit L- hole > H- shaft

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: 34 :

03. Ans: (c)

ME – GATE_ Postal Coaching Solutions

06. Ans: (c) 0.050

Sol: Hole = 40 0.000 mm ,

min. clearance = 0.01 mm, tolerance on shaft = 0.04 mm , Max. clearance of shaft = ? 0.01 = L.hole – H.shaft 0.01 = 40.000 – H.shaft

Sol: D  18  30  23.2

i  0.45 3 D  0.001 D  1.3 IT8 = 26i = 26 × 1.3 = 33.8 = 34 m = 0.034 mm 0.034

Hole size  25H 8  25  0.000

 H.shaft = 40.000 – 0.01 = 39.99mm H.shaft – L.shaft = 0.04 L.shaft = 39.99 – 0.04 = 39.95 Max. clearance = H.hole – L.shaft = 40.05 – 39.95 = 0.10 mm 04. Ans: (d) Sol: Xmax = 50.02 – (37.985 + 9.99) = 2.045 Xmin = 49.98 – (38.015 + 10.01) = 1.955 X = Xmax  Xmin= 0.09 Dimension X = 2 ± 0.045

When, t = 0.01 mm D = 30.01 + 20.01 = 30.03 mm = 30.05 + 2 0.01 = 30.07 mm When, t = 0.015 mm D = 30.01 + 20.015 = 30.04 mm = 30.05 + 2 0.015 = 30.08 mm mm

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i = 1.86 microns = 1.9 microns IT8 = 25i = 47.5 microns Tolerance = 0.0475 mm F.D = –5.5 D0.41 = – 5.5 × 63.240.41 = 30 Microns = 0.03 mm H. shaft = 60 – F.D = 60 – 0.03 = 59.97 mm L. shaft = H. shaft – Tolerance

08. Ans: (c) Sol: Case (i) 25H7

t t = 0.01 to 0.015mm

D  30

Sol: D  50  80  63.24 mm

= 59.97 – 0.047 = 59.923 mm.

05. Ans: (c) Sol: D

0.08  0.03

07. Ans: (a)

L.L = 25.00 U.L = 25.021 Case (2) 25 H8 UL = 25.033 Case (3) 25H6, UL - ? (UL)H8 (UL)H7 = (UL)H7  (UL)H6 25.03325.021 = 25.021  (25 + x) x = 0.009  (UL)H6 = 25.009

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: 35 :

09. Ans: (d)

(ii) Ans: (c) xmax = 250max– (60min+(30/2)min+ymin+(25/2)min)

0.01

Sol: A = 25.2 0.02

= (250 + 0.2) (60 +15+121.55+12.5)

B = 30.4  0.01

= 41.15mm xmin = 250min –(60max+(30/2)max + ymax + (25/2)max)

C = 32.7  0.02 Tmax = Lmax  Amin  Bmin – Cmin = (118 + 0.08)  (25.2  0.02)  (30.4  0.01) – (32.7 – 0.02) = 29.83 = 30

0.17

Tmin = Lmin  Amax  Bmax  Cmax = (118  0.09)  (25.2 + 0.01)  (30.4 + 0.01) (32.7+ 0.02) = 29.57 Tmin = 30 T = 30

= (2500.2)  (60.2 + 30.025/2 + 123.45 + 25.025/2)

= 38.625 mm Tolerance on X = Xmax – Xmin = 2.525 mm 12. Sol: L Hole = BS = 65mm H Hole = BS + Tolerance = 65.05mm (i) Ans: (c)

0.43

Allowance = (L.L)hole  (H.L)Shaft  0.09 = 65 – (H.L)shaft

0.17  0.43

 (H.L)shaft = 65  0.09= 64.91 mm Tolerance = (HL)shaft  (LL)shaft

10. Anc: (c)

 0.05 = 64.91 – (LL)shaft

11. (i) Ans: (d) Sol: Let the vertical distance between the holes is ‘y’ 2450.05

25 30

 (LL)shaft = 64.86 mm 0.09

Shaft = piston = 65

 0.14

(ii) Ans: (a) (L.L)hole = 65 mm

x

60

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(Tolerance)hole = (HL)hole  (LL)hole y

2500.2

 0.05 = (HL)hole – 65  (HL)hole = 65.05 mm

0.2  0.0

0.05

y y = 245sin30 sin30 = 245 ymax = 245maxsin30max = (245 + 0.05)sin(30 +15/60) = 123.45 ymin = (245 0.05)sin(30015/60) = 121.55 ACE Engineering Academy

Hole = Bore = 65 0.00 (iii) Ans: (b) Max Clearance = 65.05 – 64.86 = 0.19mm

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: 36 :

13. Sol: Amax = 15max + 30max = 15.06 + 30.1 = 45.16

ME – GATE_ Postal Coaching Solutions

min. interference = difference between min. material limits = L.shaft - H.hole = 50.026 – 50.025 = 0.001 mm

Amin = 15min + 30min = 44.84 A = 45 ± 0.16. = A ± ∆A Bmax = Amax – 20min = 45.16 – 19.93 = 25.23 mm

15. Sol:

C =100±0.1 9.9  0.025

14.9  0.025

Bmin = Amin – 20max = 44.84 – 20.07 = 24.77 mm B ± ∆ B = 25 ± 0.23.

15  0.05

x

14. (i) Ans: (a) , (ii) Ans: (a), (iii) Ans: (a), (iv) Ans: (c)

Let C = center distance between holes Cmax = max. Outer distance of pins – sum of min rod holes.

Sol: H

H. Limit

0.025 50

L. Limit

Hole  50 H. Limit

9.925 14.925  2 2 = 112.525 mm

p 0.026

B.S =50

L. Limit 0.042

Shaft  50 0.026

L.hole = B.S = 50 H.hole – L.hole = Tolerance = 0.025 mm h.hole = L.hole + Tolerance = 50.025 mm max. interference = difference between max. material limits = H.shaft – L.hole = 50.042 – 50.00 = 0.042 mm ACE Engineering Academy

 9 .9   14.9  Xmax = 100 max       2  max  2  max

 100.1 

0.025  0.000

0.042

10  0.05

 9 .9   14.9  Xmin = 100 min       2  min  2  min

9.875 14.875  2 2 = 112.275 mm

 99.9 

 15   10   C max  X max         2  min  2  min 

 14.95 9.95   112.525     2   2 = 100.075 mm

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: 37 :  15   10   C min  X min         2  max  2  max 

 15.05 10.05   112.525     2   2 = 99.725 mm 0.075

 C  100 0.275 16 Sol: For the given conditions

14.875 9.875  2 2 = 112.475 mm

X  100.1 

 15.05 10.05  C  X   2   2 C = 99.925 mm Because C is lying in between the limits, the assembly is possible. 17. Ans: (a) Sol: GO size = max. material limit of hole = 20.01 mm NOGO size = min. material limit of hole = 20.05 mm

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19. Ans: (c) 20. Ans: (b) Sol: To calculate exactly the data was not given in the problem. But for shaft “h”, H – Shaft = 25.000 L – Shaft = less than 25. And h7 → 7 indicates IT 7 not 7 microns.

21. Ans: (b) Sol: W max = P max – (Q min + R min) = 35.08 – (11.98 + 12.98) = 10.12 mm W min = P min – ( Q max + R max) = 34.92 – (12.02 + 13.04) = 9.86 W = 9.86 to 10.12 = 9.99 ± 0.13 mm.

7.2 Angular Measurements 01. Sol:

Ans: (a) Sine bar

Slip gauges



18. 0.03

Sol: Hole = 20 0.00

Min. interference = 0.03mm, Max. interference = 0.08 mm 0.03 = L.shaft – H.hole L.shaft = 0.03 + 20.03 = 20.06 mm 0.08 = H.shaft – L.hole H.shaft = 0.08 + 20.00 = 20.08mm shaft  20

0.08  0.06

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Given sine bar length = 200 = l Angle =3256 = 32.085 Slip gauge height = h say sin  

h 

sin 32.085 0  

h 200

 h = 106.235

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: 38 :

02. Ans: i -(b), ii-(a) Sol: l = 50 , L = 500 50  0.08 200  200 

0.08  0.32 50

06. Ans: (d) 07. Sol: (i) Ans: (b) sin =

h = h + 0.32 = 28.87 + 0.32 = 29.19 h 29.19 Sin     82332 L 200

h2  h1 w

h2  h1 = 100sin30 = 50 h2 = h1 + 50 = 75

03. Ans: (d)

(ii) Ans: (d)

sin(30) 

04. (i) Ans: (c)

 dh dw  (ii) (a) d = tan30    w  h

08. Sol:

7.3 Taper Measurement 01. Sol: 2.5

10    1000 3600 180 = 0.04845 mm/m. ACE Engineering Academy

90 250

  = 21.2 deg

 d  d1  (h 2  h 1 )   2   2  sin = w



h2

Ans: (a) L = 250mm, d = 20mm h = 100 – (d/2) = 100 – 10 = 90 mm

sin  

0.002 = 0.001 2

05. Ans: 0.048 mm/m Sol: Gradient of spirit level = Sensitivity specified in mm/m

h1

 h2 = 75.0025 + 0.005 = 75.0075 mm

0.001  180  = tan30  0   3600 = 5  125   

(c) dh = 0.002 & Calculate (d) dh = 0.005

h  25 100.005

 h = 75.0025mm

h L h = sin30o  125=62.5 mm

Sol: sin  =

(b) dh =

ME – GATE_ Postal Coaching Solutions

30 O2

50

/2

15

O1

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: 39 :

sin  / 2 

d 2  d1 2h1  h 2   d 2  d1 

sin  / 2  

30  15 252.5  30  15



O1O 2  r1  r2 =75

O2A = h1 + r1 – r2 – h2 = 70 + 50  30  25 = 65 D  50  75 2  65 2  25

15 = 1/6 105  15

= 112.4165 mm

 = 19.2

04. Sol:

02. Sol: tan  / 2 

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5 8.66

O2

d = 25

42

  = 60 O1

35

A

O2

105=5

O1A =

O1

25 2  17 2  18.33

D = r + O1A + r = 25 + 18.33 = 43.33 05. Sol:

36.345=31.34 36.34 40

B A

50

03. Sol:

12.5mm

M

70 mm 30 mm

C



C

h1+r1=O2A+r2+h2

O2

D

X

O1 d1 = 100 m





h2

h1

12.5mm

O

A

 +  = 4550 + 2910 = 75 Diameter = O1C+O1A+O2D  ACE Engineering Academy

d1  2

O1O 2 2  O 2 A 2



d2 2

  37.5  2

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: 40 :



75   = 37.5 –  = 37.5 – 2910 2 = 820

le OBC sin 37.5 =  OB 

ME – GATE_ Postal Coaching Solutions

If D = 0, h = 0 D = 1, h = 1 1 1   Sin      2  2 1  1 3    2   19.47   =38.94  

BC OB

12.5 BC   20.533 sin 37.5 sin 37.5

le OAB

3  Sol: Tan     2  28.54

OA OB

cos 820 =

08. Ans: (a)

 OA = OB cos 820 = 20.316 mm



X = M – (OA + R) = 110.89 – (20.316 + 12.5) = 78.074 mm 06. Sol: d2 – d1 = 10 ;

2

3

3

15.54 + 8 + 5 = 28.54

 h2 – h1 = 12.138

d 2  d1  sin     2  2  (h 2  h 1 )  d 2  d1 

 = 88.9

 3   Tan 1  = 6 2  28.54 

 Taper angle ( )  6 0 2 Included angle = 120

Error = 90 – 88.9 = 1.1 09. Ans: (c) 07. Sol:

Sol: tan  

10   = tan-1(1/3)    18.434 30 10mm

10mm

h  D2 

D

2

2



30mm 10mm

D

  2 Sin    2 hD Sin

 2

D 2h  D

Z=40

Z=10 Z=0

10 – (10/3)

 D  2 2h  D

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: 41 :

Distance at Z = 0, 10   D 0  210  10 tan 30  210   3  = 6.67  2 = 13.33 mm

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p   De = M  d  tan  2 2  3.5   = M  2   tan 30  2   = 29.366 – 3.010366 = 26.355 mm

1sec



1

04. Ans: (a) Sol: VED = De  VC

With probe diameter compensation Dactual  13.334  2  r sec 

VC  P cos

 2

 0.0131P 1   2 

= 13.334 + 2 ×(1 sec 18.435)

P = pitch error

= 15.442 mm.

1, 2  flank angle errors in deg

7.4 Screw Thread Measurements

1  71  0.11667 2.04103 2 = 91 = 0.15 2.618103 P = 0.004

01. Ans: (d) Sol: Major diameter = s + (R2 R1)

= 35.5 + (11.8708 9.3768) = 37.994 mm

= 30.5 + (15.3768  13.5218) = 32.355 mm

(metric thread)

Virtual correction VC = (0.004  cos30) + (0.0131  VC = 0.01569 VED = De + VC = 30.6651 + 0.01569 = 30.6807 05. Ans: (a)

03. Ans: (a) p  Sol: best wire diameter, d =   sec  2 2  3.5   60  =  sec  = 2  2   2  M = 30.5 + (12.2428  13.3768)

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 = 60

3.5(0.11667 + 0.15) )

02. Ans: (a) Sol: Minor diameter

= 29.366 mm

De = 30.6651

06. Ans: (d) R 2  R1  Sol: Sin     2  M 2  M1  R 2  R 1 

=

1.4434  0.8660 22.06  20.32  1.4434  0.8660

 = 59.5566 = 593323 Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally

: 42 :

ME – GATE_ Postal Coaching Solutions

07.

12. Ans: (a)

p   Sol: De = M   d  tan  2 2 

Sol:

p   Deff = M –  d  tan  2 2  = 16.455 – 1.155.tan30 = 14.7226 mm

2 M = 14.701 + (1.155+ tan30) =16.433 2 08. Ans: (d) Sol: Lead = pitch no of starts Pitch =

lead 3 = =1.5 no of starts 2

09. Ans: (d) Sol: Rollers will not used to measure pitch diameter. p  Best size diameter d =   sec  2 2  2   60  =   sec  2  2  = 1.1547 = 1.155

7.5 Surface Finish Measurement 01. (i) Ans: (c) Sol: Rt = max. peak – min.valley

= 4 2  18 = 24 (ii) Ans: (c) Sol: CLA(Ra) = (h1+h2+h3+……+h10)/n

=

(iii) Ans: (b) Sol: Peaks 35 40 35 42 35 Valley 25 22 18 25 23

10. Ans: (d)  Sol: V.C =  P.cos    0.0131 P(1+ 2) 2

300 = 30 10

Rz = 

= 0.2 cos30 = 0.346

peaks  valleys no of peaks

(35  40  35  42  35)  (25  22  18  25  23) =15 5

(iv) Ans: (b) Common data Q 11 & 12 Sol:

11. Ans: (a) p  Sol: Best size diameter, d =   sec  2 2  2   60  =   sec  =1.155 mm 2  2 

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RMS =

h 12  h 22  h 32  ........  h 2n = 33 n

(or) RMS = 1.1  R a=1.1  30=33 (v) Ans: (c) Sol: If Ra value from 18.75 to 37.5 international grade of roughness is given by N11.

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: 43 :

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02. Ans: (c) Sol: Ra =

=

A  w

Chapter‐ 8

1 1000  HM VM

480  480 1 1000   0.8 100 15000

= 0.8

01. Ans: (a) Sol: Pitch of lead screw = 5mm 1 rev = 5mm 1mm = 1/5 rev 200mm = 1/5  200 = 40rev = 40 360 = 14400 deg.

03. Ans: (d) Sol: Rt =

0.05 =50  m tan 45

04. Ans: (c) Sol:

40

50

Am = 0.105

2.5

Am act = 0.105 0.01 2.5 = 0.08



A m act

 Distance travelled /pulse

03. Ans: (b) Sol: For 1 rev of motor 360 are required

(10 3  2.5)  0.04

 360 pulses are required

2

0.08 1  = 0.8 3 2.5  10  0.04 1000



02. Ans: (b) Sol: Pitch of lead screw = 5mm, BLU = 0.005mm Length of travel = 9mm No.of pulses = L/BLU = 9 / 0.005 = 1800 pulse.

10

K=

Advanced Machining Methods Numerical Control (NC)



When motor is rotated by 1 rev  lead screw will rotate by 1 rev When Lead screw is rotated by 1 rev

05. Ans: (c )

06. Ans: (c)

07. Ans: (a)

3.6mm distance is travelled by axis In total For 360 pulses  360 deg of motor  1 rev of motor 1 rev of lead screw 3.6mm of linear movement of axis 360 pulses = 3.6mm 1 pulse = 3.6/360 = 0.01mm = 10 microns

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: 44 :

04. Ans: (b) Sol: 10V = 100 rpm = 100  5 = 500 mm/min That is for 500mm/min = 10V 1mm /min = 10/500 3000mm/min = 10  3000 / 500=60 V 05. Ans: (c) Sol:

10. Ans: (d) Sol: Appropriate answer but the correct answer is N05 X5 Y5 N10 G02 X10 Y10 R5 Because in CNC part program we are not suppose to indicate information about one axis more than once in one block. Common Data 11 & 12 11. Ans: (b) & 12. Ans: (a)

75

Sol: A, Stepper motor  200 steps / rev

centre

55

ME – GATE_ Postal Coaching Solutions

 200 pulses /rev

(50,55)

Pitch = 4mm, no. of starts = 1, 50

Gear ratio = N0/Ni= 1/4 = U

70

F = 10000 pulses per min

06. Ans: (b)

200 pulses  1 rev of motor

07. Ans: (b)

 1/4 rev of lead screw

08. Ans: (a) Sol: G02–circular interpolation clockwise G03 – circular interpolation counter clockwise

09. Ans: (c) Sol: because the tool has to travel from P1 to P2 in clock wise. Y

P2 = (10, 15)



 Center (15, 15)

= 1/4  4  1 mm linear distance. = 1mm linear distance 1 pulse = 1/200 = 0.005mm = 5 microns = 1 BLU Feed = BLU  pulse /min = 0.005  10000 = 50mm/min For changing BLU = 10 microns = 0.01mm  gear ratio has to be reduced to 1/2 Feed = BLU  pulse /min



P1 = (15, 10)

 Pulses per min = feed / BLU = 50/0.01 = 5000

X

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: 45 :

13. Ans: (b) Sol: Tie comes between B & C, but in C the value of x-is large and in B the value of x-is small. 14. Ans: (a) Sol: Given coordinates (0,0) to (100, 100) Mean, L = 100, depth, d = 2 m Diameter, D = 10 APC = =

d D  d  210  2  = 4

16. Ans: 60 Sol: In the combined movement, the tool is moving for 50mm with a speed of 100mm/min. whereas in the same time tool is traveling x-axis by only 30mm. hence, For 50mm 100mm/min For 30mm 

100  30  60mm / min 50

17.   Ans : (a) Sol: Because diameter of milling cutter is 16mm, the radius is 8mm. the dotted line indicates cutter center position, which is shifted by 8 mm all around the rectangular slot

104 104 Time/slot = = fN 50 = 2.08 min = 124.8 sec  120

(–8,58) S

15. Ans : (c) Sol:

(0,50)

(100,50)

(0,0)

(100,0)

R

(108,58)

CNC drill table X axis

pulses Pulse generator

Production Technology

Driver

Stepper motor

BLU = the distance traveled by the table for one pulse of electrical energy input to the motor.

(–8,–8)

P

Q

(108,–8)

If the given shape is rectangular hole, then the answer is (8,8), (92,8), (92,42), (8,42), (8,8)

Hence 200 pulse = 1 revolution of motor = 1 revolution of lead screw = 4mm That is 1 pulse = 4/200 = 1/50 = 0.02mm, hence BLU does not depends on the frequency of pulse generator. But if the speed of the table means it will get doubled. ACE Engineering Academy

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: 46 :

Chapter‐ 9 NTM, Jigs and Fixtures 01. Ans: (b) Sol: For Rough machining i.e. stock removal the electrolyte should have high electrical conductivity, called passivity electrolyte, where as for finish machining the electrolyte should have low electrical conductivity called non–passivity electrolyte will be used. 02. Ans (c) 03. Ans: (b) Sol: In ECM MRR  gram atomic weight of material MRR  Current density MRR 

1 dis tan ce between tool and work

MRR  Thermal conduction of electrolyte. 04. Ans: (c) Sol: In EDM the mechanism of MR is due to melting and vaporization associated with cavitation and also erosion & cavitation or spark erosion and cavitation 05. Ans: (c) Sol: EDM, ECM and AJM are used for producing straight holes only but in LBM by maneuvering or bending laser gun slightly it is possible perform the Zig – Zag hole.

ACE Engineering Academy

ME – GATE_ Postal Coaching Solutions

06. Ans: (b) Sol: In EBM Vacuum is provided to avoid the dispersion of electrons after the magnetic lense, but this vacuum is giving an addition function of providing efficient shield to the weld bead. 07. Ans: (b) Sol: out of all the NTM’s EDM will give large MRR and EBM will give very small MRR. 08. Ans: (d) Sol: Relative motion between tool and work piece is not necessary. 09. Ans: (d) Sol: The high thermal conductivity of the tool material will have high electrical conductivity hence the heat generated with in the tool is low and what ever heat generated it will be distributed easily therefore tool melting rate reduces and tool wear reduces. Where as due to specific heat of work material, the rise in temp of W.P is faster and more amount of MR is possible. 10. Ans: (a) Sol: In ECM MRR  gram atomic weight of material  Current density 

1 dis tan ce between tool and work

 Thermal conduction of electrolyte.

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: 47 :

11. Ans: (b) Sol: I = 5000 A A = 63, Z = 1, F = 96500 MRR 

AI 5000  63  3.264 g / sec .  ZF 1  96500

12. Ans: (a) Sol: A = 55.85, Z = 2, F = 96540 Specific resistance = 2Ω-cm Voltage = 12V Inter electrode gap = 0.2 mm Resistance

Production Technology

16. (i) Ans: (a) , (ii) Ans: (c) Sol: D = 12mm, t = 50mm, R = 40 ,

C = 20 F, Vs = 220V, Vd = 110V  Vs  Cycle time = R.C ln    tc  Vs  Vd   220  = 40 2010–6 × ln    110 

 554  10 6 sec  0.55 milli sec Average power input = W E =  t c

Sp. Resis tance  Inter electrode gap R Suface area



2  10  0.2  0.01 20  20

I

V 12   1200A R 0.01 AI 55.85  1200  ZF 2  96540  0.3471 g / sec

  

 0.5  CVd 2  tc 

  

= 218 W = 0.218 kW 17. Ans: (c) Sol:

MRR 

13. Ans: (b) Sol: Given w = 1 + (2  0.5) = 2 t =5, f = 20 mm/rev MRR = wtf = 2.5.20 = 200 mm/min 14. Ans: (d) 15. Ans: (a) Sol: As the thermal conductivity of tool material is high the heat dissipation from the tool is taking place and if the specific heat is high, it needs large amount of heat for raising the temps of tool material up to MP. ACE Engineering Academy

A

B

If D = Dmin = 59.9 X1 = distance between center of shaft and 59.9 corner of V – block  2  34.583 sin 60 60.1 X 2  2  34.698 sin 60 Error in depth = 2(X2 – X1 ) = 0.223 mm

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: 48 :

18. Sol: Resolving the force “F” into Horizontal F Sin  100

………. (1)

ME – GATE_ Postal Coaching Solutions

20. Sol: (a) Fixed rectangular block and marble V – clamp.

F Cos   100  100  200 …… (2) (1) 100  Tan  (2) 200

O1 O2 · ·

1   Tan 1    26.565 2 F

100  223.6 kg Sin 

Clamping

30

Taking the moments about vertical axis xF Cos   100  30  100  30  100  20 x = 10 mm.

30.025

Positional error = 30.025 – 30 = 0.025 (b) Fixed V – block and movable rectangular block

19. Sol: P

X2

X1 O2

O1

Clamping

30 30.025

A

Q

O1

O2 4

O1 O 2  4  3 2

2

=5

3 A

O1O 2  5  x 2  x 2 x = 3.5 Block of uniform thickness is preferable because of balanced condition. ACE Engineering Academy

x1 

30  34.64 Sin 60

x2 

30.025  34.66 Sin 60

Positional error = x2 – x1 = 0.0298mm The positional error is mainly depends on the fixed element. So when fixed V – block and marble V – block is used, the positional error is remains same as (b). Out of the 3 cases, case (a) is giving lower positional error, hence preferable.

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IM & OR Solutions for Vol – I _ Classroom Practice Questions 14. Ans: (b)

Chapter- 01 PERT & CPM 01. Ans: (a)

02. Ans: (a)

15. Ans: (c) Sol: D = 36 days, V = 4 days Z

36  36 0 4

03. Ans: (a)

04. Ans: (a)

05. Ans: (c)

06. Ans: (c)

07. Ans: (b)

08. Ans: (b)

16. Ans: (c)

09. Ans: (b)

10. Ans: (a)

Sol: cp  Va  b  Vb  c  Vc  d  Vd  e

 P(z) = 50%

 4  16  4  1  5

11. Ans: (b) 12. Ans: (b) Sol: T0 = 8 min, Tm = 10, Tp = 14min, To  4Tm  Tp

Te 

6



8  4  10  14 62  = 10.33 min 6 6

13. Ans: 70% (check the correct answer) Sol: Take 4 – 3 , Te = 6 days Critical path = 1-2-4-3 = 5 + 14 + 6 = 25 days

critical path  V1 2  V2  4  V4  3

17. Ans: (b) 18. Ans: (c) Sol: The earliest expected completion time, Critical path : A-B-C-D-F-E-H

 5 + 4 + 8 + 5 + 8 = 30 days 19. Ans: (d) Sol: Critical path : 1-3-4-6 = 20 days z

24  20 4  2 2 4

P(z) = 97.7%

 2  2.8  2 = 3.979 2

z

2

2

Due date  critical path duration critical path

z

27  25  0.5026 3.979

 P(z) = 70 % ACE Engineering Academy

20. Ans: (d) 2

2

 t  t   22  10  Sol: Variance =  p o     4  6   6  21. Ans: (a)

22. Ans: (b)

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: 52 :

ME – GATE _ Postal Coaching Solutions

23. Ans: Sol:

2 4

C(7) 1

11 18

B(8) A(4)

0 0

4 4

1

H(8)

3

2

12 12

E(9)

D(6)

G(12)

5

6

7

33 33

41 41

21 21

Path duration 1-2-4-6-7 = 4 + 7 + 15 + 8 = 34 1-2-3-5-6-7 = 4 + 8 + 9 + 12 + 8 = 41 (days) 1-2-5-6-7 = 4 + 6 + 12 + 8 = 30 2 4 4

C(7)

C(12)

A(10)

F(15)

D(9)

3

B(5)

4

Critical path : 1-2-3-4 = 10 + 12 + 9 = 31 days

cp  V1 2  V2  3  V3 4 2

2

 2 5 5     12        3 3  3

2

 7

25. Ans: (a)

4 11 18

26. Ans: (b)

TF+7

TF + 7 = 18 – 4

27. Ans: (c) Sol:

 TF = 14 – 7 = 7 24. Ans: Sol:

28. Ans: (c) Activity

Time estimated

Te 

To  4Tm  Tp 6

Standard deviation 

Tp  To 6

5  4  10  15  10 6

15  5 5  6 3

B

2  45  8 5 6

82 1 6

C

10  4  12  14  12 6

14  10 2  6 3

D

6  4  8  16 9 6

16  6 5  6 3

A

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29. Ans: (b) 30. Ans: (d) 1

T K

2

T K

3

T K

4

T

5

K

Given each activity having time mean duration ‘T’ and standard deviation ‘K’. Total time estimate Te = 4T Variance of the path (var)CP = R2 + R2 + R2 + R2 = 4R2

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: 53 :

Standard deviation of CP=  (var) CP

IM & OR

33. Ans: (c) 34. Ans: (c) Sol:

20

8

12 Min=Te3CP = 4T 6K

CP =

5

Max=Te+3CP = 4T+6K

8

7

14

22 27

32 34

EST

4K 2

CP = 2K

6

9 15

Range of overall project duration likely to be in 4T + 6K and 4T – 6K

12 LFT

i.e., 4T  6K

Total Float)6-1 = TF)6-7 = 27 9 12 = 6 Free float)6=7 = 28 9 12 = 1

Common data Questions for Q.31 & Q.32 31. Ans: (b) 32. Ans: (b) Sol:

35. Sol:

Paths

duration

Paths

Duration

AD

22

1-2-4-5 = (AEF)

8+9+6=23

ACE

41CP

1-2-3-4-5=(ADF)

8+9+6=23

BE

20

1-3-4-5 (BDF)

6+9+6 = 21

1-4-5 (CF)

16+6=22

Highest time taken paths are AEF and

ADF Critical path’s are AEF and ADF

Critical paths are ‘2’. Possible cases to crash A by 1 day that cost = 80 F by 1 day that cost = 130 E and D by 1 day that cost = 20 + 40 = 60 ACE Engineering Academy

7

9

2 A 10 1

D 12

C 16 B 5

3

E 15

4

Var cp  Var A  Var C  Var E   2A   C2   2E  2 2  2 2  12 = 4 + 4 + 1= 9  CP 

Var CP

 9=3

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: 54 :

Minimum completion time = 32 days Maximum completion time = 50 days

ME – GATE _ Postal Coaching Solutions

38. Sol:

6

5 Normal distribution curve

5 1

Te  3 cp Te  41

36. Sol: A 2

5

3

F 3

4

Paths

Duration

1-2-3-6-7-8

25

1-2-3-5-7-8

28

1-2-4-5-7-8

26

No. of day’s can crash

Extra cost/cost saved

1-2

43 = 1

250/day

2-3

5 3 = 2

500/day

3-5

84=4

50/day

for

2+1=3

5-7

75=2

300/day

1-2-4-5-6 AEF

2+3+2 = 7

7-8

42=2

400/day

1-3-6 BD

4+2 = 6

1-3-4-5-6  BEF

4+3+2 = 9

Among all the option the minimum cost slope option is 3-5, which can be reduced by 4 days, at a cost of 50/day

 CP is BEF 37. Sol:

F 3 B

1

Normal duration

1-2-6  AC

Highest Duration is ‘9’.

A

7

Paths

Possible activities crashing

2

4 3

4

7

“Crashing on critical path”

6

D

B

E

4

D

2

ACE Engineering Academy

6

G

5

7

L

I 8

The difference between longest path and next longest path is the maximum duration we can do crashing. Only if the duration is available in the activity taken for crashing.

K

J

H

8

 1-2-3-4-5-7-8 is the critical path

E

4

4

5

4

=50

C 1

4

7 8

4

2 2

3

2

Te  3 cp

=32

(EFT) 1 (LFT)

4

7

 The Critical path can be crashed for ‘2’

9 M 10

days only N

11

 Crash Cost = 2  50 = 100

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: 55 : 39. Sol: Activity

Cost slope



Crashing possibility (NT – NC)

CC  N C (Rs/week) N T  CT

1-2

150

1 week

2-3

-

-

2-4

50

2 week

2-5

-

-

3-4

30

3

4-6

40

1

5-6

25

2

IM & OR

From the option crash 3-4 by 2weeks by crashing 3-4 by 2 weeks the project duration becomes 11 weeks. Crashing cost = 2  30 = Rs. 60 Net savings by means of crashing = 2100 – 60 = Rs. 140

1

1

2

3

3

4

5

6

3

Path

Duration

1-2-4-6

11

1-2-3-4-6

11

1-2-5-6

10

4 4

5 4

3

6

3

5

Path

Duration

1-2-3-4-6

13

Critical path

1-2-4-6

11

Sub-critical path

1-2-5-6

10

Crashing possibility from the network = critical path duration – sub critical path = 13 – 11 = 2 weeks To reduce the project duration by 2 weeks Option Crash cost Crashing possibility 1-2

150

1 week

2-3

-

-

3-4

30

3 week

4-6

40

1 week

ACE Engineering Academy

2

5

Network diagram 3

3

2

4

Indirect cost = 100/week

3

3

3

Crashing possibility from the network = 11 – 10 = 1 week To reduce project duration by 1 week Option

Cost slope

Crashing possibility

1-2

150

1 week

4-6

40

1 week

3-4 & 2-4

30+50 = 80

1 week

Among the best option, crash 4-6 by 1 week, the project duration will become 10 weeks Crashing cost = 140 = 40 Net savings by crashing (4-6) = 100 – 40 = 60 3

3 1

3

2

2 4

5 4 5

2

6

3

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: 56 :

Path

41. Sol:

Duration

1-2-3-4-6 10 1-2-4-6

10

1-2-5-6

10

ME – GATE _ Postal Coaching Solutions

a d

Option

Cost slope

1-2

150

3-4, 2-4 , 5-6

30+50+25 = 105

f

b

To reduce by project duration by 1 week

e c

As crashing cost is more than indirect cost/week = further crashing is not economical Optimum project duration = 10 weeks Total cost of the project (with crashing) =

42. Ans: Sol:

2

A(8)

E(10)

B(4)

1

4

C(6) 3

D(5)

direct cost + indirect cost/week  project duration + crashing cost

(a) critical path :

= 945 + 10010 + 302 + 401 = 2045 Total cost without crashing = 945 + 100  13 = 945 + 1300 = 2245 40. Ans: Sol:

2

A(4)

1

B(3)

3

E(1)

F(1)

4

5

G(3)

A-E

8+10 = 18

A-C-D

8+6+5 = 19

B-D

4+5 = 9

6

H(2)

Critical path : 1-2-3-4-5-6 = 4 + 2 + 1 + 0 + 2 = 9

1-2-4-6

= 4 + 4 + 3 = 11  CP

1-2-3-4-6 1-3-5-6

= 4 + 2 + 1 + 3 = 10 = 3+1+2=6

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Duration

(b) To reduce the project by 1 day the available option is crashing ‘C’ by 1 day

D(4)

C(2)

Path

Option

Crashing possibilities (NT – CT)

A

8–8=0

C

6–5=1

D

5–5=0

By crashing activity C we can reduce the project duration by 1 day.

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: 57 :

IM & OR

Network diagram E(10)

2

C(5)

A(8) B(4)

1

3

Path

Duration

A-E

8+10 = 18

Linear Programming

D(5)

A-C-D 8+6+5 = 19 B-D

Chapter- 02

4

4+5 = 9

Further crashing is not possible due to “AC–D” critical path.  Minimum duration of project = 19

01. Ans: (d) 04. Ans: (d)

02. Ans: (d) 05. Ans: (b)

03. Ans: (c) 06. Ans: (a)

07. Ans: (a) Sol: Zmax = x+2y, Subjected to

4y  4x  –1……… (1) 5x + y  10 ……….. (2) y  10 …………… (3) x and y are unrestricted in sign x y  1  1   1      4  4  x y  1 (2)   2  10 y (3)  1 10

(1) 

43. Ans: (c) Sol:

Path

Duration

AB

7+5=12

CD

6+6=12

EF

8+4=12

(2)

(1)

12

(3)

10

Three critical paths, number of activities to be Crashed are 3

8  y 6 4 2 10 8 6 4

2

2

4

6

8

10

12  x

2 4 6 8 10

Only one value gives max value, then solution is unique. ACE Engineering Academy

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: 58 : 08. Ans: (b) Sol: Zmax = 3x1+2x2

Subjected to 4x1+x2 60……… (1) 8x1+x290 ……… (2)

ME – GATE _ Postal Coaching Solutions

09. Ans: Sol: Let, x1 be the number of ash trays x2 be the number of tea trays Production to be maximized Z = 20x1 + 30 x2 From the table given, constrained are

2x1+5x280 …….. (3)

10x1 + 20x2  30000

x1, x2  0

15x1 + 5x2  30000

(1) 

Fixed daily cost =Rs. 45000

x1 x 2  1 15 60

(2)  (3) 

x2

x1 x  2 1 11.25 90 x1 x 2  1 40 16

6000 5000

15x1 + 5x2  30000

4000 3000 2000

90 80 70 60 50 40 30

x2

1500A 1000

(2)

0(0,0)

O

(1)

(3)

20 10 11 20

B

30

40 50 60 70 x1

1000

Z B(1800,600) points for feasible solutions are 4

ACE Engineering Academy

2000

3000

4000

5000

x1

From the graph, common feasible region is OABC O(0,0) , A(0,1500), C(2000,0) B would be obtained by solving the constraints. B(1800 , 600) A(0,1500)

From the above graph the No. of corner

C

10x1 + 20x2  30000

C(2000,0)

200+301500 = Rs.45000 201800+30600 = Rs.54000 202000+300 = Rs.40000

Zmax = Rs. 54000 at B Profit = Zmax – Fixed daily cost = 54000 – 45000 = Rs.9000

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: 59 :

IM & OR

Common feasible region is OABCDO O(0,0), A(0,5), D(7.5,0) B is point of intersection of lines

10. Sol: Zmax = 60x1 + 50x2

x1 + 2x2  40

s.t

6x + 9y  54 ,

3x1 + 2x2  60 x1 x 2 x1 x 2  1 ,  1 40 20 20 30

5x + 13y  65 Solving this B = (3.55 , 3.64) C is the point of intersection of the lines

x2

6x+ 9y  54 ,

40 30

(0,30)

20

(0,20)

10x + 7.5y  75 Solving these, C = (6,2) Graphically solving :

(10,15)

10

y 10

(Zmax)(10,15)

20

(20,0)

30

40

x1

(40,0)

= 6010+5015 = 1350 /-

11 9 7

11. Sol:

5

Type of Products machine A B P 10 7.5 Q 6 9 R 5 13

Total time available

75 54 65

Profit for product, A = Rs. 60 per unit Profit for product, B = Rs. 70 per unit Let, x = number of A type products y = number of B type products  Maximization problem Zmax = 60x + 70y Constraints are, (in times) 10x + 7.5y  75 6x + 9y  54

10x + 7.5y  75

A B

3

6x + 9y  54 5x + 13y  65

C

1 0

1

3

Points A (0,5)

B (3.53,3.64)

5

7

D

9

11 13

x

Z=60x+70y

600+705 = 350 3.5560+703.64 = 464.8

C (6,2)

606+702 = 500

D (7.5,0)

7.560+070 = 450

O (0,0)

060+070 = 0

Zmax = 500 at C(6,2) A type products = 6 , B type products = 2

5x + 13y  65 ACE Engineering Academy

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: 60 : 12. Sol:

ME – GATE _ Postal Coaching Solutions

13. Sol:

Tables

Chairs

Availability

Wood

30

20

300

Labour

5

10

110

Profit/unit

8

6

x

y

Zmax = 8x + 6y Subject to x y   1 ----- (1) 10 15 x y 5x + 10y  110 ,   1 ------ (2) 22 11 x,y0 30x + 20y  300 ,

Demand

Products

Maximum available

Chairs (x1)

Tables (x2)

Wood

1

2

200

Chairs

1

-

150

Tables

-

1

80

Profit/loss

100

300

Zmax = 100x1 + 300x2 Subject to x1 + 2x2  200 x1  150 and x2  80

y 24

14. Sol:

20 16

Products Demand

12 B(0,11)

4 A(10,0) O(0,0)

A (x1)

B (x2)

Raw material

1

1

850

Special type of buckle

1

-

200

Ordinary buckle

-

1

700

Time

1

½

500

Profits/unit

10/- 5/-

C (4,9)

8

4

8

12

16

20

24

x

“C’ is the intersection of (1) and (2) Solve equation (1) & (2) for x,y We will get x = 4, y = 9 Z = 8x + 6y Z0 = 0 ZA = 810 + 60 = 80 ZB = 80 + 611 = 66 ZC = 84 + 69 = 86 Solution is optimal at (c) Zmax = 86 at x = 4 , y = 9 ACE Engineering Academy

Maximum available

Constraints : x1 = No. of belts of type ‘A’ x2 = No. of belts of type ‘B’ Zmax = 10x1 + 5x2 s.t x1 + x2  500 x1  500 , x1 +

1 x 2  500 , 2

x2  700 x1 , x2  0

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: 61 :

IM & OR

Zmax = 3x + 5y

x2

ZA = 3  1500 + 5  0 = 4500

1000

ZB = 3  0 + 5  600

= 3000

ZC = 3  1000 + 5  500 = 5500

700

ZD = 3  800 + 5  600 = 5400

600

400

Y 2000

300

1500 

500

200

1000 

100

100

200

300

400

500

600

Zmax = (100) + (5500) = 2500 15. Ans: (c) Sol: Let, P type toys produced = x , Q type toys produced = y

x1

D (800,600)   C (1000,500)

600  B(0,600)

O

500

1000

 1500 A(1500,0)

Q

Time

1

2

2000

Raw material

1

1

1500

Electric switch

-

1

600

Profit

3

5

x1 + 4x2  4 -------- (2)

x

y

x1 , x2  0

x y  1 2000 1000

x  y  1500

;

x y  1 1500 1500

y  600

;

y 1 600

ACE Engineering Academy

16. Ans: (c) Sol: Zmax = x1 +1.5 x2 Subject to

2x1+3x2  6 ------- (1)

x1 x 2  1 3 2

Zmax = 3x + 5y

x, y  0

X

C does not exist in answer. Hence, Zmax is at D, i.e., Zmax @ D = 5400

P

x  2 y  2000 ;

 2000

x1 x 2  1 4 1 Let, “c” in the intersection of (1) and (2) Solve (1) & (2) for ‘c’.

It follows, x 1 

12 2 ; x2  5 5

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: 62 :

ME – GATE _ Postal Coaching Solutions

But feasible region is ABCDEA (∵ x1 , x2 > 0) A(2,0) B(0,4) C(0,6) E(3,0) D can be obtained by solving

y 4 3

x1  3 & x1 + x2  6

2

 x1 = 3 and x2 = 3 and D (3,3)

 12 2  C  ,   5 5

B(0,1) 1 O(0,0)

1

3

2

A(3,0) 4

x

Zmax

Zmax = x1 +1.5x2 Z0 = 0 ZA = 3+ 1.5 0 = 3 ZB = 30+1.51 = 1.5 12 3 2   =3 5 2 5 Problem is having multiple solutions and it is Optimal at (A) and (C)

12

x1  3

11

2x1 + x2  4

10

8

Z=9

Z=6 6

3

02+16 = 6

E(3,0)

32+01 = 6

D(3,3)

32+13 = 9

19. Ans: (b)

20. Ans: (d)

21. Ans: (a) Sol: Zmax = 4 x1 + 6 x2 + x3 s.t

4 x1

6 x2

1 x3

0 s1

2 0 4

-1 0

3 0 1

1 0 0

6 B

D

2 1

G(0,6)

B0 5 0

min Ratio –5

C

5 4

18. Ans: (a)

cj  s v 0 s1 zj cj - zj

x1 , x2  0

Z=12

7

02+14 = 4

x1, x2, x3  0 2 x1 – x2 + 3x3 + s1 = 5 Zmax = 4x1 + 6x2 + x3 + 0 s1

Subjected x1 + x2  6

9

B (0,4)

2 x1 – x2 + 3x3  5

Zmax = 2x1 + x2

x2

22+10 = 4

Zmax = 9 at D (3,3)

ZC =

17. Ans: (a) Sol:

A(2,0)

Z=3 0

1

2

A

E 3

ACE Engineering Academy

4

5

6

7

8

9

10

x1

EV

Entering vector exists but leaving vector doesn’t exist as minimum ratio column is having negative values. It is a case of unbounded solution space and unbounded optimal solution to problem.

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: 63 :

IM & OR

22. Ans: (a) Sol: 10 7

Cj 

1

2

0

0

x

s

4

1

0

Common Data

S1

0

35 6

2 3

0

1

10 7

x1

1

1 3

1 3

0

0

1 3

 107 3

 107 3

0

0

107 3

110 3

113 3

0

0

107 3

s2

10

1350

0

Zmax = (2  100) + (5230) + (020) = 1350/-

Cj

23. Ans: (b) Sol: Solution is optimal; but Number of zeros are greater than the number of basic Variables in Cj – Zj(net evaluation row) hence multiple optimal solutions.

0

Cj  Zj row, hence the problem is optimal.

26. Ans: (d) Sol:

Minimum ratio column has all negative values, so can not decide outgoing variables. Problem has unbounded solutions.

2

As it contain 0’s and ‘ve’ values in

 10 3

0

27. Ans: (a)

28. Ans: (a)

6

4

0

0

0

M

B0

CB

SV

x1

x2

s1

s2

s3

A1

0

S1

0

1

x

14

S3

0

1

x

5

6

x1

1

2 3 1 3 1 3

0

0

5 3 1 3 2 3

0

x

8 48

0 0

Zj

6

4

0

2

0

x

Cj Zj

0

0

0

2

0

x

Min Ratio

As the No. of zeros greater than No. of basic

24. Ans: (b) Sol:

3

1 2

25. Ans: (a)

0

Cj

0

 15 2

1

7

0

5

4 9

10

4

 16 3

17 9

Cj Zj

Cj Zj

 21 4

0

10 7

1

7 3

x4

Zj

5

 14 9

0

x3

2

Min Ratio

x1

x2

7

B0

S V

CB

Zj

variables in CjZj row, hence it is a case of 2

5

0

0

0

x2 x3 s1

s2

s3

B0

CB

SV x1

2

x2

1 4

1

0

1 2

1 4

0

100

5

x3

3 2

0

1

0

1 2

0

230

0

S3

–2

0

0

2 1

1

20

ACE Engineering Academy

multiple solutions or alternate optimal solution exists. If Non basic variable x2 is having a zero evaluation at optimality, with that variable if we enter and performs simplex procedure in alternate solution to the problem is obtained as follows.

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: 64 : Cj

6

4

0

0

0

M

s3

A1

B0

Min Ratio

CB

SV

x1

x2

s1

s2

0

S1

0

1

2 0 3

x

14

14 5/3

0

S2

0

0

x

5

15

x1

1

1 3 1 3

1

6

5 3 1 3 2 3

0

x

8

12

48

0

Zj

6

4

0

2

0

x

Cj Zj

0

0

0

2

0

x

ME – GATE _ Postal Coaching Solutions

Common Data for Questions 33. Ans: (d) Sol: Given, Zmax = 5x1 + 10x2 + 8x3 Subjected to

3x1 + 5x2 + 2x3  60  material 4x1 + 4x2 + 4x3  72  Machine hours 2x1 + 4x2 + 5x3  100  labour hours x1, x2 , x3  0 3x1 + 5x2 + 2x3 + s1 = 60 4x1 + 4x2 + 4x3 + s2 = 73 2x1 + 4x2 + 5x3 + s3 = 100 Zmax = 5x1 + 10x2 + 8x3 + 0s1 + 0s2 + 0s3

Entering vector column

Minimum ratio having row will become entering vector row. Hence the alternate solution is 42 x2 = , x1 = 12 5 Entering vector x2 and leaving vector s1 because of s1 row had minimum ratio. 29. Ans: (c)

30. Ans: (c)

31. Ans: (a) 32. Ans: (c)

Cj  C

subjected to 5y1  10, y1  2 , Wmax = 100 3y1  5 , y1  5/3 , Wmax = 250/3 y1 , y2  0

1 0

8

0

0

B

10

x2

1 3

1

0

1 3

1 6

0

8

8

x3

2 3

0

1

1 3

5 12

0

10

0

S3

8 3

0

0

1 3

1

18

Zj

26 3

1 0

8

5 3

0

160

Cj Zj

11 3

0

0

5 3

0

Cj  Zj x2 Cj  Zj x3

s1

s2

3

11

0

0

11 2

0

0

s

B0

x1

x2

x

0

S V

Sol: Zmin = 10x1+x2+5x3+0S1

Dual , Wmin= 50y1

5

Min Ratio

3

2 3 2 3



2

2

17 12

10

0

LL=2 UL=1 0

4

0

LL=4 UL=2

102= 8 10+10 =20 84=4 8+2=1 0

 Zmax = 250 / 3

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: 65 :

In Cj Zj row all elements are negatives or zeros, hence the solution is optimal and unique.. Basic variables are: x2 = 8 , x3 = 10 , s3 = 18 i.e., production of B = 8 units C = 10 units 18 labours hours remained unutilized Non Basic variable x1 = 0, s1 = 0, s2 = 0 Resource materials and resource machine hours are fully utilized.

IM & OR

If 1 m/c hr decreases, production B increases 1 5 and production C decreases by 6 12 If 1 unit of A produces, contribution

by

decreases by

11 , production B decreases by 3

1 2 , production C decreases by . 3 3 34. Ans: (a) Sol: If 3 kg material increases, contribution

increases by 3 

In (Cj Zj) row at optimality, the values under s1, s2 and s3 columns represents the shadow prices. So, If 1 kg material increases, contribution 2 . 3 If 1 kg material decreases, contribution

increases by

2 . 3 If 1 kg material increases, then production B 1 and production C decreases 3

1 3 If 1 m/c hr increases, contribution increases by 5/3. If 1 m/c hr decreases, contribution decreases by

5 by 3 If 1 m/c hr increases, production B decreases by

1 5 and production increases by . 6 12

ACE Engineering Academy

= 2 Rs

35. Ans: (a) Sol: Present profit = 160  160 

5  12 = 140/3

36. Ans: (b) Sol: New production of B

1 1   = 8 – 12   = 8 + 12   6  6  

decreases by

increases by

2 3

8 + 2 = 10 units 37. Ans: (c)

 1  1 Sol: = 10 +  3   = 10   3   = 10 1 = 9 3    3 38. Ans: (a) Sol: If 1 unit of A produces, contribution

decreases by

11 3

39. Ans: (a)

 11  Sol: 160   6   = 138 3 

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: 66 :

If RHS value of 2nd constraint increases by 1 unit then

40. Ans: (a) Sol: Production of B , 3 

1 =1 3

Production of C, 3 

2 =2 3

ME – GATE _ Postal Coaching Solutions

s2

Common data 41 & 42 41. Ans: (b) , 42.Ans: (b) Sol: Basic variables x1 = 20 , x2 = 10 Non-basic variables

z-row

2

x1

-1

x2

2

From the table z increases by 2 units, x1 decreases by 1 unit x2 decreases by 2 units,

s1 = 0  first constraint is fully consumed. s2 = 0  second constraint is fully consumed.

If RHS value of 1st constraint decreases by 10 units then z decreases by 10 units,

x3 = 0 (unwanted variable) x1

x2

x3

s1

s2

RHS

z-row

0

0

2

1

2

110

x1

1

0

1

1

-1

20

x2

0

0

0

–1

2

10

The new objective value , Zmax = 110- 10 = 100

s1 z-row

1

x1

1

x2

–1

If RHS value of 1st constraint increases by 1 unit then From the table z increases by 1 unit, x1 increases by 1 unit, x2 decreases by 1 unit,

ACE Engineering Academy

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: 67 :

IM & OR

09.

Chapter- 03 Forecasting 01. Ans: (d)

02. Ans: (a)

Sol: At ,  = 0.2

Fmay = 100 + 0.2 ( 200 -100) = 120 Fjune = 120 + 0.2 ( 50 – 120) = 106 Fjuly = 106 + 0.2 ( 150- 106) = 114.8 Time Demand Forecast

04. Ans: (a)

100  99  101 3 = 100 4 period moving average

Sol: 3 period moving avg =

102  100  99  101 = 100.5 4 5 period moving average =

99  102  100  99  101 = 5 = 100.2 Arithmetic mean =

101  99  102  100  99  101 = 100.33 6

05. Ans: (a) Sol: Dt = 100 units , Ft = 105 units

 = 0.2 Ft+1 = 105 + 0.2 (100 – 105) = 104



April

200

100

May

50

120

June

150

106

July

-

114.8

2 n 1

n 1 

2 2  1  9 period  n  0.2

10. Sol: In, Jun, July, Aug, Sep demand is Stable

In Oct, Nov, Dec – demand is Fluctuating FJan =

327  339  355 = 340.33 units. 3

Last ‘3’ months average is forecast for next month The inflation start only from October hence considering last 3 months data was highly

06.

Ans: (c)

significant

Sol: Dt = 105 , Ft = 97,  = 0.4

Ft+1 = 97 + 0.4 (105 – 97) = 100.2 07. Ans: (c) Sol: Ft+1 = Ft + a (Xt - Ft) 08.

Simple exponential  = 0.1 FJan = FDec + (DDec  FDec) = 307 + 0.1(355307) = 311.8

Ans: (c)

ACE Engineering Academy

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: 68 : 11. Sol: Simple exponential method

 = 0.2 , FJan = 175,

ME – GATE _ Postal Coaching Solutions

13. Sol:

DJan = 200 DFeb = 170

Month Demand Forecast

Ffeb  FJan  D Jan  FJan  = 175+0.2 (200175) = 180 Fmarch = FFeb + (DFeb  FFeb) = 180 + 0.2(170 180) = 178 12. Sol: Linear Regression model:

24

78

25

65

26

90

27

71

28

80

29

101

30

84

31

60

2

73

(x)

y (Rs)

xy

x

32

1

450

450

1

33

2

550

1110

4

3

625

1875

9

4

650

2600

16

5.

750

3750

25

6.

775

4650

36

x=21

y=3800

xy=14450

|x2=91|

=

Forecast For month-8 y8 = 408.3+64.28(8) = 952.5 ACE Engineering Academy

2  0.2 9 1 + ……….+ (1-)n (Dt-n)

= 0.273 + 0.20.860 + 0.2(0.8)284 + 0.2(0.8)3 101 + 0.2(0.8)480 + 0.2(0.8)571

+

0.2(0.8)690

7

xy = ax + bx2  xy = ax + bx2

Forecast for month – 7, y7 = 408.3+ 64.28(7) = 858.26

74

F32 = (Dt) +  (1-) (Dt – 1) +  (1-)2 (Dt-2)

y = na + bx  y = na + bx 3800 = 6a + 21b …….. (1) 14425 = 21a + 91b…….. (2) Now, solve (1) and (2) for a, b a = 408.3, b = 64.28 Forecast equ. yc = a + bx yc = 408.3+64.28x

66.563

8

0.2(0.8)  65 + 0.2(0.8) 78 = 66.563 F32+1 = F32 +  (D32 – F32) = 66.563 + 0.2 (73 – 66.563) = 74 14. Sol: n = 20,

(y y )2 = 2800

x = 80,

y = 1200,

x2 = 340,

y2 = 74,800

xy = 5000 y = a + bx

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+

: 69 :

 y = na + bx

17. Ans: (a) Sol:

1200 = 20a + b(80)….. (1) xy = ax + bx2  xy = ax + bx2 5000 = a(80) + b(340)……(2) Solve (1) and (2) for a, b a = 20, b = 10 Standard error Syx =

 (y  y

r=

C

)2

n2

800  = 6.67 20  2

EV UEV  1 UEV TV

 1

IM & OR

Period Di Fi

|(DiFi)|

1

10 9.8

0.2

2

13 12.9 0.3

3

15 15.6 0.6

4

18 18.5 0.5

5

22 21.4 0.6 |DiFi| = 2.2

18. Sol: Deviation = Di – Fi

(y  yc )2 800  1 = 0.84 2 ( y  y) 2800

MAD =

As ‘r’ closer to 1 i.e., good correlation 15. Ans: (a)

=

7.5  18  0  28.  12 6 70 = 11.66 6

Tracking signal =

16. Ans: (b) Sol:

=

Period

Di

Fi

(Di Fi)2

14

100

75

625

15

100

87.5

156.25

16.

100

93.75

39.0625

(DiFi) =820.31 F15 = F14 + (D14 F14)

19. Ans: (c)

= 75 + 0.5(100 75) = 87.5 F16 = F15 + (D15 F15) = 87.5 + 0.5(100 87.5) = 93.75 Mean square error (MSE) = = ACE Engineering Academy

 24 = 2.05<4 11.66

If tracking signal < 4 – No significant deviation in data If tracking signal > 4 – significant deviation in data

2

 (D

Cumulative deviation MAD

i

 Fi ) 2

20. Ans: (d) 21. Ans: (d)

n

820.31 = 273.13 3 Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally

: 70 :

ME – GATE _ Postal Coaching Solutions

EJ - EARLY JOB , OS - ON SCHEDULE TJ - TARDY JOB

Chapter- 04 Sequencing & Scheduling

Minimum total cost = 57  60 = 3,420 01. Ans: (a) Sol: SPT rule Job

Number of jobs which fail to meet due date are 2.

Process time Completion time (days)

1

4

4

3

5

9

5

6

15

6

8

23

2

9

32

4

10

42

Ci

125

Average Flow Time = =

Job Ti 2 5 2 2 3 1 4 4 4 6 9 3

4 2 1 5 3

2 3 5 6 8

Ci

2 5 10 16 24  Ci = 57

ACE Engineering Academy

Di 15 21 17 12 24 5

 Ci = 63

125 = 20.83 6

03. Sol: SPT rule is used for minimizing mean flow time ti

Ci 2 4 7 11 15 24

 Ci n

02. Ans: (a) Sol: According to SPT rule total inventory cost is minimum.

Job

04. Sol: SPT – rule minimizes average flow time EDD – rule minimizes mean tardiness

di

Ci – di

9 12 10 8 20

-7 -7 -8 4

EJ EJ OS TJ TJ

Job 3 4 5 1 2 6

Ti 9 4 2 3 2 4

Ci 9 13 15 18 20 24  Ci = 99

MFT =

63  10.5; 6

MT =

19  3.17 6

MFT=

Ci – Di -13 -17 -10 -1 -9 19  Ci – Di = 49

Di Ci – Di 4 5 1 12 -15 1 17 -1 21 0 24  Ci – Di = 6

 C i 99   16.5 n 6

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: 71 :

IM & OR

 Ci  Di 6  1 n 6 Ti = Process Time Ci = Completion Time Di = Due Date MFT = Mean Flow Time MT = Mean Tardiness

C

MT =

D

SPT

LPT

E F

STACK

STACK

A

F

C

A

110=9

A

A

B

F

A

F

B

9  7=2

E

F

C

E

E

E

D

7  2 =5

F

E

D

C

C

D

E

7  6 =1

D

D

E

D

D

B

F

2  5=3

B

B

F

B

B

A

C

1  4=3

C

C

178

2

Dj Lj Tj = max of (0, Lj) 9

a

8

8

-1

0

b

7

15 18 -3

0

c

9

24 21

3

3

d

12 36 38 -2

0

e

14 50 41

9

9

f

10 60 60

0

0

 Fy

193  32.16 6 n (iii) No. of tardy jobs = 2 (c & e)

Critical ratio

(190175)/5=3Ahead of (178175)/2 = 1.5Ahead of schedule

ACE Engineering Academy

Fj

(ii) Mean flow time =

schedule B

(184175)/9 = 1on schedule

9

(i) Make-span time = 60 days

CDR  Todays date = PTR 5

(187175)/15 = 0.8 Behind

15

184

Job Tj

06. Ans: F-C-G-B-E-D-A Sol: Calendar date required (CDR) Processing time (PT) Process time remained (PTR)

190

187

07. Sol:

Note: Stack=Due Date (DD) – Processing time (P.T)

A

(205175)/17 = 1.76Ahead

17

If critical ratio is one job will be on schedule. If critical ratio is less than one job will be behind schedule. If critical ratio is greater than one job will be ahead of schedule.

(or)

PT

205

schedule

A

CDR

(181175)/3 =2  Ahead of

3

of schedule

(or)

Job

181

schedule

05. Sol: EDD

(184175)10 = 0.9 Behind

10

schedule

G

FCFS

184

(iv) Mean tardiness, T  08.



T j n



12 2 6

Ans: (d)

Sol: EDD rule can minimize maximum lateness.

The job sequence is

R–P–Q–S

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: 72 : 09.

ME – GATE _ Postal Coaching Solutions

Ans: (d)

Sol: Johnson’s rule :

Optimum job sequence III I IV II

13. Sol:

M

N

78

78

9

87

U

78

16

94

94

7

101

P

94

15

109 109

6

115

6 3 4 1 2 5

Idle

III

0

1

1

1

2

3

-

I

1

3

4

4

6

10

1

IV

4

7

11

11

5

16

1

II

11

5

16

16

2

18

-

Total idle time on machine (N) = 3

Sequence by Johnson’s Rule is: Job

In PT Out In PT Out

DENTER Tin Tout 0 1 1 3 3 8 8 12 12 22 22 28

Minimum Make Span = 30 14. Sol:

Optimum sequence:

TABULAR METHOD:

2 3 1 4

Job

12. Ans: (b) Sol: Optimum sequence is

R T S Q U P M1

PAINTER Tin Tout 1 7 7 12 12 16 16 19 22 24 28 30

A B C E D

11. Ans: (a) Sol: Optimum sequence of jobs

M2

In PT Out

In

PT Out

R

0

8

8

8

13

21

T

8

11

19

21

14

35

S

19

27

46

46

20

66

ACE Engineering Academy

32

6, 3, 4, 1, 2, 5

10. Ans: (b) Sol:

Job

46

The optimal make-span time = 115 days

Do the job 1st if the minimum time happens to be on the machine (M) and do it on the end if .it is on second machine (N). Select either in case of a tie.

Job

Q

M/C -I

M/C - II

Ti

T0

Ti

T0

A

0

2

2

12

B

2

5

12

20

C

5

12

20

25

E

12

18

25

29

D

18

27

29

30

Processing time

27

28

Idle time

3027=3

(3028=2)

%utilization

27  100 30

28  100 30

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: 73 :

A 37

GANTT CHART 2

0

5

12 C7

E6

M/C II Idle-2

A10

B8

0

Idle3 E4

C5 20

25

29

52

-

4  4 or 4 Comp X M W

PT = processing time Machine – 1

Machine – 2

Idle In PT Out In PT Out Time

A

0

2

2

2

4

6

-

C

2

5

7

7

6

13

1

D

7

6

13

13

7

20

-

B

13

7

20

20

8

28

-

E

20

5

25

28

3

31

-

Minimum time for completion of all jobs = 31 16. Sol: Optimum Sequence :

Machines

N

8

3

5

A

4

4

6

O

7

3

7

L

5

4

8

E

6

4

4

Since the condition is satisfied, we can create two virtual Machines ‘G’ & ‘H’. X = t1j , M = t2j , W = t3j Comp Machine G (X+M) N 11 A 8 O 10 L 9 E 10

Machine H (M+W) 8 10 10 12 8

Optimum sequence A L O N E

D C E F G B A Polish

Idle

In

PT

Out

In

PT

Out

D

0

4

4

4

5

9

4

C

4

5

9

9

12

21

-

E

9

6

15

21

9

30

-

F

15

9

24

30

11

41

-

G 24

7

31

41

6

47

-

B

6

37

47

3

50

-

ACE Engineering Academy

2

Sol: Condition : Max (t2j)  Min (tij or t3j)

30

A C D B E

31

50

17.

D1

15. Sol: Optimum Sequence :

Job

47

30

27 D9

12

10

Minimum flow time = 52

18

M/C I A2 B3

2

IM & OR

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: 74 :

Comp

Machine X

ME – GATE _ Postal Coaching Solutions

Machine M

Idle

In

PT

Out

In

PT

Out

A

0

4

4

4

4

8

L

4

5

9

9

4

O

9

7

16

16

N

16

8

24

E

24

6

30

Machine W

Idle

In

PT

Out

4

8

6

14

8

13

1

14

8

22

-

3

19

3

22

7

29

-

24

3

27

5

29

5

34

-

30

4

34

3

34

4

38

-

Gantt Chart : 8

14 A

W

4

8 9 A

M 4 X

A

13 16 19

10

24 N

15

20

E

27 30 N

16 O

34 N

24

O

9

5

O

L

L

L

29

22

34

38

E 30

38

E 25

38

30

35

40

(iii) % utilization :

Machine X 

30 100  78.94% 38

Machine m 

38  20 100  47.73% 38

Machine W 

38  8 100  78.94% 38

ACE Engineering Academy

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: 75 :

IM & OR

18. Sol: The given machine sequence is ‘ACB’ hence, we need to re-arrange the given data

Job

A

C

B

1 2 3 4 5

5 7 6 9 5

2 1 4 5 3

3 7 5 6 7

Machine G 5

5  5 or 3

2

Jon

Machine G (A+C)

Machine H (C+B)

1 2 3 4 5

7 8 10 14 8

5 8 9 11 10

A

B

In

PT Out In

0 5 14 20 27

5 9 6 7 5

5 14 20 27 32

1

Optimum sequence 2

Max { t2j}  min{t1j or t3j}

5 4 3 2 1

Machine H 4 3 2

Machine G

Job

20.

Optimum sequence 1

5 14 20 27 32

8 19 24 28 34

5 4 3

1

C

PT Out Idle In

3 5 4 1 2

Machine H

5 6 1 3 4

8 19 25 30 37

PT Out Idle

7 6 5 7 3

15 25 30 37 40

8 4 – – –

Ans: (c)

ACE Engineering Academy

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: 76 :

ME – GATE _ Postal Coaching Solutions

06. Ans: (d)

Chapter- 05 Queuing Theory 01.

Ans: (b)

Sol:

 = 3 per day

Sol:  =

1 = 0.25min1 4

=

1 = 0.33 min1 3

=

 0.25 = 0.75   0.33

 = 6 per day 1 1 1   day  63 3

Ws 

07. Ans: (b)

02. Ans: (c) Sol:  = 0.35 min-1,

    Pn  1        

Sol:  

1  0.1 min 1 10



1  0.25 min 1 4

 = 0.5 min-1 n

System busy  ()  8

 0.35   0.35  = 1   0.0173 0.5   0.5  

08. Ans: (c) Sol:  = 4hr1, = 6 hr1

03. Ans: (a) Sol:  = 10hr1,

 P(QS  2) =   

 = 15 hr1 Lq =

2 10 2 = = 1.33     1515  10 

04. Ans: (b)

60 = 5 hr-1 Sol:  = 4 hr ,  = 12

2

09.

Ans: (a)

Sol:

Lq 

2    

4

2    

2 42 16    3 .2     55  4  5

05.

Ans: (b)

Sol:

Lq 

 4 (–) = 2

2     

ACE Engineering Academy

2

4 4 =   6 9

-1

Lq =

 0.1   0.4  0.25

2    1     2



2 1  

10.

Ans: (c)

11. Ans: (c)

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: 77 :

b) No customer  service facility idle

12. Ans: Sol:  = 8 hr–1 ;  =

Ws 

60  12hr 1 5

P0 = 1  = 1  0.4 = 0.6 (c) & (d)

1 1 1      12  8 4

Customer being served  No one waiting         P0 + P1 = 1    1           

13. –1

Sol:  = 100 h



     = 1  1   = 1      

–1

 = 120 h

;

 100 10    120 12

= 1 – = 1 

   

2

= 1  0.16 = 0.84

P0 (no customer in the system)

2    

e) Lq =

2 1 10   12 10 5

=

14.

22 = 0.266 5(5  2)

As  >   Lq is finite

Sol:  = 8 h-1

=

IM & OR

If  =   Lq is infinite

60 1 h = 12 h-1 5

17. Ans: (c)

(a) L q 

8  1.33       12  4

(b) L s 

8   2     12  8

 = 3 hr

 = 6 hr1

 = 4 hr1

(c) Wq 

 8   0.1666     12  4

NPC/hr = 15 Rs LC/hr = 20

NPC/hr = 15 LC/hr = 12

(d) Ws 

1 1   0.25     4

LS represents non productive machining

2

2

 8 (e)     0.666  12

18. Sol:

A

LS = =

16. Sol:  = 2 hr1,  = 5 hr1

 2 a) Traffic intensity () =  = 0.4  5 ACE Engineering Academy

B 1

  3 =1 63

 = 3 hr1

LS = =

  3 =3 m/c 43

NPC/hr =115Rs NPc/hr =315 = 45 Rs LC/hr = 20/“A” should be hired

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: 78 :

ME – GATE _ Postal Coaching Solutions kd hQ1 kd hQ 2    Q1 2 Q2 2

Chapter- 06 Inventory Control 01.

Ans: (b)

Sol: EOQ 

2kd  Q1 Q 2 h

2AS CI

EOQ1  2 

Q *2  Q1  Q 2

2AS CI

Q*  Q1  Q 2  300  600  424.264

EOQ1  2  EOQ 02.

 Q  Q1  h   Q 2  Q1  kd 2  Q1 Q 2  2

Ans: (c)



Sol: (No of orders =

 2  800  50  2 = 400

45 Days

45 Days

06. Ans: (d)

2AS CI

08. Ans: (c) Sol: TC(Q1) = TC(Q2)

A 12 months 12    8) Q 45 days 1.5

Q =100

07. Ans: (a) Sol: A = 800 , S = 50/- , Cs = 2 per unit = CI

ACE Engineering Academy

EOQ1  CI   2AS        EOQ 2  CI  A  2AS  B

2AS CI

05. Ans: (b)

TICEOQ 

Sol:

11. Ans: (d)

2  900  100 = 300 2

04. Ans: (c)

Ans: (c)

(EOQ)A : (EOQ)B = 1:4

03. Ans: (b) Sol: A = 900 unit S = 100 per order CI = 2 per unit per year EOQ  ELS 

10.

T VC 

A Q S  CI. Q 2 = 8  100 

100  120 = Rs. 6800 2

12. Ans: (b) Sol: Average inventory

Q 6000  = 3000 per year 2 2 = 250 per month 

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: 79 : 13.

IM & OR

TAC 83.38  1000  500  1000  40  83.33  500  0.1

Ans: (c)

83.33

Sol: A = B  true demand

 5,02,563 / 

By mistake 40% higher demand A1 = 1.4B

In percentage =

Economic lost size , Q1 =

2BS CI



Economic lost size , Q2 =

2  1.4  BS CI

= 1.183 Q1 Actual rise = 1.183 Q1 – Q1 = 0.183 = 18.3% 14. Sol: Given, A = 1000 units/year, S = 40/I = 0.1, C = 500/2  1000  40 2AS   40 units 500  0.1 CI

a)

EOQ 

b)

A 1000 No. of annual orders =   25 Q 40

TACQ  100 TACEOQ 5,02,563  100  100.11% 5,02,000

15. Ans: (b) Sol: P = 1000 , r = 500 , Q = 1000

I max 

1000 1000  500 = 500 1000

16. Sol: Simultaneous consumption producing Model A = 15,000 units, C.I = 5/ units/year

S = 25/,

P = 100 units/day

No. of working days = 250/year 15,000 250  60 units / day

Consumption rate = r =

Frequency of ordering (or) time between orders

c)

TACEOQ



Q  time period A

EBQ = EPQ = ELS



40 360  360   14.6 days 1000 25

EPQ 

 AC  2ACSI

 1000  500  2  1000  500  40  0.1 = 5,02,000/Order per month 

TACQ

 AC 

1000 = 83.33units. 12

A Q .S  .CI Q 2

2

Q

2AS  P    CI  P  r  2  15000  25  100    5  100  60 

Q = 612.37 units

TVC EPQ

 2CSI

P  r  P

 100  60   15000  2  5  25   100 

= 1225/ACE Engineering Academy

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: 80 :

Q = tp . P tp  I max

Q 612.37  = 6.1237 days P 100

a)

EOQ  

Q  t p P  r   .P  r  P

 1000020   210000 4200 

A Q



15000  25 612.37

= 2,04,000/b)

TACQ

A Q S  .CI Q 2 2000

S = Rs. 1080/A = 20,00012 = 240000 units 2  240,000  1080 3 .6

2AS  CI

= 12000 units (b)

no. of production run =

(c)

Time interval =

192000 = 9 runs 20000

365  20  9  9 1

= 23.125 days  24 days

2

= 2,05,000/-

Sol: CI = Rs. 0.3/month = 0.312 = Rs 3.6/-

EOQ 

 AC 

TAC Q  2000  10000 20  10000 200  2000 4

17. Ans:

(a)

2  10000  200 = 1000 units. 4

(TAC)EOQ = AC  2ACSI

100

No of production runs 

2AS CI

Total annual cost at EOQ,

100  60  245

 612.77 

ME – GATE _ Postal Coaching Solutions

For 2000 orders to be economical the total annual cost for 2000 order with r% discount must be less than TAC at EOQ

TAC2000  TACEOQ r% TAC 2000  TAC EOQ r%

r  A Q  r   AC 1   S  CI1    2  100   100  Q = (TAC)EOQ r  10000 2000 2  r   AC1  4 1    200     2  100  2000  100 

 2,04,000

(d)

Total variable cost = =

2ASCI 2  240000  1080  3.6

= Rs. 43200/18. Sol: A = 10,000 units CI = 4/unit/year

ACE Engineering Academy

r   1   [2,04,000] = 2,03,000  100  r  2,03,000 203   1    100  2,04,000 204 r = 0.49%

S = 200/order C = 20/-

Preferred discount rate is r  0.49%

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: 81 : 19. Ans: (a) Sol: Maximum inventory  CC = shortageCS = 400

 Max inventory =

400 100

= 4 units/year

no. of order per year =

4000 = 10 runs 400

Total yearlycos t EOQ  AC 

2ASCI

= 4000  5  2  4000  30  1.5 = Rs. 20600/

R1  D S  100  Q1

+

Q1  R  CI1  1  2  100 

2  4000  = 4000  51   30   100  1000 +

1000 2    1.51   2  100 

= Rs. 20455/-

TCQ2 @%  4000 51 3   4000 30  20001.5 1 3  2  100  2000  100  = Rs. 20915/Among all 2% discount for ordering quantities of 1000 or more

ACE Engineering Academy

2  2000  20 = 200 units 8  0.25

The EOQ at Cu = Rs. 8/- is satisfying the Quantity range hence it is declared as an optimal order quantity.

A = 4000 units CI = Rs. 1.5

2  4000  30 = 400 units 1 .5

TC Q1@ R ,%  AC1 

21. Sol: Given: A = 2000 units/year , S = Rs. 20/-, I = 25% Cu = Rs. 8/- (Lowest with unit price) EOQ |Cu 8% 

20. Sol: Given : C = Rs. 5/unit , S = Rs. 30/order , EOQ 

IM & OR

22. 23. Sol: Annual demand (A) = 2000 units Cost per item (C) = 20/Ordering cost = 50/ICC (I) = 0.25 EOQ 

2AS  CI

TACEOQ

2  2000  50  200. units 20  0.25

 AC  2ACSI

= 2000  20  2  2000  20  50  0.25 = 41,000/Now, TAC at Q1 with discount r%

TACQ1  AC1  

r1  A Q r   S  1 CI1  1   100  Q1 2  100 

3  2000 1000 3     2000  201   50  20  50 1    2  100  1000  100 

= 2800+100+35000 = 37,900 As the total annual cost (TAC) with discount r% is less than (TAC) at EOQ, hence accept the discount and order 1000 at a time.

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: 82 : 24. Sol:

ME – GATE _ Postal Coaching Solutions

26.

Daily No. of Probability SL sales days Pi

SOR

10

15

0.15

0.15

1

11

20

0.20

0.35

0.85

12

40

0.40

0.75

0.65

13

25

0.25

1

0.25

Cos = CP = 2

=

2AS  CI

2  25  25 0 .4

= 55.9 units  56 units

Cus = SP  CP = 5  2 = 3

SL =

Sol: EOQ 

Cus Cus  Cos 3 = 0.6 3 2

SOR = 1  SL = 1  0.6 = 0.4

 Daily    Lead Time Re-order point =   demand  = 25  16 = 400 units 27. Sol: Given, Daily demand – D. D , Lead Time – L.T Re-order Level - ROL For Item A EOQ 

As SL = 0.6 falling in the range 11 to 12 sales, hence order 12 for 40 days. (Cus) = Cost of under stock (Cos) = Cost of over stock (SL) = Service levels (SOR) = Stock out risk

=

2AS CI 2  8000  15 = 2000 units 0.06

R.O.L = daily demand  Lead Time =

8000  10 = 320 units 250

For Item B

ROL = D.D  L.T

25. Sol: Cus = SP CP = 2  0.8 = 1.2

Cos = CP  Rebate = 0.8  0.2 = 0.6 SL =

Cus 1.2 = = 0.6 Cus  Cos 1.2  0.6

For 60%  Service levels QOptimum = Imin +SL (Imax  Imin) = 20000 + 0.6(24000 2000) = 22400

ACE Engineering Academy

A = 9000 units EOQ 

=

2AS CI 2  9000  40 = 2000 units 0.18

For Item C EOQ 

2AS CI

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: 83 :

300 =

IM & OR

Stock out risk (SOR) = 100  SL

2  7500  S 30

( SOR + SL =100%) = 100  84.13

S = Rs. 180/order

SOR = 15.87%

ROL = D.D  L.T

Stock out = 140 100 = 40 units

7500  LT 250 Lead Time = 7 days 210 

29. Sol: Expected demand during lead time

28. Sol:



SOR

0.2  0.25  0.3  0.25

= 112

ROL=140 Min inventory

80  0.2  100  0.25  120  0.3  140  0.25

Reorder level = 1.25  112 = 140 For Re order level > Expected lead time demand

Max Inventory

30. 3 120 3(20)=60

=120

Sol: +3 120 +3(20)=180

 = 60 units , SL =

51 = 98% 52

(Consider 52 weeks/year) SS = SF = 2.05 60 = 123

a)

SOR = 2%, For service level (SL) = 98% to be safety factor on  basis, SF = 2.05 Safety stock (SS) = SF  = 2.05  20 = 41 Re-order point (ROP) = Avg lead time demand + SS = 120 + 41 = 161

b)

Given, ROP = 140 units, SF = ? 140 = 120 + SF 20 SF = 1 ie., as SF basis is 1 will achieve service levels (SL) 84.13%.

ACE Engineering Academy

ROL = ALTd + SS = ALT  CR + SF = 500 1 + 123 = 623 units Where, CR = consumption rate ALT = Average lead time 31. Sol: Lead Time > order cycle

OC =

n 2 =

6  5 2 = 12.21

Safety stock (SS) = SF  = 1.28  12.21 = 15.67 m  16. ( For 90% SL  SF = 1.28) ROL = ALTd + SS = 40 + 16 = 56 32. Ans: (b)

33. Ans: (d)

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: 84 : 34. Ans: (d) Sol: C – Class means these class items will have very less consumption values. – least consumption values

ME – GATE _ Postal Coaching Solutions

ABC PLAN RANK

Part code

% of total cost%

Cumulative percentage

I

P08

70.5

70.5

II

P09

10.6

81.1

III

P05

10

91.1

IV

P04

2.4

93.5

V

P07

2

95.5

J  5  0.2 = 1.0

VI

P02

1.2

96.7

VII

P06

1.2

97.9

G  10  0.05 = 0.5

VIII

P10

1.2

99.1

IX

P03

0.7

99.8

X

P01

0.2

100

B  300  0.15 = 45 F  300  0.1 = 30 C  2  200 = 400 E  5  0.3 = 1.5

H  7  0.1 = 0.7  G, H items are classified as C class items because they are having least consumption values. 35. Sol: Raking of items according to their usage values Part code

Price per unit Rs

Units/year

Total cost (Rs)

% of total cost

Ranking

P01

100

100

10000

0.2

X

P02

200

300

60000

1.2

VI

P03

50

700

35000

0.7

IV

P04

300

400

120000

2.4

IV

P05

500

1000

500000

10

III

P06

3000

30

60000

1.2

VII

P07

1000

100

100000

2

V

P08

7000

500

3500000 70.5

I

P09

5000

105

525000

10.6

II

P10

60

1000

60000

1.2

VIII

Total

ACE Engineering Academy

Class A items → Nil Class B items → I, II Class C items → III,IV,V,VI,VII,VIII,IX,X 36.

Ans: (b)

4970000 100

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: 85 :

IM & OR

Evaluation of empty cells: Cell (A1) Evaluation = CA1CA4 + CC4CC1

Chapter- 7

=10  11 + 18  5 = 12

Transportation Model

Cell (A3) Evaluation = CA3  CA2+CB2  CB3 = 20  9 + 7  2 = 16

01. Ans: (c)

Cell (B1) Evaluation =127+211+184

Sol: A no. of allocations : m + n  1

 5 + 3 1 = 7

= 10 Cell (B4) Evaluation = 20  7 + 2  11= 4 Cell (C2) Evaluation = 14  2 + 1118= 5

02. Ans: (a)

Cell (C3) Evaluation = 169 +7218 = 5 If cell cost evaluation value is ‘ve’, indicates

03. Ans: (b)

further unit transportation cost is decreasing and if cost evaluation value is ‘+ve’ indicates further unit transportation cost is increases. If cost evaluation value is zero, unit transportation cost doesn’t change.

04. Ans: (b) 07. Ans: (a) Sol: No. of allocations : 5

 no. of allocations : m + n  1

 As for A3 cell cost evaluation is +16, means

m + n  1 = 4 + 3 1

that, if we transport goods to A3 the unit transportation cost is increased by 16/-.

 It is a degenerate solution 08. Ans: (a) Sol:

Common Data for Questions Q9, Q10 & Q11 : 3

2

1 10

12

Demand

7

14

18

16

15

15

3

2

1 6

50

25 11

B

4 9

1

A 10

5

10 55

3 45

2

8

4

7

25

30

50 C

ACE Engineering Academy

10. Ans: (a)

11. Ans: (b) Sol:

25

5 15

15

09. Ans: (b)

20

15

5 5

11

9

10 5

Supply

10

5

B

C

20

2

A

4

12 35

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: 86 :

ME – GATE _ Postal Coaching Solutions

No. of allocations = 6

Chapter- 8

R+C1=6

Assignment Model

As No. of allocations = R + C  1 Hence the problem is not degeneracy case. Opportunity cost of cell (i, j) is Cij  (Ui + Vj) If Cij  (Ui + Vj)  0  problem is optimal, Empty cell evaluation (or) Opportunity cost of cells: A1 =  12, A2 =  19, B2 = 8 B4 = 12,

C3 = 3,

C4 = 12

From the above as A2 has opportunity cost ‘19’ indicates unit transportation cost is decreased by 19/By forming loop A2, A3, B2, B3 it is observed that to transport minimum quantity is 25 among 25, 30, 35.  The reduction in the transportation cost is 25  19 = 475 12. Ans: (c) 13.

Ans: (c)

ACE Engineering Academy

01. Ans: (a)

02. Ans: (c)

03. Ans: (a)

04. Sol: Step-1: Take the row minimum of substract it from all elements of corresponding row

1 0 8 0

0 2 5 6

2 2 0 2

3 1 1 4

Step – 2 : Take the column minimum & substract it from all elements of corresponding column.

1 0 8 0

0 2 5 6

2 2 0 2

2 0 0 3

Step – 3 : Select single zero row or column and assign at the all where zero exists. If there is no single zero row or column. Then use straight line method. A

B

C

D

1

1

0

2

2

2

0

2

2

0

3

8

5

0

0

4

0

6

2

3

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: 87 :

1–B: 2–D: 3–C: 4–A:

7 8 2 5

Step – 4 : Draw the minimum number of horizontal and vertical lines necessary to cover all zeroes at least once.

Total cost = 22 05. Sol:

IM & OR

Take the above Table J1

J2

J3

J4

A

B

C

D

C1

5

0

10

7

1

10

5

15

13

C2

0

6

5

14

2

3

9

8

18

C3

8

5

0

0

3

10

7

2

3

C4

0

6

2

3

4

5

11

7

9

(i)

0 6 5 6

10 5 0 2

8 15 1 4

Step – 2 :

5 0 8 0

L3

L1

Step – 1 :

5 0 8 0

L2

0 6 5 6

10 5 0 2

7 14 0 3

Step – 3 5

0

10

7

0

6

5

14

8

5

0

0

0

6

2

3

It may be noted there are no remaining zeroes and row – 4 and column – 4 each has no assignment. Thus optimal solution is not reached at this stage. Therefore, proceed to following important steps. ACE Engineering Academy

Mark row – 4 in which there is no assignment (ii) Mark column 1 which have zeroes in marked column. (iii) Next mark row 2 because this row contains assignment in marked column 1. No further rows or columns will be required to mark during this procedure. (iv) Draw the required lines as follows. (a) Draw L1 through marked column 1 (b) Draw L2 and L3 through unmarked row (1 and 3) Step – 5 : Select the smallest element (2). Among all the uncovered elements of the above table and substract this value from all the elements of the matrix not covered by lines and add to every element that lie at the intersection of the lines L1, L2,and L3 and leaving the remaining element unchange.

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: 88 : J1

J2

J3

J4

C1

7

0

10

7

C2

0

4

3

12

C3

10

5

0

0

C4

0

4

0

5

ME – GATE _ Postal Coaching Solutions Step – 1:

9 13 35 18

It may be added that there are no remaining zeroes and every row and column has an assignment. Since, the no. of assignment = no. of row or column

Step – 2:

 The solution is optimal The pattern of assignment at which job has been assigned to each contractor. J2 J1 J4 J3

5 3 3 7 181000=18000

Minimum amount = Rs. 18,000/06. Sol: Here no. of rows  no. of column

 The algorithm is not balanced so add one

0

6

9

0

4

7

0

0

26

0

9

0

9

10

14

0

Here the operator – 4 is assigned to dummy column.

Contractor Job Amount (Rs)1000

C1 C2 C3 C4

26 15 0 27 6 0 20 15 0 30 20 0

 He is the idle worker. 08. Sol:

Ans: (c) S1

S2 S3

S1 S2 S3

P 110 120 130

0

Q 115 140 140

0 15 5

R 125 145 165

0 10 20

P

5

0

10 20

Q 0

25 25

R

20 40

0

Row Transaction

0

0

0

Column Transaction

0

0 10 0 0

5 15

dummy column. Operates Machine A

B

1

9

26 15 0

2

13 27 6

3

35 20 15 0

4

18 30 20 0

ACE Engineering Academy

C

Dummy 0

P-S2  120 Q-S3  140 R-S1  125 Total = 385

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: 89 : 09. Sol: Assignment problem is the special case of the transportation problem A

B

C

D

1

10

8

10

8

2

10

7

9

10

3

11

9

8

7

4

12

14

13

10

IM & OR

Step (4): Now in the third row, if we select (3,B) then it would not be possible to get assignment in the 3rd column . So assign (3,C) cell.

Step (1): Select the small element in a row and subtract it from all other numbers in that row.

A

B

C

D

1

2

0

2

0

2

3

0

2

3

3

4

2

1

0

4

2

4

3

0

Step (2): Now in columns, subtract the small number from all other elements in that column.

A

B

C

D

1

0

0

1

0

2

1

0

1

3

3

2

2

0

4

0

4

2

B

C

D

1

0

0

1

0

2

1

0

1

3

3

2

2

0

0

4

0

4

2

0

Step (5): In the remaining assignment, if we assign (1, A) then other assignment would be (4, D). If the assignment is (1, D) then other assignment would be (4, A) and in the both cases, total cost is same. So assign (1,A) and (4,D)

A

B

C

D

1

0

0

1

0

0

2

1

0

1

3

0

3

2

2

0

0

4

0

4

2

0

Step (3): Now select the single zero cell in a row if possible and assign that cell and cross off other zero corresponding to that cell’s row and column. Here (2,B) with single zero. A B C D 1 0 0 1 0

ACE Engineering Academy

A

2

1

0

1

3

3

2

2

0

0

4

0

4

2

0

Assignment is 1-A, 2-B, 3-C, and 4-D Optimal cost = 10 + 7 + 8 + 10 = 35 Euros.

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: 90 :

ME – GATE _ Postal Coaching Solutions

Chapter- 9 PPC & Aggregate Planning

01. Ans: (d)

02. Ans: (b)

03. Ans: (b) Sol:

Months

Month 1

Month 2 20

RT

90

1

24

10

Month 3

Unused capacity

20

24 10

26

28

Capacity Available

(1)

20

OT RT

100

OT

20

20

22

24

36

2

100 20

20 24

RT

80

OT

30

80

3

RT

90

100

130

10

40

110

OT Level of planned production in overtimes in 3rd period is ‘30’. RT = Regular time OT = Over time ACE Engineering Academy

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: 91 : 04.

Ans: (b)

05.

Ans: (b)

06.

Ans: (d)

IM & OR

07. Sol:

Demand for Supply from Beginning inventory 1 2 3 4

Regular Overtime

Total Capacity Available (supply)

Period 1

Period 2

Period 3

Period 4

Un used capacity

200

0

5

10

15

-

200

700

60

65

70

75

0

700

75

80

85

300

300

0

700

300

300

70

Regular

500

Overtime

70

60

65

200

75

80

200

Regular

500

65

0

700

200

75

100

300

Regular

700

60

0

700

Overtime

300

70

0

300

Overtime

60

70

70

900

500

200

1900

700

4200 4200

Total cost = (700 60) + (500 60) + (200 70) + (200  60) + (500 65) + (200 75) + (700 60)+ (300 70) = Rs 2,08,500/-

ACE Engineering Academy

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: 92 :

ME – GATE _ Postal Coaching Solutions

08. Sol:

Period1

Period2

Period3

Period4

Capacity Unused Available capacity (supply)

Beginning Inventory

150

0

2

4

6

-

150

1 Regular

900

25

27

29

31

-

900

Overtime

150

30

32

34

36

-

150

Subcontract

200

35

-

-

-

100

Demand for Total Supply from

-

300

600

25

27

29

-

600

Overtime

125

30

32

34

-

125

Subcontract

175

35

-

-

125

-

700

-

150

2 Regular

700

25

27

Overtime

100

30

50

Subcontract

35

3 Regular

32

-

300

-

300

-

300

800

25

-

800

Overtime

200

30

-

200

Subcontract

250

35

50

4 Regular

1400

900

800

-

300

1200+100 575

4975 4975

Total cost = (900 25) + ( 150  30) + (200 35) + ( 600 25) + ( 125 30) + (175 35) + (700 25) + (100 30)+(50 32) + (800  25) + (200 30) + (250 35) = Rs 1,15,725/-

ACE Engineering Academy

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: 93 :

IM & OR

Chapter- 10 Material Requirement & Planning

01. Ans: (b)

02. Ans: (c)

03. Ans: (d)

04. Ans: (c)

05. Ans: (c)

06. Ans: (c)

07. Ans: (b)

08. Ans: (b)

09. Sol:

A  1  10 = 10 B  2  10 = 20 C  (1  2 10) + (3  4  2 10) = 260 D  (4  2 10) = 80 E  (3  4  2 10) + (2  2 10) + (410) = 320

10. Sol: 0 Level X = 1

100

1100

1st Level

2  1100 D-2

E2 21100

3rd Level

221100

ACE Engineering Academy

B1

C1

1100

1100

2nd Level

J2

A1

F-1

G1

H3

11100 11100 31100

J2

F-1

221100 11100

K1 121100

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: 94 :

ME – GATE _ Postal Coaching Solutions

11. Sol:

Order Quantity = 200 LT = 3 Weeks

Week 1

2

Project required Receipts On hand inventory Planned order release

40

85 10

3

100 15 5 200

200

4

5

6

7

8

60

130 110 50 170

200

200

200

145 15

105 55 85

200

(On hand inventory)t 1st week = 140+0-40 = 100 2nd week = 100+0-85 = 15 3rd week = 15 + 0  10 = 5 4th week = 5 + 200  60 = 145 5th week = 145 + 0  130 = 15 6th week = 15 + 200  110 = 105 7th week = 105 + 0  50 = 55 8th week = 55 + 200  170 = 85  Order before 3-weeks 12.

Ans: (c)

13. Sol:

X

B1

A1

C1

Net required

D2

A = (1  1  20 10) = 10 C = (1  1  20110 10) = 0 B = 1  20 15 = 15 D = 2  1  20  2  10 10 = 10

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: 95 :

IM & OR

04. Ans: (c )

Chapter- 11

Sol: TC = Total cost

Break Even Analysis

TCA = Total cost for jig-A TCB = Total for jig-B

01. Ans: (c)

TCA = TCB

Sol: Total fixed cost, TFC = Rs 5000/-

800 + 0.1X = 1200 + 0.08X

Sales price, SP = Rs 30/-

0.02X = 400

Variable cost, VC = Rs 20/Break Q 

even

production

per

month

X 

,

TFC 5000 = = 500 units SP  VC 30  20

400 400   100 = 20,000 units 2 0.02

05. Ans: (d) Sol: Sales price – Total cost = Profit

02. Ans: (a)

(CP 14000) – (47000 + 1400015) =

Sol: Total cost = 20 + 3

23000

2X = 30

CP = 20

 X = 15 units When X = 10 units TC1 = 20 + (3 10) = Rs 50/-

06. Ans: (b)

TC2 = 50 + ( 1 10) = Rs 60/Among both, total cost for process is less

07. Ans: (a)

So process-1 is choose. 08. Ans: (c) 03. Ans: (c ) Sol: In automated assembly there are less labour,

so variable cost is less, but fixed is more

09. Sol:

X

Y

because machine usage is more. In job shop

S1 = 100

S2 = 120

production, labour is more but machine is

F1 = 20,000

F2 = 8000

less. So variable cost is more and fixed cost

V1 = 12

V2 = 40

is less.

P = q(S – V) – F

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: 96 :

ME – GATE _ Postal Coaching Solutions

P1 =q(100 – 12) – 20,000 P2 = q(120 – 40) – 80,000

Chapter- 12

P1 = P2

Network Analysis

88q – 20,000 = 80q – 80,000 12000 = 8q

01. Sol:

 q = 1500

SQ

11.

Ans: (b) dij

12.

Ans: (c)

13.

Ans: (d)

Q

G

P

Sol:

Standard machine Automatic tool machine tool F1= F.C.

30  200  Rs.100 60

2×800 = Rs.1600 = F2

V.C =

20  200  Rs.73.33 60

5  800  Rs.66.67 60

1600  100  225 volts 73.33  66.67 If greater than 225 units then automatic machine tool is economic.

R dij

SR

q

The are two path to from P to Q  The length of shortest path from P to G = min { SQ + dQG, SR + dRG}

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Fluid Mechanics & Hydraulic Machinery Solutions for Vol – I _ Classroom Practice Questions F   A

Chapter- 1 Properties of Fluids

W sin 30 

AV h

100 1  0.1  V  2 2  10 3

01. Ans: (d)

V = 1m/s

02. Ans: (a) Sol: The gap between two co-axial cylinder is

Common data Q. 07 & 08

very narrow. Therefore velocity profile can 07. Ans: (c)

be assumed linear.

Sol: D1 = 100mm , 03. Ans: (c)

D2 = 106mm D 2  D1 2

Radial clearance, h 

04. Ans: 1



Sol:  = 1.25 stoke S = 0.8

106  100  3mm 2

L = 2m

=

 = 0.2 pa.s

= 1.25  10-4  800

N = 240rpm

2

= 0.1 Ns/m = 1 Poise 05. Ans: 100

0.2  1.5 V Sol:    = 100 N/m2  3 h 3  10 06. Ans: 1



2N 2  240 = 60 60

 = 8 r 0.2  8  50  10 3  = 83.77N/m2   3 h 3  10 08. Ans: (b)

Sol: WSin30

30o

W ACE Engineering Academy

Sol: Power, P 

22 Lr 3 h



2  82  0.2  2  0.053 3  10 3

= 66 Watt Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally

: 100 :

17. Ans: (c)

09. Ans: (c) Sol:

ME _ GATE_Postal Coaching Solutions

 30

18. Ans: (a) Ans: In fluids, the deformation changes with time

18

so rate of deformation is far more important

Slope = constant

than deformation itself.

6

It deforms continuously because fluids 1

resist shear forces under dynamic condition. 0

1

3

5

Both the statements are true and statement

du/dy

II is the correct reason of statement I.

 Newtonian fluid

10. Ans: (d) Sol:

19. Ans: (d)

du  dy

Ans: Viscosity in liquids decreases and in gases

decreases with rise in temperature.

u = 3 sin(5y) du  3 cos5y   5 = 15cos(5y) dy  y0.05  

du dy

Ans: Blood is a pseudoplastic fluid. So statement

I is wrong. y 0.05

= 0.5  15 cos5  0.05 1  = 0.5  15  cos   0.5  15  2 4 = 7.53.140.707  16.6N/m2  y0.12  0.5  15 cos5  0.12 

21. Ans: (d) Sol: Free surface is subjected to surface tension

force in the plane of surface. It can resist small tensile loads. 22. Ans: (b) Sol: V = 0.01 m3

 = 0.75  10–9 m2/N

3   = 7.5   cos 5   25  

dp = 2107 N/m2

 3  = 7.5   cos   5  Which is negative so zero 11. Ans: (c)

12. Ans: (d)

13. Ans: (d)

14. Ans: (c)

15. Ans: (a)

16. Ans: (a)

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20. Ans: (d)

k

4 1 1   109  9 3  0.75  10

k

dp dv / v

dv 

2  107  102  3 = 1.510–4 9 4  10

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: 101 :

The plate moves with velocity V

23. Ans: 320 Pa Sol: P 

Fluid Mechanics & HM

8 8  0.04 32  102   D 1  10 3 10 3

1

(h-y)

V

P = 320 N/m2 2

y

24. Ans: (b)

From Newton’s law of viscosity,  Conventional Questions which can be asked as objective Questions

du Let A be area of plate dy

 F1 =1  Area of plate F1  1 

01.

F2   2 

Sol:

h

F1 = F2

1 y

V A y

(i) Shear force on two sides of the plate are equal:

(h-y) 2

F

1  VA  2 VA  hy y 1 h  y  2 y

Assumption:



Thin plate has negligible thickness.



Velocity profile is linear. Because of

h 1  1 y 2

narrow gap. 

V A hy

h 1   2  y 2

Given fluid is a Newtonian fluid which obeys Newton’s law of viscosity.

y

The force required to pull it is proportional to the total shear stress imposed by the two oil

2h 1   2

(ii) The position of plate so that pull required to

layers.

drag the plate is minimum.

F = F1+F2 , Where F1 = Force on top sides of

F

plate . F2 = Force on bottom side of plate

1 VA  2 VA , V, A, 1 & 2 , h are  hy y

constant ACE Engineering Academy

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: 102 :

For minimum force,

Assumption:

dF 0 dy

–2

ME _ GATE_Postal Coaching Solutions

 –2

and hence velocity profile is assumed

–1VA(h –y) (–1) –2VAy = 0

linear.

 2 VA  VA  1 2 2 y h  y 

h  y 2 y

2





1 2

Force = shear stressArea 

1 h  1 where y is the distance of the y 2

thin flat plate from the bottom flat surface.

  VA h

Where h is the clearance (radial) 10  9.75 2

h

= 0.125cm = 1.2510–3m

h 1

No change in properties

Torque = Tangential force  radius

1 hy  y 2

y

The gap between two cylinders is narrow

1 2

Area = DL = 0.12.510-2 = 7.853910–3m2

02. Ans: 8.105 Pa. S Sol: Torque = 1.2N-m

Speed, N = 90rpm Diameter, D1 = 10cm , H = 2.5cm

D2 = 9.75cm

Fs 

  r  A h



2N 2  90   3 rad / s 60 60

Torque = Fsr

2.5 cm





rA r h



r 2 A h

  3  (0.1) 2  7.8539  10 3 1.2  1.25  10 3  4  = 8.105 Pa.s

9.75cm 10 cm ACE Engineering Academy

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: 103 :

Fluid Mechanics & HM

06. Ans: 48.147 Sol:

Chapter- 2 Pressure Measurement & Fluid Statics

Oil (S=0.86)

1.5m

Water

01. Ans: (a)

0.8m

Mercury (S =13.6)

Sol: 1 millibar = 10-3105

= 100 N/m2

0.2m

Pbottom = oilghoil + wghwater + HgghHg

One mm of Hg = 13.61039.81110-3 2

= 133.416 N/m 1 N/mm2 = 1106 N/m2

= (8609.810.8)+(98101.5)+(136009.810.2)

= 48147.48 Pa Pbottom = 48.147 kPa

1 kgf/cm2 = 9.81106 N/m2 07. Ans: (a) 08. Ans: (b) 09. Ans: (c) 02. Ans: (b) 10. Ans: 2.2

Sol:

Sol: hp in terms of oil Local atm.pressure (350 mm of vaccum)

710 mm

so ho = smhm 0.85h0 = 13.60.1 h0 = 1.6m

360 mm

hp = 0.6+1.6  hp = 2.2m of oil

Absolute pressure

11. Ans: 750 Sol: Patm+wghw = Patm+0gh0

03. Ans: (d)

1000610-2 = 0810-2 0 = 750 kg/m3

04. Ans: 27 m Sol:

P = gh 200103 = 7559.81h h = 27m

05. Ans: (c)

12. Ans: (b) Sol: h M 

s w h w2 sw  h0 h w1  h N  s0 s0

hM  hN 

9 18  3 0.83 0.83

h M  h N  13.843 cm of oil ACE Engineering Academy

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: 104 :

ME _ GATE_Postal Coaching Solutions

13. Ans: 2.125

17. Ans: 1

Sol:

Sol:

 1m

2x

hP  hG   2  2

I Ah G

x

Fbottom = g  2x  2x  x FV = gx  2x  2x

D 4  4 64  D 2  2  

FB 1 FV

22  4  2.125m 64  2

18. Ans: 785 kN Sol:

14. Ans: 30.8 Sol:

F = PA

Sol:

  9810  1   2 2 4

  22 FV = gV  1000  10  4

15. Ans: 61.6

FV = 10kN

Sol: F = PA

 x = 10

F  ghA  2  2 = 61.6 kN 4

20. Ans: (d) Sol: Fnet = FH1 – FH2

FH1   

16. Ans: 10 Sol: F  ghA

 9810  1.625 

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2m

FV = x

F = 30.82kN

F = 10kN

F  ghA = 98102104 = 785 kN

19. Ans: 10

= gh  A

 9810  2 

2x



 1.2 2  0.8 2 4



D 2 D  D 1  2 2

FH 2   

D 2 D D  1  8 4 2

2  1 1  3D = D 2    = 8  2 8

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: 105 :

Fluid Mechanics & HM

03. Ans: 4.76

Chapter- 3

Sol: FB = FB,M+FB,W

Buoyancy and Metacentric Height

WB = FB x

01. Ans: (d) Sol:

water Hg

(10–x)



bgVb= mgVfd,m+gVfd,w

2m

d

bVb = mVfd,m+wVfd,w

1.25m

4m

SVb = SmVfd,m+SwVfdw 7.6103 = 13.6102(10–x)+102x

FB = weight of body

–6000 = –1260x

bgVb = fgVfd

x = 4.76 cm

640421.25 = 1025(41.25d) d = 1.248m Vfd = 1.24841.25 Vfd = 6.24m3

04. Ans: 11 Sol:

 FB

02. Ans: (c)

1.6m

Sol: Surface area of cube = 6a2

Surface area of sphere = 4r2

T

4r2 = 6a2 2  a    3 r

FB = W + T

2

W = FB – T = fgVfd – T

Fb,s  Vs 4 3 r r 3 4 3  3  3  2  3 a r    3   



4 3 = 10 3  9.81  0.8  10  10 3 3



= 21 – 10 W = 11 kN

r 3 4 6  3  2  2 3     r  3 3  

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: 106 : 05. Ans: 1.375 Sol: Wwater = 5N

Woil = 7N S = 0.85

ME _ GATE_Postal Coaching Solutions

BM 

1 9

GM 

1 1  9 8

GM = – 13.8mm  – 14mm

W – Weight in air FB1 = W – 5

08. Ans: (b)

FB2 = W – 7 W – 5 = 1gVfd…..(1) W – 7 = 2gVfd…..(2)

W = FB bgVb = fgVfd bVb = fVfd

Vfd = Vb

  0.6  d 2  2d  1  d 2  x 4 4

W  5  1gVb W  7   2 gVb 2  1   2 gVb

Vb 

Sol:

x = 1.2d

2 1000  8509.81

Vb = 1.359110-3m3 W = 5+(98101.359110-3) W = 18.33N

GM = BM – BG d 4 d   19.2 64  d 2  1.2d 4 BG = d–0.6d = 0.4d I BM   V

GM < 0 unstable

W = bgVb 18.33  b 9.81  1.3591  10 3

09. Ans: 20s

b = 1375.05 kg/m3

Sol: T  2

Sb = 1.375  2

06. Ans: (d)

k2 g (GM)

7.722 9.81  0.6

T = 20s 07. Ans: 14 Sol: GM = BM–BG

BM 

I 3  1  V 12  3  1  0.75

BM 

4 1 3 1 , BG    12  3 2 8 8

3

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: 107 :

Fluid Mechanics & HM

Total force required to lift the plate

10. Ans: Sol:

V=0.1m/s F

= Fs + W – FV = 102.2727 + 50 – 29.7978 = 122.4749 N

Fs

Fs

W The thickness of the oil layer is same on either side of plate y = thickness of oil layer 

23.5  1.5  11mm 2

Shear stress on one side of the plate 

dU dy

Fs = total shear force (considering both sides of the plate)  2A    

2AV y

2  1.5  1.5  2.5  0.1 11 10 3

= 102.2727 N Weight of plate, W = 50 N Upward force on submerged plate, Fv = gV = 900  9.81  1.5 1.5  10–3 = 29.7978 N

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: 108 :

ME _ GATE_Postal Coaching Solutions

03.

Chapter- 4 Fluid Kinematics

Ans: (a)  Sol: V  2xˆi  yˆj 

Compare V  uˆi  vˆj

01.

Ans: (b)

Where, u = 2x, v = y  Velocity, v  u 2  v 2

02.

Ans: (a)

=

Sol: Given, u = –x,

Stream line equation in 2 – D dx dy  u v

41  1  5m / s  Acceleration, a  a x ˆi  a y ˆj 2

V (1,1) =

 2 2 a  ax  ay

dx dy   x 2y

ax 

On integration dy

u u u u u v w t x y z

= 0  2x 

  x   2y 1 1 1   dx   dy 2 y x 1 log y  log c 2

–log x =

4x 2  y 2

V=

v = 2y

dx

2x 2  y 2

1 log   log y  log c x



1 log   log y .c x



 2x   y  2x  x y

= 2x (2) + y(0) = 4x ay =

v v v u v t x y

= 0  2x

 y   y.  y  x y

= 2x (0) +y (1) =y  a  a x2  a y2 

1  y .c x 1 At (1,1) point =  1.c 1 c=1 x y 1

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4x 2  y 2

 16x 2  y 2  2 2  a 1,1  161  1

 17 m / sec 2

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: 109 :

Fluid Mechanics & HM

Common Data for Questions 04 & 05

According to the conservation of mass

04.

Ans: 0.94

Total inward flow = Total outward flow

Sol:

aLocal

=

Q1 = Q2 + Q3

V t

A1V1 = A2V2 + A3V3

  x  =  2 t 1   t   2L 

A2 = A3

   

2

V1 = 2 m/s ; V2 = 3 m/s ; V3 = 5 m/s A1 2 = A2  3 + A2  5

2

x   = 1   2  2L 

A1 = 4A2 At another instant V1 = 3 m/s

0.5   (aLocal)at x = 0.5, L = 0.8 = 2 1    2  0.8 

2

2

V2 = 4 m/s V3 = ? 2

= 2(1  0.3125) = 0.945 m/sec

A1V1 = A2V2 + A3V3 4A2  3 = A2  4 + A2  V3

05.

Ans: –13.68

12 A2 = 4A2 + A2V3 V3 = 8 m/s

2 2 Sol: a convective  v. v  2t 1  x    2t1  x   x   2L   x   2L  

2   x     x  1   2t 1    2t 21       2L     2L  2L 

At

t = 3 sec; x = 0.5 m; L = 0.8 m 2

Ans: (d)

Sol: u = 6xy – 2x2

continuity equation for 2D flow u v  0 x y

 0.5  0.5  1   a convective 2 31  2 321     2 0.8   2  0.8  2 0.8

u  6y  4x x

atotal = alocal + aconvective = 0.94 – 14.62

6 y  4x   v  0

aconvective = – 14.62 m/sec2

= –13.68 m/sec2 06.

07.

Ans: 8

V2

Sol:

V1 A1

A2

y

V  4x  6 y   0 y  V = (4x–6y) dy

V =  4xdy   6 ydy = 4xy –3y2 + c

V3

= 4xy  3y2 + f(x)

A3 ACE Engineering Academy

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: 110 :

(ar)(=45) = 3  cos450  sin450

08.

Ans: 2

Sol:

V 1  m / sec/ m  x 3

ar =

11.

V2 R

V 3 9   = 1 m/s2 R 9 9

a = a r 2  a t 2  12  12 =

2 m/sec2

dV dx

 2.5  3  3  2.5  =   = 2.75  5  2  0.1 

=

1 Q A t

1 Q  0.12 ( = 0.12) 0.4 t

12.

Ans: (b)

aTotal = (ax) ˆi +(ay) ˆj

u=

  2  x  y 2 = 2y y y

v=

    2  x  y 2 = 2x x x









u u v x y

= (2y)(0) + (2x)(2) ax = 4x ( V  r )

V V   r

ay = u

v v v x y

= (2y)(2) + (x)(0)

 3 sin  3 sin  = 3cos  sin =  3 ACE Engineering Academy

a local 

ax = u

V    t

V =  

 Q u =   t  A  t

Sol:  = x2 y2

Ans: 1.5

Sol: ar =

aLocal =

= 0.3 m/sec2

at (conv) = 13.75 m/s2 10.

3 = 1.5 m/sec2 2

Ans: 0.3

(aLocal)at x = 0 =

Ans: 13.75

a t ( conv )

2

=

1   Q aLocal =    0.4  0.1x  t

2

Sol: a t (conv)  Vavg 

1

Sol: Q = Au

V 1  3  = 1 m/s2 at = V x 3

09.

2



V= 3 m/sec

R=9 m

2

1

=3 at

ar=

ME _ GATE_Postal Coaching Solutions

ay = 4y a = (4x) ˆi + (4y) ˆj

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: 111 : 13.

Fluid Mechanics & HM

Ans: (b)

Sol: Given, The stream function for a potential

Chapter- 5

flow field is  = x2 – y2

Energy Equation and its Applications

=?

u

    x y

u

  x 2  y 2   y y

01.

Sol: Applying Bernoulli’s equation for ideal

fluid P1 V2 P V2  Z1  1  2  Z 2  2 2g g 2g g

u = –2y u=

  2y x

P1 (2) 2 P (1) 2  = 2  g 2g g 2g

    2 yx

P2 P1 4 1    g g 2g 2g

 = 2 xy + c1 Given,  is zero at (0,0)

P2  P1 3 1.5   g 2g g

 c1 = 0   = 2xy 14.

Ans: (c)

02.

Ans: 4

Sol: Given, 2D – flow field

Sol:

Ans: (c)

①

① S1

Velocity, V = 3xi + 4xyj

2mm

u = 3x , v = 4xy 1  dv du  z     2  dx dy  1 z  4 y  0 2

Z at 2,2   1  42 = 4 rad/sec 2

S2②

② V12  1.27 m , 2g

P1  2.5m g

V22  0.203m , 2g

P2  5.407m g

Z1 = 2 m ,

Z2 = 0 m

Total head at (1) – (1) =

V12 P1   Z1 2g g

= 1.27 + 2.5 + 2 = 5.77 m ACE Engineering Academy

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: 112 :

Total head at (2) – (2)

ME _ GATE_Postal Coaching Solutions

04.

Ans: 395

Sol: Q = 100 litre/sec = 0.1 m3/sec

V2 P = 2  2  Z2 2g g

V1 = 100 m/sec; P1 = 3  105 N/m2 V2 = 50 m/sec; P2 = 1  105 N/m2

= 0.203 + 5.407 + 0 = 5.61 m Loss of head = 5.77 – 5.61 = 0.16 m

Power (P) = ?

 Energy at (1) – (1) > Energy at (2)– (2)

Bernoulli’s equation

 Flow takes from higher energy to lower

P1 V12 P V2   Z1  2  2  Z 2  h L g 2g g 2g

energy i.e. from (1) to (2)

3  10

Top to bottom flow take place

5

1000  10



100

2

2  10

0

1  10

5

1000  10



50

2

2  10

 0  h L hL

= 395 m 03.

Ans: 1.5

Sol: A1 =

P = gQ.hL

 2  2 d1  0.1  7.85  10 3 mm2 4 4

  2 A 2  d 22  0.05  1.96  10 3 mm2 4 4 2 1

2 2

P1 V P V   Z1  2   Z2  h L g 2g g 2g

Z1 = Z2, it is in Horizontal position Since, at outlet atmospheric pressure, P2 = 0 Q 0.1   12.73 m / sec A1 7.85  10 3

V2 

Q 0.1   51.02 m / sec A 2 1.96  10 3

P1gauge air  g



12.732 2  10

0

P1  121.53 air .g P1  121.53  air  g = 1.51 kPa ACE Engineering Academy

P = 395 kW 05.

Ans: 51.5

Sol: Apply Bernoulli’s equation to pump

 P1

 P2 PUMP

Work in

Q = 100 lit/sec = 0.1 m3/sec V1 

P = 1000  10  0.10  395

51.022 2  10

P1 V2  Z1  1 + Work in 2g g

=

P2 V2 +Z2 + 2 + HLoss g 2g

Where work in = Head raised = 10 m Since pipes are same size V1 = V2 and Z1 = Z2 P1 120  10 3  0  0  10  003 1000  9.81 g

P1 = (12.23 + 3 10)g Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally

: 113 : 07.

P1 = (5.234)(10009.81)

Fluid Mechanics & HM Ans: 65

Sol: hstag = 0.30 m

= 51.33103 N/m2

hstat = 0.24 m

= 51.33 kPa

V = c 2gh dyna 06.

Ans: 35

V  1 2g h stag  h stat 

Sol:

 29.810.30  0.24 = 65.09 m/min

fluid, S = 0.85 d2= 120 mm d1= 300 mm

08.

Ans: 81.5

Sol: x = 30 mm Pressure difference Between A & B = 4 kPa

g = 10 m/s2 air = 1.23 kg/m3; Hg = 13600 kg/m3

A1 A 2

Q Th 

A12  A 22



C=1

2gh

A1A 2 A12  A 22

V  2gh D

S  h D  x m  1  S 

 P  2g    w 

 13600   1 h D  30  10 3   1.23 

  2 A1 = d12  0.30  0.07 m 2 4 4

hD = 331.67 m

  2 A 2  d 22  0.12  0.011m 2 4 4 Q Th 

0.07  0.011

0.07 2  0.0112

09.

Ans: 140

Sol: Q a  C d

P P h  w Pf .g P 4  10  s f  w g 0.85  1000  9.81 3

3

= 0.035 m /sec = 35.15 ltr/sec.

ACE Engineering Academy

V = 81.5 m/sec

2  9.81  4  10 3

P = 4 kPa,



V  1 2  10  331.67

Cd  C d venturie C d orifice

A1 A 2 A12  A 22

2gh

1 h 

h orifice h venture

hventuri = 140 mm

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: 114 : 05.

Chapter- 6

ME _ GATE_Postal Coaching Solutions Ans: (b)

Sol: Fx = aV( V – 0) = a V2

Momentum equation and its Applications

= 1000  1  10–4  102 = 10 N 01. Sol:

Ans: 720

P = (g + a)h = (g + 5g)h = 6gh = 6  1200  10  10 = 720 kPa

02.

06. Sol:

V

Ans: 1600

Sol: S = 0.80

V

2

A = 0.02 m

Fx = aV( V1x – V2x)

V = 10 m/sec

= aV( V – (–V))

F = .A.V2

= 2 a V2

F = 0.80  1000  0.02  102

= 2 1000  10–4  52 = 5 N

F = 1600 N 03.

Ans: (c)

07.

Ans: 6000 2

Ans: (c)

Sol:

Sol: A = 0.015 m

V = 15 m/sec

V2

V1

U = 5 m/sec

A2

F = A (V + U)2 F = 1000  0.015 (15+5)

A1

2

Q = A1V1 = A2V2

F = 6000 N

(0.2104)(1102) = 0.07  10–6  V2 04.

Ans: 19.6

V2 = 2.86 m/s

Sol: V = 100 m/sec

Fx = Q(V2 V1)

U = 50 m/sec

=1000(0.21041102)(2.860.01)

d = 0.1 m F = A (V – U)

= 5.8  104 N

2

 2 F  1000   0.12  100  50 4 F = 19.6 kN ACE Engineering Academy

=

5.8  10 4 kgf 9.81

⋍ 6  105 kgf

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: 115 : 08.

Fluid Mechanics & HM

Ans: (d)

Sol: F1 = A(Vu)2

Chapter- 7 Laminar Flow

Power (P1) = F1 u = A(Vu)2 u F2 = .Q.V  Vr

01.

Ans: (d)

Power (P2) = F2 u = AV(Vu)u

02.

Ans: (d)

P1 AV  u 2  u  P2 AV(V  u )  u

03.

Ans: (d)

= .A(V).(Vu)

1 

Sol: Q = A.Vavg

5 = 0.75 20

Q = A.

Vmax ( Vmax = 2 Vavg) 2 2

Q=

04.

  40  1.5    4  1000  2 =

 2  0.04   0.75 4

=

 4 4 3 3    = m3/sec 4 100 100 4 10000

Ans: 100000

Sol:  =

 dP r  dx 2

0 .1 dP 2  250 =  10 2

 P1 P2 = 1  105 N/m2 05.

Ans: 1.92

Sol:  = 1000 kg/m3

Q = 800 mm3/sec = 800  (10–3)3 m3/sec L=2m D = 0.5 mm P = 2 MPa = 2  106Pa =? ACE Engineering Academy

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: 116 :

P= 2  10 6 

128.QL D 4

09.



128    800  10



 0.5  10

Ans: 1600

Sol: f= coefficient of friction

 2

3 3



ME _ GATE_Postal Coaching Solutions

f = friction factor

3 4

f =

f for laminar flow, 4

f

64 Re

 = 1.917 millipa – sec 06.

Ans: 0.75

Sol: Ur = Umax

  r 2  1      R   

f 

Re = 16/f = 16/0.01 = 1600

2  U r   1      R    U max

10.

Ans: 0.32

Sol: Given:

  5 2  = 1 1       10    

 = 0.01Poise = 0.0110–1 N-s/m2

 1 3 = 1 1   = = 0.75 m/s  4 4 07.

64 16  4R e R e

D = 10 mm = 10  10–3 m V = 10 mm/s = 1010–3 m/sec L = 1 km = 1000 m

Ans: 0.08 3

Sol: Given,  = 0.8  1000 = 800 kg/m

 = 1 Poise = 10–1 N-s/m

 = 1000 kg/m3 Reynolds Number, R e 

d = 50 mm = 0.05 m

1000  10  10 3  10  10 3  0.01  10 1

velocity = 2 m/s Reynold’s Number , R e  

VD 

800  2  0.05  800 10 1

( Re < 2000)  Flow is laminar,

Re = 100 < 2000  Re < 2000, hence flow is laminar For laminar flow, h f 

For laminar, Darcy friction factor f

64 64   0.08 R e 800

VD 



32VL gD 2

32  0.01  10 1  10  10 3  10 3



10 3  10  10  10 3



2

= 0.32 m 08.

Ans: (c)

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: 117 : 11.

Q = A.Vavg = (width of platey)v

Ans: 16

Sol: For fully developed laminar flow,

hf 

32VL (  Q = AV) gD 2

Q 32 L  A   32QL hf  gD 2 AD2  g

hf 

hf 

Fluid Mechanics & HM

32QL

P  Qh L  3  10 3 

12vL B2

12  1 v  1.20

50 10 

3 2

V = 0.52 m/sec Q = AVavg = (0.2  50  10–3) (0.52) = 5.2 lit/sec

 2 D  D 2  g 4 1 D4

h f 1 D14  h f 2 D 42

Given, D 2 

D1 2

D  h f1  D  h f 2   1   2 

4

4 1

h f 2  16 h f1  Head loss, increase by 16 times if

diameter halved. 12.

Ans: 5.2

Sol: Oil viscosity,  = 10 poise = 100.1

= 1 N-s/m2 y = 50  10–3m L = 120 cm = 1.20 m P = 3  103Pa Width of plate = 0.2 m Q=? ACE Engineering Academy

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: 118 : 03.

Flow Through Pipes

h fA h fB

Sol:  = 0.4 cm2/sec = 0.4104 m2/sec

d = 8 cm = 8 102 m. critical

2000 =

Reynolds

number

for

h fA h fB

V.D  V  8  10 2 0.4  10  4

04.

5

  dB        1 .2 d B 

5

f1 = f2 ;

l e l1 l2 2l l l  +  5 5  5 5 5 5 d e d1 d2 20 d e 10

Sol:  = 8  104 m2/sec; d = 0.08 m

 de = 11.4 cm

Q = 3200   106 m3/sec 05.

Q = AV  3200   10 = 0.082  V 4 6





Mean (or) Average velocity = 2 m/sec V.D 

Ans: (b)

Sol: In parallel pipe arrangement;

h fA = h f B f A .l A .Q 2A f B l B. Q 2B  12.1  d 5A 12.1  d 5B Given dA =dB ; lA = lB, fA = 4fB

2  0.08 8  10  4

Re = 200 < 2000 (Critical Reynolds’s number for laminar flow)  Type of flow is “Laminar” ACE Engineering Academy

l1 = l2 = l

le = l1 + l2 = 2l

Ans: (a)

Type of flow = ?

Ans: (a)

Sol: Given, d1 = 10 cm; d2 = 20 cm

= 2  1 = 2 m/s

Re =

d   B  dA

5

For Laminar pipe flow; Vmax = 2Vavg

Re =

f A .l A .Q a2 12.1  d 5B  12.1  d A2 f B .l B .Q 2B

 1  =   = 0.4018  0.402;  1.2 

Average (or) Mean velocity (V) = 1 m/sec

02.



Given lA = lB, fA = fB, QA = QB

laminar flow is 2000 Re =

f .lV 2 f .lQ 2  2gd 12.1  d 5

 hf =

Ans: (d)

Lower

Ans: (a)

Sol: In pipes Net work, series arrangement

Chapter- 8

01.

ME _ GATE_Postal Coaching Solutions

 QA   QB

2

 fB    fA

QA f fB 1 1  B    = 0.5 QB fA 4f B 4 2

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: 119 : 06.

Ans: (d)

99   = 1000  10 1  2   3 

Sol: For parallel pipes

= 660  103 Watt = 660 kW

h f1 = h f 2 f 1  l1  Q12 f 2  l 2  Q 22  12.1  d 15 12.1  d 52 For given data

Q12  d1    Q 22  d 2   Q1   Q2

Fluid Mechanics & HM

09.

Ans: (b)

Sol: Q = 100 m3/sec

H = 75 m

5

1 HP = 75

2

5

  2d       25 = 32   d 

kgf  m Nm  750 sec sec

1 HP = 750 Watt = 0.75 kW Power (Theoretical) = gQH  1000  10 100 75

Q1  32 = 4 2 Q2

= 75000000 W = 75000 kW

07.

0.75 kW = 1 MHP

Ans: (c)

75000 kW = – ?

Sol: de = n  5 .d 2



30 = 2 5 d 2

75000  100000 MHP 0.75

 d = 22.73 cm Select near higher size i.e. 25 cm

10.

Ans: (c)

Sol: pump = 08.

Ans: (b)

Sol: Power transmitted by the pipe,

P = gQ(Hhf) For maximum power transmission, the H condition is hf = 3 P = gQ = gQ

H  H   3  2H 3

ACE Engineering Academy

pump =

Fluid power Shaft power gQH  h f  Pshaft

Given H = 10 m Q = 0.1 m3/sec hf = 5 m 1=

1000  10  0.1  10  5 PShaft

 PShaft = 15000 W = 15 kW

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: 120 : 11.

Ans: (c)

h f1

d   2 h f2  d1

Sol: V1

V2

Given d2 = 2d1

h f 2 = 32 h f1 = 32 h

Losses due to sudden expansion, 13.

2g

V2 = 1 2g

 V2 1  V1 

  

5

5

d2

V1  V2 2

  

d h f1  2   1  5 1      h f 2  d   2  32    

d1

hL =

ME _ GATE_Postal Coaching Solutions

Ans: (b)

Sol: K = 2  109 N/m2

2

Given  = 965 kg/m3 k 2  109 ⋍ 1440 m/sec   965

C=

By continuity equation, Q = A1V1 = A2V2 V A d  2  1   1 V1 A 2  d 2 hL =

V12 2g

hL =

9 V12  16 2g

 1 1    4

2

 1      2

2

2

Ans: (b)

Sol: d1 = 100 mm = 0.10m

v1 = 5m/sec d2 = 200 mm = 0.20m Q  A1V1 

Q = A2.V2 V2 

Ans: (d)

f .l.V 2 f .l.Q 2  Sol: hf = 2gd 12.1  d 5

for same discharge 1 d5

ACE Engineering Academy

Q 0.0392   1.25m / sec A 2   0.20 2 4

Head loss due to expansion hL 

Keeping parameters constant, Except “d”,

hf =

  0.10 2  5 4

Q = 0.0392 m3/sec

hL 9  V12 16 2g 12.

14.

hL

(V1  V2 ) 2 2g

2  5  1.25 

2  9.81

 0.717m

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: 121 : 15.

Fluid Mechanics & HM

Ans: (b)

Sol: Pipes are in parallel

Chapter- 9

Qe = QA + QB ------- (i)

Elementary Turbulent Flow

hLe = h L A  h L B Le = 175 m

01.

Ans: (c)

f e L e Q e2 f A .L A Q 2A f B L B Q 2B   12.1D 5e 12.1D 5A 12.1D 5B

02.

Ans: (a)

2 0.020  150  Q 2A = 0.015  200  Q B  5 12.1 0.08 12.1 0.1

03.

Ans: (d)

04.

Ans: 2.4

fe = 0.015

QA = 1.747 QB -------(ii)

Sol: Given: V = 2 m/s

From (i) Qe = 1.747 QB + QB

f = 0.02

Qe = 2.747 QB --------(iii)

Vmax = ?

0.015  1752.747Q B  0.015  200  Q 2B  5 12.1 D 5e 12.1 0.08 2

De= 116.6 mm

Vmax = V(1 + 1.43 f )



= 2 1  1.43 0.02

≃ 117 mm



= 2  1.2 = 2.4 m/s

B

summit A h = summit height

A

C

C SDatum

05. Head difference

Ans: (c)

Sol: Given data:

D = 30 cm = 0.3 m Re = 106 f = 0.025

Fig. Siphoning Action

Thickness of laminar sub layer,  = ?  =

11.6 V*

Where V* = shear velocity = V

f 8

 = Kinematic viscosity

ACE Engineering Academy

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: 122 :

Re =

V.D 

 =

V.D Re

07.

Ans: 1.66

Sol: k = 0.15 mm

 = 4.9 N/m2  = 1 centi-stoke

VD 11.6  Re  = f V 8

k 0.15  10 3  '  11.6       V* 

o 4 .9 = = 0.07 m/sec  1000

V* 

11.6  D  = f Re 8 =

ME _ GATE_Postal Coaching Solutions

 = 1 centi-stoke

11.6  0.3 10 6 



0.025 8



= 6.22 105 m = 0.0622 mm 06.

Ans: 25

08.

Sol: Given:

 =  kg/sec Flow rate m

hL= 10 m

 = 0.001 N-s/m2

=? For any type of flow, the shear stress at  dP R  dx 2

 = 1000 kg/m3 fD 

64 …………. for laminar Re d

fD = 0.316 Red–0.25 …. for turbulent

gh L R =  L 2

=

Ans: 480

d = 5 cm = 0.05 m

D = 0.1 m

=

0.15  103  0.905 11.6  10 6 0.07

Sol: Given:

L = 100 m

wall/surface  =

1 10 4 stoke   10 6 m 2 / sec 100 100

g = 10 m/sec2

gh L D  L 4

  AV    m

1000  9.81  10 0.1  100 4

= 24.525 N/m2

  1000 

  d2  V 4

 0.052  V 4

V = 1.6 m/sec

= 25 Pa ACE Engineering Academy

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: 123 :

Re 

VD 1000  1.6  0.05   0.001

11.

 Flow is turbulent  fD = 0.316 ReD–0.25 = 0.316 (80000)–0.25 = 0.0187

fQ 2 12.1d 5

Q

36  0.6 m 3 / sec 60

R  2 log 10    1.74 4f  K

fLV 2 fLV 2   g  2gD 2D

1  300   2 log10    1.74 4f   3 

0.0187  1  1.6  1000  2  0.05  478 Pa / m  480 Pa/m 2

4f = 0.03 f = 4f = 0.03 0.03 1000  0.6 2 hf  = 11.61 m 12.1 0.6 5

Ans: 20%

Sol: Since, Discharge decrease is associated

Power = wQhf

with increase in friction.

= 9.81  0.6 11.61 = 68.35 kW

df dQ  dQ   2   2  f Q  Q

12.

 2 10  20%

10.

hf 

1

Pressure drop (P1 – P2) = hf  g

09.

Ans: 68.35

Sol: Power lost per one km length = wQhf

= 80000 > 2000 (ReD)



Fluid Mechanics & HM

Ans: 50

Sol: 50 times Dia (or) 7% of Re

Ans: 4.1

Sol: Rectangular duct = 400 mm  250 mm

Non circular section Re =

VD H 

Where, DH = Hydraulic discharge = 4 DH 

4  0.4  0.250  0.307 m 20.4m  0.250

Re 

VD H 20  0.307  4.1  105  4 0.15  10 

A P

= 4.1 lakh x = 4.1 ACE Engineering Academy

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: 124 :

R e at x 0.6 m 

Chapter- 10 Boundary Layer Theory 01.

 am 

Ans: (c)

Sol: R e Critical

Assume water properties

K.x Re

04.

R e1

x1

1m

(At given distance ‘x’)

(x1 + 1)

 x A  B

1 256 16   = 1.6 100 10 2

Sol:

x=

Ans: 7.5

05. x = 0.6 m Leading edge

Flat plate

air = 15  10–6m2/sec Blasius constant, k = 5

ACE Engineering Academy

2  3

x1 x1  1

5x1 = 4 = x1 = 80 cm

lam

U .x Re   

x1 x 1  1

x 4  1 9 x1  1

Air, U= 4m/sec

 at x = 0.6 m

B

A

R e2 1  R e1 2

03.

 7.5 mm

A = 2 cm B = 3 cm

Ans: 1.6

Sol:  

160000

Sol:

6  x critical 1 10 6

1

5  0.6

Ans: 80

xcritical = 0.08333 m = 83.33 mm 02.

4  0.6 U x = 160000 =  15  10 6

 am at x  0.6m 

U x   critical  critical

5  10 5 

ME _ GATE_Postal Coaching Solutions

Ans: 1.5

Sol:  = 

du dy

(Newton’s law of viscosity) =

d  y  u m  1.5  dy  

 =   um  1.5 

1 

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: 125 :

 = 1.5

u m 

08.

0



y  y    1   dy    0

Ans: 2

Sol:  



  1  1    ydy   2  y 2 dy 0   0

1 



1 x

1  2

 x

1  2 1 3    2 3  2  

1  42 2

07.

Ans: (0.5)

Sol:

U y  U   U *   1  U 0

 dy 





1 ydy  0 

1  y2   = y 0      2 0   1  2        2  2 2

*  / 2 1   0.5   2  ACE Engineering Academy

09.

Ans: 3

Sol:

U y  U  * ? 

 y   1  dy  0

0

     2 3 6

 /6 1   0.167  6  



  dy 



1  y2  1  y3      2    2 0   3 0

x2 x1



U  U  1  dy U  U 

Sol:   

By comparing, K = 1.5 06.

Ans: (0.167) 

u m 

=K

Fluid Mechanics & HM

 u   dy *   1  0  u 

=





y

 1  8  dy 0



y2 = y 2 0 = 

   2 2

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: 126 :

 



0





0

u u

1.226  0.15  10 4  6 = = 0.015 N/m2 3 7.33  10

 u  1   dy  u 

1  2

y y 1   dy 8 

x2   x L  x 1  X L / 2

 /2 3  /6 

12.

Ans: 22.6

Sol:

Drag force, FD =

*

10.

L = 3.0 m,  = 0.15 stokes

U L   

R ex L  

U = 2 m/sec

UL 6 1   4 10 5 4  0.15 10

Re 

Hence,

We know that

= 22.57 milli-Newton 13.

 du  Sol:      dy 

Ans: 1.62 

Sol: m  AU   B   U  (∵  = L)

U U  y 







1 1  m ab  B   .U  2 2

m bc 

.U  

=

 x L 

.v U     v   at x  L

ACE Engineering Academy



m ab  m bc  m cd

On differentiating 

1.328 1.328   2.09  10 3 5 Re 4  10

1 D.F, FD =  2.09  10 3  1.2  1.5  3  2 2 2

Kx 4.64L 4.64  1  7.33 mm     x L Re Re 4  10 5 Ans: 21

UL 23   4  10 5  0.15  10 4

CD 

Since, Re (x = L) < 5  105

11.

1 CD..AProj. U 2 2

B = 1.5 m,  = 1.2 kg/m3

Ans: 7.33

Sol: R e  x L 

L

 2  0.015 N / m 2 = 21 milli Pa

     2 3 6

Shape factor =

L / 2

 x  L / 2  2 x  L



y 2 y3   2 3 0 

ME _ GATE_Postal Coaching Solutions

1  1.2  1 1.5  10 3  30 2

= 1.62 kg/minute

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: 127 :

Fluid Mechanics & HM

03. Ans: 0.054 Sol: Given data:

Chapter- 11

V = 8 m/s

Force on Submerged Bodies

D = 0.06 m  = 1.2 kg/m3

01. Ans: 8

 = 1.6104 m2/sec

Sol: Drag power = Drag Force  Velocity

W=?

P = FD  V P = CD 

Re=

AV 2 V 2

For flow over sphere; CD = 0.5 1000 < R e < 1105

3

PV

P1  V1    P2  V2 

W = FD

3

P1  V    P2  2V 

V.D 8  0.06  = 3000  1.6  10  4

W = CD 3

AV 2 2 1.2 

W = 0.5

P2 = 8P

 0.062  82 4 2

W = 0.5  0.108 = 0.054 N

02. Ans: 4.56 m

04. Ans: 4

AV 2 Sol: FD = CD. 2

Sol: Given data:

l = 0.5 km = 500 m

 ( D) 2  V 2 W = 0.8 1.2  4 2

d = 1.25 cm VWind = 100 km/hr Air = 13.4 N/m3

(Note: A = Normal (or)  projected Area = D 2 ) 4  10 80  9.81 = 0.81.2 (D) 2  2 4

2

 = 1.4105 m2/s CD = 1.2 for R e > 10000 CD = 1.3 for R e < 10000

D = 4.56 m

ACE Engineering Academy

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: 128 :

ME _ GATE_Postal Coaching Solutions

 = 45o L

T = 25 N Cable

VWind = 54 km/hr

d

= 54  VWind

5 = 15 m/s 18 WKite

Towers

ACE INDIA

 100  5  500  V.L  18  Re   1.4  10 5 

A=1m2 Effective F

VWind = 45

Note: The characteristic dimension for

electric power transmission tower wire is “L” R e = 992 106 > 10,000

T=25 N 45

T=25N

FD = Tcos45o

AV 2 FD = CD 2

CD

 13.4  2 L  d V  9 . 81  = 1.2  2 5   13.4  1.2   500  0.0125100   18    9.81  = 2

2

AV 2 = 25  cos45o 2

 12.2  2 CD   115 1  9.81   25  2 2  CD = 0.126 Resolving forces vertically

= 3952.4 N

FL = WKite + Tsin45o

= 4 kN

C L AV 2 = 2.5 + 25sin45o 2

Sol: Given data:

WKite = 2.5 N A = 1 m2 ACE Engineering Academy

WKite=2.5(N)

Resolving forces horizontally

 CD = 1.2

05.

F

 12.2  2 CL  115 25  9.81  = 2.5 + 2 2 CL = 0.144 Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally

: 129 : 06. Ans: (a)

Fluid Mechanics & HM

08. Ans: (c)

Sol: Given data:

C D 2 = 0.75 C D1 (25% reduced) Drag power = Drag force  Velocity C AV 2 P = FD V = D V 2 AV 3 P = CD 2

CDV3 = constant = C C D2

 C D1   0.75C D 1 

   

1

V   2  V1

3



  

Velocity = 5.6 kmph = 1.56 m/s sea water = 1.025  = 1.67  106 m2/s Power required =? Drag power = Drag force  Velocity

3



C D AV 2 V 2

0.7  1025  (1.56) 3   45 2

V2 V1

= 60.766 kW

V2 = 1.10064V1 % Increase in speed = 10.064%

10. Ans: 318 Sol: Width = 3 m

Height = 0.8 m

07. Ans: 0.1875

Velocity = 50 kmph = 13.89 m/s

Sol: Given:

= 1.25 kg/m3

FD = 300N  = 0.8  1000 = 800 kg/m3 L=2 m

CD = 1.1 Drag force F 

D = 80 mm = 0.08 m V = 5 m/s CD = coefficient of drag V 2 FD  C D . A 2 300  C D 

Sol: Area = 45 m2

CD = 0.7

Keeping , A and power constant C D1

09. Ans: 60

C D AV 2 2

1.1 1.25  3  0.8  13.89 2  2 = 318.33 N

800  52  0.08  2 2

 CD = 0.1875 ACE Engineering Academy

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: 130 :

ME _ GATE_Postal Coaching Solutions

09. Ans: (b)

Chapter- 12

Sol: According to Froude’s law

Dimensional Analysis 01.

Ans: (c)

02. Ans: (b)

04. Ans: (c)

05. Ans: (b)

06. Ans: (c) Sol: L r 

03. Ans: (b)

Tr  L r tm  Lr tp

tp 

tm 10  Lr 1 / 25

tp = 50 min

1 30

VP = 15 km/h

10. Ans: (a)

5

2

6

2

0 = 1.5110 m /s

Sol: VP = 10 m/s dia = 3m

w = 1.0210 m /s

Vm = 5 m/s

Reynolds law,

Fm = 50 N

Vr 

r Lr

Fp = ? 5

Vm 1.51  10   30 3 15  10 1.02  10 6 60  60

Acc to Froude’s law:- Fr  L3r (But Lr is not given) P  V 2 

Vm = 1850.5 m/s

F A

AV2 =F 07. Ans: (c)

Now scale ratio:

08. Ans: (a) Sol: L r 

Fm Vm2 A m  m    FP VP2 A p  p

1 16

2

QP = 1024; Qm = ? Q r  L5r / 2 

Qm  1    Q p  16 

Reynolds law

Qm 1   QP  16 

5/2

5/ 2

2



50  1   5        A  L2r FP  10   10 



(∵same fluid) FP = 20000 N

Qm = 1 m3/sec ACE Engineering Academy

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: 131 :

Fluid Mechanics & HM

11. Ans: (a)

14. Ans: (a)

Sol: L = 100 m

Sol: L r 

VP  10 m / s ,

Lr 

1 100

am = 0.013

1 25

As viscous parameters are not discussed follow Froude’s law Acc to Froude

1 am  L r  6 ap

ap 

am

L r 

Vr  L r Vm  Vp

1 6



0.013  1     100 

1/ 6

ap = 0.028

1 25

15. Ans: (a)

1 Vm   10 = 2 m/s 5

Sol: L r 

1 9

yp1 = 0.5 m , yp2 = 1.5 m

12. Ans: (c)

L 1 Sol: Stilling basin L r  m  L p 20 hr = Lr

qm = ? ,

qp = ?

2q 2P  y1p .y 2 p y1p  y 2 p  g

hm 1  h p 20

2q 2P  0.5  1.5  0.5  1.5 9.81

hp = 200.20

2q 2P  0.51.52 9.81

hp = 4 m

q p  2.71 13. Ans: (c) Sol: Lr = 1 : 25 ,

Fm = 5 N , Fp = ? Fm  L3r Fp

qr 

qm  L3r / 2 qp

1 qm    9

3/ 2

 q p = 0.1 m3/s/m

Fp = 78.125 kN

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: 132 : 16. Ans: (c)

ME _ GATE_Postal Coaching Solutions

TP = (1260) + 24

Sol: For distorted model according to Froude’s

law

= 744 min Tm = 0.0516  744 ≃ 40 min

Qr  LHL

3/ 2 V

LH = 1:1000 ,

19. Ans: (d)

LV = 1:100

Sol: Froude number = Reynolds number. 3

Qm = 0.1 m /s Qr 

r = 0.0894

1  1    1000  100 

3/ 2



0.1 Qp

If both gravity & viscous forces are important then  r  L r 

QP = 105 m3/s 17. Ans: (a) Sol: LH = 1:1000 , LV = 1:100

qm = 0.1 m3/sec qP = ? qr = (LV)3/2 3/ 2 qm  1    q P  100  QP = qm100 = 0.11000 = 102

Qr    q r   LH   3/ 2  L H .L V     LH     L3 2  V    

m p

3/ 2

3

 Lr

Lr = 1:5

18. Ans: (b) Sol: L H 

LV 

1 , 150

Tr 

LH LV

1 60

Tm 1 60   TP 150 1 Tm = 0.0516 TP The actual time interval between two successful high tides in a sea  12 hour 24 min ACE Engineering Academy

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: 133 :

Fluid Mechanics & HM

Where D = diameter of wheel

Chapter - 13

N = speed of turbine = 600 rpm

Hydraulic Machinery 01.

H = Head available of pelton wheel turbine = 300 m

Ans: 1000

Sol: T = Moment of momentum of water in a



turbine = Torque developed = 15915 N-m

  D  600  0.41 2  9.81  300 60 D = 1.0 m

Speed (N) = 600 rpm 2NT 60

04.

Ans: (b)

2    600  15915 60

05.

Ans: (b)

Power developed = =

= 1000  103 W = 1000 kW

Sol: P = 8.1 MW = 8100 kW

H = 81 m 02.

N = 540 rpm

Ans: 4000

Sol: Q = 50 m3/sec

Specific speed NS =

H = 7.5 m Turbine = 0.8 Turbine = 0.8 =

=

Pshaft Pshaft  Pwater gQ(H  h f )

Pshaft 1000  9.81  50(7.5  0)

=

N P (H)

5

4

540  8100

81 4 5

540  90 = 200 243

60 < NS < 300 (Francis Turbine)

Pshaft = 2943103 W = 2943 kW =

03.

2943 HP = 4000 HP 0.736

Ans: 1

DN  k u . 2gH 60

ACE Engineering Academy

07. Ans: (b) 08. Ans: (a)

Sol: We know that

U

06. Ans: (a)

09. Ans: (d)

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: 134 :

ME _ GATE_Postal Coaching Solutions

10. Ans: (d) 12. 11.

Ans: 72

Sol: Given P1 = 100 kW

Ans: 1000

H1 = 100 m and H2 = 81 m

Sol: Given Np = 500 rpm

We know that

Dm 1  Dp 2

 P   P       H 3 / 2    H 3 / 2  2   

We know that  ND   ND       H m  H P Given H is constant  

N m Dp  Np Dm



P2 100  3/ 2 100 803 / 2

P2 = 71.55 kW ≃ 72 kW  New power developed by same turbine = 72 kW

Nm 2 500

 Nm = 1000 rpm

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Heat Transfer Solutions for Vol – I _ Classroom Practice Questions Chapter‐ 1

2K

Conduction

r3

r2

K r1

01. Ans: (b) Sol: Case (1): Higher thermal conductive material is inside and low thermal conductive material is outside

Q2 = K

r3

r2

2K r1

2LT  20  ln  ln 30  10   20 K 2K

Q2 = 1.116(2KL(T)) ……… (2) From (1) and (2) Q1 > Q2 Means that, lower thermal conductivity

Q1 =

2LT r r ln 2 ln 3 r1 r2  K1 K2

material should be used for inner layer and higher thermal conductivity material should be used for outer layer so that heat transfer will be lower  insulation is effective.

Let, r1 = 10 mm, 02.

r2 = 20 mm, r3 = 30 mm Q1 =

2LT  20  ln  ln 30  10   20 2K K

Q1 = 1.32(2KLT) …….. (1) Now, Case (2): Higher thermal conductivity

material outside, lower thermal conductivity material is inside. ACE Engineering Academy

Ans: (a)

30 =15mm 2 r2 = 15+25 = 40 mm r3 = 40 +25 = 65 mm

Sol: r1 =

5K r1

r2

K

T r3 Q1  r  r  ln 2  ln 3   r1    r2  2KL 2KL T2K    1.466T2K   40   65  ln  ln   15    40  1 5

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: 138 :

ME – GATE _ Postal Coaching Solutions

04. Ans: (c) Sol: dx

K r1

Q2 

r2

5K r3

x

2K T   0.927.T2K   40   65  ln  ln   15    40  5 1

% decrease in heat transfer 1.466  0.927   100  36% 1.466

Heat conducted from ice = heat removed to form 1mm thick ice over 200mm ice block

 kA kA

03. Ans: (b) Sol: ri = 0.0025 cm

dT    Adx  LH x x2

220 = 10  335  10   xdx 3

ri

V/30cm = 3.6V I = 0.5A

r0

0

Ti = 167 C T0 = 150C

3

x1



10 3  335  10 3 40

0.201

 xdx

0.200

0.201

HG = HT Through H2 tube =VI Ti  T0 r  n o   ri  2K  0.3

 0.5  3.6 

dT    V  LH dx

kdT = –LHxdx

ro = 0.125 cm

 VI 

00C

335  10 6  x 2     40  2  0.200 

335  10 6  0.2012  0.200 2    40 2  

 1680 sec

167  150  0.125  ln    0.0025  2K  0.3

 K = 0.22 W/m.K

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: 139 :

Heat Transfer

05. Ans: (c) Sol: H = 4m,

Ti = 30C,

H.T from inside air to inside wall = H.T from

L = 10m,



21  10.17 10.17   23  δ 1 1  1.2A 23A 8A

30  10  A  3600

h0 = 4W/m2K

δ 1 33.17   1.2 23 86.64

1  1   hi k h0

20  40  3600  3840 kJ

86.64 

1 0.115 1   2.5 1.15 4

33.17  δ  0.407 m δ 1  1.2 23

Common Data for Q. 08 and Q. 09

06. Ans: (d) Sol:

inside wall surface to outside air

T0 = 10C, K = 1.15 W/mK

hi = 2.5W/m2K, Q

 = 0.115 m

K1 1 r1  ,  0.8 K 2 2 r3

08. Sol:

r1

r2 – r1 = r3 – r2

r3

Due to steady state H.T.

Tmax

r2

T1 = 160C

T2 = 120C

Q1 = Q2 4K1r1r2 T1  4K 2 r2 r3 T2   r3  r2  r2  r1 

x L=0.02m

T1 K 2 r3 2    2 .5 T2 K1r1 0.8

QG = 80 MW/m3 = 80  106 W/m3 T1 = 160C,

07. Ans: (b) Sol: To avoid condensation in the building, the

T2 = 120C 2

k = 200 W/m K

inside wall temp should be greater than or

For 1-D steady state with heat generation

equal to DPT

equation is

10.17oC hi = 8 Ti = 21

2

ho = 23W/m K K = 1.2 

ACE Engineering Academy

d 2T QG  0 k dx 2

T = –23oC

Q Q d 2T dT  G    G x  C1 2 dx k k dx T

 QG 2 x  C1 x  C 2 ------ (i) 2k

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: 140 :

ME – GATE _ Postal Coaching Solutions

Where, C1 & C2 are constants that can be

11. Ans: (a)

evaluated by boundary conditions.

Sol: T = 100,

At, x = 0 , T = 160C

K = 20 W/mK,

160 = C2

RC = 1.016 K/W

At x = L = 0.02 m , T = 120C , 120  

d = 2cm

80  10 6 0.022  C1  0.02  160 2  200

C1 = 2000  QG 2 T x  2000 x  160 ------ (ii) 2k To get maximum temperature dT 0 dx Q   G  2 x  2000  0 2k 2000  k 2000  200  QG 80  10 6 = 5  10–3 m = 5 mm

x

Q 

Sol:

By putting the value of x = 510–3 m in eq.(ii) 2 80  106  5  10 3  2000  5  103  160 2  200





T R1  R C  R 2 100     0 . 03   2   1.016   20    0.02 2   4  

 9.42W

12. Ans: (b) Sol: Q =

09.

T

3 cm

Ti  T0 r2  r1 4Kr1r2

Ti at radius 12.5 cm = Q  = 30  103

= 165C

r0  ri  T0 4Kri r0

0.15  0.125 + 400 4  70  0.15  0.125

= 45.47 + 400 = 85.47 10. Ans (a) Sol: Q =

=

T Total Resistance 130  30 0.004  0.002  0.002

= 12500W = 12.5 kW ACE Engineering Academy

13. Ans: (d) Sol: Critical radius of cylinder, rc =

K 0.5 = h 20

= 0.025 m = 25 mm Critical thickness = 25 – 10 = 15 mm

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: 141 :

14. Ans: (a) Sol: Critical radius, rc =

Heat Transfer

π 2 (D  L)  ρ  C P (TmP  Ti ) 4 = I 2 RL

k 0.1  h 10

 2 D    C P (TmP  Ti ) = 4 I2R

= 10 mm = 1 cm For cable as r1(=1.5 cm) > r1(=1 cm), the heat transfer decreases by adding insulation.

18. Ans: (b)

15. Ans: (c)

Sol: r2 = 5 mm, h0 = 10 W/m2K, ki = 0.04 W/mK

Sol: r1 = 1 mm ,

Max. heat flow occurs at critical radius or

K 0.175 = = 1.4103 m rc (= r2) = h 125

diameter

rc = 1.4 mm ,

dc = 2 rc = 2 

As r1( = 1 mm) < rc(= 1.4 mm) The heat transfer increases by adding insulation till r2 = rc and then it decreases. 16. Ans: (c)

Sol:

Req = R1 + R2 Q2

2L L L = + k k1 k 2

k1

2 1 1 = + k k1 k2

L1

k2 L2

hc1

glass cover

hr1

Plate  Rconv = 0

k1  k 2 2k1k 2  k= k1 k 2 k1  k 2

R c1 

1 h c1

R c2 

Insulation

1 h c2 10C

70C

17. Ans: (a) Sol: Time required for melting the rod

Heat required for melting the rod Heat generated

m C P (TmP  Ti ) = I2R  L ACE Engineering Academy

hr2

hc2

Q1

L 2L L1 + 2 = kA k1A k2 A

=

4k i 4  0.04 0.16 = = 16 mm = 10 10 h0

19. Ans: (a)

Sol: Q1 = Q2 = Q

=

=

2k i h0

R r1 

1 h r1

Rins 

R r2 

1 h r2

 k

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: 142 :

R ins 

ME – GATE _ Postal Coaching Solutions

20. Ans: (a)

0.05 1 0.05

Sol:  = (81 – 76) = 5cm = 0.05m

1 1 R C1  , R r1  3 6

Q

1 1 , R r2  25 5



R C2 

1 1  3 6  0.111 R1  1 1  3 6

k.A.T  6  0.166  0.51 0.66   204  38  6 0.05

= 1.2 kW

1 1  25 5  0.0333 R2  1 1  25 5 R3 = 0.111 + 0.0333 = 0.1443 R eq 

0.1443  1  0.126 1  0.1443

Q 70  10 60    476.19 W A R eq 0.126

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: 143 :

Chapter‐ 2 Convection

Heat Transfer

Dh =

4 A c 4  0.5  1   0.67 m P 21  0.5

 Re =

1.2  10  0.67  4.46 105 18  10 6

01. Ans: (b) 06. Ans: (d) 02. Ans: (a)

Sol: Because Re > 2000, the flow is turbulent

Nu = 0.023 Re0.8 Pr0.33

03. Ans: (a)

= 0.023 (4.46 × 105)0.8 (0.73)0.33  = 0.88

h De = 685.6 k

V = 50 m/sec

h=

04. Ans: (a) Sol: Cf = 0.004 ,

 = 2.286 10-5 ,

Cp = 1001 J/kgK Pr 

Heat transfer Q = hA Δ T = h( Pl )ΔT

Sat. Stream

C p

685.6  0.025  25.58 W / m 2 K 0.67 = 25.58 × 2(1.5) × 1 (30–20)

k

= 769 W/m

2 3

C St  Pr  f 2 h  VC P h

2 P

2 3

Sol: L0 = 320 cm,

Sat. water

Tp = 1500C,

C f  VC P 2  Pr



07. Ans: (c)

Cf

Q0 = 8 kW

2 3

Q = h.AT ( keeping others constant)

0.004  0.88  50  1001  2.286  10  1001   2   0.035   5

2 3

2

= 117 W/m K

05. Ans: (c) V D h Sol: Reynolds No (Re) = 

Because of rectangular duct, it has to be converted into equivalent circular duct. ACE Engineering Academy

T = 100C

QA Q  DL QL Q1 L1  Q0 L0  L1 

Q1 1  L 0   320  40cm Q0 8

08. Ans: (c)

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: 144 :

Heat Transfer

(ii) Q/m length = hA T

09. Ans: (b) Sol:  = thickness of hydrodynamic boundary

= 6.109    0.0251 20

layer = 0.5 Pr 

C p

k

= 9.596 W/m

 1     t  0.5mm

13. Ans: (a) Sol: Heat generated = Heat transfer by

convection

10. Ans: (a)

5  107  AC  L = 1000  As  T 11. Ans: (d)

5107  r2 L = 1000  2rL  (Ts – 75)

Sol: Qt = Qtop + Qbottom + 4  Qside

Ts = 625 + 75 = 700C

6  hA T = h1A T + h2A T + 4h3A T 6h = h1 + h2 + 4h3 h

12.

14. Ans: (b)

h 1  h 2  4h 3 6

Sol:

At No slip region

Ans: (a)

Sol: d = 25mm = 0.025 m

  Q cond  Q convection

V = 1m/sec

 dT    h  ATw  T  … (1) kA   dy  y 0

 = 0.88 kg/m3  = 2.286  10–5 kg/ms

T  Tw = 1 e3500y T  Tw

Cp = 1.001 kJ/kgK k = 0.035 W/mK Re 

Vd  962.38  2000  flow through 

pipe is laminar For constant flux condition, (i) Nu 

48  4.364 11

Nu  4.364  h

T  Tw = 1 e3500y T  Tw

T Tw = (1e3500y)(T Tw) dT  3500  e 3500 y  T  Tw  ……. (2) dy Sub (2) in (1) kA 3500e3500y(TTw) = hAs(TTw)

h 

y 0

hd k

= 0.03 3500 e3500 0

h = 105 W/m2K

4.364  0.035 = 6.109 W/m2K 0.025

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: 145 :

15.

Heat Transfer

18. Ans: (d) 5  0.01 Vd Vd = =   26.66  10 6

Sol: Re =

2

G r  2  3 g T      gT Sol:    R e2 2  v 

= 1875.46 < 2300 Hence flow is laminar

Nu 

T  constant

= 3.66

20. Ans: (c)

hd = 3.66 k h =

19. Ans: (b)

21. Ans: (b)

3.66  k d

Sol: k = 1.0 W/mK

3.66  0.1351 = 49.44 W/m2K 0.01

Re = 1500  means that the flow is laminar D = 10 cm = 0.1m For a fully developed laminar flow

16. (i) Ans: (c) , (ii) Ans: (d)

Through pipe

2

Sol: kw = 0.6 W/m K

kg = 1.2 W/m2K , Tw = 480C  T     1  10 4 K/m  y  w

(i) With constant heat flux

Nu = 4.364  constant =

As Qw = Qg

 dT   dT    k w A   k g A   dx  g  dy  w 0.6  dT   1  10 4  0.5  10 4     dx  g 1.2

4.364  1.0  43.64 W / mK 0.1

(ii) With constant wall temp

Nu = 3.66 =

But, Qw = Qconv  dT  k w A   hA(48  40)  dy  y h

h

hD k

h

hD k

3.66  1.0  36.6 W / m 2 K 0.1

0.6  1  10 4  750 W / m 2 K 8

17. Ans: (d)

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: 146 :

Heat Transfer

In case of rod of length “2L” with both ends maintained at same temp of 1000C, at the

Chapter‐ 3

center the temp gradient is zero, means that

Extended Surfaces ‐ FINS

from one side upto the center of long rod, it can be treated as short fin of length 01.

Ans: (a)

Sol: Ab = 1 m2 ,

2L  L , for which temp distribution is 2

As = 2 m2

h = 20 W/m2 K

  

 = 0.75 Tb = 50C ,

T = 30C



 Q fin

  Q without fin   Q  Q fin without fin



= 0.75  h  As  (Tb – T)

Cosh (m(L  x )) 60 Cosh (0)  Cosh (mL) Cosh (m  L)

60 1  2.063  Cosh   L  L  60 60   15 Cosh(2.063) 4

  T  40 15  T  55o C

= 0.75  20 1  (50 – 30) = 300 W

03. Ans: (d)

02. Ans: (c) Sol: T  40 o C, T0  100  C,

 100  40  60

04. Ans: (b) Sol:

30 cm 2000C

  55  40  15

1 cm

In case of rod of length “L” with insulated

10 cm

tip, the temp at the end of tip = 55C Cosh m L  x   1    Cosh (mL) Cosh mL 15 1 60   Cosh mL   4 60 Cosh mL 15

20 cm

LC = L + = 30 +

mL = Cosh-1 (4) = 2.063 m

2.063 L

ACE Engineering Publications

m=

T=200

D 4 1 = 30.25 cm = 0.3025 m 4

4h = kd

4  15 = 9.607 65  0.01

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: 147 : T1  T cosh mL C  x 1   Tb  T cosh mL C 

Heat Transfer

x  x2 k  2  k2 = k 1  2  x1 k1  x1 

T1  20 cosh 9.6070.3025  0.1  200  20 cosh 9.607  0.3025

2

2

2

1  20  k 2  k 1    200     50W / mK 2  40 

T1  20 = 0.389  180  T1 = 90 T2  20 cosh mL C  x 2   200  20 cosh mL C

06. Ans: (b)

T2  20 cosh 9.6070.3025  0.2   200  20 cosh 9.607  0.3025 T2 = 49.93  50

Sol:

T  T cosh mL  x   Tb  T cosh mL

If, x = L T  T 1 = Tb  T cosh mL

( cosh0 = 0)

05. Ans: (d) Sol:

40cm

T=250C

T (1) Tb = 500 At x = 0

long fin

=25 mm

60at x  L

(2) x2 = 20cm

T

Given, k1 = 200 W/mK, k2 = ? x1 = 40 cm,

x2 = 20 cm

1   0 e  m1x1

60  25 1  500  25 cosh(mL) Cosh(mL) = 13.57

 2   0 e m2x 2

mL = 3.299  L =

1 = 2 0

L=

e  m1x1   0 e  m 2 x 2

m1x1 = m2x2 x 2 m1  x1 m 2

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  m   

4h   kd 

3.299 = 4h kd

3.299 m

3.299  0.3894 m 4  15.7 35  0.025

= 38.94 cm  39 cm

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: 148 :

07. Ans: (c) Sol:

  40    5  10 3  400   5  10 3 2 Q 4  130  30  tanh 8.94  0.10125



20 mm T= 200

h = 8.5 W/m K hPAK   b

10.

P = d =   0.02 = 0.0628 m

Ans: 31   480  10 3 W Sol: Q gen

  A= d 2 =  0.02 2 4 4

Tb = 70C , T = 30C kAl = 170 W/mK ,

= 3.14  104 = 0.000314

h = 12 W/m2K

0

b = 100 20 = 80 C

 Q hpkA s  b tanh mL sin gle fin 

  8.54  0.0628  0.000314  400  80 Q loss

L = 1210–3 m

08. Ans: (c)

P = 21.410–3 m

T  T T0  T 1  T0  T  = ln  x  T  T 

A = 0.49  10–6 m2  Q  0.0158 W  15.81  10 3 W

4h kd

sin gle

 Total heat generated = n  Q dissipated in sin gle fin

4  17.45 1  125  28  ln  3 k  0.02 100 10  91  28   k = 187.37 W/m.K 09.

Ans: (b)

5 Sol: Effective length, Lc = 100   101.25 mm 4 Lc = 0.10125 m   hpkA tanh mL  Q b c

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12  2  1.4  10 3  20.08 170  0.7  0.7  10 6

m

= 20.76 W

m=



  5W Q

2

Sol: emx =

4h 4  40 = 8.94  kd 400  5  10 3

m

K = 400 W/mK

Qloss =

Heat Transfer

n

480  10 3  30.36  31 15.81 10 3

11.

Ans: (a)

12.

Ans: (c)

13.

Ans: (b)

14.

Ans: (c)

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: 149 :

Heat Transfer



Chapter‐ 4 Transient Heat Conduction 01.

 x = 0.25

Sol: D = 1 mm = 10–3m

03. Ans: (c)

 = 8400 Kg/m3

K = 25 W/mK,

Sol: T0 = 530 0C,

2

h = 560 W/m K

G = 400 J/kgK,

8400400103



According to the Newton’s law of cooling the rate of cooling is directly proportional to difference in temperature.

 = 4.6 sec

 dT  (TT) d

Ans: (b)

dT   K T  T  d

 dT   kA   dx  x 0 2

4

dT = 50 + 24x + 45x2 60x3 dx d 2T = 24 + 90x 180x2 2 dx d 3T = 90360x dx 3 dT d

Cooling rate or heating rate to be maximum or minimum

d  dT   0 dx  d 

 dT   K  d T  T0 T  T 0 T

3

T = 50  50x + 12x + 15x  15x

Cooling rate is

m 0.5 m = V= V  9000

=

e– = 0.01

Sol: Q entering

Time = 10 sec

T = 430C

5606

 0.01  e

T = 300C

m = 500 gm

 hA  VC

Given, t – t0 = 0.01 (tb – t0)

02.

 90360x = 0

Ans: (c)

t  t0 e tb  t0

d  d 2T  d 3T  2   0  3  0 dx  dx  dx



 T  T ln   T0  T



  = K 

T  T = eKt T0  T

430  30 = e10K 530  30  K = 0.0223/sec

Now next 10 sec means that at t = 20 Temp – ? T  30 = e0.0223(20) 530  30 T = 350.090 C

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: 150 :

04. Ans: (a)

Heat Transfer   hA  T  T  Exp    Ti  T  VC 

Sol: For lumped analysis to be used

Bi  0.1

  hA  150  90    Exp 450  90  VC 

hL C a = 0.1 (LC for solid cube is ) k 6

  = 7.4 min

Where, ‘a’ is a side of cube) ha max = 0.1 6k

amax =

08.

Ans: (a)

Sol: T = 65 + 80 r – 425 r2

0.1  6k h

0.1 6  206 = = 4.944 m 25

ri = 0.25 m

ro = 0.4 m

L – 1.5 m

k = 5.5 W/m.K

 = 0.04 m2/hr

d 1 d  dT    .  r.  dt r dr  dr 

05. Ans: (b) Sol: No dimensions given and given as large

mass, so consider it has semi infinite body.

1 d  dT   d   0.004     ri  ri dr  dr   dt  inside

 dT   3.5 0C/cm , T  100  5  95    dx  At x  0



= 0.016 (80 – 1700 ri)

0.41 2 m /sec  = 0.41 m /hr = 3600 2

(Q)At x = 0 = 

= 0.016 [80 – 1700 (0.25)] = – 5.52C/ hr

 kA T

 dT    kA    dx  At x 0

100  5 0.41   3600

0.004 d ri 80  850 ri  0.25 dr

1 d  dT   d    .  ro .    ro dr  dr   dt  outside

= 350   = 206 sec

06. Ans: (d)



 d 80 ro  850 ro2 ro dr



0.004 80  1700  0.4 0.4





= – 6C/hr 07. Ans: (d) Sol: Lumped analysis is valid ACE Engineering Publications

09. Ans: (d)

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: 151 :

Heat Transfer

10. Ans: (c)

(FO)A = (FO)B

Sol: V = 60 km/hr = 16.67 m/s

       2   2  L     c A  Lc B

Time = 6 sec , Aeff = 300  10–4 m2 1 (KE)associated = mV 2 2 1 2   1500  16.67  = 208.416 kJ 2 (KE)associated/brake

208.416   52.104 kJ 4

 k   A     c  A 

 k   B     c  B 

Rate of (KE)dissipated / brake 52.104   8.683 kW 6

40  2  10 5 2  106

20  1  10 6 7 2  10

2  10 5  2 1 10 6  B  0.42 0.12

 B = 2.5 hours

After 6th sec   8.683 kW Q   kAT0  Ts   kAT Q t t

K = 55 , A = 300  10–4  = 1.24  10–5 ,

t=6 T = 80.45C 11.

Ans: (c)

12. Ans: 2.5 (range 2.4 to 2.6) Sol: It is given in the question that at all times the

surface is remaining at temperature of ambient. ACE Engineering Academy

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: 152 :

Heat Transfer

Q12 

Chapter‐ 5 Radiation





A1 b T14  T24  27.32 W  1 A1  1   1  1 A 2   2 

01. Ans: (b)

04.

Sol: T1 = 800 K, T2 = 500 K

 b T14  T24 Sol: 1000  1 1  1 0.5 0.5

1 = 0.8, 2  0.6

Q 12 



 b (T14  T24 )

Ans: (b)





1 1  1 1 2



  b T14  T24  3000

8

5.67  10 (800  500 ) 1 1  1 0.8 0.6 4

4

= 10.26 kW/m2

 2  0.25

02. Sol: 1 = 2 = 0.8, T1 = 400K, T2 = 600K



3000 3000   600 W / m 2 2  4 1 5

05. Ans: (c)

s = 0.005

1

Q1s = Qs2







b T  T  T T  b 1 1 1 1  1  1 1  s 2 s 4 1

4 s

Ts = 527K q  Q1s 

03.



 b T14  T24   Q12 1 1  1 0.5 0.25 







4 s

4 2



b T  T  14.7 W / m 2 1 1  1 1  s 4 1

4 s

1 3  10   0.3 1 10 3 06. Ans: (b)

Ans: (b)

Sol: A1 = 20cm2,

1 1 2   2 Q 0.5 0.5 0.25 Sol:  1 Q 1 1  1 0.5 0.5

A2 = 100m2

T1 = 800K,

1 = 0.6

T2 = 300 K,

2 = 0.3

1

Sol: F21+F22 = 1

F22 = 0 F21 = 1

2 0.5

b = 5.67  10-8W/m2K4

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: 153 :

A1F12 = A2F21

1 2  1.5  0.375 4 4

=

F11+F12 = 1  F11 = 1 0.375 = 0.625

S = 0.05

Given

 DL  2  D 2 A2 4 F12   2 A1 4R    0.5  0.5  0.5 2 2  2 4  0.5 1

Heat Transfer

2 = 1 = 0.8   2   2   2  1 = 79  1  1  N   0.05   0.8   0.8  N=3 10.



Sol: Error =

=

07. Ans: (d)

1  C   TC4  Tw4 h









1 0.94  5.67  10 8 300 4  2984 5



= 2.2790 C (Error)

Sol: T1 = 10000 K, T2 = 5000K

1 = 1, 2 = 0.7 Irradiation of body 1 = 0.7  5.67  10–8  5004 + 0.3  5.67  10–8  10004 = 19.5 kW/m2

11. Ans: (c) Sol: % Reduction = 75%  Q S 1– = 0.75  Q W .S

 Q 1 S = 0.25 =  4 Q WS

08. Ans: (d) Sol: T1 = 1000 K, T2 = 500 K

1 = 0.7 , 2 = 0.8 Q12 





 b T14  T24 = 31.7 kW/m2 1 1  1 1  2

12.

Ans : (a)

Tc = recorded temperature

 Q 1 S   Q W.S 79

= 20 + 273 + = 293 K Tw = wall temperature

QWS = 79QS E b1  E b 2 79E b1  E b 2  = 1 1 2  1 1  1   1  N  1 1  2 1  2  S  ACE Engineering Academy

n=3

Sol:  = 0.9

09. Ans: (c) Sol:

1 1  n 1 4

= 5 + 273 += 278 K T = fluid temperature = ? h = 8.3 W/m2 K

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: 154 :

Qconv = Q radation



h T  Tc     b T  T 4 c

4 w

Heat Transfer

dpipe = 20 cm , pipe = 0.8 , Lp = 1 m



T  Tc  0.9  5.67 10 8  293

 278 4 8.3

4

Q  A pipe



T = 301.59 K = 28.59C



As room area is very large compared to

13. Ans: (b)

pipe, hence

Sol: F11+ F12 + F13 + F14 = 1

F14 = 1 – (0.1 + 0.4 + 0.25)



Ap A wall

 Tp4  T4 Q  1 d p L p  pipe

= 1 – 0.75 = 0.25 A1 F14 = A4 F41 F41 



 TP4  T4 1 Ap  1  1 11    1  pipe A wall   wall 

A1 4  F14   0.25  0.5 2 A4

0



Q  d p  Tp4  T4  pipe Lp

14. Ans: (c)

=   0.2  5.67 108(6734 3034) × 0.8

Sol: Let Gs = Solar constant = 1400 W/m2

= 5.6 kW

L = distance between sun and earth 4 π L2 G s  A s σ b Ts

16. Ans: 702 K

4

 4 π R 2 σ b Ts

4

  0.8  10 6  4    0.063 = 723.82 Sol: Q g 3

2

Qgenerated = Qconvected + QRadiated

L G T   . s  R  σb 4 s



723.82  h O .A S (TS  313)   sphere  A S TS4  T4 2

 1.5  1011  1400   Ts  A   5802 K 8  8  7  10  5.67 10

723.82  10  4  0.06  T5  313  0.9 



2



5.67  10 8  4  0.06  Ts4  2784 2



 TS = 702 K

15. Ans: (b) Sol:

(2)

T = 300C

17.

Ans: (d)

Sol: Qloss = 0.8    0.275  5.67  10–8 (7734 –

(1)

3034) = 13.7 kW

Tp = 400 + 273 = 673K

ACE Engineering Publications

18.

Ans: (d)

19.

Ans: (c)

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: 155 :

Heat Transfer

04. Ans: (b)

Chapter‐ 6

Sol:

200C oil

Heat Exchangers

100C T 70C

water

01. Ans: (a)

20C

Sol: Cc = 4180  2 = Cmax

Ch = 1030  5.25 = Cmin A = 32.5m2, U = 200 W/m2C UA 200  32.5 NTU    1.2 C min 1030  5.25

 0 C 0 200  100  m  w C w 70  20 m  0C0  m  WCW  2m

Now  0 C 0 200  T   m  w C w T  20  m

 200T = 2T40

02. Ans: (c) Sol: NTU = 0.5, Ch = Cc = C = 1

For counter flow NTU 0.5 0.5 1      0.33 1  NTU 1  0.5 1.5 3 03.

Parallel flow Heat exchanger

Ans: (d)

Sol: Given, C = 1, NTU = 2

c =

NTU 2  = 0.667 1  NTU 3

P =

1  e 2 NTU 1  e 2 ( 2 )  = 0.49 2 2

 c 0.667  = 1.36  p 0.49

T = 80C 05. Ans: (c) Sol: AMTD = 1.05 LMTD

     θ1  θ 2   θ1  θ 2    1.05    2   ln θ1    θ    2   θ1   θ1   1   1  θ   1.05 θ  θ 2 θ2  2 2   2   θ1         ln θ       2   x 1 x 1   1.05 2  ln x 

by trail and error x = 2.2

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: 156 :

06. Ans: (c) Sol:

Heat Transfer

NTU =

150C 80C

=

60C

UA 1000  5   1.25 C min 4000

1  exp 2  1.25 C C Tc 2  Tc1   2 C min Th1  Tc1 

0.4589 = 25C Parallel flow Heat exchanger

4000Tc2  15 4000102  15

Tc2 = 40+15 = 550C 09. Ans: (b)

1 = 15025 = 125C 2 = 8060=20C (T)max = 15080 = 70C (T)min = 60 25 = 35C    2 125  20 LMTD  1   57.29  1   125   n   n  20     2 NTU 

T max LMTD



150  80  1.22 57.29

Sol: Counter flow H.E. fluid A  hot fluid

Th1 = 420oC,

mh = 1 kg/s

TC1 = 20oC,

mc = 1 kg/s

 = 0.75,

C Ph =1000 J/kgK

C PC = 4000 J/kgK Because exit temp of both of the fluids is not given, effectives-NTU method is used Ch = 11000 = 1000 = Cmin CC = 1 4000 = 4000 = Cmax

07. Ans: (b) Sol:  

C

NTU 1  NTU

NTU 0.8   NTU  4 1  NTU

C min 1000 1    0.25 C max 4000 4

  0.75 

0.75  08. Ans: (b) Sol: Ch = CC = 1  4000 = 4000,

Th1 = 1020C TC1 = 150C,

U = 1000 W/m2K,

A= 5m2, parallel flow H.E 2 =1 exp (–2NTU) ACE Engineering Publications

C h (Th1  Th 2 )



C min Th1  TC1



1000(420  Th 2 ) 1000420  20 

Th 2  420  300  120 0 C

0.75 =



 C T  T  4000T  20  T C C TC 2  TC1 min

0.75 =

h1

C1

C2

1000420  20 

C2

 95 0 C

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: 157 :

Heat Transfer





m C C PC TC 2  TC 1  UA m

10. Sol:

1500  418780  30 A  3600 = 0.707 m2 2000  61.66

65C

65C

20C

12.

Sol: Q = 250103 W

Q = CW(Tc0Tci)

 w C w TC 0  TCi  250  10 3  m 250103 =

mw 

7500  4180(TC0  20) 3600

TC0 = 48.50 1 = 65  20 = 45

65

20

?

250103 = 2.0844180(Tc2 – 20) Tc2 = 48.69C 1 = 65 – 20 = 45 2 = 65 – 48.69 = 16.31

11. Ans: (a) 1500 mc   0.417 kg / sec 3600

Th1 = 120C,

C PC = 4187 J/kg K

TC 2 = 80C,

U = 2000 W/m2K constant,

the

LMTD 

1   2  28.269  1  ln   2 

UA  LMTD = 250 103

Because the hot fluid is steam and whose remains

65

Q = mwCpw (Tc2 – Tc1)

  2 LMTD  1  28.4  1  n    2 

temp

7500  2.084 kg / s 3600

U = 1250 W/m2.K

2 = 65  48.5 = 16.5

Sol: C1 = 30C,

Ans: (d)

flow

is

 A

250  10 3  7.07m 2 1250  28.269

immaterial 120  120 30  80 1 = 120 – 30 = 90, 2 = 120 – 80 = 40  m 

90  40  61.66 0 C  90  n    40 

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ACE Engineering Publications

Heat Transfer

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Thermodynamics Solutions for Vol – I _ Classroom Practice Questions 04. Ans: (c)

Chapter- 1

05. Ans: (b)

Basic Concepts 01.

06. Ans: (c)

Ans: (a)

Sol: Given 3

 = 1kg/m

P = 600mm Hg

g = 10 m/s2

dP 

du = f (T, v)

h

gdh gC

Mountain

P = 750mm Hg

dP g C g 750  6001  101.325 kPa   1  10 dh = 2000 m dh 

02.

Ans: (a)

Sol: Volume

 Extensive property

Density

 Intensive Property

Pressure

 Point function

Work

 Path function

Energy

 Point function

03.

08. Ans: (d) Sol: LFP = Lower fixed point

UFP = upper fixed point

C  LFP 0  300  UFP  LFP 100  300 C0 0  300  100  0 100  300 C = 1500 C 09. Ans: (d) Sol: When molecular momentum of system

absolute zero values.

Sol: Assertion is true

Reason is false. is

not

the

correct explanation, in this

07. Ans: (b)

becomes zero, the pressure reaches its

Ans: (c)

Reason

1 1  Sol: du = Cv (T2 – T1) + a     v1 v 2 

the

tip

of

thermometer has only contact with Hot gases; it has no contact with cold walls. ACE Engineering Academy

Cold Walls Hot gas Cold walls

10. Ans: (b) Sol: Mole fraction of N2 =

0.3 =

n N2 Total no. of moles

n N2 1

n N 2 = 0.3

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: 162 :

m N 2 = n N 2  molecular weight = 0.3  28 = 8.4 kg Mass fraction of N2 =

ME – GATE _ Postal Coaching Solutions

Rmixure =

m N2

m N 2  R N 2  m CO 2  R CO 2

=

Total mass

m N 2  m CO 2 56  0.2969  176  0.1889 56  176

= 0.215 kJ/kg

8.4 = 0.233 = 36

13. Ans: (b) For CO2

Sol: For N2

11. Ans: (d) Sol: m N 2 = n N 2  (molar weight) = 2 28

= 56 kg m Co 2 = n Co 2  (molecular weight) = 6  44 = 264 kg m N 2  m Co 2 = 320 kg Mass fraction = =

m Co 2

n1 = 3

n2 = 7

P1 = 600 kPa

P2 = 200 kPa

n = n1+ n2 = 10 Pf = 300 kPa  n1 PN 2 =   n1  n 2

 3 Pf =    300  10  

= 90 kPa

m Co 2  n N 2

264 = 0.825 320

14. Ans: (d) Sol: V = 80 L

n = n N 2  n CO 2 12. Ans: (a) Sol: m N 2 = n N 2  (molar weight)

= 2  28 = 56 kg

mCO2 = n Co 2 (molecular weight) = 4 44 = 176 kg R N2 =

R 8.314  Molecular weight 28

= 0.2969 kJ/kg

RCO2 =

R 8.31 = molecular weight 44

= 0.1889 kJ

ACE Engineering Academy

=

5 5  28 44

PV = n R T 5   5    P n 28 44  = =  RT V 80 P1V1 = n1 R T

5   5   5      P n 1  28   28 44    = R T V1 V1 80  V1 = 48.88 L

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: 163 :

15. Ans: (b)

T2  P2    T1  P1 

Sol: For Argon:

m = 3 kgs Molar weight = 40

R 8.314  M(   1) 5  40  1 3 

(C V ) CO 2

4 3

8.314 4  44  1 3 

= 0.566 kJ/ kgmole K (CV)mix =

m Ar  C V Ar  m CO 2  (C V ) CO 2

=

m Ar  m CO 2

3  0.31  6  0.566 36

= 432.6 kJ 16. Ans: (b)

n2 = 7,  =

Ar = 70%

Adiabatic process,  = ACE Engineering Academy

5 (monatomic) 3

5 3

5  8.314 R 3 (CP)Ar =  = 20.78   1  5  1   3 

(CP)Mix = =

n He C P He  m Ar (C P ) Ar n He  n Ar

3  20.79  7  20.79 3 7

= 20.79

P

V1 V2  T1 T2

T2 =

Sol: P1 = 1.2 MPa, P2 = 0.2 MPa

5 3

For Ar:

QS = m  (C V ) mix  (T ) = (3 + 6)0.4806(350 250)

=

5  8.314 R 3 (CP)He = = 20.78    1  5  1   3 

= 0.4806 kJ/kgK

He = 30%,

= 328.7 K = 55.70 C

n1 = 3,

Sol: For He:

For CO2:

R  = M  1

5 1 3 5 3

17. Ans: (d)

= 0.311 kJ/kg mole K

m = 6 kgs, Molar weight = 44 ,  =

 1 

 0 .2  T2 = 673    1 .2 

5  = (monoatomic) 3 (CV)Ar =

Thermodynamics

V2  T1 V1

1

V

2

2V

V

= 2 323 = 646 K QS = n(CP)mix (T) = (3+7) 20.79 (646 – 323) = 67 MJ

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18.

19.

Sol: Given

Sol:

ME – GATE _ Postal Coaching Solutions

PA = 500 kPa DA = 100 mm P0 = Patm = 100 kPa

Tyre

DB = 25 mm

V = 36000 cc P = 15 atm

Pump V1 = 600cc P1 = 1 atm

mp = 15 kg g = 9.81 m/sec2

Air compressor pump, V = 600 c.c

B DB=25mm

P1 = 1 atm

PO = 100 kPa g

DA=100mm

Tyre volume, V1 = 36000 c.c Tyre pressure, P = 15 atm,

Pump

V = 36000 c.c

A

nP1V1 = PV (Isothermal) F0

FP

n=

15  3600 = 90 strokes 1  600

FA

20.

Net force, Fnet = FA F0 FP

Sol: At ground

= PAAA P0A0 FP FNet  PA  A A   Patm A A  A B  

mp g

Balloon volume, V1 =

1000

     2 Fnet   500  0.1   100 0.12  0.0252  4 4     15 9.81  1000



= 3.925 – 0.73 – 0.14 FNet = 3.055 kN.

P1 = 72 cm of Hg. In the air, R = 3r 4 Balloon volume, V2 = 27  r 3 3 Temperature constant  Isothermal  Process P1 V1 = P2 V2

FNet PB  Area of ' b'

PB 



4 3 r 3

P2 = 72 r3/27r3 = 2.67 cm of Hg

3.055 3.055   6224 kPa  2 Ab 0.025 4

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: 165 :

21.

22.

Sol: (1) By taking constant volume V = C

Sol:

Air

20cm

P2

V

Thermodynamics

Hg

P1

19cm

T

Patm =76cm of Hg

For isothermal process: T = c Case: 1

 P1V1 = P2V2 V P2 = P1  1  V2

P1 = 76 – 19 = 57 cm of Hg

  

Case: 2 (when open end is up)

As per fig we can say T1 > T2

Patm =76cm of Hg

V   P1 > P2........  1  1  V2 

Air

X2

P1

V2

Hg

19 cm

P2

V

P2 = 76 + 19

V1

 P2 = 95 cm of Hg It is Isothermal process

T

P1V1 = P2V2

(2) For isochoric : V = C

 P1A X1 = P2AX2

P1 P2  T1 T2

 X2 

P1X1 57  20   12 cm P2 95

T1 > T2 T  P2 =  2  T1

 P1 

 P2 < P1

23. Sol: Given relation t = aln(K) + b

K is 1.83 at ice point  t = 00 C K is 6.78 at steam point  t =1000 C

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ME – GATE _ Postal Coaching Solutions

As per given relation, t = aln(K) + b

24.

By condition 1  0 = aln (1.83) + b

Sol: Constant volume thermometer

0 = 0.60 a + b

V=C

By condition 2  100 = aln (6.78) + b

tP

100 = a 1.914 + b  a = 76.10

t = aP + b

b = 45.66

t = 0.273P  273.22

 By putting value of a & b

At t = 0C, P = 1000

t = 76.10 ln K – 45.66 (K = 2.42 given)  t = 21.60 C

At t = 100oC,

P = 1366

0 = 1000a + b …….. (1) 100 = 1366a + b …… (2) By solving, a = 0.273, b = 273.22 t = 0.273(1074) 273.22 = 19.90 C P = 1074 mm

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: 167 :

Thermodynamics

02. Ans: (d)

Chapter- 2

Sol:

1.4

P

n=

1.3

Work and Heat

n=1

01. Ans: (c)

n=0, P=C

Sol: Heat engine cycles as shown in fig

n=0

n=1, T=C

VQ = QR, PQ = QS,

n=1.3

UP = PR = RT P

n=, V=C

x

x

v

W

V

S

y

y1

Q y P

R x

x

03. Ans: (b) Sol: Process 1 2 3 dQ(kJ) 300 0 100 dW(kJ) 300 250 100 Heat supplied QS = 300 kJ

Work of compression, WC = 100+250 = 350 kJ

x v

Work interaction for ‘WVUR’= 48Nm

Wnet = WE  WC = 550 – 350 = 200 kJ  thermal 

Area WVUR = 2x 2y = 48

Wnet = 0.67 Heat supplied

Work ratio =

 xy = 12 From similar les PQR and PST QR ST  PR PT

Wnet 200   0.36 WE 550

04. Ans: (c)

y y1   y1  2 y x 2x Work interaction for le ‘PST’ 1 2x y1 = 1 (2x)(2y) = 2xy 2 2 = 2  12 = 24 N-m

ACE Engineering Publications

4 0 250

Work of expansion, WE = 300 + 250 = 550 kJ

T

U

=1.4, PV =C

Sol: (A) W.D in polytropic process

=

(P1 V1  P2 V2 ) (n  1)

(B) W.D in steady flow process    v dP (C) Heat transfer in reversible adiabatic process = zero (D) W.D in an PV P V  1 1 2 2  1

isentropic

process

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: 168 :

05. Ans: (d)

For isothermal process, PV = C

Sol: Given

log (P) + log(V) = log(C)

For isothermal process P1V1 = P2V2  P1V1 =

P1  0.55 10

m1 = 1 For adiabatic process

 V1 = 0.055 m

PV = C

For adiabatic process  1

 log P + log V = log C

 2

P1 V = P2 V  P1 V11.4   V

log(P) = log(C) – log(V) Compare it with y = C + mx

3

1.4 2

ME – GATE _ Postal Coaching Solutions

 log (V) = logC – log V

P1  V21.4 10

Compare with y = C + mx m2 = 

 10  (0.055)

1.4

 m1 < m2

3

V2 = 0.284 m

10. Ans: (d)

06. Ans: (b) 07.

Sol: The ratio of Cp/Cv for a gas with n degrees

Ans: (a)

or freedom is equal to 1 

Sol: Assertion is true

Reason is true

2 n

11. Ans: (c)

Correct Explanation

Sol: Cp ; CV ;  ; values are constant for ideal

08. Ans: (d)

gases

Sol: No. of degree of freedom in diatomic

Cp ; Cv values increase with temperature for Real gases.

molecule = 5 No. of degree of freedom in monoatomic

Cp Cv

molecule = 3 No. of degree of freedom in Triatomic

CP Cv

molecule = 6 or 7

 

09. Ans: (a) Temp.

Sol: The slope of log P – log V graph for a gas

for isothermal change is m1 & for adiabatic

Where

changes are m2 if gas is diatomic gas

temperature.

ACE Engineering Publications

Ideal gases

as

“”

value

decreases

with

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: 169 :

12. Ans: (c) Sol: P = 0.75 hp = 0.5 746 = 373 Watt   0.65

373 0.65 = 573.85 W= 0.574 kJ/s

Energy supplied from fan =

13. Ans: (a) Sol: The minimum power supplied 1 1  V2  V 2   Q = m 2 2 1 =  1.15  3  12 2  248.4 W 2 14. Ans: (b)   70m 3 / s ; h = 65m Sol: Q

gen = 0.85 ; Poutput = ?  gh  1000  70  9.81  65  gh  Q m = 44635 kW = 44.64MW Actual power output = 0.85  44.64 = 38MW 15. Ans: (d) Sol: P = 75 hp = 56kW P 56  63.56kW Pactual   efficiency 0.88 Annual electricity consumption = 63.562500 kW-hr So the annual electricity cost = 63.5625000.06 = $9534 16. Ans: (b) Sol: P = 320W In 30 days the refrigeration runs is 1  30   7.5 days 4 ACE Engineering Publications

Thermodynamics

So total consumption of a month in Watt-hr is = 320  7.5  24 = 57600 W-hr = 57.6 kWhr So the electricity cost per month = 57.6  0.09 = $ 5.184 17. Ans: (a) Sol: P = 2kW = 2000 W

 gh Pm  gh  2000 = Q

2000 0.820  10 3  9.81  30

   Q

 8.28 10 3 m 3 / s = 8.28 Lit/sec 18. Ans: (c) Poutput Sol:   Pinput  gh  gh  Q Poutput  m

  gh  Q   P Q

 18  10 3  211  10 3  3798Watt 

3798  100  75.96 % 5000

19. Ans: (a) Sol: m 1  6kg ;

P1  3 atm

T1 = 40C = 313K, P2 = 2.2 atm P1 V1  m1 RT1 P2 V2  m 2 RT2

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: 170 :

ME – GATE _ Postal Coaching Solutions

As the tank is rigid, the volume is constant P1 m1T1  P2 m 2 T2



Spring compression

 T2 

m1  P2  T1 P1

 T2 

6  2.2  313 = 460 K = 187C 3 3

Piston Expansion So spring compressed 0.2m

l2 = 0.2m

20. Ans: (a) Sol: Work done in isothermal process

Find: Total W.D

Here rigid cylinder so area will be same

V  P1V1 n 2 V1

Al2 = 3 Al1

 0.1   500  0.8  n     831.77 kJ  0.8 

l2 = 3l1 

l2 = 0.6m

 V2 = 0.03 m3 21.

Total work

Sol: Given Pa = 0.1 MPa = 100 kPa

= [work of gas + (work of spring due

V1 = 0.01m3

to expansion by heating]

A1 = 0.05 m3

(Due to expansion of piston, spring is

V 0.01  l1 = 1  A1 0.05

compressed)

 l1 = 0.2 m

= PdV +

V2 = 3V1

1 Kx 2 2

= P (V2 –V1) +

1 (25)(0.2)2 2

[ constant pressure process] =100 (0.03 – 0.01) + 0.2m

Total work = 2.5 kJ

0.6m 0.2m

gas

ACE Engineering Publications

1 (0.2) 2 (25) 2

gas

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: 171 :

22.

Thermodynamics 1W2 =

Sol: Given

mw = 500 kg

V  P1V1ln  2   V1 

 3  = 100  3.53 ln    3.5 

3

Total tank volume V= 4m

N2 Water

3m3

N2

1m3

Water

 W = –54.42 kJ

3.5m3

Here ‘–ve’ sign indicates compression of N2

3

0.5m

so system volume decreases 23. Sol: Total volume V = 50m3,

P1 = 1 bar, T1= 250C Air

P1 = 100 kPa mw = 500 kg, w = 1000 kg/m3 For state-1, V =

500 = 0.5m3 1000

35m

For state-2, mw = 1000 kg, V = 1 m3 For state 1 ‘N2’:

P1 = 100 kPa V1 = Total volume – vol. of water = 4 –0.5 V1 = 3.5m

state 2 (‘N2’):

V2 = Total volume – vol. of water

W.D by pump = (work of lifting water + work of compression of air) =

= 4 –1 V2 = 3m3 Condition: Isothermal process

 From eqn  P2 

3 3 VTank  (50)=37.5m3 4 4

Vair = 50 –37.5 = 12.5m3

3

For

Vwater =

=

P1V1 = P2V2

P1V1 100  3.5  V2 3

 P2 = 116.67 kPa

ACE Engineering Publications

V  m w gh  P1V1n  2  1000g c  V1 



 w Vw gh 12.5  100(50)n 50 1000g c

 (1000)(37.5)(9.7)(35)  12.5   100(50)n  1000  1  50 

Wpump = 19662.72 kJ

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: 172 :

24.

ME – GATE _ Postal Coaching Solutions WS 4.175  Time 600

 Power =

2

Sol: A = 0.12 m

= 6.95  103 kW = 6.95W

P1 = 1.5 MPa = 1500 kPa P2 = 0.15 MPa = 150 kPa

l = 0.3 m

Now, PS =

1

P

T=

2NT 60

60PS 60(6.95)  2N 2  3.14  840

2 V1

V2

= 0.079 Nm T = 0.079 Nm

V

1 1W2 = (P1  P2 )Vs , Vs = V2 V1 2 1  1W2 =  1500  150  (0.036) 2 ( Vs = Al = 0.12 0.3 = 0.036m3)  1W2 = 29.7 kJ

26. a   Sol: Given  P  2  (V  b)  mRT V   P

a mRT  2 V (V  b)

P=

mRT a  2 ( V  b) V V

V

2 2  mRT a   2  dV W.D   P dV    ( V  b) V  V1 V1 

25. Sol:

v2

PAtm=101.325 kPa d=0.4m

d = 0.4m, W = 2kJ

W.D = mRTn V2  b   a  1  1  V b  1   V2 V1 

time = 600 sec. ,

 V  2 1  W.D  mRT n [V  b]  a     2  1  v1 V2 V1

V1 = 1m3,

m = 10 kg,

l = 0.485 m

V2 = 10m3,

N = 840 rpm WNet = Piston expansion work – WStirrer

T = 293K, 4

a = 15.7  10 Nm4 =157 kNm4

2 = PdV –Ws

b = 1.07  10–2, R = 0.278 kJ/kg.K

2 = (101.325) Al – WS

W.D = (10)(0.278)(293)

2 = (101.325)   0.42  0.485  WS 4

WS = 4.175 kJ ACE Engineering Publications



 10  (1.07  10 2 )    15710  1 ln  2 1 ( 1 . 07 10 )    

W.D = 1742 kJ Hyderabad |Delhi |Bhopal |Pune |Bhubaneswar |Bengaluru |Lucknow |Patna |Chennai |Vijayawada |Visakhapatnam |Tirupati

: 173 :

27. Sol: Given Diameter of piston (D) = 0.15m

Thermodynamics 1W2

=

P1V1  P2 V2 n  1

I.P. = 4kW = 4  1000W

( PV1.2 = C = Polytropic process)

Speed (N) = 216 rpm

 Vs = V1 − V2= V1 −

Spring constant (k) = 25106 N/m3 Length of indicator diagram (ld) = 0.1  Stoke (L) Let Area of indicator diagram = (ad)



 2 4 d  s = V1 4 5

 V1 =

 Mean effective pressure (pm) =

ad k ld

p LAN [as 4 stroke engine] and I.P.  m 120

 I.P.   ad 

ad  k L A  N  ld 120

6

2

25  10    0.15  216

= 5.51 10-3m3  V2 =



1 1 V1 = 5.51  10 3 5 5

 V2 = 1.10  10-3m3

m2

In polytropic process 1W2

=

1W2 =

P1 V1  P2 V2 n 1



28.

1W2

= 1.04kJ

Sol: d = 0.15m, l = 0.25 m

For Power

PV

1.2

2

Vs= v1−v2 ACE Engineering Publications

= 1.04  2  V



( compression)

= 1W2  No. of cylinder 

=C



0.55  0.77 = −1.04kJ 0.2

=

1



101.325  5.50  10 3  699 1.10  10 3 (1.2  1)

1W2

P

1.2

 P2 = 699 kPa

= 5.03  10-4m2 = 503 mm2



P Vn  5.51   P2 = 1 n1 = 101.325   V2  1.10 

I.P  l d  120 kLA N

4  0.1  120  4  1000

 5 2  0.15  0.25  4 4

By given condition P1 V1n = P2 V2n

2   I.P  0.1L  120  4  area A  D  = k  L    D 2  N and   0.41L d  

=

1 4 V1 = V1 5 5

compression stroke sec 500 60

Power = 17.32 kW

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: 174 :

ME – GATE _ Postal Coaching Solutions

29. Sol: P 20

Chapter- 3

1

First Law of Thermodynamics 2

PV = C 5

01. Ans: (a)

2

3

V

0.05

Sol:

0.1

(2) h2

P1 = 20 bar, V1 = 0.05 m3,

V2 = 0.1 m3

kJ kg

(h2 h1)=30 kJ/kg

P1 V12 = P2 V22 dW  90kJ / kg dm

2

 0.05  P2 = 20    = 5 bar  0.1  Net work = 1W2 + 2W3 + 3W1

h1

= (Work)polytropic + (Work)P=C

kJ kg

(1)

dQ = 40kJ/kg dm

+ (Work)V = C =

P1 V1  P2 V2  P2 (V3  V2 )  0 n 1

= 50  25 + 0

dW = 90kN.m/kg = 90 kJ/kg (h2-h1) = 30kJ/kg. dQ =  40kJ/kg dm

W.D = 25 kJ

Heat rejected  dQ = ‘Ve’ According to Steady flow energy equation,  dQ   dQ   dW  h1      h2     dm  air  dm  w  dm  dW  dQ   dQ      h 2  h 1    dm  dm  w  dm  air

= 30  90 (40) =  20kJ (ve sign indicates heat is rejected from the system)

ACE Engineering Publications

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: 175 :

Thermodynamics

(d) dW = ve

02. Ans: (b) Sol: List – 1

List -2

dQ = 0 (Adiabatic)

Joule Thomson co-efficient - T / P h Cp for monatomic gas

-

CP  CV for diatomic gas -

(dU/dT)v For monatomic  = 1.33,

dU = +ve

5/2 R

05. Ans: (c)

R

Sol:

C

CV 5 Cp = R 2

Pressure

A Volume

03. Ans: (d) 1 R  dT vdP   Sol:    = dT  dP T  T   T MdT  NdP = 0 M = 0, P

B

N =0 T

So, it is an exact differential.  It is a property of system 04. Ans: (c)

D

Along A – B – C:

QAB WA-B = UA B 180 – 130 = UB UA UB  UA = 50 kJ Along A – D – B:

QAB WA-B = UA B QAB = 50 + 40 = 90 kJ 06. Ans: (a)

Sol: (a) dW = 0 (Rigid)

dQ = ve

07. Ans: (e)

dU = ve

Sol: m = 60 kg ,

P = 200 kPa

T = 250C ,

dQ = 0.8 kW dt

dQ dW = dU dU = ve (b) dQ = 0 (Insulated) dW = + ve (Expansion)  dQ  dW = dU 0 dW = dU

t = 30 min = 1800 sec Well sealed = control mass (Non flow process) dQ dW = dU dQ dW t  t = mCvdT dt dt

dU = ve (c) dW = 0 (Free expansion) dQ = 0 (Insulated)

1800[0.8 (0.12)] = 60(0.718)(T 25) T = 63.40C

 dU = 0 ACE Engineering Publications

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: 176 :

ME – GATE _ Postal Coaching Solutions

11. Ans: (d)

08. Ans (d) Sol: Q = 2000 W

t = 15 min = 900 sec

Sol: ‘He’ is a monatomic gas. He =

m = 75 kg

T

P=10 atm

2

Well sealed = control mass (Non flow

5 3

P1 =1 atm

process) dQ dW = dU

1

dQ dW t  t = mCvdT dt dt

T1 = 298 K For minimum temperature condition at exit,

 dQ dW  t   = mCVdT dt   dt

the compression has to be isentropic.

900 [0(2)] = 75(0.718)dT

T2  P2  T1  P1

dT = 33.420C 09. Ans: (c)

  

 1 

 10  T2 = 298   1

Sol: P1 = 0.25 kW ( ve sign indicates it is a

power consuming device)

 5 1   3   5   3 

= 748.5 K = 475.540 C

P2 =  0.12 kW P3 = 1 kW

12. Ans: (d)

P4 =  0.05 kW

Sol:

2 T

3MPa

Temperature = constant

1

(Isothermal process) dQ = dW

( dU = 0)

0.2MPa

dQ = (0.25+ 0.12 +1 + 0.05)  3600

s

= 5112 kJ/hr

T1 = 1023 K Argon is a monoatomic gas.

10. Ans: (c) Sol: 1 = 1000 C,

2 = 200 C,

Ar =

 w = 2 kg/s m  a = 3 kg/s m

Molecular weight of Argon, M = 40

Heat gained by water = heat lost by air 2  4.187(T215) = 31.005 (10020) T2 = 390 C ACE Engineering Publications

5 3

C PAr

5  8.314 R = = 3 = 0.5196 M  1 5  40    1 3 

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: 177 :

P T2 = T1  2  P1

  

 1 

 0.2  = 1023    3 

0.4

Thermodynamics

14. Ans: (a) Sol:

V2.=1500cm3

= 346.28 K = 73.280 C

 C PAr (T1T2) Power (kW) = m

Helium

= 0.51965(1023 346.28) =

Vc = V1 = 15 cm3

1758.1 = 1.758 MW 1000

VCO =V2 = 1500 cm3 P1 = PHe  20 atm

13. Ans: (b)

T1 = 400C

5 Sol: ‘He’ is monoatomic gas,  = , M = 4, 3

 dW  0 (Free expansion)

5  8.314 R 3 (C P ) Air   M (   1) 5  40    1 3 

dQ = 0 ( due to insulation) By 1st law dU  dQ  dW  dU  0  C V dT  0  dT  0  T= constant

= 0.5196 kJ/kgK 5  8.314  R (C P ) He  3 M (   1) 5  4    1 3 

Temperature = constant (Isothermal)  P1 V1  P2 V2  20  15  P2  1500  P2  0.2 atm

= 5.196 kJ/kgK mixture =

5 3

15. (i) Ans: (c), (ii) Ans: (c)

CP mix = 0.5(CP)He + 0.5(CP)Ar

Sol:

= 0.50.519 + 0.55.19 = 2.857 P T2 = T1  2  P1

  

(ii) Ideal gas stored in Rigid insulated Tank.

 1 

 100  = 1200    1000 

Total volume of Tank Vf  3m 3 State : 1

0.4

= 477.72 K

 a CP mix (T1T2) Power (kW) = m

= 0.32.857(1200477.72)

Tank has two compartments. State : 2

Partition

between

two

compartments

Ruptured

= 619.05 kW ACE Engineering Publications

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: 178 :

nf =n1 +n2 Pf = ? Tf = ?

ME – GATE _ Postal Coaching Solutions

T1=300K

T2.=1000K

0.040C V Tf  300   0.24C V Tf  1000  0  C V 0.28Tf  241.2  0

V1=1m3

V2.=2m3

 Tf  900K

P1=0.1MPa

P2.=1MPa

Again  PfVf = nf R Tf

PV  mRT

= 700 kPa

m   PV  n RT  n   M 

 Pf = 0.7MPa

By equation PV  n RT

16. (i) Ans: (a) , (ii) Ans: (b) , (iii) Ans: (b)

 P1 V1  n 1 RT1

Sol:

P

0.1MPa  1m  n1   8.314  300K RT1

3

P1 V1

=

0.1  10 3 kPa  1m 3 8.314  300

 n 1  0.040 moles

For n 2 

P2 V2 RT2



1  10  2 8.314  1000

nf = n1 + n2 = 0.04 + 0.24 = 0.28 3

Vf = V1 + V2 = 3m

Here rigid & Insulated tank given dW  0

 Rigid  V  C  dW  0  By 1st law dQ  dW  dU  C V dT  0 Here dU 1  dU 2  0

3

P3

PV=C P1=140kPa 2

1

3

= 0.24 Moles

 dQ  0,

n f RTf 0.28(8.314)(900) = Vf 3

 Pf =

RT   PV  m  R  R M   M

V1=0.028m3. V2

Given Process 1 2: (P = C),

P1 = 1.4 bar, V1 = 0.028m3, W12 = 10.5kJ Process 2 – 3: (PV = C),

U3 = U2 Process 3 – 1: (V = C),

U1U3 = 26.4 kJ Process 1 – 2: (Constant pressure)

 n 1C V dT  n 2 C V dT  0

 Q12  W1  2 = U1  2

 n 1C V Tf  T1   n 2 C v Tf  T2   0

Given

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V

1W2

= 10.5kJ = P(V2 V1)

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: 179 :

W1  2 = P1 (V2 V1)

Thermodynamics

17. Ans: (b)

 10.5 = 140(V2  0.028)

Sol: For process 1 – 2 : (P = C)

 V2 = 0.103m3 U3 = U2 & U1  U3 = 26.4kJ

 1W2 = P1(V2 V1)

 U1  U2 =  26.4kJ

 1W2 = 100 (0.30.003)

 U2 U1 = 26.4 kJ.

 1W2 = 29.7kJ



1Q2

= 1U2 + 1W2



1Q2

= 26.4 + 10.5



1Q2 = 36.9 kJ

By Ist law 1Q 2

 0  29.7 = E2  E1 ( 1Q2 = 0)

Process 2 – 3:(Isothermal Process)

29.7 = E2 0

 Q2-3  W2-3 = U2-3

E2 =  29.7 kJ

Hence T = C  2U3 = 0  Q 2-3 = W 2-3 = P2V2ln

V3 V2

 0.028  = 1400.103ln    0.103  

2W3

 1W 2 = E2 E1

By Process 2 – 3 2Q3 2Q3

 2W3 = 2E3 = E3  E2  P(V3 V2) = E3 E2

105 100(0.060.3) = E3 (29.7)

=  18.79 kJ

Process 3 – 1:(constant volume) 3W1

E3 = 110.7 kJ 18. (i) Ans: (a), (ii) Ans: (a)  1 = 0.01kg/sec Sol: m

=0

h1 = 2952 kJ/kg

 dQ  dW = dU

 2 = 0.1kg/sec m

h2 = 2569 kJ/sec



3Q1

= 3U1 = 26.4

 3 = 0.001kg/sec m

h3 = 420 kJ/kg



3Q1

= 26.4 kJ

V1 = 20m/sec

V2 = 120m/sec

For checking answer Q = 1Q2 + 2Q2 + 3Q1 Q = 8.28 kJ W = 1W2 + 2W3 + 3W1

Fluid 1

Engine

Fluid 2

 W = 8.28kJ

Fluid 3



Fluid 4

Q = W (First law proved)

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: 180 :

1) Mass balance

ME – GATE _ Postal Coaching Solutions

 air enter = 80 kg/hr. m

Mass entering = mass leaving

h1 = 84 kJ/kg

1 m 2m  3 m 4 m

h2 = 59 kJ/kg

 4 = 0.109 kg/sec m

 dQ   630  4  2520kJ / hr    dt  person

(2) Energy balance

 dQ  =?    dT  R cooler

  V12  V22  dQ   m 1 h 1    m 2 h 2   2000  2000  dt  

By steady flow energy equation

dW  3h 3  m  4h 4  m dt

dQ   dQ   1 h 1   m     dt  person  dt  RCooler

dQ =0 dt

Here

dW   dW   2 h 2   m     dt  fan  dt  bulb

2 2 0.01 2952  20   0.12569  120 



2000 

80  84 2520  dQ     3600 3600  dt  RCooler

2000 



= (0.001  420) + (0.109  h4) + 25  h4 = 2401 kJ/kg



 dQ      1.91 kW  dt  RC

19. Ans: (a) Sol:   80 m

  80 kg / hr m

h1 = h  84 kJ air kg

80  59  (0.66) 3600

R

kJ hr

kJ h2=59 kg

Heat is removed from a cooler is 1.91 kW 20. Sol: (i) Ans:(c)

m = 1.5 kg P1 = 1000 kPa

 dW  = 0.182 = 0.36 kW    dt  fans

P2 = 200 kPa V1 = 0.2m3 V2 = 1.2 m3

 dW    = 3  0.1 = 0.3 kW  dt  Bulb

P = a + bv

 dW  =  0.36 + ( 0.3) = 0.66 kW    dt  Total

1000 = a + 0.2b ……… (i)

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u = 1.5PV  85 200 = a + 1.2b …….. (ii)

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: 181 :

Thermodynamics

By solving

For Maximum internal energy

b =  800 a = 1160

du =0 dv

P = 1160  800V

 a + 2bv = 0

V2

v2

V1

v1

(ii) 1W2 =  PdV   (1160  800V)dV 1.2

=  (1160  800V)dV 0.2

= 1160(1.20.2)400(1.22  0.22) u = (1.5 PV  85) kJ/kg = (1.5P

 a  1160 3  m = 0.725m3 2b  2  800

umax = (11600.725)  (800  0.7252) 85 = 335.5 kJ/kg   umax Umax= m = 1.5  335.5

= 600 kJ (iii)

v=

V  85) kJ/kg m

= 503.25 kJ 21. (i) Ans: (a), (ii) Ans: (b) , (iii) Ans: (a)

= 1.5P

V  85 m

Sol: h1 = 3000 kJ/kg ,

= 1.5P

V  85 1.5

V1 = 60 m/s ,

= (Pv  85) kJ/kg u1 = P1V1  85

= 1000  0.2  85

h2 = 2762 kJ/kg V2 = ?

A1 = 0.1 m2 , v2 = 0.498 m3/kg v1= 0.187 m3/kg

= 115 kJ/kg u2 = P2V2  85

= 200  1.2  85 = 155 kJ/kg dQ 0 dt

u2  u1 = 40 kJ/kg dU = m(u2  u1)

Applying steady flow energy equation

= 1.5  40 = 60 kJ

h1 +

dQ  dW = dU dQ = 60 + 600 = 660 kJ u = Pv  85

 3000 +

V2 (60) 2 2  2762  2 2000 2000

 V2 = 692.5 m./s

= (a + bv)v  85 = av + bv2  85 = f(v)

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V22 V12 dW dQ  h2   2000 dt 2000 dt

  m

A 1 V1 A 2 V2  v1 v2

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: 182 :

  m

ME – GATE _ Postal Coaching Solutions

0.160  m  32.08 kg

sec

0.187 Find, A2 =?



A V mV2   2 2  A2  m v2 v2

 A2 

 1002  0  dW = 0.422512   2000   dt

32.08  0.498 692.5

dW  112.51 kW dt

23. (i) Ans: (a) , (ii) Ans: (b) Sol: CP = 2.093 +

 A2 = 0.023 m2

41.87 0 J/ C t  100

22. Ans: (a)

P = 1 atm

Sol: Given:

V1 = 2000 cm3

T1=00 C T2=1000C

P1=1.2 MPa,

P2 = 20kPa , T1=1880C,

V2= 2400 cm3

h1=2785kJ/kg,

h2 = 2512kJ/kg

Here CP = J/0C form it should always in J/kg

V1=33.33 m/s,

V2 = 100m/sec.

0

Z2 = 0m , Z1= 3m ,  = 0.42 kg/sec m

C form

So CP × kg = J/0C  Sp. Heat × mass = Heat capacity

1

T2

100 0 C

T1

00 C

dQ   C P dt  dW dt



41.87   2.093  t  100  dt

 2.093t 0  41.87 lnt  1000 100

100

= [2.093(100)]+[41.87ln(200)  41.87ln(100)] dQ  0.29kW dt

= 209.3+[2218-192.81] 2 (i) dQ = 238.32J

Applying steady flow energy equations  Zg V 2  dQ  h 1  1  1   m 1000 2000  dt   V22 Z g  dW  h 2  m  2  2000 1000  dt  2  39.81 33.33   0.42 2785     0.29 1000 2000  

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Here constant pressure is given  1W2 = P1 (V2 – V1) = Patm (V2  V1) = 101325(2400  2000) 106 1W2

= 40.53J

 dQ  dW = dU  dU = 238.32  40.53 dU = 197.79J

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: 183 :

Thermodynamics

24. Ans: (a)

=

Sol: n = 1 mole

CV = 20.785 J/mol K

Applying Steady flow energy equation

T1 = 300 K

 h1  m

P = 50 W t = 120 sec

dW dQ  h2  m dt dt

h 2  h1 

dQ = 0 (Insulated non flow process)

 35  26   1877 kJ / kg of mixture 0.0325

dQ  dW = dU

26. (i) Ans: (d) , (ii) Ans: (a) , (iii) Ans: (d)

0 (P t) = nCVdT

Sol:

P

50  120 = 120.785(T2 300)

T2= 2

T2 = 588 K For Ideal gas,  P2 =

P1 V1 P2 V2  T1 T2

T2 V1   P1 T1 V2

P2 = 196 kPa = 0.196 MPa 25. Ans: (b) dW dQ = 26 kW, = 35 kW dt dt  f kg / hr  m kg = bsfc = Brake Power (kW ) kWhr

0.3 =

 f = 7.8 kg/hr m  a 14 m  f m 1

 f = 14  7.8 = 109.2 kg/hr  a = 14  m m

 =m  a m  f = 117 kg/hr m ACE Engineering Publications

1(T1)

T3=

1 T1 2

V

V1 V2  T1 T2

3 T V2 T2 4 1 3     = 0.75 V1 T1 T1 4 1Q2 1Q2

1W2 = 1U2

 P(V2 V1) = (U2U1)

1Q2 = CV(T2 T1)+ P(V2 V1) T  V  = CVT1  2  1  PV1  2  1 T  V   1   1 

f m 26

AFR =

3

3 T1 4

Process 1 – 2: (P = C)

588 =  100 300

Sol:

117 = 0.0325 kg/s 3600

3 3 = CVT1   1  PV1   1 4  4  1 =  C V T1  PV1  4 =

1 C V T1  RT1  4

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: 184 :

=

1 T1 R  C V  4

For 2–3 process

=

C T 1 T1C P  P 1 4 4

  150 2W3 = U3U2 = U3 U1

2Q3  2W3

= 2U3 = 200

2W3

Process 1 – 2: 1W2

ME – GATE _ Postal Coaching Solutions

= 350 kJ

For 3 – 4 process

= P(V2 V1) V   RT1 = PV1  2  1  V  4  1 

3Q 4

 3W 4 = 3U 4

3U4

= (U4 U3) = (U4 U1)(U3U1) = (U1U4)(U3U1) = 50 200 = 250

Process 2 – 3: (V = Constant) 2W3 2Q3

= 3U4 250 ……… (1)

=0 = 2U3 = U3 U2

For 4 – 1 Process

= CV(T3  T2)

4Q1

 4W1 = 4U1 = U1  U4

T 3T = CV  1  1  4  2

4Q1

 300 = 50

4Q1

= 350 kJ

1 = CVT1   4

Q = 1Q2 + 2Q3  3Q4 + 4Q1

= 1Q3

3Q4

= 100  150 500 + 350 = 200 kJ

 C V T1 4

( Q = W) W = 100 350 250 + 300

= 1Q2 + 2Q3  C P T1  C V T1  = 4 4

=

= 200 kJ

 T1 C P  C V  = (CV + CP) T1 4 4

27. Sol: For 1–2 process 1Q2

 1W2 = 1U2

 100  100=U2  U1  U2=U1 ACE Engineering Publications

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: 185 :

Thermodynamics

04. Ans: (b)

Chapter- 4

Sol:

T1 = 900K

Second Law of Thermodynamics

W = 50kW

01. Ans: (b)

H.E

02. Ans: (c)

T2 = 300K

Sol: Given: H.E  H.E 

T

T1  T2 Q1  Q 2  T1 Q1

40kW=Q1



Power kW  = Q S kW 

H

1200  300 40  Q 2   1200 40

Q2 =?

 Q2 = 10 kW



T

03. Ans: (c) Sol: Given: Electric power generating station

means H.E

C 

Q1=36  108 kJ/h

= 0.8

T1  T2 627  27   0.67 T1 900

W= 400MW

Sol: COP R 

Q2= ? T2

Q1  Q 2 W  Q1 Q1

 Q2 = Q1 – W

1   E 1  0.75  = 0.33 E 0.75

06. Ans: (d) Sol: Given

 E  0.4

36  10 8 (MW )  400MW = 1000  3600 = 1000 MW  400MW

H.E

Q2 = 0.6Q1

E 

Q3

Q1

Q 2  Q 4  3Q1

KJ  400MW Q2 = 36  10 8 h

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3  75000 3600

05. Ans: (b)

H.E

Q2 = 600MW

50,000

 > c  Not possible

T1

 H .E 

W kW   kJ  kg   f  m   C.V   sec   kg 

Q1  Q 2 Q1

Q2

W

R

Q4 = Q3+W = Q3+0.4Q1 Q4

 0.4Q1  Q1  Q 2 = W  Q 2  0 .6 Q 1

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: 186 : Q 2  Q 4  3Q1

10. Ans: (c) Sol: 1 Q 2 1 W2  1 U 2

 0.6Q1+Q4 = 3Q1  Q4 = 2.4Q1 Q3 = Q4  W = 2.4Q1  0.4Q1 = 2Q1 (COP)R 

ME – GATE _ Postal Coaching Solutions

Q3 2Q 1  5 W 0 .4 Q 1

07. Ans: (a) Sol: Assertion is true

Reason is true and reason is the correct explanation.

=

P1 V1  P2 V2  C V T2  T1  n 1

=

R T1  T2   R T2  T1  n 1  1

=

R T1  T2   R T1  T2  n 1  1

 1 1  = R T1  T2      n  1   1    1  n  1  = R T1  T2     n  1  1 

08. Ans: (a)

=

n R T  T  n  1  1 1 2

=

  n   R T1  T2    1 n  1

09. Ans: (d) Sol:  1 

2 

3 

W 8 .2   0.328 Q1  1500     60  W 8.75   0.328 Q 2  1600     60  W 9 .3   0.328 Q 3  1700     60 

W 9.85 4    0.328 Q 4  1800     60 



W W = n  Q1 Q1

W = 0.328 

2000 kJ / sec  10.94kJ / sec 60

ACE Engineering Publications

 n   W  =     1  = Heat transfer for polytropic process 11. Ans: (a) Sol: COP = 3.2

COP 

NREkW  WC kW 

Net refrigeration effect =

=

 a kg / sec  C pa kJ / kgK T  m

WC kW 

Va  C pa  T  t WC kW 

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: 187 :

COP 

 

C pa T  AL  t WC

Thermodynamics

14. Ans: (d) Sol: COP 

200  2.4  1.2  1.00532  22 3.2  3600 WC

NRE NRE  WC 680kWhr 

NRE 680  3600

 NRE = 680  3600  1.4

WC = 0.5kW

= 3427 MJ/yr

Electricity bill = WC(kW)no. of hours 

cos t kWhr

= 0.5 10  0.1 = 0.5

15. Ans: (d) Sol: NRE = 4kW

W = 1kW Q1 = 5kW ref

12. Ans: (a) Sol: COP 

3.1 =

NREkW  WC kW 

Q2 = Q1 – W = 1kW

10 4.187  10 3  23  6  3600 WC kW 

WC = 197 W

Net effect = Q1 – Q2 = 5 – 4 = 1 kW (heating) 16. Sol:

13. Ans: (a) Sol: COP = 3.2 , m = 1200kg , P = 5kW

 3.2 

Q3

Q1

NREkW  COP  WC kW  =

T3 = 243 K

T1 = 473 K

E

W

m a  C pa  T a

Q3+W

WC kJ 

12000.71822  7  WC

WC = 4169kJ  Time  =

WC kJ  WC kW 

834 4169kJ  =  13.5 min 5kJ / sec  60

ACE Engineering Publications

R

T2 = 303 K

T1  T2 W  T1 Q1 473  303 W   W = 0.359Q1 473 Q1 COP R  T3  Q 3 T2  T3 W Q3 243  303  243 0.359Q1

E 

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: 188 :



ME – GATE _ Postal Coaching Solutions

 E  0.4   Carnot

Q3  1.45 Q1

T1  T2 W = T1 Q1

0.4 

Q1  0.68 Q3

0.4 

17.

1000  300 1000

=

W Q1

 W = 0.28Q1

Sol: For minimum collector area condition the

efficiency

is

maximum

and

Q2 = Q1 W

maximum

= Q1  0.28Q1 = 0.72Q1

efficiency is Carnot efficiency.

C 

Q2 = 0.72Q1 Q3 = 2Q2 + W

W T1  T2  Q1 T1

= 1.44Q1 + 0.28Q1

T1 = 363K

Q3 = 1.72Q1

Q1

1 kW 363  293  Q1 363

 T3  1.72Q1   0.5 0.28Q1  T3  300 

W

 Q1 = 5.1857 kW

T3 = 326 K T2 = 293K

If Q1 = 50 kW Q3 = 2Q2 + W

Q1 (kW ) 5.1857 Area    9.93 m 2 2 1880 C d (kW / m ) 3600

= 2 0.72Q1 + 0.28Q1 = 1.72  50 = 86 kW

18.

19.

Sol:

Sol: (a) T1=1000K

T1 = 303K

T3 = ? Q3= 2 Q2+W

Q1 = 50kW

Q1 W

R

W= Q1– Q2 E

Q2=mLH

HP

Q2

2Q2 Q4 T2 = 300K

ACE Engineering Publications

T2 = 273 K

Latent heat of ice = 335 kJ/kg

COP R



Q2 T2  W T1  T2

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: 189 :



Thermodynamics

1000  335 273  W 303  273

 0.15 

 W = 36813 kJ

 0.15 

1 kW/hr = 3600kJ Electrical work 

36813 =10.22 kW hr 3600

1 kW hr = 3600 kJ 

T1 = 800K

5701 1.58 kW hr / day 3600

Electricity bill

Q1 = 1000 kJ WC

= No. of unit/day  No. of days  cost/kWhr = 1.58  30  0.32 = 15.168/-

Q2=400 kJ T2 = 400 K

21.

800  400 = 0.5 800

Sol:

W Q1

=

W WC 0.5 = C = Q1 1000

E

W 0.3 = Q1

WCarnot = 500 kJ WAcutual = 1000  400 = 600 kJ

W=0.3Q1

R Q2=1 MJ

W = 0.3Q1

As Wact > WCarnot

COP R

Hence claim is not justified

5

 Q1 

20. Sol:

8400 275 = 303  275 W

 W = 5701 kJ

(b)

C =

T2 Q  2 T1  T2 W

Q2 W

1 MJ  0.667 MJ 5  0.3

T1 = 303K Q1 = Q2 + W W=Q1 – Q2

22. Sol:

R

T1 = 294 K Q1=60000 kJ/hr

Q2 W

T2 = 275 K

HP

Q2 = 20  420 = 8400 kJ/day

COP actual  0.15 COP max ACE Engineering Publications

T2 = 263 K

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: 190 :

COP H.P 



ME – GATE _ Postal Coaching Solutions

333.43  10 3 273  W 310  273

Q1 T1  W T1  T2

 W = 45190.147 kJ

60,000 294  W 294  263



W

W = 6326.5 kJ/hr

45190.147  12.55 kW 3600

From (1)

WHP 6326.5   0.1054 WEH 60,000

Q1 = 130.759 kW

With heat pump, initial investment is high

Q2 = Q1–W

but running cost is less. With electrical heater

Q2 = 118.20 kW

the initial investment is less but running cost

Q4 = Q3+W =

is high.

333.43  10 3 W 3600

= 92.62 + 12.55 = 105.17 kW Q3 92.62   0.708 Q1 130.75

23. Sol: (a) T1=343 K

T3=273 K

W

= 118.2 + 105.17 = 223.37 kW

R

Energy of freezing water 92.36   0.706 Energy of heat engine 130.759

Q12

Q2 T2= 310 K

  m

reservoir = Q2 + Qref

Q3

Q1 E

Total heat rejected to lower temperature

24.

1000  0.277 kg / s 3600

Sol: (a)

T1 = 293K

 latent heat = 0.277×333.43 Q3 = m = 92.36 kW W T1  T2 E   Q1 T1

 W = 0.096 Q1 -----(1) Q3 T3  W T2  T3

ACE Engineering Publications

W

HP Q2

 343  310  W   Q 1  343 

(COP) R 

Q1

T2 = 263 K

Q 1  0.65 T1  T2  = 0.65(293  263)

= 19.5 kW

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: 191 :

COP HP 

T1 Q  1 T1  T2 W

Thermodynamics

25. Sol:

T1

293 19.5   293  263 W W = 1.98 = 2 kW

Q1 W=0.27 Q1

E Q2

(b)

T2

T1

Q12 =Q3+W

Q 1

W

HP

R

Q3

Q2 = 0.65 (T1293) T3

T2 = 293K

Q2 = 0.65(T1  T2) = 0.65(T1  293)

COP R

(House)



Q2 T2  W T1  T2

0.65 T1  293 293   1.99 T1  293

Q2 = Q1 – W = Q1 – 0.27 Q1 = 0.73 Q1 (COP)HP = 4

COP HP



T1 = 323 K = 500 C Up to 500 C outside temperature, the temperature of room can be maintained 200 C.

W  W = 0.27 Q1 Q1

E = 0.27 =

4

Q 12 W Q 12 0.27  Q 1

Q12 1.08 Q1 Total heat supplied to house = Q 2  Q 12  0.73 Q1  1.08 Q1 1.81 Q1

Q 2  Q 12  1.81 Q1

ACE Engineering Publications

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: 192 :

ME – GATE _ Postal Coaching Solutions

04. Ans: (a)

Chapter- 5

Sol:

Entropy

T

P1 A

B C

01. Ans: (c)

P2

 dh  Sol: C p     dT  p

S

05. Ans: (b)

Tds = dh – vdp

Sol:

As P = c , dp = 0

T

So, Tds = dh

1

 Tds  Cp     dT  p

3

 s  C p  T.   T  p

1

2

2

3 S

P

1 , 1 2

02. Ans: (c)

2

Sol: Motor power =5 kW.

3 3

V

T = 200C = 293 K Due to friction, there is heat between brake and

shoe

and

heat

is

transferred

Q Power Time 5  3600  = T T 293

dS = 61.4

2  3  constant volume 21  31  constant pressure

surrondings. (dS) sur =

to

12, 12  constant temperature

kJ K

3 1, 3111  Entropy constant 06. Ans: (c) 07. Ans: (c) Sol: Sgen = (S2– S1 ) 

03. Ans: (b)



dQ T

Q Q  1600 1600       T2 T1  400 800 

800K 1600kJ 400K

 2 kJ / K ACE Engineering Publications

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: 193 :

08. Ans: (b)

11. Ans (b)



Sol: Clausius inequality 

Sol:

dQ 0 T

dQ Q1 Q 2 600 450     T T1 T2 1000 300

dQ  0.9 kJ / K <0  T Q1  Q 2 Q1



T

100 K1

S=C V=C 2 P=C 1 3

600 KJ H.E

4 T=C

450 KJ 5

600  450 =0.25 = 25% 600

P

09. Ans: (b) Sol:

A V=C

P=C T=C S=C

hfg =1800 kJ/kg

T

S

300 K

 Irreversible cycle 

Thermodynamics

B C D V

500K

s1

s2

Slope of constant volume curve is more than that of constant pressure curve in T-S

sf

diagram. Similarly slope of adiabatic curve is

sg s

2.6 kJ/kg

more than that of isothermal curve in P-V diagram.

sg  sf 

h fg Tsat

s g  2.6 

12. Ans: (c)

1800 500

sg = 6.2 kJ/kg.K

Sol: Tds = du + Pdv.

This process is valid for any process, reversible (or) irreversible, undergone by a closed system.

10. Ans: (c) Sol: i) Temperature measurement is due to

Zeroth law of thermodynamics. ii) Entropy is due to Second law of thermodynamics iii) Internal Energy is due to first law of

13. Ans: (c) Sol: (dS)system = 0

(dS)surr = 0 (dS)univ = 0

thermodynamics ACE Engineering Publications

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: 194 :

ME – GATE _ Postal Coaching Solutions

14. Ans: (b)

19. Ans: (d)

Sol: m s  100kg

Sol: Find the cycle of thermodynamic.

T1 = 285 + 273 = 558 K

(dS) system  m(s 2  s1 ) = 100(0.1) = 10

T2 = 5 + 273 = 278 K

kJ (dS) system = 10 K (dS)surrounding = (S2 – S1) = – 5



kJ K



(dS)universe = (dS)sys – (dS)surr = 10 5 (dS)uni. = 5 kJ/K > 0  irreversible process



1000 492 = 1.79 – 1.76  558 278

dQ  0.022  0 T

 It is an impossible cycle.

15. Ans: (c) Sol: Area on T-S graph gives amount of heat

20. Ans: (d) Sol: Q = T + T2

supplied.

dS  

16. Ans: (c) Sol: T

dQ Q1 Q 2   T T1 T2

Q T  T 2   T T

=  ln T=C

4

3

P

B

V=C

S=C

T=C C D S=C

2 P=C 1

B V=C

21. Ans: (b) Sol: T = 273 + 30 = 303 K  = 55 106 W dQ

A P=C

S

T2  2T2  T1  T1

V

17. Ans: (d)

As heat is removed at constant temperature,   55 dQ = dS = = 0.18 MW/K T 303

Sol: For irreversible process entropy change never

be equal to zero as it always increases. Entropy once created can not be destroyed. 18. Ans: (a) Sol: For finding the final properties during an

adiabatic mixing process, use the 1st & 2nd law of thermodynamics. ACE Engineering Publications

22. Ans: (c ) 23. Ans: (c) V Sol: (dS)Isothermal = mRln  2  V1

  

P = mRln  1  P2

  

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: 195 :

Thermodynamics

24. Ans: (d)

Q1 Q 2 Q 3 =0   T1 T2 T3

25. Ans: (a)

Q 6 8   3 =0 800 600 100

dQ Q Q   T T2 T1

Sol: Sgen = (S2 – S1)–  =

 Q3 = 2.08 kJ

600 600   0.11 W / K 278 293

W = Q1 + Q2 Q3 = 6 + 8 2.08 = 11.92 kJ th =

26. Ans: (a) Sol: T1= 273+15 = 288K,T2=288.2K

(S2 – S1)– 

dQ  Sgen T

28. Sol: (dS)mixing = (dS)separation

Here dQ = 0 T  S2–S1 = dS = mC p ln 2  T1

In isothermal process, T = C

  

dQ = dW dQ=T(dS)mixing

 288.2  = 8  4200  ln  = 23 W/K  288 

Mole fraction of N2= X N 2 = 0.79 X O 2 = 0.21

dSmixing

27. Ans: (d) Sol:

T1=800 K 6kJ=Q1

W 11.92   100 = 85% QS 8  6



= 18.314[0.79 ln 0.79+0.21 ln 0.21]

T2=600K

= 4.2727

Q2=8 kJ

E



 nR X N 2 n (X N 2 )  X O 2 n (X O 2 )

dW = dQ= T(dS) = 3004.2727=1281 kJ Q3

29. Sol:

T3=100K

I=10A

Q = W For a reversible cycle,  

Q =0 T

dQ =0 T

Tatm= 27C

R = 30 t=1sec

mw = 10 gm, Cpw = 0.9  103 With work transfer there is no entropy change so entropy change of resistor is equal to zero.

ACE Engineering Publications

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: 196 :

dSRe sister

0

dS1  mC P n

Q I 2 Rt 10 2  30  1     10 J / K Tatm Tatm 300

dssurrounding

dS 2 

= 10 J/K When it is insulated:

 m i L ice  10  335   12.27 J / K T2 273

dS 3  mC Pice n

Tatm=300K

T3  263   10  2.1n   T2  273   0.78J

R = 30 t=1sec

(dS)system= dS1+dS2+dS3 = 2.9612.270.78

Heat gained by wire = work done = I2Rt

= 16.01 J/K

mw Cpw (T2Tatm) = I2Rt 10  103(0.9103)(T2 300) = 102301

31.

T2 = 633 K

Sol: Given CP = a + bT.

(dS)wire =

T dQ = mwireCPwire  n 2 T T1



 10  10 3

 



 633   0.9  10 3  n   300 

 6.720 J / K

2

T2

T2

1

T1

T1

Q   dQ   mC p dT   m(a  bt ) dT  ma[T]TT12 Q = a{T2 –T1 }+ S2

T2

(dS)universe = 6.720 + 0 = 6.72 J/K

S1

T1



b 2 T2  T12 2

 dS   mC P

(dS)surrounding = 0



dT T T2

 dS  (S 2  S1 )   ma  bT 

30. Sol:

T2  273   10  4.2  n  T1  293 

 2.96 J / K

(dS)universe = (dS)resistor +(dS)surrounding

I=10A

ME – GATE _ Postal Coaching Solutions

T1

Water

T1 = 20C = 293 K

T   (S2-S1)= amln  2   bmT2  T1   T1 

T2 = 0C = 273 K

CP = a + bT

 Water

dT T



25.2103 = a + b500 ……….. (1)

Ice T2 = 0C=273K 

30.1  103 = a + b 1200 ….. (2) By solving

o

Ice T3 = –10 C = 263K

a = 21700,

b=7

m = 10g, CP = 4.2 J/gK ACE Engineering Publications

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: 197 :

Q = 21,700(1200 500) +

7 (12002 5002) 2

Thermodynamics

33. Sol: H2O T1 = 00C = 273K

= 19355, 000J

 H2O T2 = 164.970 C = 437.97K

 1200  dS = 21700    71200  500  500 



= 2897.6716 J/K

Steam T3 = 164.970 C=437.97K (dS)1 

32. Sol: (a): Given Carnot cycle

T1=623K

 T2     T1 

(dS)1 = 1.97 kJ/K

Q1

(dS)2 =

W

Q m  L.H 1  (2066.3)   T T 437.97

 (dS)2 = 4.71 kJ/K

Q2

(dS)Universe = (dS)1 + (dS)2

T2=300K

 Carnot

dQ  mCpw ln T T1



 437.97  = 1  ( 4.187 ) ln    273 

ds = 1.44 kJ/kgK

E

T2

T  T2 623  300 =  1 = 0.518 T1 623

W = Q1

 (dS) Universe  6.68 kJ/K 34. Sol: Case.1:

W = Q1 = T1dS   W = 623 1.44 0.518 = 464.7 kJ

Copper block m = 600g, CP= 150 J/k, T1 = 1000 C Lake Temp = 80 C = T2

Find: (dS)Universe

(b) Given

If CP is in J/K means mass is included and it

Power = 20 kW  kJ  kg   s  Power (kW )  m   W   sec   kg   s  0.043kg / sec m  0.043  3600 kg/hr

 s  154.8 kg / hr m

ACE Engineering Publications

is known as heat capacity.) (dS) Cu block = mCP ln

T2 T1

 281  = 150 ln    373   (dS) Cu block  42.48J / K

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: 198 :

dSH O  mCP ( T2  T1 )

T   dSBlock 2  C n  f   T2 

T2

2

 150

ME – GATE _ Postal Coaching Solutions

(373  281)  49.11J / K 281

 50  273  150 ln    25.22 J / K  273 

 (dS) Universe  (dS) Cu block  (dS) H 2 0

 (dS) Uni  21.58  25.22

=  42.48 + 49.11

 (dS) Uni  3.64 J / K

(dS)Universe = 6.626 J/K

(dS)Universe > 0 So it is an irreversible process.

Case – 2

Same block, T2 = 80C = 281 K As it is dropped from certain height, so there is change in potential energy. Work is done by the block

35. Sol: 200K

300K Q2

Q3

Change in entropy due to work interaction = 0

E

 (dS)block = 0 (dS) H 2O 

Q mgh 0.6  9.81  100   T2 T2 281  2.09J / K

400K Q1=5 MJ

W = 840 kJ

As per first law Q  W

 (dS) Uni  2.09 J / K

 Q1  Q 2  Q 3  840  5000  Q 2  Q 3  840

Case - 3

Now by joining two blocks find (dS)uni 0

0

T1 = 100 C, T2 = 0 C Heat lost by block -1 = Heat gained by block -2 C(T1 Tf) = C(Tf  T2) T  T2 Tf  1  50 0 C 2 T   ( dS )block 1  C ln f   T1   50  273   150n   21.58 J / K  373  ACE Engineering Publications

 Q 2  Q3  4160 -------(1) For reversible engine Q1 Q 2 Q 3   =0 T1 T2 T3 

Q 5000 Q 2   3 0 400 300 200

 Q 2  4980 kJ

 2Q 2  3Q3  7500 -------(2) By solving (1) & (2) Q3 = 820 kJ, Q2 = –4980 kJ

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: 199 :

36.

Thermodynamics

dSH O

Sol:

2

dT dQ T1 =  m w CP w T2 T T

T  = mCP ln  1   T2 

(a) Heat engine reversible Q1 = 420 kJ/cycle, Q2 = 210 kJ/cycle

(dS)Reservoir = 

0

T1 = 327 C = 600 K T2 = 270C = 300 K

Q T1

=

 m w C Pw  (T1  T2 ) T1

(dS) Universe  (dS) H 2O  (dS) surr

For a reversible engine

  T   T  T2 = mwCPw ln 2    2   T2   T1

dQ Q1 Q 2  T  T1  T2 = 0 

T

= T21

  

  373   373  273  = 1(4.187)  ln   = 0   273   373 

Q1 Q 2 420 210    0 T1 T2 600 300

(dS)universe = 0.183 kJ/K

 Reversible cycle. (b) Q 2  105kJ / cycle

(b)

Q Q1 Q 2 420 105       0.35  0 T T1 T2 600 300

T1 =323K

T1 =373K

Q

Hence Impossible cycle. T1 =273 K

(c) Q 2  315kJ / cycle 

dQ Q1 Q 2 420 315     T T1 T2 600 300 = – 0.35 kJ/K <0 (Irreversible cycle)

37. Sol: (a)

Q= mwCPw(T1T2)

m=1kg water ACE Engineering Publications

By providing one more reservoir at 323 K 1st stage,(dS)universe,1st  T  T  T2  (ds) universe1  m w C Pw ln 1   1  T2  T1

  

  323   323  273 (dS) univ,1st stage 1 4.18 ln     273   323  = 0.056 kJ/kgK

Reservoir T1 =373 K

H2O T2 =273 K

T1 =323 K

 (ds) univ, 2 nd stage   373   373  323   1  4187 ln     323   373  = 0.041 kJ/K

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: 200 :

ME – GATE _ Postal Coaching Solutions

(dS)universe  dSuniv , 2 nd stage  dSuniv ,1st stage

Chapter- 6

= 0.041 + 0.056 = 0.097 kJ/kgK

Availability

(dS)uni = 0.097 kJ/K (c) From above problem, when compared to singe stage heating in a two stage heating entropy is halved. As the no. of stages of heating goes on increasing, entropy change of universe are decreasing. This way we can heat the fluid with almost no change in entropy of universe.

01. Ans: (b) Sol: Whenever certain quantity of heat transferred

from a system available energy decrease 02. Ans: (d) Sol: Irreversibility is zero in the case of

Reversible process 03. Ans: (d) Sol: A. irreversibility  Loss of availability

B. Joule Thomson exp  Throttling process 04. Ans: (a) Sol: Availability is the maximum in theoretical

work obtainable Availability

can

be

destroyed

in

irreversibilities 05. Ans: (b)

06. Ans: (b)

07. Ans: Sol: Given Ideal gas

n = 1 k mol. P1 = 1 MPa, P2 = 0.1 MPa , T1 = 300 K Tf = T1 T = constant  isothermal process ACE Engineering Publications

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: 201 :

Thermodynamics

For isothermal process,

09.

p  s2– s1 = mR ln  1   p2 

Sol: Given,

P1 = 6 MPa, T1 = 4000C

P2 = 5MPa, P0 = 0.1 MPa, T0 = 200C Availability:  = h –T0S

p  = n R ln  1   p2 

For state 0 – 1

(AE)1 = 1 – 0 = (h1–h0) –T0(s1–s0) = (3177.2–83.96)–293(6.5408– 0.2966)

For Non-flow process

= 1263.68 kJ/kg

Wmax = (u1– u2) –T0(s1–s2)

For state 0 -2

Wmax = –T0(s1– s2) ( T = constant)

(AE)2 = 2 – 0 = (h2 – h0) –T0(s2–s0)

p  Wmax = T0 n R ln  1  = T0(s1 s2)  p2 

= (3177.2 – 83.96) –293(6.63–0.2966) = 1237.5 kJ/kg Loss in availability = 1 – 2

= T0(s2 s1)

= 1263.68 – 1237.5

 1  = 300 (1) (8.314) ln    0.1  Wmax = 5743 kJ

Loss in availability = 26.18 kJ/kg 10. Sol:

08. Sol: Given, P1 = 3 MPa, T1 = 3000C

P2 = 20 kPa, T0 = 300 K 2nd law efficiency for turbine is  turbine 

W Q1  Q 2

h1  h 2  h1  h 2   T0 s1  s 2  2993.5  2609.7 2993.5  2609.7   300 6.539  7.9085

  0.48  48%

ACE Engineering Publications

500kJ

720

280

But, W = h1 – h2



835

Loss in A.E = T0 Sgen Q Q = 280     T2 T1 

 500 500  = 280    720 835  Loss in A.E = 26.77 kJ

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: 202 :

ME – GATE _ Postal Coaching Solutions

11.

14.

Sol: m = 1000 kg, T1 = 1200 K, T2 = 400 K

Sol: (a) If mixed and operated (Case I):

CPsteel = 0.5 kJ/kgK ,

For 1st unit

T0 = 300 K

Available Energy = Q – T0 (s2 – s1)  T = mCPsteel T1  T2   T0 ln 1 T  2 

T1 = 1273 K For 2nd unit m2 = 5000 kg,

  

T2 = 873 K m3 = m1 + m2 = 10000 kg.

=1000(0.5) (1200  400)  300 ln 1200    400  



m1 = 5000 kg,

3 

 A.E = 235.208 MJ

T0 = 298 K

Unavailable Energy(UAE) = T0 (s2– s1) T  = T0 mCPsteel ln  1   T2 

T4 = 313 K AE = Q – T0ds T = m3CP(T3 – T4) – m3CPT0 ln 3 T  4

= 300  1000 (0.5) ln(3)  U A.E = 164.79 MJ



12.

  

10000  1 1073  313  10000(1)  298 ln 1073  1000 1000  313 

= 3928.60 MJ

Sol: For Reversible Non-flow process

Wuseful =(u1–u2)– T0(s1– s2) + p0(v1-v2)

T1  T2  1073K 2

(b) Not mixed and operated separately (Case II):

= [(h1–p1v1) – (h2 – p2v2)

For 1st unit

+ p0(v1–v2) – T0(s1 – s2)

m1 = 5000 kg, T1 = 1273 K

= [(2993.5 – (30000.08114)) – (2706.7 – (2000.8857)]

nd

For 2 unit m2 = 5000 kg,

+ (100 (0.08114 –0.8857) –300 (6.539 – 7.1217)

T2 = 873 K

 Wuseful = 316.49 kJ

m3 = m1 + m2 = 10000 kg. T0 = 298 K, T4 = 313 K

13. Sol: Loss in available Energy in pipe  p  (E)lost = mRT0    p1   0.1p1  = 3  0.287  300    p1 

= 3  0.287 300 (0.1) (E)Lost = 25.83 kJ ACE Engineering Publications

T  AE1 = mCP (T1–T4) – mCP  T0  ln 1   T4  

5000 5000  1273   1 1273  313   1 298  ln  1000 1000  313 

= 2709.63 MJ  T2    T4 

AE2 = mCP(T2 – T4) – mCP  T0  ln = 1271.85 MJ

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: 203 :

Thermodynamics

Total AE = AE1 + AE2 = 3981.48 MJ (AE)Case II > (AE)Case I

T3 =573 K

i.e., due to mixing (which will be irreversible

Q1

causes reduction in available energy.

E

Reduction = 52.88 MJ

W Q1 –W

 Case II is preferred

T2 =298 K

15.

Now it is done by diffusion process

Sol:

 T  a C Pa T3  T2   T0 ln 3 AE2 = m  T2 

T1 =1273 K

Q1 E

 573  = 4.92(1)[(573298)(298ln   )]  298 

W Q1 –W

AE2 = 394.7

T2 =298 K

Loss in A.E = AE1  AE2 = 753.18 – 394.7 Loss in A.E = 358.48 kW

Maximum work obtained if reaction products could be directly used in heat engine. Maximum work (A.E1) = Q– T0dS   T  A.E1= mCp (T1  T2 )  T0 ln  1    T2    

5000 3600

16. Sol:

  1273    (1273  298)  298ln  298     

A.E1 = 753.18 kW

4oC

25oC Resistance heater

Now as per given

  

Q=54000 kJ/h

Heat gained by air = Heat lost by gas  g C Pg T g ma Cpa(T)a = m 

Second law efficiency,

Cop act Wrev or Cop rev Wact



 II 



Cop rev 

5000 ma  1(300 – 25) = (1273  298) 3600 ma = 4.92 kg /sec

TH 298  TH  TL 21

(Cop)act = 14.19 ACE Engineering Publications

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: 204 :

Cop act



ME – GATE _ Postal Coaching Solutions

Heating effect Work input

Chapter- 7 Properties of Pure Substance

For resistance heater, the entire work in heating is lost as heat  Cop act

54000  54000

(Cop)act = 1  II 

1 14.19

II = 7.04 %

01. Ans: (d) P T Sol: 2   2 T1  P1

  

n 1 n

02. Ans: (b) 03. Sol: Given Non flow process & adiabatic system

m = 1kg at P1 = 700 kPa, T1 = 3000 C, v1= 0.371 m3/kg ,

h1 = 3059 kJ/kg

Due to Paddle work T2 = 400C,

v2 = 0.44m3/kg,

P2 = 700 kPa,

h2 = 3269 kJ/kg

At P1 = 700 kPa from pressure Table Tsat = 164.950C  T1 > Tsat so it is in super heated steam

state. u 1  h 1  P1 v1  3059  700  0.371

= 2800kJ/kg u 2  h 2  P2 v 2  3269  700  0.44

= 2961kJ/kg It is a non flow process P=C 1

W2  P(v 2  v1 )  7000.44 - 0.371

Ws = 1W2 = 48.3kJ/kg For non flow process u1  Q  u 2  W ACE Engineering Publications

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: 205 :

Thermodynamics

W= WS+WP = Q + (u1 – u2)

05.

Wp = Q + (u1 – u2) – WS

Sol: Given , P1 = 8 bar, T1 = 2500C

= 0 + (2800 – 2961) – 48.3 2

1

= –209.3 kJ/kg

CV

P3 = 5 bar 3 TV

04. Sol: Given Non flow constant volume process

P1 = 1.5MPa

x1 = 0.9

V = 0.03m3

P2 = 500kPa

P1=8 bar T1 = 2500C

P2=7 bar T2 = 2000C

P4=0.1 bar x4 = 0.9

v1  x 1 v g  0.9  0.132  0.1188 m /kg 3

h 1  h f  x 1 h fg  845  0.9  1947  2597.3kJ/kg u1 = h1 – P1v1 = 2597.3–(15000.1188) u1= 2419.1 kJ/kg , V = 0.03 m3 V 0.03 (i) Mass of wet steam =   0.253 kg v1 0.1188

First check steam is in which state for section 1 P1 = 8 bar , T1 = 2500C Tsat = 170.40C at 8 bar T1 > Tsat  super heated state From steam table,

h1 = 2950 kJ/kg v1 = 0.2608 m3/kg

V = C (Rigid vessel) v1 = v2 = x2vg2

s1 = 7.403 kJ/kgK For section 2 :

0.1188 = x2(0.375) (ii)

P2 = 7 bar , T2 = 2000C

x2 = 0.317

Tsat = 164.970C

h2  h f 2  xh fg 2  640  0.317  2109

T2 > Tsat  super heated state.

h2 = 1308.55 kJ/kg

From steam table, h2 = 2845.2kJ/kg

u2 =h2 – P2v2 = 1308.55 – 5000.1188 = 1249.15 kJ/kg (iii) H = m(h2 – h1) = –1288.75m = –326.054 kJ

(iv)

4

s2 = 6.893 kJ/kgK For section 3 :

P3 = 5 bar , Tsat = 151.83 C Throttling section

U = m(u2–u1) =0.253(1249.15–2419.1)

In throttling enthalpy remain constant so it is

U = –295.997 kJ

also known as isenthalpic device

For constant volume dV = 0, 1W2  0

(Throttle valve)

1

Q 2  mu 2  u1  = U

U = –295.997 kJ ACE Engineering Publications

hg = 2748.5 kJ/kg, sg = 6.8203 kJ/kgK  h2 = h3 = 2845.2 kJ/kg

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: 206 :

So from h3 we can say

ME – GATE _ Postal Coaching Solutions

ii) Temperature drop through throttle valve:  (T)TV = T2  T3 = 200  195.17

h3 > hg at P = 5 bar At state 3 it is in super heated state.

(T)TV = 4.830C

Tsat = 151.83 C, P3 = 5 bar T

h (kJ/kg)

s

iii) Work output of engine: h3 + Q = h4 + WE

151.83  2748.1  6.8207

 WE = h3  h4

195.17  2845.2  s3 = ?

Assuming Q = 0, adiabatic and very fast

 2855.8  7.0610

200

process.

dT = – 48.17, dh = –107.7, ds = –0.2403 dT = ?,

 WE = 500.5 kJ/kg

dh = –97.1 ds = ? dT =

dh  dT dh

iv) Entropy change during throttling: s3 – s2 = 7.0369  6.893

 97.1  48.17 151.83 – T3 =  107.7

= 0.1439 kJ/kgK Entropy change in engine:

T3 = 195.17C ds =

( Q = 0)

s4  s3 = 7.4008  7.0369

dh  97.1  ds   0.2403  107.6 dh

s4  s3 = 0.3639 > 0 At both sections in throttling and engine

= 0.216 kJ/kgK

entropy change > 0, so both are irreversible

6.8207–s3 = –0.2163

process.

 s3 = 7.0369 kJ/kgK At 0.1 bar, hf = 191.81 kJ/kg

06.

hfg = 2392.1 kJ/kg

Sol:

h 4  h f 4  x h fg 4  191.81  0.9  2392.1

h

 2344.70kJ/kg s 4  s f 4  x s fg 4  0.649  0.9  7.50

1

= 7.4008 kJ/kg.K i)

2

Heat lost in pipe: h1 + Q = h2 + W  h1 + Q = h2

s

( W = 0)

 Q = h2  h1 = –104.8 kJ/kg ACE Engineering Publications

P1 = 3 MPa = 30 bar, h f1 = 1008.41 kJ/kg, h fg = 1795.7 kJ/kg P2 = 0.1 MPa, T2 = 1200C

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: 207 :

Thermodynamics

From steam tables

At 1 bar, 1000C, h = 2776.6 kJ/kg

At P2 = 0.1 MPa,

A 1 bar, 1300 C, h2 =?

Tsat =99.610C and T2 = 1000C,

P2 = 100 kPa,

T2 > Tsat ( superheated)

T2 >Tsat (Superheated state)

h = 2676.2 kJ/kg

100C 2776.6 kJ/kg

At

Tsat = 99.61C

1300C  h3 = ?

P2 = 0.1 MPa and T2 = 1500C

150C 2675.8 kJ/kg

h = 2776.4 kJ/kg dT = 50C 100.8 kJ/kg

At P2 = 0.1 MPa and T2 = 1200C,

dT =20C  x

h2 = ?

x=

1000C  2675.8 kJ/kg

h2 = 2776.6 – 40.32

0

dT = 50 C  dh = 100.8 kJ/kg

h2 = 2736.28 kJ/kg

0

dT = 30 C  x = 60.48 kJ/kg



m

h2 = 2716.12 kJ/kg = h1

(For throttling process)

3.4 kg 3.4   0.0113 kg/s 5 min 300

By steady flow energy equation

 If dryness fraction is x



 m  h1  Q  h2  W m

 h1 = h f1 + xh fg1



Q h1  h 2  m

1008.41+ x(1795.7) = 2716.12  x = 0.95

2  2559.28 kJ/kg 0.0113 h 1  h f1  xh fg1  444.36  x  2240.6  2736.28 

07. Sol:

100.8  20  40.32kJ / kg 50

1500C  2776.6 kJ/kg

(1) 1.25 bar

1 bar

1300C

x = 0.9439

(2)

1300C

08. Sol: V1 = 3 m3

m = 5 kg 2 kW capacity

From steam tables At 1 bar, 1500 C, ACE Engineering Publications

h = 2675.8 kJ/kg

 v1 =

3 = 0.6 m3/kg 5

P1 = 200 kPa = 2 bar,

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: 208 :

v1  v f 1  x ( v g1  v f 1 )  x1 =

ME – GATE _ Postal Coaching Solutions

09.

v1  v f 0.6  0.001061  v g  v f 0.8857  0.001061

Sol: Given condition: adiabatically mixed

1 3

= 0.677

4

h1 = hf +x1hfg = 504.71 + 0.67702201.6 = 1995.19kJ/kg u1 = h1 p1v1 = 1995.19 2000.6

P4 = 1kPa

= 1875.19 kJ/kg

2

As it is a closed rigid tank, volume of the

P1= 2 MPa = 20 bar

steam remains constant

T1 = 3000C ,

 1 = 3 kg/min m

3

v1= v2 = vg2 = 0.6 m /kg v (m3/kg)

P1(kPa)

300



P2 = ?  325



h (kJ/kg)

0.6058

 2724.9

0.6

 h2 = ?

0.5619

 2728.6

P = 25

v = – 0.0439, h = 3.7

P = ?

v = – 0.0058, x = ?

Section At 1, h1 = 3024.2 kJ/kg

s1 = 6.788 kJ/kgK P2 = 2 MPa = 20 bar T2 = 4000C  2 = 2 kg/min m

At Section 2,

h2 = 3248.4 kJ/kg

x = 3.303

s2 = 7.1292 kJ/kg-K

P2 = 300+3.303 = 303.3 kPa h2 = 2725.38 kJ/kg u2 = h2 – P2v2 = 2725.38 – 303.30.6 = 2543.4 kJ/kg

 3 m 1 m  2 = 5 kg/min m

Energy balance : 





m1 h 1  m 2 h 2  m 3 h 3

h3 = 3113.88 kJ/kg

V = C, 1 W2  0 1

Mass balance :

Section At 3, P3 = 20 bar

Q 2 1 W2 1 U 2

 Heat supplied, 1Q2 = m(u2  u1)

hg = 2798.2 kJ/kg h3 > hg (superheated)

= 5(2543.4 – 1875.19)

3000C  3024.2 kJ/kg  6.7604 kJ/kgK

= 3341.05 kJ

T3 = ?  3114 kJ/kg  s3 = ? 3500C  3137.7 kJ/kg  6.9583 kJ/kgK T = 500C h = 113.5  s = 0.1979

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: 209 :

Thermodynamics

 2 A = 0.0054m2  4  d

s3 = 6.918 kJ/kgK dT = 10.440C

d = 0.083 m = 8.3 cm

T3 = 339.560C Rate of increase of the entropy of the universe

10. Sol: Given separating & throttling calorimeter

 3s 3  m  1s 1  m  2s 2 s gen  m = (5  6.9186) – (3 6.788) – (2 7.1292) = 0.0294 kJ/min.K As system is perfectly insulated (ds) surrounding = 0

P1 = 15 bar = P2 , m1 = 0.55 kg T1 = 198.30 C = T2 , m2 = 4.2 kg P3 = 1 bar, T3 = 1200 C  h3 = 2716.3 kJ/kg (1)

(ds)universe = (ds)system+(ds)surr

(2) TV (3)

CV

= 0.226 kJ/minK m2=4.2kg

h 3

mw=0.55 kg h

4

2 s

At 4,

3

1

s3 = sf4 + x (sfg4)

6.917 = ( 0.106) + x (8.87) x = 0.77

s

h4 = hf4 + x hfg4 = 1942.29 kJ/kg v4 = x vg4 = 0.77129.19 = 99.47 m3/kg

(Dryness fraction)separator, x1 =

(V) Nozzle exit  44.72 h  44.72 h 3  h 4  1530.69 m / s AV m v1 

5 A  1530.69  60 99.47

=

m2 m1  m2

4.2  0.88 4.2  0.55

For throttling

h2 = h3

h f 2  x 2 h fg 2  h3 844.55 + x2 (1946.4) = 2716.2 x2 = 0.9616

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: 210 :

ME – GATE _ Postal Coaching Solutions At state 3:

Mass of vapour = mv= x2m2 = 0.964.2

h 3  h f3  xh fg3  561.43  0.95  2163.5

= 4.032 kg (Dryness fraction)Boiler =x=

= 2616.75 kJ/kg

mv  0.8488 m total

Mass balance: s m w m 3 m

As quality of steam at boiler is < 90% so

w m 3 5 m

only throttling calorimeter can not be used.

Energy balance:      m s h 1  m w h f 2   m f  m w h 3  

11. Sol: Given

5  3064.8   m w  167.5    5  m w 2616.75

At, P1 = 5 bar = 500 kPa,



T1 = 3000C , h1 = 3064.2 kJ/kg









At,

 m w  0.9147

0

P3= 3 bar , T2 = 133.5 C hf2 = 561.4 kJ/kg

12. Ans: (d)

hfg2 = 2163.5 kJ/kg

Sol: At P = 1 atm, hfg = 2256.5 kJ/kg

T3 = 400C

Power =

w m (2)

 w  h fg m

=

time

0.5  2256.5 18  60

= 1.05 kW (3)

(1)

13. Ans (c) Sol: Heat lost by steam = Heat gained by ball

ms  hfg = mb  Cpb (T2 T1) 1 m

ms 2256.6 = 5 1.8 (10025)

m 3

ms = 0.299 kg = 299 gm

P1= 5 bar, T1= 300C, P3 = 3 bar, x3 = 0.95

14. (i) Ans: (b), (ii) Ans: (a), (iii) Ans: (c)

At state 1, T1= Tsat

Sol:

h1 = 3064.8 kJ/kg,

P

2 P=C 3 V=C

s1 = 7.46 kJ/kg K At state 2:

1

h2 = (hf)40C =167.5 kJ/kg ACE Engineering Publications

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: 211 : i)

Thermodynamics

Piston cylinder device (Non flow process)

15. Ans: (d)

T2 = 7000C, P3 = 1.2 MPa, T3 = 12000C

Sol: QS = 380 kW At 5 MPa, Tsat = 263.990C = T1

At point -2; (Superheated state)

At 2 MPa, Tsat = 212.420C = T2

3

u2 = 3474.5 kJ/kg , v2 = 0.37294 m /kg

=

At point - 3:

u3 = 4465.1 kJ/kg, v3 = 0.56646 m3/kg 2W3

T1  T2 263.99  212.42 = = 0.096 263.99  273 T1

QS  Q R QS 380  Q R 0.096 = QR QR = 343.5 kW

= P(v3 v2)

=

= 1.2  103 (0.56646 0.37294) = 232.224 kJ/kg 2Q3

 2W3 = 2U3

2Q3

= 2W3 + u3 u2 = 232.224 + (4465.13474.5) = 1222.824 kJ/kg

16. Ans: (d) Sol: m = 5 kg

For process 1  2:

T 3

v1 = v2 = 0.37294 m /kg

3MPa

2

1

At the beginning; It is in dry and saturated state. Hence by looking the vg column of the

s1

s2

steam table for a value of 0.3729 m3/kg, we

s

At 3 MPa, s2 = sf = 2.6656 kJ/kgK

will get pressure P = 500 kPa and ug =

s1 = sg = 6.1869 kJ/kgK

2561.2 kJ/kg

dS = m(s2 s1) ii) For 1  2 ; V = C 1W2

= 5 (2.6656 6.1869)

= 0, 1Q2 = u2 u1

= 17.7kJ/K

= 3474.5 2561.2 = 913.3 kJ/kg iii) Total heat transfer = 1Q2 + 2Q3

= 913.3 + 1222.824 = 2136.124 kJ/kg

17. Ans: (a) Sol: T 8MPa

iv) Total work done = 1W2 + 2W3

0.1MPa

= 0 + 232.224 = 232.224 kJ/kg

1 2 s

 = 3 kg/sec m ACE Engineering Publications

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: 212 : By applying steady flow energy equation to turbine

 = m   h1 + Q  2h 2 + W m

20. Ans: (d) Sol: At P = 10 MPa, Tsat = 311C

 = 0 (For adiabatic) Q

T = 500C

 = m  (h 1  h 2 ) W

T > Tsat  superheated state At 10 MPa and 5000C

At 8 MPa & 5000 C, h1 = 3398.3 kJ/kg

v = 0.03279 m3/kg

At 0.1 MPa, h2 = hg = 2675.5 kJ/kg  = 3(3398.3 2675.5) = 2168.4 kW W

18. Ans: (c) Sol:

ME – GATE _ Postal Coaching Solutions

Mass =

V 3 = 91.49 kg  v 0.03279

21. Ans: (a)

T 8MPa

T1= 500C

Sol:

T

T1= 500C

4MPa

0.2MPa

T2= 300C

0.5MPa

T2= 250C

s

s

0

At 8 MPa and 500 C, s1 = 6.726 kJ/kgK 0

At 0.2 MPa and 300 C, s2 = 7.8491 kJ/kgK

m(s2 s1)  

Q = Sgen T

m(s2 s1) = Sgen

Sgen = 18(7.8491 6.7266) = 20.22 kW/K 19. Ans: (a) Sol: At 200 kPa, vf = 0.001061 m3/kg,

vg = 0.8857 m3/kg v = 300 m3, P = 200 kPa mf + mv = m

h1 = 3445.2 kJ/kg 0

At 0.5 MPa and 250 C, h2 = 2960.7 kJ/kg  = 1350 kg/hr = 0.375 kg/s m By applying steady flow energy equation to  = m   h1  Q  h2  W m

 0.375(3445.2  2960.7) 25 = W  = 156.68 kW W 22. Ans: (d) Sol: For throttling h1 = h2

At 1.4 MPa & 90 , h1 = 319.37 kJ/kg At 0.6 MPa from

vf + vv = v (0.25m0.001061)+(0.75m0.8857) = 300

ACE Engineering Publications

At 4 MPa & 500 C,

turbine

Q    0 for adiabatic process    T  

m = 451.44 kg

0

Super heated table R134a H = 319.37 kJ/kg Pump is , P = 0.6 MPa at 80C

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: 213 :

Thermodynamics

09. Ans: (d) and (a)

Chapter- 8

Sol: For same maximum pressure in Otto &

Air Cycles

Diesel cycle

 Diesel   Otto 01. Ans: (a)

And rk Diesel  rk Otto

Sol: 1–2  Isothermal

31  Adiabatic process

10. Sol:

02. Ans: (d) Sol: For Maximum specific output in case of Otto

v4  rc  1.615 v3

cycle, the temperature of working fluid at the end of compression and expansion should be equal

re 

3

P

v1  rk  12 v2

v=c

Qin

T

exp

2

4

2

v =c

comp

1 V

03. Ans: (c)

pv1.35  C, n  1.35

3 T 2 = T4

pmax = p3 = p4 = 54p1

4

4

Qout

1

rk 12   7.43 rc 1.615

3

T

S

P

2

3

4

5

5 2

04. Ans: (c)

05. Ans: (b)

1

1 S

06. Ans: (d)

Process 1-2 :

Sol: For equal rk & heat rejected

p1 v1n  p 2 v n2

otto   Dual   Diesel 07. Ans: (c) Sol: Otto cycle

08. Ans: (c) Sol:  Carnot  Stirlling

ACE Engineering Publications

v p2  p1  1  v2

V

n

  = p1 12 1..35 = 28.64p1 

v  p1v1  28.64p1  1  p1v1  p 2 v 2  12   1W2  n 1 1.35  1 = – 3.96p1v1

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: 214 :

11.

Process 2-3:

V = C,

2

W3  0

Sol: T1  350K , P1 = 2 bar

Pmax = P3 = 6 bar

Process 3-4:

P=C 3

ME – GATE _ Postal Coaching Solutions

V1 V4   1.4  rk  re V2 V3

W4  p 3 v 4  v 3  v  = p 3 v 3  4  1  0.615p 3 v 3  v3 

54p1v1 v  W4  0.615p 3  1   0.615  12  12 

3

RT1 0.287  350 m3   0.5022 V1  P1 kg 200 Process -12 P1 V1  P2 V2 P2  P1 

= 2.76p1v1 Process 4-5:

p4 v  p5v n 4

V2 

n 5 n

4

V1 0.502   0.358 1 .4 1 .4

Process – 2 – 3 :

1.35

v   1  p5 = p 4  4   54p1    7.43   v5 

V1 = 21.4 = 2.8 bar V2

 3.6p1

p 4 v 4  p 5 v 5 54p1  1.615  v 2  3.6p1v1  n 1 n 1 1.615 v1  3.6p1v1 54p1  12 = = 10.47p1v1 1.35  1

W5 

Wnet 1W2  2W3  3W4  4W5  5W1  3.96p1v1  0  2.76p1v1  10.47p1v1

P2 P3  T2 T3

 T3 

P3 6  T2   350  750K P2 2 .8

V QS = 3 W4  RT3 ln 4  V3

  

= 0.287  750 ln 1.4 = 72.41 kJ/kg Q R 3W4  RT1 ln

= 9.27 p1v1

V4  0.287  350 ln 1.4 V3 = 33.79 kJ/kg

Mean effective pressure =

Wnet Wnet 9.27p1 v1   v swept volume v1  v 2 v1  1 12

= 10.1p1

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th  Mep =



QS  Q R  53.35% QS

Wnet 72.41  33.79 = swept volume V1  V2 

72.41  33.79 = 268.2 kPa 0.502  0.358

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: 215 :

T3, T1 are constants hence work done

12. Sol:

Thermodynamics

depends only on rk

P

3

2

So, s=c

dW  C V T1   1rk 2  C V T3 1   rk1  1 drk

4

s=c

1

=0 1

V

 rk opt

V1  15  V1 = 15V2 V2

rk 

= compression ratio at which the work is maximum

6.5 V1  V2   0.91V2 100

V3  V2 

T2 = T1(rk)

V3 = 1.91V2

 th  1   1

1

  rk

.  1

1

1     2  T  1  T2  T1  3     T1   

V3  1.91 V2

rc 

rc  1 rc  1

T4 

T3

rk  1

13. T

When

3 Tmax

2 1

Tmin S

W  C V T3  T2   T4  T1    T  C V  T3  T1rk 1   31  T1   f rk  rk  

ACE Engineering Publications

1   1   T3  2 1   

 T3 T1

  

T3 T1

we

are

operating

at

optimum

compression

Wnet = QS – QR



T3

of expansion = temperature at the end of

T3

rk  1

 T3 T1

compression ratio, the temperature at the end

4

T2 = T1 rk 1

 1

   T1   T2 = T4 =

T4 



 1.911.4  1  1    60.8% . 1.4  15 0.4  1.91  1 

Sol:

1

 T  2  1  Tmin  21      1     T3   Tmax 



Wopt = CV[T3T2T4+T1]

  C T  T  2 T T  C  T  T 

= C V T3  T3 T1  T3 T1  T1 V

3

1



3 1

2

V

3

1

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: 216 :

14.

ME – GATE _ Postal Coaching Solutions

D = 10 cm ,

Sol:

P

L cos  TDC 2 L/2

L=2r

L = 14 cm

FVO 2

FVC 3

Le

4

40

1

EVO BDC

L = 14 cm ,  = 40 , D = 10 cm Effective stroke length Le  L / 2 

L cos  2

 2  D L e   102  12.36  970.75cm3 4 4 3 VC = 157 cm 

rk effective 

1  th  1    rk

VC

  

 1

L L  cos  = 7+7cos 60 = 10.5 cm 2 2

Lf 

L L  cos 1 2 2

Vs eff

Effective stroke volume

Vs eff  VC

Le 

 7  7 cos 20 0 = 0.42 cm

= 7+7cos 40 = 12.36 cm

Vs eff

970.75  157   7.18 157

 1   1    7.18 

V

 2  D L e   10 2  10.5 4 4 3  824.6cm 

VC = 40.2 cm3

rk effective 

Vs eff

 VC

VC

 21.51

Volume corresponding to fuel cutoff

0.4

V3  V2 

  2 D L f   10 2  0.42 4 4 = 32.98 cc

= 0.54 or 54% V3  40.2 = 32.98.

15.

V3 = 73.18

Sol:

Lf

rc 

TDC FVC 20 L/2

Le

L

 th  1 

60

 1

EPC BDC ACE Engineering Publications

V3 73.18   1.82 V2 40.2 1

.  1

rk

rc  1 rc  1

1

1.4  21.51

0.4



1.821.4  1 1.82  1

= 66.5%

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: 217 :

Thermodynamics

T3  T2 = 700C

16. Ans: (c) Sol: C  1 

C 

300  0.85 2000

T3 = 700 + 1097.46 = 1797.46C

work done Heat supplied

0.85 

19. Ans: (b) Sol: v1 

600 QS

rk 

QS = 705.8 kW dS 

dQ 705.8   0.35 kW/K T 2000

1 1   0.833 m 3 / kg 1 1.2 V1 2.2   8.46 V2 0.26

V2  mep 

17. Ans: (c)

0.834  0.098m 3 / kg 8.46 Wnet 440  598.8 kPa  V1  V2 0.833  0.098

Sol: V1 = 3L, V2 = 0.15L

Compression ratio rk 

V1 3   20 V2 0.15

rc 

V3 0.30  2 V2 0.15

20. Sol: Qs = 1500 kJ/kg,

P1=100 kPa,

T1 = 27C = 300 K rk = 8 =

1 1 r  1  Diesel  1    1  c  rk rc  1 1 1 21.4  1 =1     0.6467 1.4 20 1.41 2  1

= 64.67%

V1 8 V4   V2 1 V3

(Cv)air = 0.72 kJ/kgK For process 1- 2

P1 V1  P2 V2

V  P2  P1  1  V2



  = 100  81.4 

= 1837.9kPa 18. Ans: (a)

P

V 2 .5 Sol: rk  1   10 , V2 0.25

7393.57 kPa

T1 = 20C = 293K He   

1837.9 kPa

5 3

T1 = 100 kPa 5

3

2

4 s=c

402.2 kPa 1

1

T2  293  10 3  1370.46 K ACE Engineering Publications

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: 218 :

ME – GATE _ Postal Coaching Solutions

T

Chapter- 9

3

2772.54K 1206.8K 689.2K

2

Psychrometry

4

01. Ans: (b) Sol:

T1 = 300K

1 S

P T2  T1  1  P2

  

1  

 100   300     1837.9 

11.4 1.4

1

DBT

= 689.2K QS = Cv (T3 – T2) = 1500 kJ/kg  0.72  (T3 – 689.21) = 1500 T3 = 2772.54 K P P  2  3 T2 T3

 P3  P2 



2

T3 2772.54 = 1837.9  689.2 T2

During sensible cooling of air DBT decreases, WBT decreases, h decreases and  = constant, DPT = constant, R.H increases 02. Ans: (c) Sol:

2

P3 = 7393.57 kPa

1

DBT

Process 3  4

During adiabatic saturation process DBT

 P3 V3 = P4 V4 

V  1  P4 = P3   3   7393.47    8  V4   P4 = 402.2kPa  T4 = 1206.8K 1 1  Otto  1   1  1.41  1 8 rk 

W W = 847 kJ/kg 1500

ACE Engineering Publications

decreases, WBT = constant, h = constant,

1.4

T3 V3 1  T4 V4 1

 Otto = 0.56 =

 

specific

humidity

()

increases,

DPT

increases, relative humidity increases. 03. Ans: (a) Sol: T1

20oC

T2

40oC

Tcoil

45oC

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: 219 :

By pass factor =

Tcoil  T2 45  40  Tcoil  T1 45  20

Thermodynamics

09. Ans: (c) Sol: L.H.L = 0.25 S.H.L

BPF = 0.2 04. Ans: (b) Sol: When warm saturated air is cooled, excess

S.H.F =

SHL SHL  LHL

S.H.F =

S.H.L  0.8 1.25  S.H.L

moisture condenses but relative humidity remains unchanged

10. Ans: (d) Sol:

05. Ans: (b) Sol:

RH=100%

B C  

A 

  WBT

BC AC

By pass factor =

DBT

Specific humidity will never 100%

11. Ans: (b)

06. Ans: (a)  P  Psat Sol:      atm  Patm  Pv

12. Ans: (d)

 Pb  Ps     Pb  Pv

Sol: 1 

2 

07. Ans: (c) Sol: Case (A): Moist air is adiabatically saturated Case (B): Moist air is isobarically saturated  W.B.T

isobaric

During chemical dehumidification Enthalpy & W.B.T remains constant, specific humidity decreases, dew point temperature decreases and relative humidity decreases.

 adiabatic

Case-(A): Adiabatically saturated  W.B.T

13. Ans: (c)

Case-(B): isobarically saturated  D.P.T

14. Ans: (a) 08. Ans (b)

Sol: Tcoil is greater than dew point temperature but

Sol: For dehumidification, the coil temperature

should

be

less

than

the

dew

point

less than dry bulb temperature hence it is sensible cooling.

temperature of the incoming air. ACE Engineering Publications

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: 220 :

15. Ans: (c) Sol: Temperature of water spray is greater than

dew point temperature hence it is a heating process and water molecules are mixing with air hence it is humidification.

ME – GATE _ Postal Coaching Solutions Pv  0.012 95  Pv

  0.622

PV = 1.798 kPa 

Pv 1.798   0.567  56.7% Psat 3.1698

16. Ans: (b)

19. Ans: (c)

Sol: Heat is absorbed so it is absorption or

Sol: Tsat = 30C  Psat = 4.2469 kPa

chemical process in which WBT remains

Patm  90 kPa kPa ,  

constant & DBT increases.

PV  0.75  4.2469  3.185 kPa

17. Ans: (d)

Pa = Patm – Pv

Sol: Tsat = 25C   = 100%

Pair  86.815 kPa

Psat = 3.1698 kPa Ptotal = 100 kPa 

ma 

PV P  1  V  3.1698 Psat Psat

  0.622

Pa V 86.815  40   39.93 kg R a T 0.287  303

20. Ans: (c)

Pv = Psat PV 3.1698  0.622  Patm  PV 100  3.1698

 0.0204  =

PV  75% Psat

kg Vap kg dry air

  75% 

PV Psat

21. Ans: (d) Sol: Tsat = 20C, Psat = 2.339 kPa

18. Ans: (c) m kg of Vapour 0 .6 Sol:   v   0.012 ma 50 kg of dry air

ACE Engineering Publications

Patm = 96 kPa

Pa = Patm  PV = 96  3.185 = 92.8 kPa

m V  100  0.0204  2.04 kg

Psat = 3.1698 kPa

Psat = 4.2469 kPa

PV = 0.754.2469=3.185 kPa

mv ma

Tsat  25 0 C ,

Sol: Tsat = 30C,

 = 0.5 

PV  PV  0.5  2.339  1.169 kPa Psat

Corresponding to partial pressure of vapor whatever the saturation temperature is the Hyderabad |Delhi |Bhopal |Pune |Bhubaneswar |Bengaluru |Lucknow |Patna |Chennai |Vijayawada |Visakhapatnam |Tirupati

: 221 :

Thermodynamics

temperature at which the water vapor present

23. Ans: (a)

in air starts condensing and this beginning of

Sol:

2

this condensation is called as dew point temperature.

1

At 5C  0.8725 T2 - ?  P2 = 1.169

150C

10C 1.2281

Ptotal = 90 kPa, Tsat = 15C, Psat = 1.7057 kPa

1.2281  0.8725 10  5  1.169  0.8725 T2  5

1 

T2 = 9.16 C

Psat1

 0.75

1  0.622 

1



300C





 C P a T1  0  1 h fg 00 C  C PV T1  0

 0.622 

 1.005  30  0  0.0232500  1.8830  0

= 0.7(52.9988.94) = 25 kW

Patm  PV2

kg of Vap 2.377  0.618 90  2.377 kg of dry air

 a  2  1  mv  m



 40.0168  0.0089   0.03kg of vapor

 1.00515  0  0.0152500  1.8815  0

Total heat load = ma (h2h1)

Pv 2

Mass of vapour added ,

h 2  C Pa T2  0   2 h fg 0 0 C  C P V T2  0

= 52.99 kJ/kg of dry air

PV 2  PV 2  2.377 kPa Psat 2

 2  0.622 

Cooling load on coil h1



kg of vapour kg of dry air

Tsat = 25C, Psat = 3.1698 kPa,  = 0.75

2=0.015

= 88.94 kJ/kg of dry air

PV1 1.278  0.622  Patm  PV 1 90  1.278

 0.0089

1=0.023

2

150C

PV1

PV1 = 1.278 kPa

22. Ans: (d) Sol:

250C

24. Sol: Psat = 3.166 kPa,

Tsat = 250 C = 298 K

 = 0.74 =

Pv P  v Psat 3.166

Pv = 0.74  3.166 = 2.34 kPa, ACE Engineering Publications

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ME – GATE _ Postal Coaching Solutions

At 2.34 kPa , Tsat = 200C = 293 K = DPT

26.

Corresponding to saturation temperature

Sol: At T1 = 20C

whatever the saturation pressure, the water vapor starts condensing.

Psat = 2.337 kPa, 1 = 0.65

1 

dT = 298 293 = 5 K

PV1 Psat

 PV1  1Psat  0.65  2.337 = 1.52 kPa

6.5 K drops in temperature = 1000 m 5 K drop temperature =

5  1000 6.5

= 769 m 25.

1  0.622

PV1 Patm  PV1

 0.622   0.0096

Sol: T1 = 20C

1.524 100  1.524

kg of vapour kg of dry air

1 = 80%

RH=65%

 3  2  1  1  1 4 4

2

Psat = 2.3339 kPa 1  0 . 8 =

Pv1  Pv1  1.87 kPa Psat

1  0.622 

PV1

2 

Patm  PV1

Kg of vapour Kg of dry air

3  0.0118  0.008 4

 0.622

PV 2 Patm  PV2

 PV2  1.45 kPa  T2 = 12C

ACE Engineering Publications

45C

20C

1.87  0.622  100  1.87

 0.0118

1

 0.008

T2 = 45C,

2 = 100 %

PV2  Psat  9.557 kPa

 2  0.622   0.066

9.557 100  9.557

kg of vapour kg of dry air

1

2 T2 = 45C 2 = 100%

T1 = 20C 1 = 65% m =1kg t=40 min =2400sec

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Mass of water absorbed by air

Thermodynamics

 2 

w m  a   2  1  m

1  a 0.066  0.0096 m 2400

0.6224.241 101  4.241

 2  0.0272

= 0.0073 kg/sec

kg vapor kg Dry air

Pa = Patm – Pv 2

 a = 0.0073  3600 = 26.6 kg/hr m

Pa = 101 – 0.8525 = 100.147 kPa 



Pa v = m a RaT

27.

a  m

Sol: Given air enters to cooling tower 

V  10 m3 / sec

12.116 kg / sec 0

P1 = Psat = 1.705 kPa, T1 = 15 C

Mass or water vapor removed

0

  2  1  = m

T2 = 30 C = Tsat

1 = 50%

P2 = 4.241 kPa

= 12.116 (0.0272 – 0.00529)

Patm = 101 kPa,

= 0.265

Rair = 0.287 kJ/kg K

1 

PV1 Psat

Pa v 100.1475  10  R aT 0.287  288

PV1

 0.50 

kg sec

= 0.265  3600 = 954 kg/hr

1.705

28. Sol: h f ADP  46.2kJ / kg of dry air

PV1  0.8525 kPa

2 = 3 = 0.008 kg/kg of dry air

Now 1  0.622

Pv1 Patm  Pv1



0.6220.8525 101  0.8525

kg vapour  1  0.00529 kg dry air

h1 = 58 kJ/kg h2 = 32 kJ/kg h3 = 39 kJ/kg v1 = 0.873 m3/kg

Now for T2 = 300C Psat = 4.241 kPa 1

 Pressure at saturation is equal to Pv 2  2 

PV2 Psat

2 = 0.622

2

3

 Pv 2 = 4.241 kPa

 Tcoil TADP

Pv 2 Psat  Pv 2

ACE Engineering Publications

1

30C 18.5C DBT DBT (Apparatus Dew point)

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ME – GATE _ Postal Coaching Solutions

1 = 0.0105 kg/kg of dry air

29. Ans: 0.02

2 = 0.008 kg/kg of dry air

Sol: m  m a  m v1

3 = 0.008 kg/kg of dry air

m m  1  v1 ma ma

T1 = 30C,

TWBT = 20C

T2 = 18.5C

R.H = 60%

TADP = 11C

hf = 46.2 kJ/kg

Mass of water removed from air f m  a 1   2  m

m  1  1 ma m v1  1  m a  0.0110  0.1kgs / sec mv2 = 0.1 kgs/sec

= 3.818(0.0105 – 0.008)

Total mass of vap or after mixing

–3

= 9.54510 kg

 m v  m v1  m v 2  0 .1  0 .1

At 11C from the steam table whatever is the

 0.2 kg / sec

saturated liquid enthalpy is the enthalpy of water coming out of the coil. h f  46.2 kJ / kg a  m

Specific humidity of mixture



 V 200  v1 60  0.873

m v 0.2 kg / sec  m a 10 kg / sec

= 0.02 kgvap/kgd.a

 3.818kg / sec

 

 : Cooling load Q

 m  a h1  Q  ah2  m  f hf m  ah2  m  a 1   2 h f m

 m  a h 1  h 2   m  a 1   2 h f Q = 3.818(58–32) –3.818(0.0105–0.008)46.2 = 98.8 kW

 

 Heat supplied in the heater Q 1

 m  ah2  Q  ah3 m 1  m  a h 3  h 2   26.726 kW Q 1

ACE Engineering Publications

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: 225 :

Thermodynamics

h3  h2 = WHP h3 613 = 3

Chapter- 10 Rankine Cycle

h3 = 616 kJ/kg QS = h1 h3 = 3514 616

01. Ans: (d)

= 2898 kJ/kg

Sol: Assertion is false and Reason is true

Steam rate or specific steam consumption =

1 kg 3600 kg  WT  WC kW sec WT  WP kW.hr

08. Ans: (d) PW vdp  Power input Power input

Sol:  =

For Carnot cycle as pump work is very high

0.15 5000  200 60 0.75 = Pi

so specific steam consumption is very high. For Carnot cycle the mean temperature heat addition is greater than Rankine cycle,

Pi = 16 kW

so C > R 09. 03. Ans: (d)

02. Ans: (d)

Sol: At P = 70 bar , Tsat= 285.880C At P = 0.075 bar, Tsat = 40.290C

05. Ans: (a)

04. Ans: (b)

T =

06. Ans: (c)

(Tsat ) boil  (Tsat ) cond n 1

=

07. Ans: (b) Sol:

n = number of feed water heaters = 7

T

T = 30.690C

1

Now for T1 = Temperature of (FWH)1

3

(Tsat)boiler − T = 285.88 30.69

2

= 255.190C S

h1 = 3514 kJ/kg WHP = 3 kJ/kg h2 = 613 kJ/kg, WLP = 1 kJ/kg ACE Engineering Publications

(285.88  40.29) 7 1

And the corresponding pressure is P1 = 4322.7 kPa. T2 = Temperature of (FWH)2 = (Tsat)boiler −2× T = 285.88 − (2  30.69)

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: 226 :

(T2)

FWH

= 224.50C and the corresponding

pressure is P2 = 2549.7 kPa Temp of (FWH)3 = 285.88 330.69 0

ME – GATE _ Postal Coaching Solutions

h2 = h f 2 + x2 h fg

2

= 191.83 + 0.82  2392.8 = 2153.92 kJ/kg

= 193.8 C Similarity we can find the rest of the temperatures and corresponding pressure.

Process (3-4)

s3 = s4 = 0.6393 kJ/kgK Wp = h4 – h3 = Vf 3 (Pboil – Pcond)

10.

= 1.01 10-3 (7000 – 10)

Sol: At T1 = 500C and 7 MPa

= 7.05 kJ/kg

h1 = 3410.3 kJ/kg

h4 = 7.05 + 191.83 = 198.88 kJ/kg

s1 = 6.7975 kJ/kgK

WT = h1 – h2 = 3410.3 – 2153.92

At 10 kPa

= 1256.4 kJ/kg

h3 = h f 3 = 191.83 kJ/kg

Heat supplied, QS = h1 – h4

s3 = s f 3 = 0.6493 kJ/kgK

3410.3 – 198.88 = 3211.5 kJ/kg

v3 = v f 3 = 1.01  10-3 m3/kg

Wnet = WT – WP = 1256.4 – 7.05 = 1249.35 kJ/kg

s g 2 = 8.1502 kJ/kgK Work ratio =

T 1 4

3

th =

7MPa

10kPa

Heat rate =

s

Process (1-2)

s1 = s2 = 6.7975 kJ/kg K < 8.1502

kJ kgK

So point 2 is in wet state s2 = sf2 + x2 s fg 2

3600  9254 kJ/kW-hr  th

 s= m

30  10 3 = 24.01 kg/sec 1249.35

Boiler capacity is mass flow rate of steam expressed in kg/hr

6.7975 = 0.6493 + x2 (8.1502 – 0.6493)

ACE Engineering Publications

3600 = 2.8816 kg/kW-hr Wnet

 s  Wnet = 30  103 Power = m

Q = 0, s = c

x2 = 0.82

Wnet 1249.35   38.9% QS 3211.5

Steam rate =

2

Wnet 1249.35 =  0.99 WT 1256.4

= 24.01  3600 kg/hr = 86436 kg/hr

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: 227 :

(As s2 > sg2 it is in superheated state)

 s (h2 – h3) Condenser load = m

= 24.01 (2153.92 – 191.83) = 47109.7 kW (T)w = rise in temperature of water Heat gained by water = condenser load  w Cpw (T)w = 47109.7 m

(T)w =

Thermodynamics

47109.7 = 5.630 C 2000  4.186

Tm1= mean temperature of heat supply h  h 4 3410.3  198.83 = 1   522C s1 s 4 6.7975  0.6493

s2 = 6.5966 kJ/kgK

  T2  = 6.5865 + 2.1ln   (273  179.91)  T2 = 455.09 K h2 = h g 2 + C Pvapour  (T2 – Tsat) = 2778.1+2.1[455.09–(273 + 179.91)] = 2782.67 kJ/kg State – 3

P3 = 1 MPa T3 = 5000 C h3 = 3478.5 kJ/kg

11.

s3 = 7.7622 kJ/kg K

Sol: State – 1

State – 4

P1 = 10 MPa T1 = 5000 C

P4 = 10 kPa

h1 = 3373.7 kJ/kg

T4 = 45.810 C

s1 = 6.5966 kJ/kgK

h4 = ? s4 = s3 = 7.7622 kJ/kg K



T 1

5000

s 4 = s f 4 + x4 s g  s f 4 4

3

10MPa 6



= 0.6493 + x4(8.1502 0.6493) x4 = 0.94  (wet state)

2

1MPa



h4 = h f 4  x 4 h g 4  h f 4 5

4

10kPa

= 191.83 + 0.94 (2584.6 – 191.83) s

State – 2

= 2441.03 kJ/kg State – 5

P2 = 1 MPa T2 = ?



P5 = 10 kPa

0

Tsat = 179.91 C , h2 = ?

s1 = 6.5966 kJ/kg = s2 T s2 = s g 2 + C PVapur  ln  2  Tsat ACE Engineering Publications

  

h5 = (hf)w = 191.83 kJ/kg s5 = s f 5 = 0.6493 kJ/kg K v5 = 0.0010 m3kg

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: 228 :

ME – GATE _ Postal Coaching Solutions

State – 6

P6 = 1 MPa

1

T

S6 = 0.6493 kJ/kg

7

Wp = v f5 (P6 – P5) 5

= 0.0010  (10103 10) = 9.99 kJ/kg Wp = (h6 – h5)

6MPa

6

2

20kPa

3

4

h6 = Wp + h5

0.4MPa

s

= 9.99 + 191.83 = 201.82 kJ/kg WT = (h1 h2) + (h3 h4) = (3373.72782.67) + (3478.52441.03)

=1628.5 kJ/kg

State - 2: (wet state)

P2 = 0.4MPa h2 = ? T2 = ? s2 = 6.7193 kJ/kg

Wnet = WT WP = 1628.5  9.99 = 1618.51 kJ/kg W 1618.51 Work ratio = net  = 0.99 WT 1628.5

Heat supplied QS = (h1 h6) + (h3 h2)

s g 2 = 6.8959 kJ/kg K As s2 < s g 2 this is in wet state



s2 = s f 2  x 2 s g 2  s f 2



6.7193 =1.7766 + x2(6.8959 – 1.7766)

= (3373.7201.82)+(3478.5 2782.67)

x2 = 0.96

= 3867.71 kJ/kg

h2 = h f 2 + x2 h g  h f 2 2

th =

Wnet 1618.51  = 0.418 = 41.8% Q S 3867.71





= 604.74 + 0.96 (2738.6 – 604.74) = 2653.24 kJ/kg State - 3: (wet state)

12. Sol: State - 1: (super heated)

P1 = 6 MPa T1 = 4500 C h1 = 3301.8 kJ/kg s1 = 6.7193 kJ/kgK

P3 = 20 kPa h3 = ? s2 = s3 = 6.7193 kJ/kgK s g3 = 7.9085 kJ/kg K 6.7193 = 0.8320 + x3(7.9085 – 0.8320) x3 = 0.83

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: 229 :



h3 = h f3 + x3 h g3  h f3



Thermodynamics

Energy balance: mh2

= 251.4 + 0.83 (2609.7 – 251.40) = 2208.8 kJ/kg State - 4: (saturated)

1 kg

h6

P4 = 20 kPa

h5

(1-m)

h4 = h f 4 = 251.40 kJ/kg Energy in = Energy out

State - 5: (compressed)

(1m) h5 + mh2 = 1h6

P5 = 0.4 MPa

 m (h2 h5) = h6 h5

State – 6: (saturated)

h6  h5 h2  h5

m=

P6 = 0.4 MPa h6 = hg6 = 604.74 kJ/kg



s6 = 1.7766 kJ/kg

604.74  251.786  0.146 kg/sec 2653.24  251.786

WT = (h1 – h2) + (1m) (h2 – h3)

State – 7: (compressed)

= (3301.8–2653.24) + (1 – 0.146)

P7 = 6 MPa

 (2653.24–2208.8)

High pressure pump work

= 1028.12 kJ/kg

WHP = v f6 (P7 – P6) =1.084  103 (6  103 – 0.4  103) = 6.0704 kJ/kg

WP = WHP + WLP = 6.0704 + 0.386 = 6.456 kJ/kg

h7 = h6 + WHP Wne = WT – WP = 1028.12 – 6.456

= 604.74 + 6.0704

= 1021.66 kJ/kg

= 610.08 kJ/kg Low pressure pump work

Qs = h1 – h7

WLP = v f 4 (P5 – P4) 3

= 3301.8 – 610.07 = 2691.73 kJ/kg 3

= 1.017  10 (0.4  10 – 20) = 0.386 kJ/kg

th =

h5 = h4 + WLP

W net 1021.66  = 0.379 = 37.9% Qs 2691.73

= 251.40 + 0.386 = 251.786 kJ/kg The mass of steam m1 extracted from turbine at 0.4 MPa ACE Engineering Publications

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: 230 :

ME – GATE _ Postal Coaching Solutions

05. Ans: (d) Sol: T

Chapter- 11

3 (Tmax)

Gas Turbines

4

3

01. Ans: (a)

Tactural

2

(T)max

02. Ans: (a) 1

Sol: During regeneration process, turbine work

S

and compressor work remain unchanged and only heat supplied decreases so thermal efficiency increases.

P2 = 400 kPa, T1 = 283 K

03. Ans: (d)

T3 = 1273K

04. Ans: (d) Sol:

3 4

2

S

P1 = 100 kPa,

P2 = 400 kPa

T1 = 298 K,

T3 = 1473 K

400 4 100

T2  T1 rP  T4 

T3

rP 

 1 

T31 = 450 + 273 = 723 K T4 

1

 1 

400 5 80

rP 

3

T

rP 

P1 = 80 kPa,

 298  4

0.67 1.67

T3

rP 

 1 



1273

5

0.4 1.4

 804K

Effectiveness of heat exchanger.



T act T max



T3'  T2  100 = 77% T4  T2

06. Ans: (c)

 519.7K

 844.61K

The maximum temperature up to which we can heat the compressed air is turbine exhaust temperature and this will happen when effectiveness of the heat exchanger must be unity.

Sol: Tmin =T1 = 20 + 273 = 293 K,

Tmax= T4= 900 + 273 = 1173 K

 = 1.3, rp = 6 (B)ideal regeneration = 1 

 1 Tmin rp   Tmax

1.3  1  293  = 1    6 1.3  1173 

= 0.62 or 62%

i.e. T3'  T4  844.61K  5730 C ACE Engineering Publications

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: 231 :

07. Sol: Whenever we reheat, we reheat to the same

temperature until unless mentioned in the

Thermodynamics

P2 = 4 = rp P1 The temperature after isentropic compression

problem. Whenever, we intercool to the same temperature, if there are infinitely large number of reheats and if there are infinitely large number of intercoolers then reversible adiabatic expansion becomes isothermal

expansion

and

T2 = T1 (rp )

= 300  4

1.4 1 1.4

= 446 K

The isentropic efficiency of the compressor C T  T1  Ws  c  P 2 Wactual C P T2  T1 

c =

reversible

adiabatic compression becomes isothermal compression and thermal efficiency of

 1 

445.8  300 T2  300

 0.8 =

Brayton cycle becomes equal to Ericsson

 T2 = 482.5 K

cycle.

  th Braton   th Ericsson Ericsson cycle is an ideal gas turbine cycle Ideal ()Ericsson = ()Carnot

T3  P3    T4  P4 

08.

T4 

3

T

4 2

2

ηT  S

Caloric value (CV) = 42000 kJ/kg

T = 0.85 , c = 0.8 Tmax = T3 = 8750C = 1148K P1 = 1 atm T1 = 300 K

at

exit

of

T3

4

0.4 1.4



 1 

1148 4

0.4 1.4

T4 = 773 K

4

1

temperature

Process 3-4: Q=0, S=C

T ()Carnot = 1  min  76% Tmax

Sol:

T2 = actual compressor

Wact CP T3  T4   WS C CP T3  T4 

T4  1148  0.851148  773 T4  829K

WT = CP(T3 – T4 ) = 1.0051148  829  = 320.32 kJ/kg WC  CP T2  T1   1.005483  300  184 kJ/kg QS  C P T3  T2   1.0051148  484 = 668.325 kJ/kg

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: 232 :

09.

Wnet =WT – WC = 320.32 – 183 = 137.32 kJ/kg Work ratio =

Sol:

Wnet  0.428 WT

Back work ratio 

Air rate =

Tm1 

Tm2 

1

3600 kJ  17560.97 η th kW

At rp opt , T2 =T4 = T1T3  600 K



WC = CP (T2 – T1) =1.005(600300) = 301.5 kJ/kg



h 4  h l C P T4  T1 = =520.4 K s 4  s1  T4   C P  ln  T1   f ) (m

m a Cpa, T2

S

Tmin = 300 K

h 3  h 2 CP T3  T2 =768 K  s 3  s 2  T3  C P  ln   T2 

CV, comb

T1=300K

Tmax = 1200 K

3600 kg  26.22 Wnet kWhr



T3=1200K

4

2

WC  0.571 WT



3 T

W  th  net  20.5% QS

Heat rate =

ME – GATE _ Postal Coaching Solutions

0  f ) m Cpa, T3 a (m

WT = CP (T3 – T4) = 1.005(1200 – 600) = 603 kJ/kg th, (rp)opt = 1

= 1

Tmin Tmax 300  50% 1200

Energy balance:  a CPa T2  m  f CV ηcomb  m  a CPa T3 m

a  m m  f CV ηcomb  a CPa T3 CPa T2  m f f m m

AFR CPa T2  CV ηcomb  (AFR)CPa T3 AFR1482.5+420000.9 = AFR11148 AFR = 56.56:1 ACE Engineering Publications

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: 233 :

Thermodynamics



Chapter- 12

268 0.35  W 293  268  Wmin = 33 W

Refrigeration 03. Ans: (c) 01. Ans: (a) Sol:

Sol:

T

2

T

3 3

31.31C 18.77C

800KPa

Ideal

1

4

1

4

2

vapour

S

compression

means

compression starts from dry and saturated S

vapor line.

P1 = 140 kPa, P2 = 800 kPa

P1 = 120 kPa, P2 = 800 kPa

From steam table

h1 = 236.97 kJ/kg

h2 = hg = 267.29 kJ/kg

h3 = h4 = 95.47 kJ/kg

h3 = hf = 95.49 kJ/kg

Net refrigeration effect (NRE) = 32 kW  r h 1  h 4  =m

QR = h2h3 = 267.29  95.49 = 171.82

COP HP



the

QR T2  W T2  T1

171.82 273  31.31  273  31.31  273  18.77  W W = 28.54 kJ/kg

 r  0.23kg / sec m

04. Ans: (d)  r  0.193kg / sec Sol: m

s1 = s2 = 0.93 kJ/kgK After compression the refrigerant is in super heated state with entropy = 0.93 at a pressure

02. Ans: (b)

1.2 MPa

Sol: For minimum required power input condition

h1 = 251.88 kJ/kg.K

the efficiency has to be maximum and the

h2 = 278.27 kJ/kg,

maximum efficiency is the Carnot efficiency.

h3 = 117.73 kJ/kg

COP Carnot Re frigerator  ACE Engineering Publications

T2 Q  2 T1  T2 W

 r h 2  h 3   30.98 kW Heat supply = m

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: 234 :

ME – GATE _ Postal Coaching Solutions

05. Ans: (b)

08. Ans: (b)

Sol: h3 = 107.32 kJ/kg = h4 at 1 MPa

Sol:





h4 = h f 4  x h g 4  h f 4

T

2

250kPa

3 100kPa 1

107.32 = 22.49 + x (226.97 – 22.49) 4

Dry fraction of liquid, x = 0.4

S

Mass fraction of liquid = 1x = 0.6 For Helium,  = 1.67   0.2 kg/sec, T1 = 100C = 263 K m

06. Ans: (d) Sol:

2

T 3

Pressure ratio, rp 

1.2 MPa

4

0.32 MPa

T2  T1  rp 

1

h1 = 251.8 kJ/kg at 0.32 MPa h2 = 278.27 kJ/kg h  h 4 251.8  117.71 COP  1   5.07 h 2  h1 278.27  251.8 07. Ans: (a) Sol:

T4 

T3

r  p

 1 



293

2.5

0.67 1.67

= 202.87 K

 C P T2  T1  WC  m

 m 

T

0.67

 263  2.51.67

 379.84K

s

h3 = 117.71 kJ/kg at 1.2 MPa

 1 

P2 250   2 .5 P1 100

2

R T2  T1  M   1

0.2  1.67  8.314 379.84  263 41.67  1

= 121 kW 3

 WE  m

1 4 S

T3  308K T4 

T3

r  p

 1 



0.4

 280  1.4    80 

ACE Engineering Publications

0.2  1.67  8.314  (293  202.87) 4  1.67  1

= 93.3 kW

308

T4 = 58C

=

R T3  T4  M   1

Wnet = WC  WE = 121 93.3 = 27.7 kW

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: 235 :

Thermodynamics

09. Ans: (d) Sol:

WC(kW) = 0.402 28.92 = 11.62 kW 2

T

COP 

800

3

120

4

NRE 15  3.517   4.54 WC 11.62

11.

1

Sol: S

T

h1 = 241 kJ/kg

2

80C

h2 = 286.69 kJ/kg 49.31C

h4 = h3= 95.47 kJ/kg h  h 4 NRE COP  1  h 2  h1 WC

= 3.2

1200

3

2

4

45C 21.910C

5

140

1 S

T2 = 80C

10. Sol:

h 2 = 230.4 kJ/kg

T

2 3

h4 = h3  CP(T3T4) = 79.9 kJ/kg

9.6bar

h4 = h5, C PL = 1 kJ/kgK 4 2.19bar

1

h1 = 177.87 kJ/kg S

h3 = h4 = 64.6 kJ/kg ,

h1 = 195.7 kJ/kg

3

v1 = 0.082 m /kg

 r  0.2kg / sec m

s1  s2  0.7102 kJ/kgK

h4 = h3  C PP (T3 T4)

n = 1.13  r h 1  h 4  NRE (kW) = 3.517  15 = m

= 84.21 1(49.31 45)

 r  0.402 Kg / sec m

= 79.9 kJ/kg

n 1   n   n P  2 WC kJ / kg   P1v1    1  P1   n 1   1.131   1.13 1.13 9 . 6      219  0.082   1   2.19   1.13  1  

 r h 1  h 5  NRE  m

 0.2177.87  79.9 kJ / kg  19.59kW  r h 2  h 1  WC  m

 0.2230.398  177.87   10.5kW

= 28.92 kJ/kg ACE Engineering Publications

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: 236 :

COP 

ME – GATE _ Postal Coaching Solutions

NRE 19.59   1.86 WC 10.5

T W

 10 0 C

Chapter- 13 Thermodynamic Relations

Heat gained by water = Heat lost in condenser

 w C Pw TW  m  r h 2  h 4  m  w  4.187  10 = 0.2(230.9 79.9) m

01. (b)

02. (a)

03. (d)

04. (c)

05. (a)

06. (c)

07. (d)

08. (c)

09. (d)

10. (d)

 w = 0.718 kg/Sec m

s1 = s2 = 0.7102 kJ/kgK

Chapter- 14

At 800 C, h = 230 kJ/kg,

Reciprocating Compressors

s = 0.754 kJ/kgK At 700C h = 214.8 kJ/kg and s = 0.7060 kJ/kgK At s = 0.7102 kJ/kgK, h = ?

01. (b)

02. (a)

03. (a)

04. (c)

05. (d)

06. (b)

07. (c)

08. (d)

09. (a)

10. (b)

11. (d)

12. (a)

13. (b)

14. (d)

By interpolation method h = h2 = 216.2 kJ/kg  Compressor 



Ws c h 2  h 1  Wact h 2  h 1

216.2  177.87 = 0.72% 230.4  177.87

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Machine Design Solutions for Vol – I _ Classroom Practice Questions 03. Ans: (a)

Chapter- 1

Sol: G = 0.8  103 MPa

Static Loads

T1 G1  l1 J1

01. Ans: (d) Sol: t = 0.2 mm,

1 

d = 25 mm,

E = 100 GPa

M E b   I R y

= 7.62103 radian

 0 .2  100  10 3    2   b   800 MPa  25     2 

T1

04.

180 = 0.436 degrees 

Ans: (b)

Sol: b d

120 kN

T2 0.5 m

1m

t = 13 mm

120  10 3  75 (b  22)  13

T = T1 + T2  = 1 = 2

 b = 145 mm

T1l1 T2 l 2  GJ 1 GJ 2

7358  1  4905.33 Nm 1.5

7358  0.5 T2   2452.66 Nm 1.5 Maximum shear stress 16 T1 16  4905.33  10 3   = 48.8 MPa d 3   80 3 ACE Engineering Academy

P = 120 kN,

120 103  75 MPa (b  d) t

T

T1 

= 7.62103 

120 kN

02. Ans: (b) Sol:

4905.33  10 3  0.5  10 3   80 4  0.8  10 5 32

05. Ans: (c) Sol: d = 17.5mm,

 cp  330 MPa Su = 140 MPa  330  17.5 2    17.5  t  140 4 t = 10.3 mm

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: 240 :

ME – GATE _ Postal Coaching Solutions

10. Ans: (c)

06. Ans: (b) Sol:

500 mm

Sol: The reaction force acting on the pin

R  2.25 2  1.875 2  2.928 kN 1 mm

R = Pressure  (Projected area of the pin) = 6.5  (d1  l1) = 6.5  (d1  2 d1)

Bending equation : M b E   I y R

b 

( l1  2d 1 )

Ey 210  10  0.5   210MPa 500 R

d1 = 15 mm 11. Ans: (a)

07. Ans: (a)

Sol:

 I Sol: M  b y 

2.928 103 = 6.5  2  d 12

3

d = 250 mm, r = 125 mm, Pulley

210  15  13 = 525N-mm = 0.525N-m 0.5  12

Bearing 750 mm

08. Ans: (b) Sol: Force applied on the bar = 95  100  t N

Maximum stress induced Force = Minimum area 

95  100  t  100 MPa (100  5)  t

T1

T1 3 T2

T2

T2 = 900 N T1 = 900 3 = 2700 N Torque supplied to the shaft is given by T = (T1– T2)  r = (2700 – 900) 25 = 225 kN-mm

09. Ans: (c) Sol: By taking moment of force about the axis of

fulcrum 2.25  1.25 = P 15

Bending moment = T1  T2   = 1800 

750 2

750  675 kN  mm 2

P = 1.875 kN

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: 241 :

Machine Design

12. Ans: (a)

14. Ans: (b)

Sol: d = 30mm, t = 3mm

Sol: Fc = mr2

Outer diameter = 30 + 2  t = 30 + 2 3 = 36mm Speed ratio = 4:1 T

300  2    10,000  = 0.2    1000  60 

2

= 65797.36 N

60  10  10 3  190.985 Nm 2000 2  4 T   J r

Shear Stress,  = Shear stress =

Load  Double shear Area

5797.36 = 130 MPa  2 2   18 4

Maximum shear stress =

16  190.985  10 3 = 40.27MPa  36 4  30 4    36  

15. Ans: (b) 16. Ans: (b) Sol: The shear resistance of the plate

13. Ans: (b) Sol:  



Ps  2 

load area

  d2   4

300 103 = 2 

Pavg  A piston

  d 2  55 4

d = 59 mm  60 mm

A pin

  1.25  0  2 6  10  0.06  2  4 10 10 6    2 d 4

150  103 = b  15  80 b = 125 mm

d = 15mm

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: 242 :

ME – GATE _ Postal Coaching Solutions  max 

Chapter- 2 Theories of Failure

480 119.68  2  F.S 2 F.S = 4

01. Ans: (c) 0  40 Sol: Given      0  30

03. Ans: (c)

1 = 40, 2 = –30 , yt = 350 MPa



 = 40MPa,

According to maximum principal theory

Ssy S yt 1   2   2 FOS 2  FS

1 =

40  30 350  2 2  FS

 FS 

Sol:  = 60 MPa,

Syt = 330 MPa

Max shear stress theory  max 

S ut   2 2  F.S

S yt F.S 2

60  0  60  0  2 1      2 2  

350 =5 70

 30  50  80 MPa 02. Ans: (d) Sol: d = 50mm, L = 250 mm, P = 235 kN,

 80 

330  F.S  4.125 F.S

Sut = 480 MPa

04. Ans: (b) Sol: Ft = 48 kN 235 kN

Syt = 200 MPa

FS = 18 kN

FS = ?

Since bolts are made of ductile material, so 250mm

According maximum shear stress theory  max 

S yt 1   2  2 2  F.S

x =  & y = 0 and xy = 0 

235  10 3  119.68 MPa  2  50 4

ACE Engineering Academy

we can use maximum shear stress theory 

48  10 3  80 MPa 600



18  10 3  30 MPa 600 2

2

  80   max      2     30 2 2  2 = 50 MPa

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: 243 :

According to maximum shear stress theory  max   max  50 

Ssy

Machine Design

According max shear stress theory 15.09 =

F.S S yt

15.09 =

2  F.S 200  F.S = 2 2  F.S

FS =

Ssy F.S

Sy 2  FS 260  8.62 2  15.09

05. Ans: (d)

06. Ans: (c)

Sol: Given thin cylindrical shell

Sol: t = 200 MPa = 1

di = 4.6 m, t = 16 mm,

p = 0.210 MPa Syt = 260 MPa

Fs = ? pd 0.21  4.6  10 3 h   2t 2  16 pd 0.21  4.6  10 3  15.09 MPa l   4t 4  16 h = 1= 30.18 MPa

c = 100 MPa = 2 Syt = 500 MPa Tresca theory  max 

S yt 1   2  2 2  FS

200  (100) 500  2 2  FS FS = 1.666 = 1.67

t = 2 = 15.08 MPa 3 = 0

 max

07. Ans: (b)

 1   2  2   1  Max. of   2  2  2 

30.18  15.08  7.55 2 30.18  0 i.e.,  15.09 2 15.08  7..54 2

Sol:  max 

 max 

Ssy FS S yt 2  FS 2

But  max

    b   2  2  2

 55  =    (31.5) 2 = 41.81  2 FS 

S yt 2   max



284 = 3.39 2  41.81

max = 15.09 ACE Engineering Academy

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: 244 :

ME – GATE _ Postal Coaching Solutions

08. Ans: (a) Sol: FT = 20kN,

Fs = 15kN

Syt = 360 MPa,

103 360 202  3 152  A 3



 A = 273.22 mm2 = (/4) d2

Fs =3

d=?

 d = 18.65mm



FT 20 10 N / mm 2  A A



FS 15 103 N / mm 2  A A

3

09. Ans: (b) Sol: F

A

T

F

T

2

1 &  2  S yt FS

      2 2 2

 

A



  12   22   1 2



According to distortion energy theory 2

     1       2    max   R 2 2 2 2 

2

Syt = 310 MPa,

FS = 2, F = 40 kN ,

  R     2 2

d = 20 mm,

 eq   12   22   1 2

According to Distortion Energy Theory

2

S yt

2

           R    R    R  R 2  2  2  2  2

   eq     3R 2 2 2

     eq     3     2 2 2

2



F  2 d 4



40   20 2 4

127.32

2

= 127.32 MPa

 3 2 

  = 51.03 MPa

S yt



Fs 2

 20  103   15 103  360     3     3  A   A 

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  2  3 2

310  2

2

 eq   2  3 2 

FS

T=?

16T d 3

 51.03 

16T   203

 T = 80157.73 Nmm = 80.157 Nm

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: 245 :

Machine Design

10. Ans: (b)

2

Sol: 1 = 100 MPa,

127.32  127.32  2      (12.73) 2  2 

2 = 60 MPa

Maximum working stress according to

 128.58 MPa

distortion energy theory is  12   22  1 2

Minor principal stress

 100 2  60 2  100  60

2 

 87.17 MPa

2

127.32  127.32  2     (12.73) 2  2 

= –1.26 MPa According to Tresca’s theory of failure

11. Ans: (b) Sol: P = 5 kN , d = 10 cm= 0.1 m

Ssy

Torque, T = 5  103  0.5

FS

Syt = 425 MPa



S yt 2  FS

1   2 2



425 128.58  1.26  FS 2 FS = 3.27

Bending moment M = 5  103  2.5 = 12500 Nm Maximum shear stress 16T 16  2.5  10 3  = d 3   (0.1) 3 2

= 12732395 N/m = 12.73 MPa Maximum bending stress 32M 32  12500  b  d 3   (0.1) 3 = 127323954 N/m2 = 127.32 MPa Major principal stress

b



12. Ans: (a) Sol: d = 7.5 cm

T = 420 N-m

Syt = 3800 N/cm2

32M 32  250  10 2  b  d 3   (7.5) 3 = 603.6 N/cm2 

16T 16  420  10 2  d 3   (7.5) 3

= 507.03 N/cm2 2

1 

603.6  603.6  2     (507.03) 2  2 

= 891.82 MPa

2

    b   2 1  2  2 

M = 250 Nm

2

2 

603.6  603.6  2     (507.03) 2  2 

= –288.2 MPa ACE Engineering Academy

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: 246 :

According to maximum shear stress Theory

1   2 2



Ss y



FS

S yt

2  FS

 Te = 1313.5 Nm  max 

13. Ans: (c)

16Te  104.5MPa d 3

Applying MSST

10103 N

 max 

S yt 2  FS

 FS 

0.5m 5103N

M2  T2

 1250 2  403.49 2

 F.S = 3.22

Sol:

16  Te d 3

 max 

Te = Equivalent Torque =

3800 F.S

891.82 + 288.2 =

ME – GATE _ Postal Coaching Solutions

5103N M=1250 N-m

S yt 2   max

=2

14. Ans: (a) Sol: Syt = 200 N/mm2

d = 40 mm,

P = 30 kW , N = 710 rpm,

L = 500 mm, W = 10 kN, Syt = 420 MPa FS = ? (MSST) Power =

FS = 2.5 d =2 b S yt

2NT 60

FS

= b =

200 = 80 MPa 2.5

60  P 30  10 3  60   T 2N 2  710

bd 3 b2b  I= = = 0.66b4 12 12

 T = 403.49 N-m

Maximum Bending moment,

M

M = 5 1500 + 5  500

10 500   1250 N m 2 2

= 10000 103N-mm

2

   max   b    2  2  2

 32M   16T    3   3   d  2   d   max 

16 d 3

M2  T2

ACE Engineering Academy

3

10 7 d M 80 = y =  4 I 2 0.66b 2

80 =

10 7 0.66b

4



2b 2

 b = 57.42 mm

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: 247 :

15. Ans: (b)

Machine Design

For M = 5 kN-m and T = 6 kN-m

Sol: x = 100 MPa

y = 40 MPa,

 = 40 MPa

Te =

maxFS = constant 2

=

5 2  6 2 = 7.81 kN-m

100  40  100  40  2     40 2 2  

1 = 70 +

30 2  40 2 = 120 MPa

2 = 70 

30 2  40 2 = 20 MPa

=

16  7.81  FS 16  14.14  1.5 = d 3 d 3

FS = 2.7

According Distortion Energy Theory 12   22  1 2 

S yt FS

120 2  20 2  120  20 =

360 FS

FS = 3.23 16. Ans: (b) Sol: T = 10 kN-m

M = 10 kN-m Equivalent torque, Te = 10 2  10 2 = 14.14 kN-m max =

16Te 16  14.14 = 3 d d 3

According to Maximum shear stress theory max =

S sy FS

16  14.14 Ssy = 1.5 d 3 Ssy =

16  14.14  1.5 108.02 = d 3 d3

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: 248 :

ME – GATE _ Postal Coaching Solutions

Se = 0.5 Sut

Chapter- 3 Fluctuating Loads

= 0.5440 = 220MPa Se = 0.670.850.90.897 Ke  Se Kf = Actual stress concentration modifying factor

01. Ans: (c)

Kf = 1 + q(Kt 1)

Sol: Given:

b = 50mm ,

d = 10mm

t = 10mm ,

 = 62.5 MPa

= 1 + 0.8(1.37) = 2.096 Ke = Stress concentration modifying factor 1 1 = 0.48 = 2.096 Kf

=

Area, A = (bd)t = (50 – 10)10 = 400mm2

F max = A

 Se = 48.63MPa

For completely reverse load m = 0

F = max  A = 62.5  400 = 25000 N

a 

F = 25 kN  a =

02. Ans: (b) Sol: Given:

Su = 440 MPa

q = 0.8

Ka = 0.67

Kb = 0.85

Kc = 0.9

Kd = 0.897

Kt = 2.37

F.S = 1.5

16 10 3 50  10t

400 N/mm2 t

a m 1   S e S ut F.S

a =

    Here m  0  S ut  

Se 48.63 400   F.S 1 .5 t

 t = 12.3mm

t = 12mm Goodman’s equation 03. Ans: (b)

a m 1   S e S ut F.S

Sol: F = 50 kN,

Sut = 300 MN/m2

S 'e = 200 MN/m2 , Kt = 1.55, q = 0.9 Se = Endurance strength of standard specimen under ideal conditions. Se = Modified endurance strength Se = Ka Kb Kc Kd Se ACE Engineering Academy

M=? Kf = 1 + q(Kt – 1) = 1 + 0.9(1.55 – 1) = 1.495 Se =

1 ' 200 Se =  133.779 Kf 1.495

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: 249 :

Machine Design

Similarly 37.5mm

M

25mm

M

50 kN 3.75mm

Mean stress, m = =

F  2 d 4

=

50 kN

F A

2m =

32 M 32 M  3 d (25) 3

According to Goodman’s equation a m 1   S e S ut FS

32M 101.85  1 3 300 (25) 133.779  M = 135.5 N-m

175  25 = 100MPa 2

150 = 75MPa 2 According to Soderberg’s equation a m 1   S e S ut F.S

Here, Se = Ka Kb …. S'e

=

1  250 =135 N/mm2 1.85

According DET  S yt    12   22  1 2 meq    F.S 

 meq =

Sol: Given:

1 = 50MPa to +150MPa 2 = 25MPa to 175MPa

Sut = 500MPa , Se= 250MPa Kt = 1.85 1 max = 150 MPa, 1min = 50MPa

= 1a =

12m   22 m  1m  2 m

= 86.6MPa

04. Ans: (b)

1mean =

 2 min = 25MPa

2a =

50  10 3  101.85 MPa  2 (25) 4

Stress amplitude,  a 

 2 max = 175MPa,

aeq =  12a   22a   1a 2 a

= 90.14MPa Substituting these values in Soderberg’s equation 90.14 86.6 1   135 500 F.S F.S = 1.2

1 max  1 min 2

150  50 =50 MPa 2

150  50 = 100MPa 2

Linked Answer Questions 05 & 06

05. Ans: (a)

&

06. Ans: (c)

Sol: Given:

Sut = 630MPa ACE Engineering Academy

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: 250 :

ME – GATE _ Postal Coaching Solutions

Se = 0.22 Sut for 106 cycles

07. Ans: (a)

Sf= 0.9Sut at 103 cycles

Sol: Given that 1 = 0.85, 2 = 0.12, 3 = 0.03

L = 25,000 cycles

L1 = 128798,

1 = 0.5 2 = 0.3 3 = 0.2

Sf1 = 225MPa

L2 = 37770,



1  2  3 1   = L1 L 2 L 3 L



0.85 0.12 0.03 1    128798 57770 4865 L

L3 = 4865

Life of component

Sf2 = 145MPa

L = 62723 million rev. We know that

08. Ans: (a)

A = Sf LB = 0.9 Sut 103B …….. (1) A = 0.22 Sut 106B ……… (2) By solving (1) & (2), we get A = 2319, B = 0.203 2319 = Sf L0.203

Tmax = 2 kN-m Tmin = –0.8 kN-m Ssy = 225 MPa, FS = ? (Soderberg) Sse = 150 MPa

2319 = 225 L01.203 4

L1 = 9.310 cycles A = 2319,

Sol: d = 50 mm

B = 0.203

2319 = Sf L0.203 2319 = 145 L02.203 L2 = 8.04105 cycles

Ta 

2  (0.8)  1.4 kN  m 2

Tm 

2  0.8  0.6 kN  m 2

16 Tm 16  0.6  10 6  24.446 MPa m   (50) 3 d 3

1    L1 L2 L3 L

16 Ta 16(1.4) 10 6 a    57.04 MPa d 3  (50) 3

0.5 0.3 0.2 1    4 5 L3 25,000 9.3  10 8.04  10

a m 1   S e S yt FS

 L3 = 5.83103

a m 1   Sse Ssy FS

1

2

3

2319 = Sf LB 2319 = Sf (5.83103)0.203  Sf = 400 MPa

24.446 57.04 1   225 150 FS FS = 2.04

ACE Engineering Academy

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: 251 :

09. Ans: (c)

Machine Design

eq=

Sol: L1= 10 hours

 2  3 2

 meq   2m  3 2m  3  36.5 2

N1 = 9.8 hours

 63.21 MPa

N2 = 8.2 hours L2 = ?

 aeq   a2  3 a2  130 2  3(20.5) 2

According to Minor’s Equation N1 N 2  1 L1 L 2

= 134.76 MPa According to Goodman’s equation,  aeq

Se

9.8 8.2   1 10 L 2 L2 = 410 hours Common Data for Questions 10 & 11



 meq

S ut

1 Fs



134.76 63.21 1    FS  1.54 224.4 1400 FS 12. Ans: (d) Sol: Sf1 = 500 MPa

N1 = 10 cycles

5

L1 = 1  10 cycles

10. Ans: (c)

Sf2 = 600 MPa, 11. Ans: (d)

L2 = 0.4 105 cycles

Sol: max = + 130 MPa

Sf3 = 700 MPa,

min = –130 MPa

Kd 

L3 = 0.15  10 cycles

Se = Ka...... S 'e  0.76  0.85  0.897 

1 (0.5  1400) 1  0.95(1.85  1)

= 224.411 MPa  For a completely reversed, a = 130 MPa

m =

57  16  36.5 MPa 2

a 

57  16  20.5 MPa 2

ACE Engineering Academy

N3 = 3 cycles

5

1 1  K f 1  0.95(1.85  1)

m = 0

N2 = 5 cycles

1  2  3 1    L1 L 2 L L

1 =

N1 10 = N1  N 2  N 3 18

10 5 3 1    5 5 5 181 10  180.4  10  180.15  10  L L = 42352.94 Cycles For

18 cycles 

1  60 2

42352.94 cycles  ? L  L = 19.6 hrs

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: 252 :

ME – GATE _ Postal Coaching Solutions

13. Ans: (c)

Chapter- 4 Riveted Joints

14. Ans: (a) Sol: Se = 280 MPa 3

Sf = 0.9 Sut for 10 Cycles Su = 600 MPa N = 103 cycles

01. Ans: (b) Sol: Given

d  0.5 p

Sf = ?

Basquin’s equation,

p

F

A = Sf LB A = 280(106)B ……… (1) A = (0.9600)103B 3B

A = 540  10 ……… (2)

F

Tearing Area

Tearing efficiency =

pd p

 d p1   p =  p

By solving A = 1041.42 B = 0.095

= 1

0.095

 1041.42 = SfL

1041.42 = Sf (200103)0.095 Sf = 326 MPa  1041.42 = 420  L0.095

L = 1.4 104 cycles

d = 1  0.5 p

= 0.5 × 100 = 50% 02. Ans: (d) Sol: Resultant Force  F12  F22



4 2  32

= 5 kN F F1  F2   F  2F2  2  3  6 kN 2 Stress 

F 5000  2 500 = 10 MPa

F L r  r22 As r1 and r2 not given, so it is not possible to C

2 1

calculate eccentricity (L). ACE Engineering Academy

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: 253 : Common Data Question (03, 04, 05)

Machine Design

05. Ans: (c) Sol: Crushing Strength

Given d = 30 mm

t = 40MPa = 40N/mm2 P = 90mm 2

PC = d ×t ×c = 30 × 12.5 × 55 = 20625 N

s = 30MPa = 30 N/mm

Tearing Strength

t = 12.5 mm

Pt = (p – d)t × t 2

c = 55MPa = 55N/mm

= (90  30) 12.5 × 40 = 30,000 N Shear Strength

03. Ans: (b) pd Sol: Tearing Efficiency = p

PS = 21206 N ,

P = 45000 N

Strength of riveted joint

90  30 60 = = 90 90

=

Least value among PC , Pt & PS P

2 =  100 3



20625  0.458 = 45.8% 45000

Tearing = 66.67%

06. Ans: (c) Sol: Given t = 7mm

04. Ans: (b) Sol: Strength of Riveted plate = P = p×t×t

P = 90 ×12.5 × 40

PS = n ×

Shearing Resistance,

 2 d  t 4

Ps =

 302  30 4

=3×

 × d2 × s 4  2 d  60  141.4d 2 N …...(1) 4

PC = n× d ×t × c = 3 ×d ×7 ×120 = 2520d N …….. (2)

= 21206N P 21206 Shear efficiency = S  P 45000

= 0.47= 47%

ACE Engineering Academy

c = 120 MPa = 120N/mm2 n = 3 (Triple riveted joint)

= 45000 N

PS =

s = 60MPa = 60N/mm2

From equations (1) & (2) 141.4 d2 = 2520d d=

2520  17.8  18mm 141.4

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: 254 :

ME – GATE _ Postal Coaching Solutions

07. Ans: (d)

Shear strength = crushing strength

Sol: Given:

S yc  2 Ssy d  = d t 4 FS FS

t = 7mm, n=3

t = 80MPa = 80N/mm2 s = 60 MPa = 60N/mm2 c = 120 MPa = 120N/mm2 Let p = pitch of rivets, d = 18mm Tearing resistance is Pt = (p – d)t × t = (p – 18)7 × 80 = 560(p – 18) N ……. (1)

 Ps  d 2   s   4

d = 6

150 4 75  

d = 15.27 mm. 09. Ans: (c) Sol: Given

P = 2500 kPa = 2.5N/mm2 D = 1.6 m = 1600 mm d = 34.5 mm

s = 60 MPa = 60N/mm2 No of rivets, n =

D2P d 2 s

16002  2.5 34.52  60

 182  60  3 = 45804 N …….. (2) 4



From equations (1) and (2)

= 89 .6  90

560(p18) = 45804

n = 90

p = 99.79 p  100 mm

10. Ans: (b) Sol: Given:

08. Ans (a) Sol:

S yt FS Ssy FS S yc FS

= 90 N/mm2 = 75 N/mm2 = 150 N/mm2

ACE Engineering Academy

s = 100MPa = 100N/mm2 d = 20 mm,

n=4

Direct shear load on each rivet PS 

P P   0.25P n 4

PA = PB = PC = PD = PS All dimensions are in mm Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally

: 255 : FD PSD

200 D

FC

PSC

200 C 200

D

F

B

PSB

C FB

RC = PS – FC = 0.25P – 0.05P = 0.2P Resultant load on rivet D, RD = PS – FD = 0.25P – 0.15P = 0.1P RA is the maximum shear load

B

0.4P =

FA A

P=

From fig

lA = lD = 200 + 100 = 300 mm lB = lC =100 mm [ Secondary shear loads are proportional to their radial distances from the C.G ]



F P × e = B l 2A  l 2B  l c2  l 2D lB



F = B 2 l 2A  2l 2B lB P × 100 =

F 100





 2 d ×s 4

 202 100  31420 4

31420  78.55 kN  78 kN 0.4

11. Ans: (b) Sol: Tensile load (Ft) = (p – d)t  t = (60–20)  15  120 = 72000N

= 72 kN ( lA = lD & lB = lc )

2 300 

2

 2 100

2 



FA 2 l A  l 2B  l c2  l 2D lA

Shear Load (Fs) =

   d 2     20 2  90 4 4

= 28274.33 N = 28.274 kN Crushing load (Fc) = d  t  c

FB  0.05 P  FC P×e=

Resultant load on rivet C,

0.40P =

A PSA

Machine Design



= 20  15  160 = 48000 N = 48 Kn

FA = FB = 0.15 P Resultant load on rivet A RA = Ps + FA = 0.25P + 0.15P = 0.4P Resultant load on rivet B,

Load carrying capacity (F) = Minimum of (Ft,Fs& Fc) = 28.274 kN

RB = PS + FB = 0 .25P + 0.05P = 0.3P ACE Engineering Academy

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: 256 :

ME – GATE _ Postal Coaching Solutions

Linked Answer Questions 12 & 13:

Chapter- 5 Threaded Fasteners

12. Ans: (a) Sol: No. of Rivets = 2

4 Primary shear load P1 = = 2 kN 2 Per Secondary shear load P2 = 2 1 2 r1  r2 =

4 103  1.8  0.2 0.2 0.2 2  0.2 2

= 20000 N = 20 kN

01. Ans: (b) Sol: Given d = 24 mm

Fi = 2840d = 2840×24 Fi  2 dc 4

t 

Here, dc = 0.84d  dc = 0.84 × 24 2840  24

t 

 (0.84  24) 2 4

13. Ans: (b) Sol:

P1

P1

r1

02. Ans: (c)

P2

CG

P

 t  213.529 MPa

r2

Sol: Given

Q

d = 36mm dc = 0.84 d = 0.8436

P2

F.S = 1.5 Resultant load on Rivet P = P2 P1 = 18 kN

Syt = 280MPa

t =

Resultant shear stress on Rivet P

P=

18  10 3 = 159 MPa =  2  12 4

=

s yt F .s

 4



280 1.5

d c2 t

 0.84  362  280 4 1.5

= 134066 N P = 134 kN

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: 257 :

Machine Design

03. Ans: (d)

Syt = 300MPa

Sol: Given pitch = 4mm

n =8

Torque (T) = 1.4 kN-mm Work done = force distance

Fb = Load (P) =

 2502 1.2 4 = = 7363.1 N n

Forcedistance = Torque  Angle of rotation F4 = T F=

1.4  2 = 2.199 kN = 2.2 kN 4

04. Ans: (d)

t = 

Sol: Given

F = 5.3 kN , C = 0.25 , P = 9.6 kN Fb = CP +Fi = (0.25) (9.6) + (5.3) Fb = 7.7 kN

 2 D  P 4

Fb S yt  A b F.S

7.36 10 3 300  Ab 5

 Ab = 122.66 mm2 07. Ans: (d) Sol: Given,

D = 500mm

05. Ans: (c)

n=8

Sol: Km = 4Kb

P = 20 bar = 2 MPa

C=

Km = 3Kb

Kb =0.2 Kb  Km

c

To open the joint

Kb 1   0.25 K b K m 4

To avoid leakage

(1–C)P = Fi P 1 1 = = 1.25  Fi 1  C 1  0.2

Load (P) = Pr A 2 =

06. Ans: (b)

 4

5002 8

= 49 kN

For leak proof joint Fm  0

Sol: Given

D = 250mm

Fi = (1 C) P

Pressure = 12bar = 1.2 MPa

Fi = (1 0.25) 49 = 36.75 kN  37 N

F.S = 5 ACE Engineering Academy

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: 258 : Linked Answer Q 08 & 09

ME – GATE _ Postal Coaching Solutions

CP 

08. Ans: (d)

CP 

Sol: Syt = 650 MPa

A = 115 mm2

S yt FS

 A b  Fi

650 115  59800 1

CP = 14950

6

Km = 1.710 N/mm , 5

P

2

Ecu = 1.0510 N/mm

14950  59800 N = 59.8 kN  60 kN 0.25

Esteel = 2105 N/mm2 Fi = 0.8SytA = 0.8 650 115 = 59800 N For bolt, A E P Pb Kb = b  = b b lb  b Pb .l b AbEb

=

115  2  10 5 = 5.75 105 N/mm 20  20

Where, lb = t1 + t2 = 20 + 20 = 40 mm Stiffness factor C =

Kb = 0.25 Kb  Km

Linked Answer Q 10 & 11

10. Ans: (b) 11. Ans: (a) Sol: Given

d = 20mm ,

Syt = 630MPa

Se = 350MPa,

F.S = 2.5

Core area of bolt = 2.45cm2 = 245mm2

m = 180MPa Soderberg’s criterion a  m 1   Se S yt F.s

09. Ans: (a) Sol: Safe external load that can be applied safely

on the joint (1C)P  Fi = 0

 a 180 1   350 630 2.5

(1– 0.25)P = 59800 N

a 1 180   350 2.5 630

P = 79733 N = 79.733 kN

a = 40MPa

For strength

For calculating maximum & minimum

t =  Fb 

Fb S yt  A FS S yt  A b

CP  Fi 

FS S yt  A b FS

ACE Engineering Academy

values of varying loads.

max =  mean   a  180  40  220 MPa Pmax = maxArea = 220245 = 54 kN

min =  mean   a  180  40  140 MPa Pmin = minArea = 140245 = 34 kN Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally

: 259 :

12. Ans: (b)

Machine Design

8.498103 =

Sol: Load (p) = Pressure  Area

 P = 2  250 2 = 98174.77 N 4

d = 12.74mm  13mm

C = 0.1 , n = 12

14. Ans: (a)

For leak proof joint (1C)P Fi = 0

Sol: n = 4

Fi =

S yt  2 d   4 2  F.s

0.9  98174.77 = 7363.10 N 12

P = 10 kN Syt = 400N/mm2 FS = 6

Fi = 2840d

dc = 0.8d

 d = 2.5 mm

Using Rankine theory PA = CL2 (tensile load)

13. Ans: (b) 2

Sol: F.S = 3 , Syt = 400 N/mm , P = 5 kN

Direct shear load PS1

250

=

PL L2 2 L  L22

=

10  550  325 = 8 kN 2 75 2  325 2

5kN

75

PS2

75

PS3



Pdirect =



2 1





P  2.5 kN = PA = PB 4

Bolt ‘A’ is subjected to maximum load

PS1

Rankine Theory

 Total Tensile load on bolt = PA+ PA

5 Ps = =1.67 kN 3

= 8+ 2.5 = 10.5kN

Secondary shear Load, PS1 =

5  250  75  8.3kN 2 75  02  752

Resultant Load (R) = =

PS2  PS21

1.67 2  8.32

= 8.498 kN S  R  d 2  sy [Syt = 2  Ssy] 4 F.s ACE Engineering Academy

t 

F

 2 dc 4



S yt FS

10.5 400   2 6 dc 4 dc = 14.16 d

dc  17.7  18mm 0 .8

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: 260 :

ME – GATE _ Postal Coaching Solutions

According to Rankine Theory

15. Ans: (c) Sol: Given

n=4,

1 

P = 5 kN 2

Syt = 380N/mm



dc = 0.8d

F.S = 5 ,

PtA = KL2 = (Tensile)

S yt FS

2292.22 380  A 5

 A = 30.16 mm2 =

=

PL  L2 2L  L22 

 dc = 6.196 mm



5  250 375 2 75 2  375 2

d=

2 1





 2  dc 4

6.196 = 7.745 mm 0.8

= 1.6 kN Pshear =

16. Ans: (a)

P 5  = 1.25 kN 4 4

Sol: No of bolts = 4

Direct shear load, PSA = PSB =

Primary shear force (P1) =

P = 1.25 kN 4

Secondary shear force

Bolts at ‘A’ is under maximum bending

P2 =

Rankine Theory

 A=

F e r1 r  r22  r32  r42 2 1

r1  r2  r3  r4  75 2  75 2 = 106.06 mm

PSA 1.25 10 3  A A

P2 =

 2 dc 4

 = Least angle between P1and P2

1.6 10 3 t   A A

2

=0.3535F N

weakest. So design for bolt B and C.

t     t    2xy 2  2

Resultant shear force 2

1.6  10 3 1  1.6 10 3     1.25 10 3  2A A  2  2292.22  N / mm 2 A

ACE Engineering Academy

4  106.06

Here bolt B and C are the most loaded and

2

1 

F  150  106.06

 75    = Cos 1   = 45 106 . 06  

Pt A

1 

F N= 0.25 F 4





2

R= P12  P22  2 P1  P2 cos  = 0.25F2  0.35F2  2  0.25F  0.3535Fcos45

= 0.5589F

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: 261 : P1





P2B

A

P2A

B

r1 r4

P1

75mm

r2 P1

D

P2C

S yt 0.5589F S sy = = π FS 2  FS  20 2 4

0.5589F 200  π 2 2  20 4

 F = 56 kN

Fi A

4.5  10 3 36.6

m = 140MPa For calculating Factor of safety Sut = 630 N/mm2 Syt = 380N/mm2 Kf = 3 Reliability = 50% Se’ = 0.5 Sut = 0.5630

Common Data Question (Q17& Q18)

Se = 315 N/mm2

  S  Fi ut  A Sa =    1   S ut K  S  f   e 

17. Ans: (a) 18. Ans: (d) Sol: Given

      

3   630  4.5  10 36.6 =   630  3  1   315  

Km = 3Kb Pmin = 0 Pmax = 5 kN Fi = 4.5 kN

     

= 72.4MPa

M8 d =8 A = 36.6mm2

Factor of safety =

CP 2A

Here, C =

m = a + = 17+

Resultant shear stress

a =

0.25  5 10 3 = 17.07MPa 2  36.6

C

P2D



 a =

Kb K  b = 0.25 K b  3K b 4K b

a = 17MPa

75 P1

r3

=

Machine Design

 72.4  S a  F.S i.e 17   a 

= 4.258

Kb Kb  Km

ACE Engineering Academy

F.S = 4.26

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: 262 :

ME – GATE _ Postal Coaching Solutions

04. Ans (a)

Chapter- 6

Sol: Given: d = 60mm , s =10mm,

Welded Joints

  70MPa

=

01. Ans (b) Sol: Given: s = 10 mm ,

=

 = 80 MPa P = 0.707 s  l  

T 2r  0.707s 2

T

=

= 0.70710  10  80 = 5.6kN

2 

 =

02. Ans (c) Sol: Given , P = 400 kN ,

P = 2 0.707  s   

d  0.707  s 4

2.83T sd 2

70    10  (60) 2 T 2.83

Ssy FS

= 2797460 N-mm  T = 2.797 kN-m

400 1000 = 2 .70710 80 l 400000 1.414  10  80

l = 354 mm The nearest answer is option (c)

05. Ans (a) Sol:

t = 10 mm d = 15  103 mm S yt

= 85 MPa

03. Ans: (b)

FS

Sol: S = 10 mm, P = 4 kN/cm

 l = h =

Ptransverse = 0.707  S l 

Ssy FS

4 kN  1 cm 180 180 kN = = 45 cm = 450 mm 4

l = 175 mm

pd = 1 4t

According to Rankine Theory 1 =

 l + 100 + l = 450

2

s = Size of the weld

 = 80 MPa

l=

T T T r = r = 3 J 2r t 2r 2 t

S yt FS

pd = 85 4t p  15  10 3  = 85 4  10  p = 0.226 MPa

ACE Engineering Academy

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: 263 :

Machine Design

06. Ans: (a) Sol: Given:

t

D1

P = 340kN = 340000N Ssy FS

D1t

= 80MPa,

Equate (1) & (2)

s =15mm P = 0.707s l 

Ssy

P

Ssy  2 D 2  D1 t 4 FS

P=

205  4  1.7675  110 (200) 2

FS

340  10 3  0.707  15  l  80

l = 400 mm length of weld adjusted on both

P = 3.9857 MPa

sides i.e., 200 mm on each side.

08. Ans: (a)

07. Ans: (b)

1

Sol:

Sol:

2

Weld S

45

D0

D2

D1

b

D1 = 205 mm, D2 = 200 mm D0 = 210 mm, Ssy FS s=

b/2

A

rmax

b

b

1 =

210  205 = 2.5 mm 2

t = 0.707 s = 0.707 2.5 = 1.7675 mm Force = Pressure  Area  2 D 2 ………. (1) 4

F = D1t 

r1

Primary shear stress

= 110 MPa

=P

P

e

Ssy FS

ACE Engineering Academy

……….. (2)

P b t2

Secondary shear stress 2 =

T  rmax J

 T = P  e = P b  

b  = 1.5 Pb 2

 l2  J = A  r12   2  12 

A=bt Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally

: 264 :

l=b

 T = 2r2  t 

b t  r1 =    2 2

Ssy FS

2

= 2  25  (0.707 6) 140  T = 2332161 N-mm

b ( b >> t) 2

=

ME – GATE _ Postal Coaching Solutions

= 2332.161 Nm

r1 = distance between two centroids. rmax = distance from centroid to maximum distance on weld  b2 b2  2b 3 t  J = b t     2 = 3 12 4  2

rmax =

    1.5Pb  b 1.59P  =   3  2b t  bt 2    3 

Resultant Shear stress =

2

11. Ans: (a)   75 N / mm 2 ,

s = 10 mm

P = 200 kN ,

a = 145 mm

P = 200  10 3 N b = 55mm P =   0.707 s  l 200  10 3  75  0.707  10  

12   22  21  2 cos 

l=

2

 P   1.59P   P  1.59P       2   cos 45  bt  2   bt   2bt  bt 

=

10. Ans: (a)

Sol: Given:

2

b b b     = 2 2 2

T 2 =  rmax J

Linked questions (Q.10 &Q.11)

 = Angle between 1 and 2

l = 377.18mm la =

2

= P  1   1.592  2  1 1.59  cos 45 bt  2 

=

2

=

1.975P bt

377.18  55  103.72 mm (145  55)

P =   0.707  s   a  =140 MPa ,

s = 6 mm

= 75  0.707  10  103.7

d =50 mm ,

r = 25 mm

= 54986.9 N

We know that 

l b ab

For calculating force carried by top weld

09. Ans: (a) Sol: Given:

200  10 3 75  0.707(10)

T 2 r 2 t

ACE Engineering Academy

= 54.9 kN P = 55 kN

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: 265 :

Machine Design

03. Ans: (b) Sol: Given

Chapter- 7 Sliding Contact Bearings

p = pressure in MPa d = dia of shaft = 10mm

01. Ans: (b)

d1 = Dia of bearing = 14mm

Sol: Given:

l = length of journal = 16mm

Load W = 3 kN

W = load = 88N Projected Area = d1 × l

d = 40 mm 2

W=p×A

p = 1.3 MPa = 1.3 N/mm Pressure (p) =

88 = p× 160

W ld

l=

W pd

=

3000 1.3  40

p = 0.55 MPa 04. Ans: (d) Sol:

l = 57.69 mm

 57.69 = 1.44 1.45  d 40

Sol: Given:

300mm

d

A

500mm

d2 = 12mm 2

p = 0.6 MPa = 0.6N/mm  2 d 1  d 22 4





 75 2  12 2 4



B RB

d = 25 mm

l = 500 mm

W = 2.2 kN

a = 300 mm

P= ? MB = 0



A = 4304.77 mm2 Axial load = p ×A = 0.6 × 4304.77 N = 2582.862 N P = 2.58 kN ACE Engineering Academy

l

  1.5 d

d1 = 75mm

A=

2.2 kN l

RA

02. Ans: (a)

Area =

88 160

p=

RA  500 = 2.2  300 RA = 1.32 kN RB = 2.2 kN –1.32 = 0.88 kN Bearing pressure, P

RA 1.32  103   1.408 MPa d 25  1.5  25

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: 266 :

ME – GATE _ Postal Coaching Solutions

(Cd) = 2  0.075 = 0.15 mm

05. Ans: (a)

= 0.15  103 m

Sol: Given:

N1 = 300 rpm

d = 75 mm ,

N = 1000 rpm

2

p1 = 1.4 MPa = 1.4 N/m  = 0.06 Pa  sec ,

N2 = 400 rpm

Heat dissipated by bearing =90 kJ/min 90 kW = 1.5 kW 60

p2 = ?

H=

1 N 1  2 N 2  p1 p2

Heat generated at the bearing = 1500 W

Since, same oil is used  is same I . e. 1 = 2

V=



N1 N 2  p1 p2

=

f = coefficient of friction

400  1.4 300

p2 = 1.87 MPa

Load (W) = 9000N Heat generated = f .V.W 1500 = f (7.85) (9000) f=

06. Ans: (b)

1500 7.85  9000

f = 0.021

Sol: Given: Eccentricity ratio,  = 0.8

=1

  0.15  1000 60

V = 7.85 m/sec,

300 400  1.4 p2 p2 =

dN 60

d 150   1000 c d 2  0.075

h0 C

h0  1  0 .8 C

Pressure (p) =

h0  0 .2 C

p=

Load ld

9000 = 0.267MPa 0.15  0.225

According to Mckee equation 07. Ans: (a) Sol: d = 150 mm = 0.15 m

L = 225 mm = 0.225 mm Load (W) = 9kN = 9000 N C = 0.075 mm, Diametral clearance ACE Engineering Academy

 N  d  f = 0.326   p  C d

  + 0.002 

0.0212 = 0.326   1000 6 1000  0.002  0.267  10 

 = 0.0157 Pa  sec

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: 267 :

Machine Design

08. Ans: (a)

Mckee’s equation

Sol: Given:

 N  d  f = 0.326     K

d = 50 mm ,

l = 75 mm,

p = 2MPa ,

N = 500 rpm

C = 11.6 W/m2oC ,

f = 0.0015 Tr = 28 oC

 p  c d 

 0.018  3600  0.15   0.0235  0.326    0.002  9000 / 0.3  0.15  C 

W W   Here p    A d   

Heat lost in friction = f× W×V  dN  = (f) (p × l × d)    60  = 0.0015×2×50×75×

C = 7.369 × 104 m

  0.05  500 60

C = 0.736 mm The nearest answer is option (c) i.e. 0.76mm

= 14.72 Nm/sec 14.72 = CA ( Ts Tr)

10. Ans: (a)

14.72 =11.6×0.05×0.075×8(Ts 28)

Sol: Given:

Ts = 70.20C

l = 120 mm = 0.12 m N = 1500rpm

09. Ans: (c)

W = 45 kN

Sol: Given:

ZN = 20 × 106 p

l = 300mm = 0.3 m d = 150mm = 0.15 m

Z = Dynamic viscosity

Load (W) = 9000N N = 60rps = 60 × 60 rpm = 3600rpm  = 0.018 Pa  sec Power lost = heat generated = 6kW = 6000W Here, W = p × A = p ×d × l dN V= 60 Heat Generated = f W V 6000 = f × 9000 × f = 0.0235 ACE Engineering Academy

d = 100mm = 0.1 m

  0.15  3600 60

d  1000 Cd Mckee’s equation  ZN  d  f = 0.326   p  C d

   K 

f = 0.326 (20 × 106) (1000) + 0.002 f = 0.00852 V=

dN 0.11500  = 7.85 m/s 60 60

Heat generated = fWV Heat generated = 0.00852 × 45 × 103 × 7.85 Hg = 3 kW

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ME – GATE _ Postal Coaching Solutions

13. Ans: (a)

Linked Answer Question (11 & 12)

Sol: Given

11. Ans: (a)

W = 150 kN ,

N = 1800 rpm

12. Ans: (c)

d = 300 mm = 0.3 m

Sol: Given:

p = 1.6N/mm2 = 1.6 106 Pa

d = 100mm = 0.1m

Cd = 0.25,  = 20 103 Pa-sec

l = 150 mm = 0.15 mm

K = 0.002

W = 4.5 kN = 4500N N = 600rpm  =18.5×103 kg/ms = 0.0185 kg/m s

  N  d  f = 0.326  p  C d 

    K   

3    300   = 0.326 20  10  1800  0 . 002     6 

Cd = 0.1





 = 0.4 Sommerfeld Number = Here pressure (p) = =

 N s   p

 d   C d

   

2

Heat generation = 0.01  150 dN = 0.01  150    0.3  1800 = 2748.7 kJ/min 14. Ans: (a) N = 600 rpm ,

Sommerfeld no “S”  600  0.0185    2  60   100  =   0.3  10 6  0.1  = 0.617

h0  Cd     2 

h 0.4 = 1  0  0.1     2  h0 = 0.03 mm

ACE Engineering Academy



Sol: d = 60 mm = 0.06 m

P = 0.3 MPa

Eccentricity ratio,  = 1 

 0.25 

= 0.01

W W  A ld

4500 = 30  104 N/m2 0.15  0.1

1.6  10

P = 120 kPa

f = 0.05 For foot step bearing Tf  

2  F  r 3 2   0.05  120  10 3   0.06 2  0.03 3 4

Tf = 0.339N-m P



2NTf 60

2  600  0.339  21.29 60

P = 21.3W

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Machine Design

For Ball bearing, K = 3

Chapter- 8 Rolling Contact Bearings 01. Ans: (d)

 48.545  L10 =    P 

6305 series bearing have higher load carrying capacity than 6205 bearing. Hence among the given option 16.2 kN is greater than 10.8kN.

L 50  48.545    5  P 

L50 =

60 NL H 60  500  6000 = 6 10 10 6

= 180 million rev L 180  48.545   L10 = 50   5 5  P   48.545  36 =    P 

02. Ans: (b) Sol: Given: 6210 bearing

3

L10 =

Sol: 6205 bearing

C = 10.8 kN

3

3

3

P = 14.7 kN

C = 22.5 kN L = 27 million rev

Linked Answer Question (04 & 05)

6 – series – Ball bearing

04. Ans: (a)

C L10 =   P

3

05. Ans: (c)

K = 3 for Ball bearing

 22.5  27 =    P  P3 =

3

11.39  10 27

Fa = 1.5 kN Cs = 1.5 N = 1000 rpm

3

X = 0.56 Y = 1.4, V = race rotation factor = 1

P = 7.5 kN

Equivalent load (P) = (XVFr + YFa)Cs V for most bearings = 1

03. Ans: (b) Sol: Given:

Sol: Fr = 2.5 kN

C = 48.545 kN L = 6000 hrs N = 500 rpm

C L10 =   P

K

ACE Engineering Academy

P = [(0.56  1  2.5) + (1.4  1.5)]1.5 P = [11.4 + 2.1]1.5 P = (3.5)(1.5) P = 5.25 kN c L10 =   P

K

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: 270 :

K = 3 for Ball bearing

K = 3 for ball bearing LH =

40 hrs 52 weeks   5 years week yr

= 10,400 hrs L= =

ME – GATE _ Postal Coaching Solutions

 270  5 3   576  7 3   216  33    270  576  216   1

60  N  L H 10 6

 33750  197568  5832  3 =   1062 

60  1000  10,400 10 6

 237150  =   1062 

L = 624 million revolutions C L10 =   P

13

P = 6.067 kN

3

C L=   P

 C  624 =    5.25 

13

K

3

 16.6  L=    6.067 

C  8.545 5.25

3

L = 20.5 million rev

C = 44.86 kN

08. Ans: (b) Sol: P = 2.5 kN

Linked Answer Question (06 & 07)

n1 = 400,

06. Ans: (c)

N1 = 400  0.3 = 120

07. Ans: (a)

P2 = 5 kN n2 =900

Sol: Given C = 16.6 kN

2 = 0.7

% of element time =  N1 = 1n1 = N2 = 2n2 = N3 = 3n3 =

N2 = 900 0.7 = 630

30  900 = 270 100 40  1440 = 576 100 30  720 = 216 100

ACE Engineering Academy

Lh = 8

1K

hours day  365 6 years day year

Lh = 17520 hrs and L10 = L10 =

N = 270 + 576 + 216 = 1062  N 1 P13  N 2 P23  N 2 P33   P =   N N N   1 2 3  

1 = 0.3

60nL H 10 6

60  120  630)  17520 10 6

= 788.4 million rev.

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: 271 :

 N P 3  N 2 P23  Pe =  1 1   N1  N 2 

1

Machine Design

W = weight of pulley = 1 kN

3

Resultant Radial load of shaft

R = 4.61 kN = RA + RB Take MB = 0

Pe = 4.75 kN C L=   P

RA  500 = R  300

3

RA =

 C  788.4 =    4.75 

3  1.52  12

=

1

 1202.53  63053  3  =     120 630   

3

4.61  300 500

RA = 2.766 kN, RB = 1.8436 kN

C = 43.9 kN

Equivalent load P = [XVFr + FaY]

Linked Answer Question (09 to 12)

= (0.56  1  2.76) + (1.5  2)

09. Ans: (b)

P = 4.546 kN

10. Ans: (a)

Dynamic load rating

11. Ans: (b)

K

C L10 =   , P

12. Ans: (a) Sol: Given:

L10 =

T1 = 3 kN T2 = 1.5 kN Fa = 2kN

W

LH = 5000 hrs

200

X = 0.56

60  400  5000 =120 million rev 10 6

 C  120 =    4.55 

500

[K = 3 For Ball bearing ]

3

C = 22.44kN

Fr

Y = 1.5 RA

RB

T1 T2 ACE Engineering Academy

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: 272 :

ME – GATE _ Postal Coaching Solutions

T = nW R

Chapter- 9 Clutch Design

400 = 4(0.5) (W) 0.125 W = 1600 N  Four springs exert axial load,

01. Ans: (b)

Load per spring =

Sol: Given,

1600 = 400 N 4

W = 1000N, n = 2 r1 = 150mm = 0.15mm

Linked Answer Question (03 & 04)

r2 =100mm = 0.1mm

03. Ans: (b)

 = 0.5

Sol: N = 1000 rpm,

r r Mean Radius (R) = 1 2 2

150  100  2 R = 125mm

2 = 240   = 120  = 0.2,

p = 70 kN/m2 T=

60P r r = Wnrm = Wn  1 2  2N  2 

T=

6020  1000 = 191 N-m 21000 

Torque Transmitted, T = nWR (For both sides effective n = 2) = 2  0.5  1000  125 = 125000 Nmm

191 103 N-mm = 0.2  Wn  150 Wn = 6366.19 N [ Wa = Wn sin  ] Wa = 1323.60 N

T = 125 Nm

b

02. Ans: (c) Sol: Given ,

rm = 150 mm, P = 20 kW

 = 0.5



Wn

r1 = 150mm = 0.15m r2 = 100mm = 0.1m T = 0.4 kNm = 400 Nm n1 + n2 = 5, n = No. of pairs of contact surface n = n1+ n2 1 = 5  1 = 4 R=

r1  r2 0.15  0.1 = = 0.125m 2 2

ACE Engineering Academy

Wa

Force required for engagement Wae = Wa + Wncos = 1323.60+[0.26366.19 cos12] Wae = 2.56 kN

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: 273 :

07. Ans: (a)

04. Ans: (b) Wn = p  2rmb

Sol:

Machine Design

Sol: Given

D = 300mm

6366.19 = 70 103 2    0.15  b

b = 100mm

 b = 0.0964 m = 96.4 mm

 = 0.2 Linked Answer Questions (5 & 6)

 = 100

05. Ans: (a)

p = 0.07 MPa = 0.07 N/mm2 N = 500 rpm

06. Ans: (a) Sol: P = 10 kW

b

T = 100 N-m n=2

r1 r2 

pmax = 0.085 MPa d1 = 1.25d2

r1

r2

r1 = 1.25r2  = 0.3 W r1  r2  T =  n for uniform wear 2 =

We know that W = 2C(r1r2)

 2Cr1  r2 r1  r2  2 2

= 2pr(r1r2) (  C = p  r)

[ W = 2C(r1  r2), C = p1r1 = p2r2] 100 = (0.3)2(0.085)(r2)( r  r ) 2 1

2 2

3

10010 = (0.3)2(0.085)(r2)[(1.25r2)  r ]

r2 = 104 mm, d1 = 260 mm W = 2C(r1 r2) = 2(pmax)(r2)(r1r2) = 2(0.085)(104)(130  104) W = 1.44 kN

ACE Engineering Academy

(r1  r2 = bsin)

= 20.07150 (100 sin100) W =1146 N

2

r1 = 130 mm, d2 = 208 mm

= 2prbsin

2 2

Wn =

1146 W = sin  sin 10

Wn = 6599 N Force required for engagement Wen = Wn(sin+ cos) = 6599(sin10+0.2 cos10) Wen = 2445 N

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: 274 :

08. Ans: (b)

ME – GATE _ Postal Coaching Solutions

Linked Answer Question (11 & 12)

Sol: Tmax = 140 N-m

d1 = 220 mm, d2 = 150 mm Pmax = 0.25 MPa

11. Ans: (b) Sol: rm = 125 mm,

 = 0.3

 = 0.2,

r r  T = W  1 2   2 

  

300 = 0.2 Wn 0.125 Wn =12 kN Wn = Pm 2rm b 12  103 = 1.5 10220.125b b = 101 mm

i) T1 = 114 N-m Slip takes place ii) T2 = 148 N-m suitable iii) T3=173 Nm

12. Ans: (c) Sol: Axial force required to hold the clutch

Linked data for (Q09&10)

Wa = Wn sin Wa = 12sin12.5

09. Ans: (d)

Wa = 2.6 kN

10. Ans: (d) Sol: N = 600/ rpm, P = 3kN,  = 0.1 r r rm= 1 2 = 100 mm 2

For uniform wear and both sides effective surfaces r r Torque (T) = 2  W  1 2 2

= 2  0.1  3  10 3  100 4

= 6 10 N-mm = 60 N-m 2  P=

P = 1.5 bar

r r T = Wn  1 2   2 

r r  = (2)C(r1r2)  1 2   2   r 2  r22 =  2 Pmax r2  1  2 

 = 12.50

600  60  =1200 W 60

13. Ans: (d) Sol: P = 5 kW

N = 400 rpm  = 0.2 rm = 600 mm  = 14.3 T

60 P 2N

60  5 103 = 120 N-m  2  400 T =  Wn . rm = 120  103 = 0.2  Wn  300 Wn = 1989.4N

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: 275 :

Wn 

Machine Design

18. Ans: (b)

Wa Sin

19. Ans: (a)

Wa = Wn. Sin = 1989.4  Sin (14.3)

Sol: r1 = 140mm,

r2 = 70 mm

Power = 20 kW, N = 200 rpm,  = 0.3

= 491 N

For uniform wear: pmax = C  r2

Common Data for Q. 14 & 15

C = p1r1= p2r2,

14. Ans: (c)

60  20  10 3 Torque (T) = = 95.5 N-m 2    2000

15. Ans: (a)

r r  T = 2  W  1 2   2 

Sol: N1 = 200 rpm,

1 =

2N 2    200 = = 20.95 rad/s 60 60

2 = 0    2 20.95 = 1  = 4.18 rad/s2 t 5

Torque T = I = 20  4.18 = 83.6 N-m  = 0.3 For uniform pressure,  r13  r23  2 T  W  2 2   n 3  r1  r2 

83.6103 =

1003  603  2  0 .3  W  2 2 2 3 100  60 

 140  70   95.5103 = 2  0.3  W    2   W = 1.515 kN W = 2 Cr1  r2  1515 = 2    C  140  70  W = 2Pmax r2(r1 – r2) Pmax 

1515 2  70  70

= 0.0492 MPa = 49.2 kPa

Pmin 

1515 = 0.0246 MPa 2  140  70 = 24.6 kPa

W = 1706.12 N Pavg 

Common data for (Q16 to Q19)

W 1515  2ravg r1  r2  2  105  70

= 32.8 kPa 16. Ans: (b) 17. Ans: (a) ACE Engineering Academy

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: 276 :

Chapter- 10

ME – GATE _ Postal Coaching Solutions

=

4  0.3  sin 45 0   sin 90 0 2

=

0.848 = 0.329 = 0.33 2.57

Brakes Linked Answer Questions (01 & 02)

01. Ans: (b)

Common Data Question (04 & 05)

Sol: MPivot = 0

04. Ans: (c)

300 500 = RN 200

Sol:

 150

RN = 750 N Ft = RN =180 N T = Ft r

B

220 N All dimensions T2 in ‘mm’

T1

 300  3 = 180     10 = 27 N-m 2   02. Ans: (a) Sol: 1=

A

2  100 = 10.47 rad/sec 60

50

100

100

T = 450 N

=?

P = 220 N – m

a = 50

b = 100

2 = 0

Mpivot = 0

Capacity to bring the system to rest from 100

(220  200) (T2 100) + (T1 50) = 0

rpm = work done = Heat generation = T

T2 100 50T1 = 220  200 ……. (1)

   2  = T 1 t 2   = 275.235 5 = 706.725 J

T = (T1 T2)r  0.150  T = (T1 T2)    2  T1  T2 = 6000 …… (2)

03. Ans: (b)

From (1) and (2) T1 = 12880 N

Sol: µ = 0.3

T2 = 6880 N

2 = 900 = /2 rad

T1 = e T2

 = 450 Equivalent coefficient of friction 4 sin    2  sin 2 1

ACE Engineering Academy

12880 = e 6880   = 0.199 = 0.2

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: 277 :

05. Ans: (a)

Machine Design

T2 = 146.17 N, T1 = 513 N

Sol:

Torque = (T1 T2)  r

 150 A

= (513 146.17)  75 103

B

= 27.5N-m

220 N All dimensions T1 in ‘mm’

T2

Common Data Question (06 & 07) 50

100

100

l

a

We know that T ln  1  T2

  = µ  

P

T2

Here, µ = 0.4, as given T ln  1  T2

  = 0.4 ×  

T ln  1  T2

  = 0.546 

(or)

r

T1



06. Ans: (b)  T1   T2

  = eµ 

 T1   T2

  = e(0.4 × ) 

 T1   T2

  = 3.51……. (1) 

07. Ans: (c)

Here when the drum rotates in anti clockwise direction. T1 will be attached to B and T2 will be attached to A. i.e. tight side and slack side tensions will be changed. Taking moments about “O” 220 × 200 + T2 × 50 = T1 × 100…..(2) By solving 1 & 2 ACE Engineering Academy

Sol: n = 14

2 = 200   = 100 a = 150 mm ,

 = 0.25

T = 4 kN-m ,

l = 1 m, d = 1m , P = ?

As end is connected directly So it is a simple band brake T1  1   tan     T2  1   tan  

n

 1  0.25 tan 10 0 =  0  1  0.25 tan 10

  

14

T1 = 3.43 ………. (1) T2 Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally

: 278 :

P  l = T1  a

ME – GATE _ Postal Coaching Solutions

09. Ans: (d)

T = (T1 T2)(r + t)

Sol: Energy absorbed, E = T  

1  4 kN-m = (T1 T2)   0  2 

   2  75  T   1  t  2 

T1 T2 = 8 kN ......... (2)

  2  500   75  T   60  0   0 .4 2      

From (1) and (2) T1 = 11.27 kN , T2 = 3.27 kN

 T = 7.16 Nm

11.27(150) P= = 1.692 kN 1

Linked Answer Question (10 & 11) Linked Answer Questions (08 & 09)

10. Ans: (c) 08. Ans: (b)

Sol: T = 800 N-m, r = 0.5 m

Sol: d = 250 mm

T = (T1 T2)  r

3

 = 7200 kg/m

 T1  T2 =

t = 20 mm  = 0.40 sec

T1 T2 = 1600 N

N = 500 rpm

But, T2 = 300 N

Energy absorbed by brake E=



1 I  22  12 2



T1 = 1900 N

 d  I = mK = At   2 2

T1 1900 = e  = e0.45  T2 300

2

2

I = 7200 

 0.252 (0.02)  0.250  4  2 2 

 = 2350

2

= 0.055 kgm2

11. Ans: (c)

N2 = 0  Stop

Sol: Pmax = 2

1  2  500   E = 0.05  = 75 J 2  60 

ACE Engineering Academy

800 0.5

T1 1900  r.W 0.5  0.03

Pmax = 126.67 kPa

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: 279 :

Machine Design

Common Data Question (12 & 13)

Chapter- 11

12. Ans: (a)

Spur Gear Tooth

13. Ans: (b) 01. Ans: (b)

Sol: Given

d = 320 mm = 0.32 m r = 160mm = 0.16m  = 0.3

Sol: Given:

Np = 1200 rpm ,

NG = 200 rpm

m=4,

C =?

C=

F = 600N Taking moments about ‘O’ 600(400 +350) − Ft (200−160) = RN(350) 600(750) − Ft(40) = RN(350)

Tp = 25

mTp  TG  2

TP N G 1200  TG =  25 = 150  200 TG N P

C=

450000 − RN(40) = RN(350) ( Ft = RN)

425  150 = 350 mm 2

450000 − RN(12) = RN(350)

02. Ans (b)

RN(350) + RN(12) = 45000

Sol: Given ,

RN =

T1 = 19 ,

T2 = 37

C = 140mm

450000 362

C=

mT1  T2  2

140 =

m19  37  2

RN = 1243N For calculating breaking torque (TB) Ft = RN m=

Ft = 0.3  1243

140  2 = 5mm 56

Ft = 372.9N TB = Ft  r = 372.9  0.16 = 59.664 TB = 60Nm

03. Ans: (c) Sol: m = 80 mm

Face width (w) = 90 mm Ft = 7.56 kN Tensile stress = 35 MPa = S Form factor (y) = ? Let CV = 1

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: 280 :

Ft = SwmyCv

Ft =

7.56103 = 35 106908y y = 0.3



04. Ans: (a) Sol: P = 9 kW ,

d = 100 mm ,

ME – GATE _ Postal Coaching Solutions

P = SwmyCv  V

20  103 = 80106(14m)m 0.0941  d p  300     60  1000 

N = 1440 rpm Ft = ?

P = Ft  V P 9  10 3 Ft =  = 1.19 kN V   0.1  1440 60

( dp = mTp) 

20  10 6  60 = 80 140.094m2106   18  m  300  m = 5.98  6

07. Ans (d) Sol: Given

05. Ans (b) Sol: P = 10 kW = 10103W

V = 600m/min d = 100mm  r = 50mm Ft =

P V

m = 8mm ,

 = 141/20

b = 80mm ,

 = 60MPa

Y = 0.12 ,

V = 3.8 m/s , P = ?

P = Ft  v Ft =  b. m. y. Cv 

10  103  60 = = 103 N 600 Ft = 1 kN 1  10 3  50 Torque = Ft  r = 1000 T = 50 Nm

= 60  80  8  0.12   Ft = 14476 N P = Ft  V = 14476  3.8 P = 55 kW Linked Answer Questions (08 & 09)

08. Ans: (a) 06. Ans: (b)

09. Ans: (b)

Sol: Given P = 20 kW

Sol: P = 11 kW ,

NP = 300 rpm

1 , 2

b = 80 MPa

 = 14

y = 0.094, Cv = 1

TP = 25 ,

w = 14 m

TG N P  =3:1 Tp N G

Tp = 18, m = ? ACE Engineering Academy

NP = 1440 rpm m = 6 mm

y = 0.1, Cv = 0.21

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: 281 :

Machine Design

Tmax = 1.5 Tmean

dp = mTp = 8(15) = 120 mm

S = 210 MPa

Ft = ?

Ft = ? , Ft =

w=?

Fr on bearing = ? dN    V   (d = mT) 60  

P Cs V

w=? Ft =

11  10 6  25  1440 60  1000 3

Ft =

Ft = 0.98 kN

=

Ft = S w m y Cv 0.98 103 = 210  w  6  0.1  0.21 w

P 500(kW )  V d p N p 60 500  10 3  120   1800  1 m      1000   60  sec

Ft = 44.2 kN

= 37 mm 11. Ans: (c) Linked Answer Questions (10 to 12)

Sol: Fr = Ft . Tan 

= 44.2 Tan (22.5) = 18.3 kN

10. Ans: (b)

Fn 

Sol: P = 500 kW

NP = 1800 rpm C = 660 mm  = 22

12. Ans: (d)

1 2

Sol:

m = 8 mm

C

w=

N mm

Sol: Steel = 120 MPa  for pinion

8TP  10TP  660 = 2

TG = 150 ACE Engineering Academy

47.85  10 3 = 240 mm 200

13. Ans: (c)

m TG  TP  2

TP = 15

200 N  1 mm width 47.85 kN  ?

TG = 10:1 TP

Fn = 200

Ft 44.2  = 47.85 cos  cos 22.5

SCI = 100 MPa  for gear Form factors For gear, for pinion (yCI)g = 0.13 Form factors (ysteel)p = 0.093

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: 282 :

Ssteel  ysteel = 120  0.093 = 11.16

ME – GATE _ Postal Coaching Solutions

16. Ans: (d) Sol: Given, P = 120 kW = 120103 W

SCI  yCI = 100 0.13 = 13  Ssteel ysteel < SCI  yCI

DP = 250mm

(Strength)pinion < (Strength)gear

N = 650rpm

So Pinion is weaker than gear

 = 200

V=

Linked Answer Question (14 & 15)

dN = 60



250  650 1000 60

V = 8.5 m/s

14. Ans (c)

Ft =

120  1000 P Cs = (1) 8.5 V

15. Ans (a)

Ft = 14.12 kN

Sol: Given

Total load on Bearing 0,

 = 20

F=

3

P = 50 kW = 5010 W N= 30rev/sec = 3060 rpm = 1800 rpm Fn = 35 N/mm dG = 400 mm FT =

Fr

P V

50  103 = 1326.295 N = 37.699





17. Ans: (a) Sol: Given

m = 8.5mm

Fr = Ft . tan

 = 200

= 483

TP = 12

Width = =

14.12  1000 cos 20

F = 15 kN

Ft Fn

Ft cos 

Ft Fn

P = 18.65 kW

1326.29  37.89 35

Torque transmitted by idler is zero

NP = 600rpm

w = 37.89 mm  40 mm

ACE Engineering Academy

P = Ft V

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: 283 :

V=

mTP N  D P N P = 60 60

 =

8.5  12  600 1000 60

V = 3.204 m/s 18.65 Ft = = 5.82 kN 3.204

Machine Design

18. Ans: (b) Sol: Given:

G.R =

TG =2 TP

w = 10 cm = 100 mm dp = 40 cm = 400 mm Stress factor for fatigue = 1.5 N/mm2 =K

1 F32 = Ft = 5.82 kN

F32r  F32t tan = 5.82Tan20 = 2.12 kN

Q=

2TG 22TP  4 =  TG  TP 2TP  TP 3

Fw = KdpwQ

F23t = 5.82 kN

Fw = (1.5)(400)(100)

F23r = 2.12 kN

4 = 80 103 3 = 80 kN

F43r = 2.12 kN F43t = 5.82 kN RX = F23t  F43r RY = 7.94 kN R=

7.942  7.942

R = 11.228 kN

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: 284 :

ACE Engineering Academy

ME – GATE _ Postal Coaching Solutions

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Theory of Machines & Vibrations Solutions for Vol – I _ Classroom Practice Questions 06. Ans: (a)

Chapter- 1 Analysis of Planar Mechanisms

07. Ans: (d) Sol: At toggle position velocity ratio is ‘zero’ so

01.

mechanical advantage is ‘’.

Ans: (c)

Sol: It is the failure of Gruebler’s equation of DOF because it does not consider the shape

08. Ans: (d)

and dimensions of the mechanism.

Sol: The two extreme positions of crank rocker mechanisms are shown below figure.

02. Ans: (c) 03. Ans: (b)

B

20

Sol: Gruebler’s Criterion

A

N = Number of links, Given  DOF = 3(8–1) – 2  10  = 21 – 20 = 1

4e2

20 04. Ans: (a)

4 e 2

R

D

 602  502  202    18.19o  cos   2  60  50  1

3 2

P

50

B

05. Ans: (c) Q

D

 502  60 2  60 2    65.37 0 4 e1  cos 1   2  50  60  C 20 60 A

P1 = Number of rotary joints

Sol:

60

4e1 50

DOF = 3(N–1) – 2P1

D

40

2.5 2.7

S

The given dimensions of the linkage satisfies Grashof’s condition to get double rocker. We need to fix the link opposite to the shortest link. So by fixing link ‘RS’ we get double

09. Ans: (a) Sol:

C 40



B

60

20

A

D 50

rocker. ACE Engineering Academy

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: 288 : Where,  = Transmission angle

 h  1  10  2  sin 1    sin    30  20  r

20 2  50 2  53.85 cm

BD =

ME – GATE _ Postal Coaching Solutions

 = 2 1 = 20.41

By cosine rule

Quick return ratio

BC 2  CD 2  BD 2 cos = 2BC  CD

QRR   180    1.2558 180  

40 2  60 2  53.85 2 =  0.479 2  40  60

12. Ans: (c)

 = 61.37

Sol:



O

10. Ans: (c)

900

A

Sol: Two extreme positions are as shown in 

figure below. Let r = radius of crank = 20 cm

O1

l= length of connecting rod = 40 cm h = 10 cm

20

OO1 = 40 cm , OA = 20 cm

 20

sin  

40

2

10

1

   30 0 S1

S2

O A 20 1   O O1 40 2

QRR =

180  2 180  60  180  2 180  60

 QRR = 2 Stroke = S1–S2 S1 = S2 

  r 2  h 2

  r 

2

 60 2  10 2  59.16cm

 h  20  10  17.32cm 2

2

2

Stroke = S1 – S2 = 59.16 – 17.32 = 41.84 cm

13. Ans: (c) Sol: The rubbing velocity is defined as the

algebraic sum between the angular velocities of the two links which are connected by pin joints, multiplied by the radius of pin.

11. Ans: (b)

A

 h  1  10  Sol: 1  sin 1    sin    9.55  60  r

ACE Engineering Academy

1

1 O

B

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: 289 :

Theory of Machines & Vibrations

Rubbing velocity at point O

Alternate Method:

Vo= radius  relative radial velocity

The position diagram is isosceles right angle triangle and the velocity triangle is similar to

= (1  2) r

the position diagram.

Vo= (1 + 2) r ( the links OA & OB are moving in

 lr PQ 2a

opposite direction) q

14.

Ans: (c)

Vqp = 3 l3 

Instantaneous Centers.

Vq = l4 4  PQ=O4Q=

15. Ans: (b) Sol:

(2,4) (I centre) ) 90 C +

Q I34

P I23

4

2

O2

x

1

O4

3

2

I14.I13

I12

2a  2a  4

2a

O4

3

2a  3  2a

 4 = 1rad/sec

O2P=O2O4=a

O2

p

3 = 1

Q

2 rad /sec

2a

Velocity (Diagram)

for the given input angle and identify the

P

45

45o

lr O4 Q

Sol: O 4 O 2 P  180o sketch the position diagram

o2, o4 o

90 12





4

O 1

I13 is obtained by joining I12 I23 and I14 I3

OC = r

3 I12 I 23 a    2 I13 I 23 2a

Velocity of slider VS = (12 – 24) 2

3 1  2 2

x r  sin (  ) sin (90  )

3 = 1 rad /sec

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= x 2

x

r sin (  ) sin (90  )

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ME – GATE _ Postal Coaching Solutions

+ve sign means 2 and 3 are moving in

VS = r 2 sin ( + )  sec   = VC sin ( + )  sec

opposite directions. At joint 3, rubbing velocity =(4+3)r

16. Ans: (a)

= (1+0)  10 = 10 cm/s

Sol:

lr to CD

At joint 4, rubbing velocity

a,d,c

= (4 – 0)  r

lr to BC b Velocity diagram

= (1 – 0)  10 = 10 cm/s 18. Ans: (a)

VC = 0 = dc  CD

Sol:

 CD = 0

C

B

75

50

Note: If input and coupler links are collinear, then output angular velocity will

A

E

D

be zero.

50 75

17. Ans: (c) Sol: In a four bar mechanism when input link and

F

output links are parallel then coupler velocity(3) is zero.  l2 2 = l4 4

Considering the four bar mechanism ABCD,

l4 = 2l2 (Given )

 4 = 2 / 2 = 2/2 = 1 rad/s

l2|| l4

50  3  2 rad / sec 75

2, 4 = angular velocity of input and output

  2 2   4 4  4 

link respectively.

CDE being a ternary link angular velocity of

Fixed links have zero velocity. At joint 1, relative velocity between fixed

DE is same as that of the link DC ( 4 ).

link and input link = 2–0 = 2

For the slider crank mechanism DEF, crank is

Rubbing velocity at joint 1 = Relative

perpendicular to the axis of the slider.

velocity  radius of pin = 210 = 20 cm/s

 Slider velocity = DE  4

At joint 2, rubbing velocity = (2+3)  r

= 50  2 = 100cm/sec (upward)

= (2+0)10 = 20 cm/s

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: 291 :

Theory of Machines & Vibrations

19. Ans: (c)

Where, r = radius of crank pin

Sol:

From the velocity diagram VAB = ab = ?

B

b ⏊AB c

||G C

a,g

⏊BC

A

oab is right angle . tan  

oa 40    = 53.13 ab 30

tan  

r2 3

(Velocity Diagram)

G

From the velocity diagram when crank is perpendicular to the line of stroke, the velocity of slider = velocity of crank and angular velocity of connecting rod is zero.   4 ,  = 90, r

oa = 2  r = 10  0.3 = 3 m/sec

Where, n  3 

2 = 10 rad/s

Vs = r2

 r 2 10 90    5.625CW  2 2 n 16 4   3

Vrb = (2 + 3)  r

Vs = 1 = r  10  r = 0.1 m

= (10+ 5.625)  2.5 = 39 cm/s

l = 4r = 0.4 m 22. Ans: (d) 20. Ans: (a)

Sol: As for the given dimensions the mechanism

Sol: Here as angular velocity of the connecting

is in a right angle triangle configuration and

rod is zero so crank is perpendicular to the

the crank AB is perpendicular to the lever

line of stroke.

CD. The velocity of B is along CD only

Vs = velocity of slider = r2

which is purely sliding component

2 = 1  2  2 = 2 rad/sec

 Velocity of the slider  AB   AB  10  250  2.5 m / sec

21. Ans: (d) a

Sol: l3 3 b

Here

900

r 2 o



90 - Vs

the

crank

Common data Question 23, 24 & 25 o

23. Ans: (d)

is

perpendicular

connecting rod

to 24. Ans : (a)

Velocity of rubbing = (2 + 3) r ACE Engineering Academy

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ME – GATE _ Postal Coaching Solutions

25. Ans: (c)

28. Ans: 0.618

Sol: Considering the four bar mechanism ABCD,

Sol:

B

AB||DC

r

 AB  2  DC  4  4 

30  6  2 rad / sec 90

A

 C

CDE being a ternary link angular velocity of

b

DE is same as that of the link CE.

l3

For the slider crank mechanism DEF, crank is perpendicular to the axis of the slider.



Slider velocity = DE  4

e f

lr to DE and EF || to AB

(Velocity diagram)

OS OP sin    OS   250mm OP 2 27. Ans: (b)

90

diagram tan  

 240     76 60 r

Vs r2  sin  sin 90  Vs 

26. Ans : (a)

180  2 2 Sol: QRR      30 o 180  2 1

r2

Refer the configuration diagram and velocity

From the above velocity diagram Vef = 0, ef = 0

90

Vs (Velocity diagram)

= 30  2 = 60cm/sec. d,g

l

90

2

r2  618 mm / sec sin 76

= 0.618 m/sec 29. Ans: (d) VP

Sol:

VQ = VP + VPQ

VPQ

VQ

Sol: Maximum speed during forward stroke

occurs when PQ is perpendicular to the line of stroke of the tool i. e. PQ, OS & OQ are

30. Ans: (a)

in straight line

Sol: For rigid thin disc rolling on plane without

 V  250  2  750   PQ

  PQ 

2 3

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slip. The ‘I’ centre lies on the point of contact.

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: 293 :

Theory of Machines & Vibrations

31. Ans: (a)

By considering the links 1, 2 and 4 as for

Sol:

three centers in line theorem, I12, I14 and I24 P R

30

A

lies on a straight line I12 is at infinity along VP

0

V

the horizontal direction while I14 is at infinity along vertical direction hence I24 must be at infinity

O

Here ‘O’ is the instantaneous centre

33. Ans: (a)

VP =   OP

Sol:

VA = R R 2  R 2  OP 2 In  OAP , cos 120 = 2R  R

A I

1m/s 1m

2R 2  OP 2 – 0.5 = 2R 2 OP =

3R

VP =

3R    3V

or

600 O

VO = V

o

120 VP =

Va = 1 m/s

60o

o

B

Va = Velocity along vertical direction VPO

3V

Vb = Velocity along horizontal direction

    VP  VO  VPO  V  OP  

So instantaneous center of link AB will be perpendicular to A and B respectively i.e at I

 3V

IA  OB  cos   1  cos 60 0 

1 m 2

IB  OA  sin   1  sin 60 0 

3 m 2

32. Ans: (d) Sol: 1

2

N

L

Va    IA

3



90

4

M

Va 1   2 rad / sec IA V2

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: 294 : 34. Ans: (a)

ME – GATE _ Postal Coaching Solutions

35. Ans: (c) C

Sol:

Sol:

E,I13 900

 40

50 B

A

D

20

B 27

2 rad/sec

C

I23

I34

36 I41

I12 A

D 50

30

(Position Diagram)  BC

I13 = Instantaneous center of link 3 with respect to link 1

 AB

As AED is a right angle triangle and the sides are being integers so AE = 30 cm and

b

DE = 40 cm

l33

BE = 3 cm and CE = 4 cm

l22

By ‘I’ center velocity method,  a,d

l44

 DC

V23= 2(AB) = 3(BE)

c

3 

(Velocity Diagram)

1 27  9 rad / s 3

Let the angle between BC & CD is . Same will

be

the

angle

between

their

Sol: Similarly, V34= 3(EC) = 4(CD)

perpendiculars. From Velocity Diagram,

 2 2 = tan   4 4

From Position diagram, tan  =  2 =  4 

36. Ans: (a)

30 40

4 40 30  tan   2   = 3 2 20 40

2 = 3 rad/sec

4 

9 4  1 rad / s 36

37. Ans: (d) Sol: Refer the figure shown below, By knowing

the velocity directions instantaneous centre can be located as shown. By knowing velocity (magnitude) of Q we can get the

Note: DC is the rocker (Output link) and

angular velocity of the link, from this we

AB is the crank (Input link)

can get the velocity of ‘P using sine rule.

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: 295 :

Theory of Machines & Vibrations

H

I

P

A

45 VQ=1m/sec

F

C

Q

D

45 65 20 70 20

VP

E

As for the three centers in line theorem all

P

the three centers should lie on a straight line

‘I’ is the instantaneous centre.

implies on the line joining of A and H. More

From sine rule

over as both the spools are rotating in the

PQ IQ IP   sin 45 sin 70 sin 65

same direction, P should lie on the same side

IP sin 65  IQ sin 70

spool running at higher angular velocity.

VQ  I Q    1

of H. Whether P belongs to bigger spool or

 

of A and H. Also it should be close to the Implies close to H and it is to be on the right smaller spool its velocity must be same. As

VQ

for the radii of the spools and noting that the

IQ

velocity of the tape is same on both the

sin 65 IP VP  IP     VQ  1 IQ sin 70

spools H = 2A

= 0.9645

AP.  A  HP  H and

AP  AH  HP  HP  AH

38. Ans: (c) Sol: Consider the three bodies the bigger spool

(Radius 20), smaller spool (Radius 10) and

Note: (i)

If two links are rotating in same directions then

the frame. They together have three I

their Instantaneous centre will never lie in

centers, I centre of big spool with respect to

between them. The ‘I’ center will always close

the frame is at its centre A. that of the small

to that link which is having high velocity.

spool with respect to the frame is at its

(ii)

If two links are rotating in different directions,

centre H. The I centre for the two spools P is

their ‘I’ centre will lie in between the line

to be located.

joining the centres of the links.

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: 296 : 39.

Ans: (b)

ME – GATE _ Postal Coaching Solutions

41. Ans: 1 (range 0.95 to 1.05)

Sol: I23 should be in the line joining I12 and I13.

Sol: Locate the I-centre for the link AB as shown

Similarly the link 3 is rolling on link 2.

in fig. M is the mid point of AB Given, VA = 2m/sec

Locus of I23 VA

B

QB A I

30o

D

A

C

30

QA

o

B

60o

Q

Locus of I23

5

45

60o

I23

I13

I12

M

VB

VA  IA.   

So the I – Center I23 will be on the line perpendicular to the link – 2. (I23 lies

VM  IM.  IM

common normal passing through the contact

VA IA

VA IM  .VA IA IA

1  sin 30 o.VA  .2  1m / sec 2

point) So the point C is the intersection of these two loci which is the center of the disc. So  2 I12 , I 23   3 I13 , I 23   2  50  1 5  2  0.1 rad / sec

40. Ans: 20 Sol: Velocity of P  r  10 m / sec

Sol:

fco = 0.4 fr fl = 0.5



300

O

X

fc = 0.4

ft = 0.2

Centripetal acceleration,

10   R R Vp

fc = r2 = 0.4 m/s2 acts towards the centre Tangential acceleration, ft = r

Velocity of Q  2R  2R 

42. Ans: (a) & 43. Ans: (b)

10  20 m / sec R

= 0.2 m/s2 acts perpendicular to the link in the direction of angular acceleration. Linear deceleration = 0.5 m/s2 acts opposite to velocity of slider

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: 297 :

Theory of Machines & Vibrations

As the link is rotating and sliding so coriolis

45. Ans: (c)

component of acceleration acts

Sol:

co

2

f = 2V = 2  0.2  1 = 0.4 m/s

Z

VB



To get the direction of coriolis acceleration,

VA

0

rotate the velocity vector by 90 in the direction of .

O

Resultant acceleration =

VB = OB  

0.6 2  0.12  0.608 m / sec 2

VA = OA  

 0.6    tan    80.5  0.1  1

VBA = VB – VA = (OB – OA)   =  (rB – rA)

Angle of Resultant vector with reference to

and direction of motion point ‘B’.

OX = 30 + =30 + 80.5 = 110.530 44.

46. Ans: (d)

Ans: (d)

Sol: As uniform angular velocity is given,

Sol:

aTA = r

A aTO = r

Tangential acceleration,  = 0

an

Centripetal acceleration, fBA = (rB2 – rA2)   from Z to ‘O’.

O

Acceleration at point ‘O’ a

 o

a

 TO

a

 TO

47. Ans: (a)

a

 TA

 TA

are linear accelerations

and a

a

 n

Sol:

with same magnitude and opposite in direction.

a

 aO  a n 

2

b

AB

c

DC

ba

cd o 2

r a

r (Acceleration diagram)

Resultant acceleration, fR = r 2 ACE Engineering Academy

aʹ

V  r 2 r r

fR

BC

Velocity Diagram

bʹ

cʹ

Acceleration Diagram

From velocity Diagram, VC = VB

l44 = l22 25  4 = 50  0.2  4 = 0.4 rad/sec

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: 298 :

From Acceleration Diagram,

l44 = l22

ME – GATE _ Postal Coaching Solutions

49. Sol:

y

25  4 = 50  0.1  4 = 0.2 rad/sec2

P Vrel = 0.5 m/s

 

 =  0.732 rad/s2  = 1 rad/s(ccw)

 = 300

48. Ans: (d) Sol:

C

D

2rad/sec 50

50

O3

B 50 2 O2

1

O

90 90

100

x

O1

90

90

1m

f C = r2 A

=30

 f resultant

t

f = r f cor = 2V

As links O1A and O2B are parallel then VA = VB  50  2 = 50  2  2 = 2 rad/sec As a O2 C and O3D are parallel links then VC = VD  100  2 = 100  1  1 = 2 rad/sec VD = r1 = 100  2 = 200 mm/sec  = 0 (given), so tangential acceleration at = r = 0 Centripetal acceleration, fc = r12 = 100  (2)2 = 400 mm/sec2

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Acceleration diagram

Radial relative acceleration, flinear = 0 Centripetal acceleration, fc = r2 =112 =1 m/s2 (acts towards the center) Tangential acceleration, f t = r = 10.732 = 0.732 m/sec2 Coriolis acceleration, fcor = 2V = 2  0.5  1 = 1 m/sec2 Resultant acceleration, f r = 12  1  0.732 

2

= 2 m/sec2  1.732   = tan 1    60  1  reference = 30 + 180 + 60 =2700

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: 299 :

Theory of Machines & Vibrations

50.

53. Ans: (b)

Sol: xB = P, yB = Ptan

Sol:

VX  Vy 

&

54. Ans: (a)

d X B   0 dt

A r

d y B   P sec 2   d  P2 dt dt cos 

l 



B

O

Acceleration along X direction aX =

d Vx  0 dt

r

l–r l

Acceleration along Y direction ay 

d Vy  dt

FP = 2 kN

l = 80 cm = 0.8 m

 P 2 cos 3  sin   2 P 2

r = 20 cm = 0.2m

sin  cos 3 

From the triangle OAB 2  2  r 2 cos   2 2

51. Ans: (d)



52. Ans: (a) Sol:

20 2  80 2  80 2 cos      82.82 2  20  80

m1 C.G L1

m2

Thrust connecting rod

L2

FT  m1 

mL 2 100  60   60kg L1  L 2 100

mL1 100  40 m2    40kg L1  L 2 100 = 60  402 + 40 602

FP 2   2.065 kN cos  cos14.36

Turning moment, T = FT  r  

I  m1L21  m 2 L22 2

2  80 2  20 2    14.36 2  80 2

FP sin(  )   r cos 

2  sin(14.36  82.82)  0.2 cos14.36

= 0.409 kN – m 2

= 240000 kg cm = 24 kg m

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: 300 : 55. Ans: (b)

ME – GATE _ Postal Coaching Solutions

58. Ans: (d)

Sol: Calculate AB that will be equal to 260 mm

L = 260 mm,

P = 160 mm

S = 60 mm,

Q = 240 mm

I

Sol:

d 2  T  f sin , cos  dt 2

Where ‘T’ is applied torque, f is inertia

L+S = 320

torque which is function of sin & cos

P+Q = 400

d T  t  f sin , cos   c1 dt I

 L+S < P+Q It is a Grashof’s chain



Link adjacent to the shortest link is fixed

T 2 t  c1 t  f sin , cos  I

 is fluctuating on parabola and @ t = 0 ,  = 0 ,  slope  0 (because it

 Crank – Rocker Mechanism. 56. Ans: (b)

starts from rest)

Sol: O2A || O4B



Then linear velocity is same at A and B.

Parabola

 2 O2A =  4  O 4 B

Fluctuation because of inertia

 8  60   4  160 4 = 3 rad/sec

t

57. Ans: (b) Sol: As the plane is horizontal  mg = 0

 = 0  I = 0 ,  = 0 (driving torque) As the link O2A is balanced so that its centre of mass falls at O2  centrifugal force = 0

59. Ans: 1 (range 0.9 to 1.1) Sol: Ft

 Frod 0.8m

0.2m



30N

5kN

(force exerted by connecting rod)

Given Fp = 5kN 30N

(reaction force)

Frod 

Fp cos 

, Ft  Frod cos 

For the given data the only force acting on

Ft = 5kN

the link is 30N at A along AB hence the

Turning moment = Ft.r = 50.2 = 1kN-m

reaction at joint O2 is 30N ACE Engineering Academy

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: 301 :

Theory of Machines & Vibrations

06. Ans: (b)

Chapter- 2 Gear and Gear Trains

07. Ans (b) Sol: For two gears are to be meshed, they should

have same module and same pressure angle.

01. Ans (a) 02.

08. Ans: (b)

Ans: (d)

Sol: Angle made by 32 teeth + 32 tooth space

m

P

= 360.

a D b

Sol:

Pitch circle

Centre distance Q

C

A

R

B

E

S 2

R = 64

O

Given Tp = 20, TQ = 40, TR = 15, TS = 20 Dia of Q = 2  Dia of R

360 2 =  5.625 64

mQ.TQ = 2mR.TR Given, module of R = mR = 2mm

  2.8125

mT 4  32   64mm R= 2 2

 mQ = 2 mR

TR 15  2  2  1.5 mm TQ 40

mP = mQ = 2mm

a = R sin  2 = 64×sin(2.81)×2 = 6.28 OE = Rcos = 64×cos(2.8125) = 63.9 mm

b = addendum+ CE = module +(OC – OE) = 4 + (64– 63.9) = 4.1 03. Ans: (c) Sol: Helix angle = 90 – 22.5 = 67.5

mS = mR = 1.5 mm Radius = module 

No. of teeth 2

Centre distance between P and S is given by RP RQ RR RT = mP

TQ T TP T  mQ  m R R  mS S 2 2 2 2

04. Ans: (a)

 40  20  15  20   1.5  2  2   2 

05. Ans: Decreases , Increases

= 45 + 35 = 80 mm

ACE Engineering Academy

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: 302 :

ME – GATE _ Postal Coaching Solutions

09.

Ans: (a)

13. Ans: 5rpm (CCW)

Sol:

N 5  T2  T4 20 15 1      N2 T3 T5 40 30 4

Sol:

N5 

N 2 1200   300 rpm 4 4

in

the

1

2

same

3

direction as that of gear 2 i.e, CCW 10. Ans: (c) Sol:

N 2 N3 N5 N6 N3 N6   N6 N 2 N 4 N5 N 2 N 4

T1 = 104 , N1 = 0,

Wheel 5 is the only Idler gear as the number

N2  Na T 104  1 N1  N a T2 96

T2 = 96 , Na = 60rpm(CW+ve),

of teeth on wheel ‘5’ does not appear in the

N 2  60 104 = 0  60 96

velocity ratio. 11. Ans: (a) Sol:

 60  8  104  = N2 = 60 1  = 5 rpm CW  96  96 

4

1

N2 = ?

= 5 rpm in CCW 14. Ans: (b)

3

2

Sol:

Z1 = 16 , Z3 = 15 , Z2 = ? , Z4 = ? First stage gear ratio, G1 = 4 , Second stage gear ratio, G2 = 3 ,

Ring 80T Sun 20T Planet 30T

Arm

m12 = 3, m34 = 4 Z2 = 16  4 = 64

TS = 20, TP = 30, TR = 80

Z4 = 15  3 = 45

NS = 100 rpm (CW +ve), NR = 0 , Na = ?

12. Ans: (b)

NS  Na  TR = NR  Na TS

Sol: Centre distance

=

m12 m  Z1  Z2   34  Z3  Z4  2 2



4  15  45  120mm 2

ACE Engineering Academy

100  N a  80 = 0  Na 20

 Na = 20 rpm CW

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: 303 :

Theory of Machines & Vibrations

15.

Ans: (a)

18. Ans: (c)

Sol:

By Analytical Approach

Sol:

1  5  T2  T4 45 40      4  5 T1 T3 15 20

arm = 80 rad/s (CCW) = –80 rad/sec 5  a  T2 T4   2  a T3 T5

1  5 6  4  5

16.

Ans: (d)

Sol:

Data given:

1 5  (80)  20 32    100  (80) 24 80 3  5 = 140 CW = 140 CCW

1= 60 rpm (CW, +ve)

19. Ans (c)

4 = –120 rpm [2 times speed of gear -1]   5 6 We have, 1 4  5 

60  5  6 , simplifying  120  5

NA T T  D B ND TC TA



N A TD 25 TD    100 50 50 100

 NA = TD

5 = –156 rpm CW

From given option (d) is correct.

5 = 156 rpm. CCW

21. Ans: (c) Sol: No .of Links, L = 4

17. Ans: (a)

epicyclic

20. Ans : (d) Sol:

60 – 5 = –720 – 65

Sol: An

2 = 100 rad/sec(CW+ve),

gear

train

is

shown

No. of class 1 pairs J1=3

schematically the gear 5 is fixed and gear 2

No. of class 2 pairs J2 =1 (Between gears)

is rotating 60 rpm (CCW, +ve) Then the arm

No. of dof = 3(L – 1) – 2J1 – J2 = 2

4 attached to the output shaft will rotate at ? T2 = 20 , T3 = 40 ,

T5 = 100

N2 = 60 CCW (+ve) N5 = 0 (fixed), N4 = ? N 2  N 4  T5  N5  N4 T2 60  N 4 100  0  N4 20

 N4 = 10 CCW ACE Engineering Academy

22. Ans: (c) Sol: Given T2 = 60

T4 = 100

N2 = 0

T3 = 20

N4 = 100rpm (ccw +ve )

Relative velocity equation N4  Na T  2 N2  Na T4



100  N a 60  100 0  Na

1.6 Na = 100 Na 

100 = 62.5 rpm (ccw) 1.6

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: 304 :

ME – GATE _ Postal Coaching Solutions

02.

Chapter- 3 Fly Wheels

Sol: Power = 20 kW ,

N = 240 rpm 

01.

=

Sol: Given

P = 80 KW = 80103W = 80,000W E  0.9 Per cycle

240  4rps 60

1 sec per rev. = 0.22 sec per rev. 4

1 cycle = 2 revolution for 4 stroke

N  300 rpm

So cycle time = 2 

C S  0.02

Cs = 0.01



2N 2  30   31.41rad / s 60 60

T

 = 7500 kg/m3  c  6MN / m

3x

2

 c  V 2  R 2  2 C R   2

6  10 6 7500  31.412

D = 2R = 1.8m

N  300rpm  5rps  0.2 Sec/rev 1 cycle = 2 revolution 4 stroke engine 

 0.4 sec Energy developed per cycle = 0.4  80  32 kJ E  E per cycle  0.9  32  10 3  0.9 E  28800 J E  I 2 C S E 2 CS I  1459.58 kg-m2 ACE Engineering Academy

T

0.5x

0

3 

2

4



x

Let torque under the compression stroke = x Torque under the expansion stroke =3x

R = 0.9 m

I

1 2   0.5 sec 4 4

Work done per cycle = Net area under the turning moment diagram = 3x – x = 2x Joule Mean Torque, Tm  

Total work done per cycle Duration of cycle 2 x  0.5x 4

So fluctuation of energy , E = 3x – 0.5x = 2.5 x Power = 20 kW = 20000 kJ/sec 1 cycle time = 0.5 sec Energy per cycle = 20000  0.5 = 10000 kJ

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: 305 :

 2x = 10000  x

Theory of Machines & Vibrations

10000 N.m 2



Fluctuation energy, E = 2.5 x 10000    12500 Nm 2

 2.5 

Workdone per cycle 4

Mean torque Tm =

2

 2N  E = I Cs = 12500= I     0.01  60  2

6 1.5  cm 4 

15.725 

17 

b 1.275 

2

 2    240   12500  I     0.01 60  



 I = 1978.93 kg-m2

Area of the triangle (expansion) 03.

=

Sol: T

H = 18 / Area above the mean torque line

9 cm2 H

18 

h b



0 0.5 cm2

1   H  9 2

2

1.7 cm2

B

1 .5 

Tmean 4

3

0.8 cm2

2

Given: 1 cm = 1400 J

1 E   b  h 2 From the similar triangles , 16.5 b h  b  18 B H 16.5 1 E   b  2 

Assume on x-axis 1 cm = 1 radian and on y-

1 16.5 16.5 = 7.56 cm2    2 18 

axis 1 cm = 1400 N-m a1 = –0.5 cm2

E = 7.56 1400 = 10587 N-m

2

a2 = –1.7 cm

N1 = 102 rpm,

2

a3 = 9 cm

N2 = 98 rpm, 2

a4 = –0.8 cm

Work done per cycle = –a1– a2 + a3 – a4 = – 0.5 –1.7 + 9 – 0.8 = 6 cm2 ACE Engineering Academy

1 

2N 1  10.68 rad / s 60

2 

2N 2  10.26 rad / s 60

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: 306 :

Actual punching time = 30/20 = 1.5 sec

1 E   I  (12   22 ) 2 I

ME – GATE _ Postal Coaching Solutions

Energy supplied by the motor in 1.5 sec is

2  E 2  10587  2 2 (1   2 ) 10.68 2  10.26 2

E2 = 2639  1.5 = 3958.5 = 3959 N-m Energy to be supplied by the flywheel

I = 26.468 kgm2 04.

during

Sol:

fluctuation of energy Power

or

the

maximum

E = E1  E2 8.52639 = 22431Nm

= 26390  3959 = 22431 N-m Coefficient of fluctuation of speed

1.52639 8.5 sec

Given:

punching

CS 

10 sec

Time

d = 40 mm,

V1  V2  0.03 V

We know that maximum fluctuation of energy (E)

t = 30 mm

22431 = m V2 CS = m (25)2 (0.03)

2

E1 = 7 N-m/mm , S = 100 mm V = 25 m/s, V1  V2 = 3%V, CS = 0.03

m = 1196 kg

A = dt =   40  30 = 3769.9 = 3770 mm2 Since the energy required to punch the hole is 7 Nm/mm2 of sheared area, therefore the

05. Sol: Given:

Total energy required for punching one hole

P = 2 kW ;

K=0.5

= 7  dt = 26390 N-m

N = 260 rpm ;

 = 27.23 rad/s

Also the time required to punch a hole is 10 sec, therefore power of the motor 26390 required =  2639 Watt 10

Actual punching time = 1.5 sec Work done per cycle = 10000 Joule per hole

Motor power = 2 kW N = 30 rpm

The stroke of the punch is 100 mm and it

 = 2 (30/60) =  rad/sec

punches one hole in every 10 seconds.

600 holes/hr = 10 holes/min  6 sec/hole

Total punch travel = 200 mm (up stroke + down stroke)

Cycle time = 6 sec

Velocity of punch = (200/10) = 20 mm/s ACE Engineering Academy

Power Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally 7.5 kJ

7.5 kJ

2.5 kJ

: 307 :

Theory of Machines & Vibrations

I 2 Cs  22 C s I 

I2 C s I  125.6  0.04  8I   2 Cs 62.82  0.02 2

08. Ans: (d)

Energy withdrawn from motor

Sol: At A, Energy = E + 100

At B, Energy = E + 100  75 = E + 25

= (10000/6) = 1666.67 J

At C, Energy = E + 25 + 89 = E + 114

Energy stored in flywheel 

10000  4.5  7.5 6

At D, Energy = E + 114  77 = E + 37

kJ

At E, Energy = E + 37 + 36 = E + 73

Fluctuation of Energy E = 7500 J

At F, Energy = E + 73  73 = E

2

E = I   = mk  m

Maximum energy = E + 114 = maximum

E k 

speed

2

Minimum energy = E = minimum speed

Where k = radius of gyration

In flywheel energy is stored in form of

7500  349.5 kg m 2 0.5  27.23  

kinetic energy. When flywheel stores energy its kinetic energy increases and when it releases energy

06. Ans: (a)

its kinetic energy decreases.

1  TP   2 Sol:  10 4

09.

Sol: Work done = 0.5+12+250.8+0.5

Tp = 80 N-m

= 23.2 cm2 Work done per cycle = 23.2100 = 2320

07. Ans: (d)

1cm

Sol: N = 1200 rpm ,

2  N  125.6 rad/sec = 60 C Cs = 0.04 , Cs= s  0.02 2  

  62.8 2

E = E ACE Engineering Academy

Ans: (d)

2

Tmean 



 100 N  m



W .D per cycle 4

2320 580  Nm 4 

Suction = 0 to , compression =  to 2 Expansion = 2 to 3, Exhaust =3  to 4 10. Ans: (c)

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: 308 : Sol: 60 A

ME – GATE _ Postal Coaching Solutions

3W E , E min  4 4 E E = Emax – Emin = 2 E max 

80 B

40 C

D

60 100

E

F 60

G

E  0.5 E 4

EA = E

14. Ans: (c) Sol:

Motor shaft

EB = E + 60 EC = E + 60 – 40 = E + 20



ED = E + 20 + 80 = E + 100 = Emax EF = E + 60

Cs = 0.032

EG = E + 60 – 60 = Emin

Gear ratio = 4 I2  Cs = I2 Cs

11. Ans: (b)

mr 2

Punching Machine

Gear box

EE = E + 100 – 100 = E

Sol: I disk 

Flywheel

4

2

C  2 C  Cs  Cs    s 2  s 16  16   

2

= 0.0032 / 16= 0.002 (by taking moment

2

mr I1  1 , Cs1 = 0.04 2

of Inertia, I = constant). Thus, if the flywheel is shifted from

I2 = 4 mr12 = 4I1

machine shaft to motor shaft when the

I C s 2  1  C s1  0.01  1% reduce I2

fluctuation of energy (E)

is same, then

coefficient of fluctuation of speed decreases by 0.2% times.

12. Ans: (d) 13. Ans: (a)

15.

Sol: Let the cycle time = t

Sol: Given E  400 N  m

Actual punching time = t/4 W = energy developed per cycle Energy required in actual punching = 3W/4 During 3t/4 time, energy consumed = W/4 ACE Engineering Academy

Ans: (a)

 = 20 rad/sec,

Cs = 0.04

We know E  I  2 C s  I 

E  25 kg  m 2 2 Cs

16. Ans: 0.5625

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: 309 : Sol: The flywheel is considered as two parts

as rim type with Radius R and type with Radius

m 2

m as disk 2

Theory of Machines & Vibrations

E 

 I 2max  2min 2

I

2E   2min

R 2

1 m R mR 2 m  R 2 , I disk       2 2 2 16 2

mR 2 mR 2 9 I   mR 2 2 16 16

19.

Ans: 104.71

N = 100 rpm Tmean 

  = 0.5625

N = 200 rpm

400   rad / sec 60

1  10000  1000 sin 2  1200 cos 2d  0



1 10000  500 cos 2  600 sin 20 

= 10000 Nm

Cs = 0.01 (  0.5% = 1%)

Power =

E  I 2 C s I

1  Td  0

 17. Ans: 592.73 kg-m2

 

2 max

  3000 2 I  31.42 kgm 2 2 2 20  10

Sol:

= 0.5625 mR2

Sol: E = 2600 J,



2

2

I Rim



2600  60 2 E   2 C s 400  2  0.01 = 592.73 kgm2



2NT 60 2    100  10000 = 104719.75 W 60

P = 104.719 kW

18. Ans: 31.42 Sol: From the T -  diagram energy is

stored into flywheel during E 

 to  2

  3000 N  m 2

From - diagram, max = 20 rad/sec min = 10 rad/sec ACE Engineering Academy

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: 310 :

ME – GATE _ Postal Coaching Solutions

07. Ans: (c)

Chapter- 4

08. Ans: (c)

09. Ans: (b)

10. Ans: (a)

Governor

Sol: r1 = 50 cm , F1 = 600 N 01. Ans: (a)

F = a + rb

02. Ans: (d)

600 = a + 50 b 700 = a + 60 b

03. Ans: (b) Sol:

h

g m 2

(M ) (1  k )   m   2  

10 b = 100 b = 10 N/cm a = 100 N

K=1 2

F = 100 + 10 r

g 21   = 24 rad/sec 2



04.

Ans: (a)

Sol:

mr2 

Stable

Mg1  k   r   mg  2 h 

k =1 2 

Isochronous

Unstable

Force

0.50

Radius

9.8 10  2 2  0.2

This is unstable governor. It can be

 = 17.15 rad/sec

isochronous if its initial compression is reduced by 100 N.

05. Ans: (a)

1  200    a 2

1 20 2  0.25  2 = 200 = 0.5  2 = 1 cm

11. Ans: (d)

12. Ans: (d)

13. Ans: (a) Sol:

(3)

Force

Sol: mr2 a =

(2)

06.

Ans: (a)

Sol:

F  mr2  a =  s   a 2 Fs = 2mr2

40

C.F=mr2

Radius

-1000

= 2 1  0.4  (20)2 = 320 N ACE Engineering Academy

10 20

(1)

-2000

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: 311 :

Theory of Machines & Vibrations

At radius, r1 = F1 < F2 < F3  As Controlling force is less suitable 1 is for low speed and 2 for high speed ad 3 is for still high speed. (1) is active after 40 cm (2) is active after 20 cm (3) is active after 10 cm At given radius above 20 F3 > F2 mr32 > mr22 3 > 2 14. Ans: (b) 15. Ans: (c)

16. Ans: (c)

17. Ans: (b) Sol: F = 14 N , r = 2cm

F = 38 N, r = 6cm 14 = 2a + b 38 = 6a + b 4a = 24 a=6 b=2  unstable governor

ACE Engineering Academy

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: 312 :

ME – GATE _ Postal Coaching Solutions

mea = m B2 e2a2

Chapter- 5

9(0.5)(0.5) = m B2 (0.5)(1.5)

Balancing 01.

m B2 = 3 kg , m B1 = 6 kg

Ans: (c)

Sol:

L

RL

RR

03.

Ans: (c)

04.

Ans: (a)

Sol: Dynamic force = (a) Static Balance

At High speed RL and RR at ‘2L’ apart

Couple =

i.e.,

W e 2 g

W e 2 a g

Reaction on each bearing =  2L

RL

W 2a e l g

Total reaction on bearing

RR

W =  e 2  g

(b) Dynamic balance

a W 2    e l   g

a 0 l 

 RL  2L = Unbalanced couple  RL  2L = RL L  R 'L  02.

RL 2

05.

Ans: (b)

06.

Ans: (a)

07. Sol:

Ans: (a)

a

mb = 6 kg, rb = 20 cm

e a2

e1 mB1

ma

ma = 5 kg, ra = 20 cm

m

Sol:

Ans: (b)

mc = ? , e2 mB2

rc = 20 cm

md = ?, c = ? , d = ? Take reference plane as ‘C’

mb

For complete balancing  mr = 0

&

 mrl = 0

m2ea = m B1 2e1+ m B2 2e2

2mdcos d – 9

couple about the plane of B

 mdcosd = 9 2

ACE Engineering Academy

225

2 =0

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: 313 :

2mdsind –5 –9

2 =0





1 59 2 2

mdsind = 

2



Theory of Machines & Vibrations

 mccosc + 10.91 cos 54.31 – 3 2 = 0 mccosc = –2.122 mc sinc +mdsind – 3 2  5  0



2

 9  1  md      2 5  9 2  = 10.91kg   2 

mc sinc + 10.91 sin54.31 –3 2  5  0 mcsinc = – 9.618

1  59 2   0  d  tan 1  2   54.31 9    2 





2

 9.85kg

9.618 2.122

c = 257.56 or 257.56 – 90 w.r.t ‘A’ = 167.56

mc cosc +mdcosd – 3 2  0 (r20)cm (l20)cm



mrcos

mrsin

mrlcos

mrlsin

–1

90

0

5

0

–5

1

3

225

–3 2

–3 2

–9 2

–9 2

mc

1

0

c

mccosc mcsinc

0

0

md

1

2

d

mdcosd mdsind 2mdcosd 2mdsind

S.No

m

A

5

1

B

6

C D

Common data Q. 08 & 09

100kg-cm

08. Ans: (a)

m1 = kg ,

m2 = 5kg ,

r1 = 10cm

r2 = 20cm,

md = ?,

rd = 10cm

10kg

mr = 100kg-cm

Resultant force 30

m1r1 = 100 kg cm m1

2

tan  c 

= 90 – 54.31 = 35.68 w.r.t ‘A’

Sol:

 2.122    9.618

mc 

30

m2r2 = 100kg cm mdrd

100kg-cm

10cm

Keep the balancing mass md at exactly

30 20cm

m2

opposite to the resultant force 5kg

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: 314 :

ME – GATE _ Postal Coaching Solutions

mdrd = 100kg-cm 10. Ans: 0.456cm, 235.26

 md10 = 100 kg-cm

Sol:

md = 10kg cm

1

m2

2

d = 180 + 30 = 210 09.

20cm 25cm

Ans: (d)

m1

 e

2

mr

Sol:

r1 = 10cm, r2 = 10cm, m1 = 52 kg 0.2m

m2 = 75kg, 1 = 0 (Reference) 2

mdrd

2 = 90, m = 2000kg , e = ?, =?

mr = 100kg-cm = 1kgm N = 600 rpm   =

me cos = m1r1 = 520

2 N  20 rad / s 60

2

me sin = m2r2 = 750 me 

2

Couple ‘C’ = mr  0.2 = 1(20) 0.2



789.56  1973.92N 0.4

2

 5202  7502

 913  e   0.456cm  2000 

Reaction on the bearing couple dis tan ce between bearing

2

 913 kg  cm

= 789.56Nm



 m1r1    m 2 r2 

m r   75    tan 1  2 2   tan 1    55.260  52   m1r1 

= 180 + 55.26 = 235.26 w.r.t mass ‘1’.

11. Ans: (a) Sol: Plane

m

r (m)

(kg) D

2 kg.m

A

-ma

B

-mb

ACE Engineering Academy



L (m) (reference Plane A)

Fx

Fy

Cx

Cy

(mrcos)

(mrsin)

(mrlcos)

(mrlsin)

0.3

0

2

0

0.6

0

0.5m

0

a

–0.5macosa

–0.5masina

0

0

0.5m

0.5

b

–0.5mbcosb

–0.5mbsinb



mb cos b 4



mb sin b 4

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: 315 :

Cx = 0 

m b cos  b  0.6 4

Cy = 0 

m b sin b 0 4

Theory of Machines & Vibrations

b  tan 1

Fy Fx

 353.553   tan 1   = 98.7   53.55 

13. Ans: 30 N

 mb = 2.4kg , b = 0

r

Sol:

Fx = 0

Crank radius

 2 – 0.5 ma cosa – 0.5 mb cosb = 0

30

= stroke/2 = 0.1 m,

m  a cos  a  0.8 2 m Fy = 0  a sin a  0 2

 = 10 rad/sec

r mb

= 6 kg

Unbalanced force along perpendicular to the line of stroke = mbr2 sin 30

 a = 0 , ma = 1.6 kg

= 6 × (0.1) × (10)2 sin 30 = 30 N

(Note: mass is to be removed so that is taken as –ve).

12.

Ans: (a)

Y

Sol: m2 r2

 X

r1 m 1

Fx  m1r1  m 2 r2 cos  2 = 20  15 + 25  20 cos135 = –53.55 gm-cm Fy  m 2 r2 sin  2 = 2520 sin135 2 = 353.553 gm-cm 2

m b rb  Fx  Fy

2

2

 mb  

Fx  Fy

14.

Ans: (b)

15.

Ans: (b)

16.

Ans: (b)

Sol:

m = 10 kg, r = 0.15 m , c = 0.6 ,

 = 60 ,  = 4 rad/sec

Residual unbalance along the line of stroke = (1 – c) m r2 cos = (1 – 0.6)10 0.15 42cos60 = 4.8 N 17. Ans: 2 Sol: By symmetric two system is in dynamic

balance when mea = m1e1a1

2

rb

m1  m

 53.552  353.5532

50 2 e a  2kg .  1 20 2.5 e1 a 1

20

= 17.88 gm ACE Engineering Academy

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: 316 :

Chapter- 6

ME – GATE _ Postal Coaching Solutions

 25 

10    2 

Cams

  = 5 rad/sec

01. Ans: (d)

02. Ans: (a)

amax =

03. Ans: (d)

04. Ans: (b)



2

05. Ans: (b) Sol:

 = 90o = /2 radian ,

L = 4 cm ,

2  = 2 rad/sec ,    90  60 3  2   3

 21  cos 120   3cm



   L      sin   2     4  2  2 sin 120  7cm / s 2 2

   L a t      2  cos  2     

4 2 2  2  2  cos120  16cm / sec 2 2

07. Ans: (a)

 2 

Sol:  = 90 ,

 = 1 rad/s

L = 6 mm ,

s

  L 1  cos    2

   3  31  cos    

cos

 0    2

 cos 2 = 0  2 cos2 –1 = 0   = 45  1   2 V

06. Ans: (a) Sol: L = 10 cm ,  = 180 =  rad ,

Vmax = 25 cm/s Vmax= 25 =

10 2  1  52 = 125 cm/sec2 2

When, s = 3mm,

  L s(t) = 1  cos    2

V(t) 

L      2 2 

L    2 

ACE Engineering Academy



L    sin 1 / 2 2 6   2  1  sin   = 6 mm/sec 2 2 2

   L  a      2  cos   0 2    

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: 317 : 08.

Theory of Machines & Vibrations

2 x 2( y  10) dy  . 0 152 52 dx

Ans: (b)

Sol: normal tangent

dy 1  15 3 x    dx ( y  10)9 3 3 2 9 2

Radial line 16.10

tan = 43.897 120

60

30

150

Then normal makes with x-axis tan1 ( 3 ) = 60o

x = 15cos ,

tan  

y = 10 + 5sin tan =

1 3

dy dy 5 cos    d  15 sin  dx  dx     d 

y 10  5sin  10  5sin 30   x 15cos  15cos 30

 = 43.897 With follower axis angle made by normal (pressure angle) = 6043.897 = 16.10o

at  = 30 , 3 2   1   = 150 tan   1 3  15  2 5

y 10  5sin  10  5sin 30 tan     x 15cos  15cos 30

09. Ans: (a) Sol:

tangent normal

Radial line 26.52 6.3

x y

96.3 26.52 83.7

 = 43.897 Pressure angle is angle between normal and

6.3 x

Let  be the angle made by the normal to

radial line = 16.10 .

the curve x = 15 cos ,

or

o

y = 10 + 5 sin at  = 30 2

2

 x   y  10   1     15   5  15 3 , x= 2 ACE Engineering Academy

 dy  9    dx  4, 2 

y = 125

tan  

dy = 4x – 7 dx

At x = 4 & y =2 ,  = tan-1(9) = 83.7

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: 318 :

ME – GATE _ Postal Coaching Solutions

The normal makes an angle  1  tan 1    6.3 with x axis  9  2   tan    26.52 4

Chapter - 7 Gyroscope

1

Pressure angle is angle between normal and radial line = 26.52 + 6.3 = 32.82 10.

Ans: (b)

01. Ans: (b)

02. Ans: (d)

03. Ans: (b)

04. Ans: (b)

05. Ans: (c)

06. Ans: (d)

07. Ans: 151.81 N.m

Sol: For the highest position the distance

Sol:

y

between the cam center and follower



p ˆj

= (r+5) mm For the lowest position it is (r –5) mm



x ˆi

So the distance between the two positions = (r+5) – (r–5) = 10 mm



z kˆ

Iw = 2.5 kg-m2 , Ie = 0.15 kg-m2 , w 

E G

G = 5.1 D = 0.65 m ,

r = 0.325 m

G.C = ? R = 30 m , VP 

V R

e 

GV R

V = 16 m/s

Gyroscope couple

  C g  p  H

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: 319 :

I = mk2 ,

Assume left turn  H  2I w   w  I e e   kˆ

  V H   2I w   I e r 

  GV  ˆ  k  r 

 = 2400 rpm = 80 rad/sec = 251.2 rad/sec

 

 

[till about –ve axis direction] 08. Ans:

p =

18  1860  0.155rad / sec 60  3600

p 

Vˆ j R

(i) Gyroscope couple   C g  H  p

 

 I  p ˆi  ˆj  mk 2   p  kˆ

Sol: N = 300 rpm

s 

2N 2  3000  = 100 ˆi rad/s 60 60

= 6000 0.452  251.2  0.155 = 47.3 kˆ kN-m

I = 47.25 kg-m2

Bow portion is raised.

2 rad  17s  2  p    ˆj  17    C g  p  H = I  p ˆi  ˆj

(ii) Pitching amplitude, A = 7.5

 



 = A sint  = 18 sec ,

 I  p kˆ

= 47.25  100 

2 17

= 5486.33 N.m 09. Ans: Sol:

m = 6000 kg ,

k = 0.45 m ,

V V  2I w  GI e ˆj   kˆ R r = 151.81  ˆi N.m

Cg 

Theory of Machines & Vibrations

f

1 Hz 18



2 rad / sec 18

Maximum angular velocity of precession, y

p = A 



p ˆj

 7.5 



x ˆi



z kˆ

ACE Engineering Academy

 

 2  = 0.0457  kˆ rad/sec 180 18

 H  I ˆi  6000  0.452  80ˆi = 30536.28 ˆi

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: 320 :

ME – GATE _ Postal Coaching Solutions

  C g  H  p



 I  p ˆi  kˆ



Chapter - 8 Mechanical Vibrations

 

 6000  0.452  80 ˆi  0.0457  kˆ



01.

= 13.955 ˆj kN-m (as the bow portion is lowered, the ship turns

Ans: (b)

Sol: T  2

towards left or port side)

 L = 62.12 mm

Maximum acceleration = A 2 2

 7.5 

L L  0 .5  2   g 9.81

  2     rad / sec 2 180  18 

02. Ans: (d) Sol:  

= 0.016 rad/sec2



    

(iii) rolling = 0.035 rad/sec p = 0 during rolling   C g  H  p  0 (No gyroscope effect)

10.

Ans: 200 (range 199 to 201)

Sol: R=100m,

p 

V rad  0.2 R sec

mg

Let the system is displaced by  from the equilibrium position. It’s position will be as shown in figure.

v = 20m/sec,

By considering moment equilibrium about the

s  100rad / sec

axis of rotation (Hinge) 

I   m g  sin     m g  sin     0

I = 10kg-m2 Gyroscopic moment = Is  p = 100.2100N-m = 200N-m

I  m 2  m   2m 2 After simplification 2m 2   2mg cos  sin   0 For small oscillations (  is small) sin  =  2 m  2   2 m g  cos  .  0 n 

ACE Engineering Academy

mg

2 m g  cos  g cos   2  2m

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: 321 : 03.

Theory of Machines & Vibrations

05. Ans: (c)

Ans: (c)

Sol: Let, Vo is the initial velocity, ‘m’ is the mass

l  2

Sol:

K

Equating Impulse = momentum mVo = 5kN  10 4 sec



 5  103 10 4  0.5 sec V0 

..

I

0.5  0.5 m / sec m

n 

E=

2

The amplitude

n

Differentiating w.r.t ‘t’

(Initial displacement )

dE   K  2 = I     2  0 2 4 dt

V0 0.5  103  X   5 mm n 100

I

Sol: Note:  n depends on mass of the system not

  

on gravity

If  n =  n =

m 2 12

m 2  K 2  0 12 4

04. Ans: (a)

 n 

1 2 1 I  Kx 2 = constant 2 2

1 1   E = I 2  K      cons tan t 2 2 2 

initial velocity,

V0

K =300N/m

By energy method

K 10000   100 rad / sec m 1

When the free vibrations are initiate with

X=

l

1

3K 0 m

 n 

m mg g ,  K  g   mg     K 

06.

K m

  n is constant every where.

3K  30 rad / sec m

Ans: (a)

Sol: K O

M 

a

A

L ACE Engineering Academy

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: 322 :

ME – GATE _ Postal Coaching Solutions

Assume that in equilibrium position mass

08.

M is vertically above ‘A’. Consider the

Sol:

displaced position of the system at any instant as shown above figure.

K

K

If st is the static extension of the spring in

a

equilibrium position, its total extension in O

the displaced position is (st + a).

r



From the Newton’s second law, we have 

A

I 0   Mg(L  b)  k ( st  a)a...(1) But in the equilibrium position MgL=kst a Substituting the value in equation (1), we

KE 



have I 0   (Mgb  ka 2 )





 I 0   (ka 2  Mgb)  0

ka  Mgb I0 2

n 

  2

I0 ka 2  Mgb

PE 

1 22 1 22 3 22 mr   mr   mr  2 4 4

1 1 Kx 2  Kx 2  Kx 2 2 2

x = (r + a)  PE = K{(r + a)}2

quantity if ka2 < Mgb. This makes the

d d KE  PE  0 dt dt

system unstable. Thus the system to vibrate

Substituting in the above equation

The time period becomes an imaginary

the limitation is ka  Mgb 2

b

ka 2 Mg

Where W = Mg 07.

1 22 1 2 mr   I 2 2

Ans: (a)

ACE Engineering Academy

3 2 m r 2   2K r  a   = 0 2 Natural frequency fn 

1 4K r  a  2 3mr 2

2

So fn = 47.74 Hz.

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: 323 : Or

Theory of Machines & Vibrations mL2   L KL2   = 0 (∵sin ≈ )    mg   9 6 9  

I A  K

K

n =

a O

10.

A

Taking the moment about the instantaneous

Sol:

centre ‘A’.

Ans: (d)

11. Ans: (c) &

2k r  a  3 2 mr 2

4k r  a  3mr 2

2

2

m eq

  

2

 X = x0 = 10cm

3 2 2 mr   2k r  a    0 2 

v x   0  n 2 0

If v0 = 0 then X = x0

mr 2 3  mr 2  mr 2 2 2

k eq

n = 5 rad/sec

X0 = 10 cm, X=

IA  + 2K (r+a)  (r+a) = 0

n 

3g K  2L m



r

IA 

L KL2 mg   6 9  = n 2 mL 9



12. Ans: (c)

Sol:

mg sin

I

mg



09.

L

Ans: (b)

Sol:

K

L  3

Kt 

L 6

O

I = mL2



The equation of motion is mL2   k  mgL  0

mg

t

Inertia torque = mL2 By considering the equilibrium about the pivot ‘O’ L L L IO  + mg sin   K   = 0 6 3 3 ACE Engineering Academy

Restoring torque = kt – mgL sin = (kt – mgL)

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: 324 : 13. Ans: 0.0658 N.m2 Sol: For a cantilever beam stiffness, K 

Natural frequency,  n 

K  m

3EI 3

3EI m 3

ME – GATE _ Postal Coaching Solutions

By taking the moment about ‘O’, mo = 0 (m 2a  2a )  (ka  a )  0  4a2 m  +ka2 = 0 Where, meq = 4a2m, keq = ka2 Natural frequency, n 

Given fn= 100 Hz  n = 2fn= 200  200 

3EI m 3

3

= 0.0658 N.m2

f 

n 1 k   Hz 2 2 4m

16. Ans: (a)

14. Ans: (d)

Sol: Moment equilibrium above instantaneous

Free body diagram m2r 

kr

centre (contact point)  k (a  d).a  d   I c 





Moment equilibrium about hinge m 2r.2r  k.r  0

K(a+d)

4mr 2  kr 2   0



2

kr k 400   2 4m 4 4mr

Ic 

C

3 Ma 2 , 2

a 

k a  d  3 Ma 2 2

n 

2 k (a  d ) 2 3Ma 2

2

15. Ans: (a) Sol:

ka O

k rad 4m sec

 n  2f

2  200. .m 3 EI 

n 

m eq

ka 2 = 4a 2 m



Flexural Rigidity

Sol:

k eq



ACE Engineering Academy

m2a 

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: 325 :

Theory of Machines & Vibrations

Damped frequency natural frequency,

17. Ans: 10 (range 9.9 to 10.1) Sol: KE 

1 1 mx 2  I 2 2 2

20  r 2  10r 2 2

KE 

 20  1   2  25  0.6  60%

x  r

m = 5kg, I

d  1   2   n

20. Ans: (a) Sol: K1 , K2 = 16 MN/m

K3, K4 = 32 MN/m

1 2 1 x 2 1 5x  10r 2 . 2  15x 2 2 2 2 r

Keq = K1 + K2 + K3 + K4

 m eq  15

PE 

m = 240 kg

1 2 kx 2

 k eq  k  1500 N / m

Keq = 16  2  32  2  106  96  106 N/m

Natural frequency k eq

n 

meq

Ke m

n =



1500  10rad / sec 15

18. Ans: (b) Sol: In damped free vibrations the oscillatory

motion becomes non-oscillatory at critical

96  10 6  632.455 rad/sec 240

n = N=

 n  60  6040 rpm 2

21. Ans: (a) Sol:

 C2lI 

damping. Hence critical damping is the smallest



damping at which no oscillation occurs in free vibration

kl

Sol:  n = 50 rad/sec =

5 m

If mass increases by 4 times k 1 k 50     25 rad / sec 4m 2 m 2

ACE Engineering Academy

Io 2

19. Ans: (a)

 n1 

O

 x3  For slender rod, I o     3   

93   83  3   33  m 2 3 3





Where, = m/3l Considering the equilibrium at hinge ‘O’.

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: 326 :

Io  + c2l   2l + kl  l = 0  ml2  + 4l2c  + kl2 = 0 Iequivalent = ml2, Ceq = 4l2c, keq = kl2 22. Ans: (b) Sol: Damping ratio,  



c eq c  c c 2 k eq m eq

24.

Sol:

ME – GATE _ Postal Coaching Solutions Ans: (a) c eq



2  k 2  m 2

2c   4 km 2  mk

Sol:

400  12 4   0.316 2 2 (400  1  10  9.81  1)  5  10  12

25. Ans: (a) Sol:

KL 

Ca 

 a

I 

m m 

By moment equilibrium I  Ca 2  KL2  K   0 

mgsin = mg mg

I = m(2l)2 + ml2 = 5ml2





c 2    m  2   m     k 2   mg  0 4 2 cl  = 5ml 2     kl 2   0 4 2

m eq

2

k 2  mg  5m 2

400   3.162 rad / s 5  10 ACE Engineering Academy



 mL2    Ca 2   KL2  K    0 3

n 

The equation motion is

k eq



I

L



c   2 k 

n 

2 (k 2  mg)  5m 2

K 

23. Ans: (a)



  

4 2c 4 2c

mgcos



2 k eq m eq

 c 2   4

n 

K eq m eq



KL2  K  mL2 / 3

1500  42.26 rad / sec 0.833

26. Ans: (c) Sol: Refer to the above equilibrium equation

Ceq = Ca2 = 500  0.4 2  80

N  m  sec rad

 C = 80 Nms/rad

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: 327 : Note: For angular co-ordinate

Unit of Equivalent inertia =

Theory of Machines & Vibrations

28. Ans: (a)

Nm  kg  m 2 rad / s 2

Unit of equivalent damping coefficient =

Sol: Given length of cantilever beam,

l = 1000 mm = 1m, m = 20 kg

Nm rad / s

m

l = 1m

Unit of equivalent stiffness = N-m/rad

25 25

Cross section of beam = square 27. Ans: (a)

W = mg

2 Sol: x  2n x  n x  0

x t  0   X , x t  0  0

l

x(t) = Ae n t cosd t   After ‘n’ complete oscillation t = n d x(t) = Ae n nd cosd n d   d 

2 d

x(t) = Ae

2 d

x(t=nd) = Ae

  2 cos d n    d  

 n n

2 n 1  2

cos 

 2 n 

= Ae

1  2

cos

x (t=0) = X

I=

1 3 25  (25) 3 bd  = 3.25 10–8 m4 12 12

Esteel = 200  109 Pa Mass, M = 20kg 3EI 3

Critical damping coefficient, C C  2 Km  1250 Ns / m

29. Ans: (1.25) Sol: Given m = 1 kg , K = 100 N/m

C  25

N  sec m

Critical damping

X = A cos (–) = A cos   2  n 

X (t= nd) = Xe



Moment of inertia of the shaft,

Stiffness, K  n n

x

EI

C C  2 Km  20

1  2

Damping Ratio 

N  sec m

C 25   1.25 C C 20

30. Ans: (c) ACE Engineering Academy

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: 328 : 31.

33.

Ans: (d)

Sol: x = 10 cm at

  1; n

4   2 2

0.693



 x0 = 2  0.1  10 = 2 cm x0 = static deflection

= 2.19 N-sec/m x0

x    1      n

  

2

2

     2    n 

34. 2

As  > n

1  0.5   2  0.1 0.5 2 2

Ans: (b)

Sol: xstatic = 3mm ,  = 20 rad/sec

2

2

So, the phase is 180 .

 2.64 cm

x static

x

2

2    2   1       2     n    n   

32. Ans:(a) Sol: m x  Kx  F cos t

x=

m=? K = 3000 N/m, F = 100 N,

35.

  100 rad / sec

3 2   20  2   1       2  0.109  20    10    10   

= 1 mm opposite to F.

X = 50 mm = 0.05 m

Ans: (c)

Sol: At resonance, magnification factor =

F X K  m 2 K F   0.1 kg 2  X 2

ACE Engineering Academy

 0.109

4 2  0.693 2

c  2 k m  2  0.109  100  1

  0.5 , n

m

   ln 2  0.693 





x At resonance x  0  10 cm 2

x

Ans: (a)

x Sol:   ln 1  x2

  0.1

At

ME – GATE _ Postal Coaching Solutions

 20   

1 2

1 2

1  0.025 40

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: 329 : 36.

Ans: (c)

Theory of Machines & Vibrations

Sol: M = 100 kg, m = 20 kg, e = 0.5 mm

K = 85 kN/m, C = 0 or  = 0 Dynamic amplitude 20  5  10 4  20  me 2 X= = 2  k  M 2  8500  100  20 

Magnification factor =

2





-4

= 1.2710 m



Sol: Given

N = 3600 rpm  = 0.15

x(t) = Xsin(t - )

m=50kg k

y(t) = 0.2sin(200t)mm

n =

K = 20 rad /sec m

 =

2  N  377 rad/sec 60

 = 200 rad/sec, –X = 0.01 mm Y = 0.2 mm

TR=

X k  Y k  m2

  1  2  n

 0.01 k  2 0.2 k  50  200

For which, X  =4

F= 8, F

k  m   c 2

2

      2   n

  

2

= 0.0162

F

K  m   C 2 2

2

Given K = 6250 N/m, m = 10 kg, F = 10 N 2

 = 25 rad/sec, X= 40×10–3

8

80  5  4   20  4 2

2

Sol: Given systems represented by mx  cx  kx  F cos t

c = 20 ,

k = 80,

  

40. Ans: 10 N.sec/m

38. Ans: (b) Sol: m= 5kg,

  1   2  n 2

 k = 939.96 kN/m



m = 250 kg K = 100, 000 N/m

Sol:

x

0.1 1 0.1

39. Ans: (c)

37. Ans:



x static

F 8   0.1 k 80

x static 

 = 20 rad/sec



x

Magnification factor =

2

 0 .1

n 

K  25 rad / sec m

t = 25t   = 25 rad/sec ACE Engineering Academy

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: 330 :

ME – GATE _ Postal Coaching Solutions

Given F = 10, n  10

 = n or K  m 2n F F C C X 10 N  sec   10 3 m 40  10  25

k  150 N / m or

X 

  0.2 X

41.

Ans: (b)

in damping up to the frequency ratio of

10 / 150

1  0.1  2  0.2  0.12 2

 0.0669 ≃ 0.07

Sol: Transmissibility (T) reduces with increase

Beyond

 1   0.1 n 10

2.

2 , T increases with increase in

44.

Ans: 6767.7 N/m

Sol: Given f = 60 Hz, m = 1 kg

damping

  2f  120 rad / sec

42. Ans: (c).

Transmissibility ratio, TR = 0.05

Sol: Because f = 144 Hz execution frequency.

Damping is negligible, C = 0 , K =?

f Rn (Natural frequency) is 128.

We know TR 

 f 144    1.125  R n f R n 128

As TR is less than 1   / n  2

It is close to 1, which ever sample for which  close to 1 will have more response, so n sample R will show most perceptible to vibration 43.

TR is negative   0.05 

K K  m2

Solving we get K = 6767.7 N/m 45. Ans: 20 (range 19.9 to 20.1) Sol: k = 10kN/m ,

Ans: (b)

X

Sol: Given Problem of the type mx  cx  kx  F cos t

for which, X  or X 

K when C = 0 K  m2

F

k  m   c 2 2

2

F/ K    1     n 

ACE Engineering Academy

  

2

     2    n 

2

  

F0 = 100 N ,

 = 0.25

 F0 / k  2

2    2     1       2   n    n   

  1 at resonance n X

F0 100  = 20 mm 2 10  0.25  103 2k

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: 331 : 46. Ans: (b)

49.

Sol: To get the equivalent inertia of disc on ‘B’

Sol:

Theory of Machines & Vibrations Ans: (b)

e = 2mm = 210-3m,

at the speed of shaft ‘A’ then their kinematic

n = 10 rad/s,

energies will be same.

N = 300 rpm

Shift the disc to shaft “A” end

I2

I2

1 1  I2  2A   I 2  2B 2 2   I2  I 2   B  A

2

   I 2 n 2 

=

2N  10 rad / sec 60

e2 me2 e2 X   2 k  m2  k  2 n  2 2     m 2

2    10  3  e 2  10    n  10    = X 2   10  2      1    1     10     n  

= 2.25 103 m = 2.25 mm

47. Ans: (c) Sol:



ma

T



50. Ans: (a) Sol: Number of nodes observed at a frequency of

1800 rpm is 2 mg

Where, a = acceleration of train T cos  = mg

n=1

T sin = ma tan =

ma mg

n=3 n-mode number

a = g tan = 9.81 tan(9.81) = 1.69 m/s2 48.

Ans: (a)

ACE Engineering Academy

The whirling frequency of shaft, f 

 gEI  n2 2 WL4

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: 332 :  gEI For 1st mode frequency, f 1   2 WL4 2

fn = n f1

ME – GATE _ Postal Coaching Solutions

51. Ans: (b) Sol: Critical or whirling speed

As there are two nodes present in 3rd mode,

c = n =

f 3  3 f 1  1800 rpm 2

1800  200 rpm f 1  9 The first critical speed of the shaft = 200 rpm

K  m

g rad / sec 

If NC is the critical or whirling speed in rpm then 

2N C  60

g 

2N C 9.81m / s 2  60 1.8  10 3 m

 NC = 705.32 rpm  705 rpm

ACE Engineering Academy

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Engineering Mechanics Solutions for Vol – I _ Classroom Practice Questions Chapter- 1 Force and Moment Systems

02.

Ans: (b)

03.

Ans: (b)

10kN

Sol: 01.

Ans: (b)

45

20kN

Sol:

20 2 kN

F2 R1 =180

30kN R = 260

(180–) 

F1x

Fx = 20 2 cos45–20 = 0 Fy = 10+20 2 sin45–30

F1

R=

Assume F1 = 2F2 (F1>F2) F1x = 2F2 R=

260 = 2

F12  F22  4F22 cos 

180 = 2

2 2

2 2

2 2

A 3m

Fx

Fy

F

0

4F22  F22  2.F2 .F2 cos180   2 2

2 2

2 2

2 2

180 = 5F  4F cos  2 2

2 2

260  180  10F 2

Y

F12x  F22  2F1x F2 cos 

260 = 5F  4F cos  2

Sol:

2 2

180 = 5F  4F cos  ------ (2) 2

Ans: (b)

4F  F  4F cos  2 2

260 = 5F  4F cos  ------ (1) R1 =

04.

Fx2  Fy2 = 0

2

2 2

 F2 = 100N,

2602 = 5(100)2+4(100)2cos

x

6m

M 0F  180 N  m M FB  90 N  m M FA  0 M 0F  180  Fx  3  Fy  0

Fx = 60N ……. (1) M FB  Fx  3  Fy  6  90

  = 63.89

Where  angle between two forces.

ACE Engineering Academy

B

603–6Fy = -90

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: 336 :

 Fy =

ME – GATE_Postal Coaching Solutions

270 6

05.

16

0

0

 dw   wdx

Fy = 45N

 F=

w

16

F F 2 x

2 y

=

 1 1   x2  w =  90 x dx = 90  1  0   1  2 0

60  45 = 75 2

16

2

Ans: (a)

= 90

Sol: M 0  0

F5–p4=0

16

= 60 (16)3/2

0

w = 3840N

F  5 – 200  4 = 0 F=

 

2 3/ 2 x 3

R  d = dw  x

800 = 160N 5

16

3840d =

 90

x .dx.x

0

06. Sol:

15

= 90  x 1.5 dx

Ans: (c)

0

3m

3m

A

16

H

 x 2.5  3840d = 90    2.5  0

2m

V V

120 N–m H

(i) H4 = 120

d = 9.6 m

2m B

08.

Ans: (c)

Sol: Moment about ‘0’

H = 30N

M0 = 100sin 603

(ii) V6 = 120 = 300

V = 20N

3 = 150 3 2

= 259.8  260N. 07.

Ans: (a)

Sol:

09.

dw

Sol: 100N

360 N/m H dx 16m

Ans: (a) 25N

200N

x A

C

B 0.9m

ACE Engineering Academy

150N

1.2m

D 0.75m

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: 337 :

Engineering Mechanics

Fy = 0

R+100+150–25+200 = 0 (upward force

Chapter- 2

Positive downward force negative)

Equilibrium of Force System

R = –425N For equibrium MA = 0 (since R = negative, resultant is

01.

downward)

Sol:

Ans: (d) Y

Let R is acting at a distance of ‘d’

X 300

P

downward.

45o

425d –1500.9+252.1–2002.85=0

Wsin30

30o

30o

Wcos30 N

W

 d = 1.535m (from A)

Resolve the forces along the inclined surface Fx = 0

Pcos45 –Wsin30 = 0 300 sin 30  P = 212.13 N cos 45

P=

02.

Ans: (a)

03.

Ans: (c)

Sol:

C B 75

60

45o o

200 A

ACE Engineering Academy

o

45o

P D

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: 338 :

ME – GATE_Postal Coaching Solutions

04. Ans : (d) FAB 120

60

T 

60

B FBC

Sol:

B Rx

45 75 200

mg

tan =

Fig: Free body diagram at ‘B’

125   = 24.450 275

Tsin = mg. Tsin24.45 = (359.81)

FCD

75 60

T  829.5 N

FBC

105

Rx = Tcos24.45 = 755.4 N

75

Ry = 0

45

P Fig: Free body diagram at ‘C’

05.

Ans: (c)

Sol:

T

2T

T

m

For Equilibrium of Point ‘B’ FAB FBC 200   sin(60  75) sin(60  45) sin(120)

T+2T+T = mg

FBC = 223.07 N

m = 4T/g

4T = mg

From Sine rule at “C”. FCD FBC P   sin(75  45) sin(60  75) sin 105

06.

Ans: (b)

Sol:

a

P = 304.71 N

T

T

223.07  sin 105 P= sin 135

W L

ACE Engineering Academy

A

B N

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: 339 :

For body, Fy = 0

Engineering Mechanics

07.

Ans: (a)

Sol: Free body diagram

N–W+T=0 N=W–T T

W–T

B

T

15 a

75

R

A RA

L

W= 100 N

Fy = 0 for entire system Apply sine rule

W T R   sin 75 sin 90 sin 165

RA + T – (W – T) = 0 RA = W – 2T

------- (1)

100 R  R = 26.79  sin 75 sin 195

For equilibrium MA = 0 T× L = (W –T) a TL = Wa – Ta TL +Ta = Wa

08.

Ans: (c)

Sol: P = 600 N

T (L+ a) = Wa T=

Wa La

RC = R

C

A

T substitute in equation (1)

=

WL  Wa  2Wa La

=

WL  Wa La

RA =

W (L  a ) La

ACE Engineering Academy

D

3m

 Wa  RA = W  2  La W (L  a )  2Wa = La

B

P = 600 N 5m

RD = R

Fy = 0 600 – RC + RD – 600 = 0

 RC = RD = R M = 0 600 ×5 = R × 3

 R = 1000 N = RC = RD

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: 340 : 09.

ME – GATE_Postal Coaching Solutions

Now, Fx = 0,

Ans: (a)

Sol:

N

RAH –Tcos = 0

P

P

RAH = 125 N  Fy = 0;

45 45 45

W

45

S

RAV – 200 –100 +Tsin  = 0

E

RVA = 50 N 11. Ans: 400 N

F

Sol:

A

Fy = 0

NA

P sin45 +Psin45 –F = 0

2.5m

F = 2Psin45

3m

1  2  =  2P   2 2 

2.5m W=600N

F= P 2

2m 10. Sol:

B P

2m

Ans: (a)

NB

FBD

Fy = 0 T A

RAH

NB – W = 0 

B C

RAV

200 N

NB = 600 N MA = 0

100 N

P3+W2 – NB 4= 0

MA = 0

P

4 N B  2W 3

8 4

P

4  600  2  600  400 N 3

Tan =

 = 63.43 Tsin ×4 (↺)–200 ×2 (↻) –100 ×6 (↻) = 0  T = 279.5 N ACE Engineering Academy

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: 341 :

Engineering Mechanics

Free body diagram for block (1)

Chapter- 3

N2

Friction 01.

W1

F2 P

Ans: (c) F1

Sol: The FBD of the above block shown T

N1

From FBD of block (2) W

Fx = 0

100 N

F

F2 = Tcos N

FBD of the block

F2 =

Y = 0  N+T–W = 0

Fy = 0

N = W–T = 981 – T

N2 + Tsin – W2 = 0

F = N = 0.2 (981 – T)

N2 = W2 – Tsin

X = 0  100 – F = 0.

N2 = 50 – 0.6 T

F = 100 = 0.2 (981 – T)

But F2 = N 2

 T = 481 N 02.

4 T = 0.8T ------ (1) 5

 F2 = 0.3(50 – 0.6T) F2 = 15–0.18 T ------ (2)

Ans: (c)

Sol: Given Tan =

From (1) & (2)

3 4

0.8T = 15 – 0.18 T

4

sin  = 3/5

 0.98T = 15



cos = 4/5

5

Free body diagram for block (2) W2

T

3

 T= 15.31 N  N2 = 50 – 0.6T = 50 – 0.6 (15.31) = 40.81 N F2 = N2 = 0.340.81 F2 = 12.24 N



From FBD of block (1) F2 N2

ACE Engineering Academy

Fy = 0 N1 – N2 – W1 = 0 N1 = N2 + W1 = 40.81 + 200 = 240.81 N

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: 342 :

F1 = N1  F1 = 0.3 240.81

WS =150 9.81=1471.5 N Fy = 0

F1 = 72.24 N Fx = 0

N1 = N2 + WS = 981+1471.5= 2452.5N

P – F1 – F2 = 0

F1 =  N1 = 0.4 2452.5= 981N

P = F1 +F2

Fx = 0  F – F1 – F2 = 0

= 72.24 + 12.24

F = F1 + F2 = 981 + 392.4

P = 84.48 N 03.

ME – GATE_Postal Coaching Solutions

= 1373.4 N =1.3734 kN 04.

Ans: (d)

Ans: (b)

Sol: Free Body Diagram

Sol:

10 cm

P

20 cm FB

Q

NB P

R S

F

35 cm

10 cm

FA

NA W = 100 N

FBD of block R : WR = 1009.81 = 981N T F2

FA = NA =

1 NA 3

FB = NB =

1 NB 3

MB = 0 –10030(↺)+ (NA20)(↻)+(Fa  12)(↻) = 0

N2

Fy = 0

– 3000 + NA  20 +

N2 = WR =981N F2 =  N2 = 0.4 981= 392.4N

1 NA  12 = 0 3

 NA = 125 N Fy = 0 NA – NB – 100 = 0

FBD of block S:

 NB = 25 N

N2 F2 F

WS F1 N1 ACE Engineering Academy

Fx = 0 1 N A  N B  3 1 = (125  25)  50 N 3

P = FA +FB =

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: 343 : 05.

Engineering Mechanics

Ans: (d)

Sol:

B

m2 = 1 kg

H

10 m

m1 = 1 kg  A

W2 = 9.81 N

N2

6m

4m W F2 8m

F2

FBD for block (1)

FBD of block 2

N2

N2

F2 P

N2 F2

F1

W1 = 9.81 N N1

F FBD of block 1

F1

Given W = 280 N ,

W1 = 400 N

Now, MB = 0

N1

From FBD of book 2, FY = 0

–W  4 (↺) + N2  8(↻) – F2  6 (↺) = 0

 N2 = W2 = 9.81 N

–280 4 +N2 8 – N2  6 = 0  N2 = 200 N

F2 = N2 = 0.3  9.81  2.943N From FBD of book 1, FY = 0

But, F2 = N2 = 0.4 200 = 80 N

 N1 = N2+W1

From FBD of block (1) Fy = 0

= 9.81 + 9.81 = 19.62 N

06.

W1

F1 = N1 = 0.3  19.62  5.886 N

N1 – N2 – W1 = 0

F = F1 + F2 = 8.83 N

N1 = N2 + W1

Ans: (d)

N1 = 600 N

Sol: Tan =

3 3  sin = 4 5 cos =

FBD for bar AB (2) ACE Engineering Academy

4 5

= 200 +400 But, F1 = N1 = 0.4 600 5

 4

3

F1 = 240 N Fx =0 P = F1 +F2 = 240 + 80 P = 320 N

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: 344 : 07.

Ans: (a)

ME – GATE_Postal Coaching Solutions

T = WAsin – FA

Sol: Given, WA = 200 N , A = 0.2

T = 200 sin – 40cos

WB = 300 N , B = 0.5

But from equation (1) T = 150 cos – 300 sin

FBD for block ‘B’.

150cos – 300sin = 200sin – 40cos

X

Y

190 cos = 500 sin

FB

B

tan = 

T

Fy = 0

NB

WB

NB = WBcos

  = 20.8o 08.

NB = 300 cos But, FB = NB = 0.5  300 cos

190 500

Ans: (d)

Sol: FBD for the block

= 150 cos

X

Y F

Fx = 0

P

T + WBsin – FB = 0 T = FB –WB sin

N W = 500

T = 150 cos –300 sin ------ (1)

45o

FBD for block ‘A’ X Y

Fy = 0 N – Wsin45 –Psin45 = 0

T FA

N=

A  WA

NA

Fy = 0

500 2



P 2

 500 P   But, F = N = 0.25  2  2 Fx = 0

NA –WAcos = 0

Pcos45 + F – Wsin45 = 0

NA = 200 cos FA = NA = 0.2  200 cos But, FA = 40 cos Fx = 0

 500 P  1   500  P cos 45  0.25  =0 2 2  2  P = 300 N

T + FA –WAsin = 0 ACE Engineering Academy

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: 345 : 09.

Ans: (a)

10.

Sol: FBD of block

Engineering Mechanics Ans: 64 N-m

Sol: FBD of shoe bar : W

800 1000N A

480

F2 C

N2

B

HB

100 FC

r

VB

VC

FBD of Drum Brake :

F1

VC

N1

FC

F1 = N1 200

F2 = N2

MB = 0

Fx = 0

VC  480 + FC  100–1000  800 = 0

N2 –F1 = 0  N2 = F1 (∵ F1 = N1)

FC = VC = 0.2 VC 480VC + 0.2VC 100 = 800000

N2 =  N1

500VC = 800000

Fy = 0

VC = 1600 N

N1 + F2 – W = 0

FC = 0.2 VC = 0.2×1600 = 320 N

N1 + N2 –W = 0 N1 +2N1 –W = 0

M = 0.2FC = 0.2×320 = 64 N-m (∵ N2 = N1)

2

N1 (1+ ) = W N1 =

W 1  2

N2 =

W 1  2

11. Sol:

Ans: (a)

 = 2 cos =

6 12

  = 60  = 360 –2

Couple = (F1 + F2)  r

 = 240 =

= r (N1 + N2)

r  W 1     1 2

2 + 2 = 180 (∵  = f)

2 = 180 – 120

 = 30 = ACE Engineering Academy

4 3

 6

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: 346 :

ME – GATE_Postal Coaching Solutions

FBD

1   6

T1 = 1000  e 

T1 = 1181.36 N  6



T2  e  T1

6



12 cm T2

1 4  3

T2 = 1181.36  e 

T1

T1

Pmax  e  T2

T2

12 cm 6

= 4481.65 N

6

  

Pmax = 4481.68  e

1    6

Pmax = 5300 N P

W = 1000

For Pmin calculation,

12.

Ans: (b)

tan =

3 4

 cos  =

4 5

sin =

3 5

 = 0.2,

Sol: Given

W > T1 W  e  T1 T1 =

1000 e



 1  6 

= 846.48 N

X

y

T1

T1  e  T2

T2 =

848.48 e

1 4   3

= 223.12 N

W2sin

F2



N2 W2 cos

W2 = W

T2  e  Pmin 223.12  Pmin = 1 

X T2

y

Fig: FBD (1)

N2

F2 F1



e 6 Pmin = 188.86 N  189 N For Pmax calculation T1  e  W ACE Engineering Academy

 W1sin

N1 W1 cos

W1 = 1000 Fig: FBD (2)

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: 347 :

From FBD (1)

13.

Fy = 0

Sol:

Engineering Mechanics Ans: (d) R

N2 –W2 cos = 0 N2 = W2 cos = W0.8

R R

R

N2 = 0.8 W

F2 = N2 = 0.2  0.8 W

2000 N

F2 = 0.16 W

Fx = 0

At equilibrium

T1 – W2sin – F2 = 0

2R = 2000

T1 = F2 + W2sin = 0.16 W +0.6W

R=

T1 = 0.76 W

2000  10,000 N 2  0.1

Taking moment about pin From FBD (2)

10,000  150  F  300

Fy = 0

 F = 5000 N

N2 +W1 cos = N1 N1 = N2 +W1 cos 4 N1 = 0.8W + 1000  5

14.

Ans: (b) 1  9.81  9.81N

Sol:

N1 = 0.8 W + 800 F1 = N1 = 0.2 ( 0.8 W+800)

0.8N

= 0.16 W +160

F

T2  e  T1

N

T2 = T1 e = 0.76 W e0.2 T2 = 1.42 W

Y = 0  N = 9.81 N

Fx = 0 T2 + F1 + F2 = W1 sin 3 1.42W+0.16W+160+0.16W = 1000  5

Fs = N = 0.1  9.81 = 0.98 N The External force applied = 0.8 N < Fs

 Frictional force = External applied force = 0.8 N

1.74 W = 440

 W = 252.87 N ACE Engineering Academy

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: 348 : 15.

ME – GATE_Postal Coaching Solutions

Ans: (b)

Sol:

Chapter- 4

T2

W

Analysis of Trusses

T1 T1

P F N Fig: FBD (1)

01. T2

Ans: (b)

Sol: At joint

FCD

200 Fig: FBD (2) Fig: FBD (3) FAD

From FBD (3)

60o

FBD

Fy = 0 T2 – 200 = 0

 T2 = 200

Fy = 0

From FBD (2)

FCD sin60 = 1000

T1  e  T2

FCD = 

T1 = T2 e = 200  e

0.3

T1 = 320.39 N

1000

1000 sin 60

FCD = 1154 N

 2

02.

Ans: (a)

Sol: At joint ‘Q’

From FBD (1)

T = FRQ

Fy = 0 N–W=0

Fy = 0

N = 1000 N

F – Tsin45 = 0

F = N

F = Tsin45

= 0.3  1000

FPQ

Fx = 0

 Fx = 0, T1 + F – P = 0

FPQ +Tcos45 = 0

 P = 620.39  P = 620.4 N

ACE Engineering Academy

Q

F

T= F 2

F = 300 N 320.39 + 300 = P

45o

FPQ =

T 2

FPQ = –F ( negative indicate compression)

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: 349 : 03.

Ans: (c)

04.

Sol:

Engineering Mechanics Ans: (d)

Sol: A

B

a A

P

B

4m

a

C

4m

a 2t HE

E

Reactions at C & D are 2t & 2t respectively A

E 2t

F

RE

RF

FAB

Fx = 0 FEB

C

D

2P

4t

2t

C

D

E 4m

FED

4t

 HE –P – 2P = 0 HE = 3P

 Fy = 0 RE +RF = 0

MF = 0 ME = 0

P × 2a + 2P × a + RE × a = 0

FAB  4 = 2t 4

RE = –4 P (downward)

FAB = 2t (compression)

RF = 4P (upward)

(Positive indicate taken direction is true, i.e

P

AB is in compression) FAC

Note: “Always all top members are in

compression”

FCD

Fx = 0

FDF

P – FCD = 0 P = FCD (Positive indicate CD in tension)

ACE Engineering Academy

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: 350 : 05.

Ans: (b)

Sol: At point ‘B’ B 60

At joint ‘P’ 10 kN

60

FPR = =

From Lami’s Theorem FBC F 10  AB  sin 240 sin 60 sin 60

Rp RP

sin 45 50 1/ 2

Fx = 0 FPT = FPR cos45

FBC = –10 kN (Tension)

FPT = 50 2 

FRS

R

Sectioned S

R

FSU h

h

1 2

FPT = 50 kN (Tension)

Ans: (d)

P

FPT

FPR = 50 2 (compression)

10  sin 240 sin 60

Sol:

45

Rp = FPRsin45o

FAB

FBC =

FPR

Fy = 0

FBC

06.

ME – GATE_Postal Coaching Solutions

T

U

h

60 kN

h

h

Q

RP

30 kN

RP

T

U

h

60

30 kN

RQ

Mu = 0 Taking moments about point ‘P’

FRS × h (↺) + 60 × h (↺) – RP × 2h(↻) = 0

RQ ×3h – 30 ×2h –60×h= 0

FRS × h + 60 h –100 h = 0

RQ ×3h = 120 h

FRS h = 40 h

RQ = 40 kN

FRS = 40 kN (Compression)

∴ RP +RQ = 60 + 30

Fy = 0

RP = 90 – 40 RP = 50 kN

ACE Engineering Academy

FSU + RP – 60 = 0 FSU + 50 –60 –30 = 0 FSU = 40 kN (Tension)

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: 351 : 07.

Engineering Mechanics

Ans: (b)

Sol:

TAE

F

PQsin45 450

PQ

A

PRsin30

P

450 TAC RA

300

PR PRcos30

PQcos45

F

y

Force in member PQ considering joint P

0

 TAE sin 45  R A 

PQ cos45 = PR cos30 PQ = 1.224 PR

 TA 

PQ sin45 + PR sin30 = F 1.224PR  0.707  0.5PR  F

F

PL 2

PL 2

 0  T AC  T AE cos 45 

x

PR = 0.732 F Now, considering joint R

PL 2

FBD at Point C:

PRcos30 PR

TEC R QR

PRsin30

C

F

y

TAC  TCD 

= 0.63F (Tensile) Ans: (a)

Sol:

 Fy  0  R A  R B  P  L 3L  M B  0  R A  3L  PL  2

 RA 

PL PL , RB  2 2

FBD at Point A: ACE Engineering Academy

TCD

TAC

 TEC = 0

QR = PR cos30 = 0.732F  cos 30

08.

0

09.

PL 2

Ans : 20 kN

Sol:

1m B

A

B

0.5m 10kN C

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: 352 :

ME – GATE_Postal Coaching Solutions

Adopting method of sections –section x-x

FAB

adopted and RHS taken

26.560 10kN 63.44

0

 2.0  0   53.13  1.5 

  tan 1 

FBC

Fy = 0 (W.r.t.RHS of the section x-x) tan  

0.5  0.5  0    tan 1    26.56 1.0  1 

V1 + F2 –V2–Fy = 0

 Fsin 53.13 = 30+3–24

From the Lami’s triangle

F = 11.25 kN (Tension)

FBC FAB 10   0 0 sin 26.56 sin 90 sin 63.44 0

10.

Force in member QS = 11.25 kN (Tension)

FAB 

10  sin 63.44  20 kN sin 26.56

FBC 

10  sin 90  22.36kN sin 26.56

11.

Ans: (c)

Sol: W

E

F

C

P

Ans: (a)

Sol:

A

F2 =3KN 3m

F1 =9KN 3m x Fx

h W

Q

P

h

R

RA =



Fy

V2

B

h

W

2W 3

RB =

4W 3

FEF

W

F

h

2m FCF FCD

T

S 1.5m

3m x

Fy = 0

1.5m V1

RA = 2W/3

W

MB = 0 W ×h (↻) –W×h (↺) –W(2h)(↺) +RA×3h(↻) = 0

V1 +V2 – 9+3 =0

MR = 0

Wh –Wh –2Wh +3hRA = 0

 V1 1.5 +3 3 –9  6 = 0

3hRA = 2Wh

 V1 = 30 kN ()

RA =

V2 = –30 +9 – 3 = – 24 kN () ACE Engineering Academy

2W 3

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: 353 :

At joint ‘B’

∴ RA +RB = 2W

RB = 2W 

2W 3

FBD

Fy = 0 ( at the joint C)

FBD = 15 kN

FCF sin45 – W + RA = 0

Fy = 0

FCF ×

1 2

 FCF =

FAB = 0

2W =0 3 13.

W 3

=

FY = 0  FAC sin45 = FBC sin60

W 2 3

FAC =

5 kN C

A RVA B

1 = 1.224  cos45  FBC + FBCcos60 1 = 0.865 FBC + 0.5FBC,

3m

3m

D

3m

E

FBC =

5 kN

RVB = 0

1 = 0.732 kN 1.365

Vertical force at ‘B’ = FBCsin60 = 0.732sin60 = 0.634 kN

MA = 0 5×3 (↻) +5 ×6 ( ↻) – RHB × 3 = 0 15 + 30 = RH × 3 RHB =

FBC sin 60 = 1.224 FBC sin 45

= 1.224 0.732= 0.895 kN

Sol:

RHB

Ans: (a)

Sol: FX = 0  F = FAC cos45 + FBC cos60

Ans: (c) RHA

FAB B

Fx = 0

FCF sin45 – W +

12.

RHB

4W 3

RB =

Engineering Mechanics

45 3

14.

Ans: (b)

Sol:

FAC F   FAC  0.8965F sin 120 sin 105 FAC = Maximum force

RHB = 15 kN

0.8965F 0.8965F Stress  100 Area 100

FX = 0 ∴ RHA + RHB = 0

F = 11.15 kN

RHA = –RHB RHA = –15 kN (Negative indicate RHA is left side) ACE Engineering Academy

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: 354 :

ME – GATE_Postal Coaching Solutions

03. Ans: (d)

Chapter- 5

Sol: x  2t 3  t 2  2t

Kinematics of Particle Rectilinear and Curvilinear Motion

01.

V

dx  6t 2  2t  2 dt

a

dv  12t  2 dt

At t = 0  V = 2 and a = 2

Ans: (b)

ct 2 2

Sol: Given x 

04.

Ans: (a)

Sol: V = kx3 – 4x2 + 6x

c = 8 m/s3

Vat x = 2 if k = 1= 23– 4(2)2 + 6(2) = 4

dx  4t 2 x  dt

a=

 dx   4t

a = 3x2(V) – 8x(V) + 6(V)

2

dt

dV dx dx dx  k.3x 2  8x 6 dt dt dt dt

= 3(2)24 –(8×2×4)+6(4)

4t 3 x=  C1 3

= 8 m/s2

At, t = 0, x= 0 05.

∴ C1 = 0

Ans: (d)

Sol: Given,

a= 6 V

3

x=

4t 3

dV 6 V dt

4(3) 3 ∴ at t = 3 sec, x =  x = 36 m 3 02.

dV  6 dt V 

2 V  6 t  C1

Ans: (a)

Given, at t = 2 sec, V = 36

Sol: V = 10 m/s

 2 36 = 6(2) + C1

S = 25 m u = 0,



C1 = 0

a=?

2 V = 6t

v2 – u2 = 2as 102 – 02 = 2(a) 25

 a = 2 m/sec2

ACE Engineering Academy

V = 9t2 But V =

ds  9t 2 dt

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: 355 :

 ds   9t

2

2 3/ 2 s  4t  C 2 3

dt

S = 3t3 + C2

At t = 1, S = 4

At, t = 2sec, S = 30 m



 30 = 3(2)3 + C2  C2 = 6 ∴ S = 3t + 6

S = 3(3)3 + 6

2 4  s 3 / 2  4t  3 3

S = 87 m

At t = 2 sec

Ans: (a)

2 3/ 2 4 s  4(2)  3 3

Sol: Given A = –8S–2

dV d 2s  2 = –8s–2 = a dt dt

 s = 5.808 m

We know that,  V dv   a ds V2    8s 2 ds 2 V 8   C1 2 S

V2 8 ∴  2 S



4t 3  2t  C1 3

dx 4t 3   2t  C1 dt 3

4 s



dv  4t 2  2 dt

v=

C1 = 0

ds 4  dt s

Ans: (c)

dv = (4t2 – 2) dt

22 8   C1 2 4



8 8 = = –0.237 m/sec2 2 2 s 5.808

Sol: Given, a = 4t2 – 2

Given, at S = 4m , V = 2 m/sec

V=

a=

07.

2



16 4 4= 3 3

2 ∴ s 3 / 2  4t  C 2 3

At t = 3 sec



2 3/ 2 (4)  4(1)  C 2 3

C2 =

3

06.

Engineering Mechanics

 4t 3    dt dx   2 t  C 1    3 

x=

4t 4 t2  2.  C1t  C2 2 3 4

s ds   4 dt

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: 356 :

t4 x=  t 2  C1 t  C 2 3

4t2 – 40t – 384 = 0

Given condition,

∴ t = 16 sec

t = 16 sec or t = –6 sec

At t = 0, x = –2 m

 –2 = C2

09.

At t= 2, x = –20 m

Ans: (b)

Sol: Take , y = x2 – 4x + 100

24  –20 =  2 2  4(2)  (2) 3

Initial velocity, V0 = 4ˆi  16ˆj

 29  C1 = 3

Vy , ay at x = 16 m

If Vx is constant

Vx = V1x =

t4 29 ∴x=  t2  t  2 3 3

Vy =

∴ at t = 4 sec

x=

dx 4 dt

dy dx dx  2x 4 dt dt dt

(Vy) = 2x (4) – 4(4)

44 29  4 2  (4)  2 3 3

Vy = 8x – 16 (Vy)at x = 16 = 8 (16) –16 = 112 m/sec

= 28.67 m 08.

ME – GATE_Postal Coaching Solutions

ay =

Ans: (b)

dV d  (2xVx  4Vx ) dt dt (∵ Vx = constant)

Sol: uA = 20 m/sec aA = 5 m/sec2 Pt “A”

uB = 60 m/sec aB = –3 m/sec2

= 2Vx ay = 2Vx2

A&B

Pt “B”

dx = 2Vx. Vx dt

(ay) x = 16 = 2×42 = 32 m/sec2

SB SA

Let SA be the distance traveled by “A” Let SB be the distance traveled by “B” SA= SB +384

10.

Ans: (c)

Sol:

1

u1 = 0 x1

h = 36

1 1 u A t  a A t 2  u B t  a B t 2  384 2 2 1 1 20 t  5t 2  60t  3t 2  384 2 2 ACE Engineering Academy

x2 2

u2 = 18 m/sec

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: 357 :

Let at distance of “x1’ ball (1) crossed ball (2)

Engineering Mechanics

Velocity loss = 20% of V =

1 1 x1 = 0(t) + gt 2 (∵s = ut + at 2 ) 2 2

= 9.81 m/sec

1 2 gt -------- (1) 2

x1 =

∴ Initial velocity for further movement in

glass = 49.05 – 9.81

1 x2 = 18( t )  gt 2 2

= 39.24 m/sec Distance traveled for 1 sec of time is given

(∵a = –g moving upward)

by

x1 + x2 = 36



1 S = ut  at 2 2

1 2 1 gt  18t  gt 2  36 2 2

1 S = 39.24(1)  (9.81)(1) 2 2

 18 t = 36  t = 2 sec

S = 44.145 m

1 ∴ x1 = (9.81).2 2 2

12.

= 19.62 m (from the top) x2 = 36 – 19.62 = 16.38 m (from the bottom) 11.

49.05  20 100

∴ x1 + x2 = 36

Sol: Vy

V0 = 100 m/sec 30o

Ans: (b)

Sol:

Ans: (a)

Vx u=0

g

t = 5sec V = u +at

60 m x=?

ax = –4 m/sec2 ,

ay = –20 m/sec2

Vx = V0 cos30 = 100 

S

Vy = V0 sin30 = 100 

V = u + at V = 0 + 9.81 (5) V = velocity with which stone strike the

ACE Engineering Academy

1 = 50 m/sec 2

1 y = Voy t  a y t 2 2

V = 49.05 m/sec glass

3 = 86.6 m/sec 2

1  60  50t  (20) t 2 2

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: 358 :

10t2 – 50t – 60 = 0 t = 6 or –1 sec

1 1  1    9.81  8.0 = 20 sin    cos  2  cos  

∴ t = 6 sec

8 = 20 tan – 4.9 sec2

1 x = V0 t  a x t 2 2

8 = 20tan – 4.9 (1+ tan2 )

x = (86.6 × 6) +

2

4.9 tan2  –20 tan +12.9 = 0 1 (4)6 2 2

tan1 = 3.28, tan2 = 0.803

1 = 73.04 , 2 = 38.76

x = 447.6 m ≃ 448 m 14. 13.

ME – GATE_Postal Coaching Solutions

Ans: (a)

Ans: (b)

V02 sin 2 Sol: Range = g

Sol: Given, V = 20 m/sec

x = 20 m, y = 8.0 m

Range is maximum when sin2 = 1 2 = 90

y

Vy

 = 45o

V

15.

Sol: Range = maximum height

Vx x

Vx = Vcos ,

Vy = Vsin

1 x = Vx t  at 2 ( ∵ a = 0 along x direction ) 2 x = Vcos t

V02 sin 2 V02 sin 2   g 2g

sin2 =

sin 2  2

 2sin cos =

sin 2  2

 tan = 4

20 = 20 cost t=

Ans: (d)

1 ------- (1) cos 

  = tan–1(4) = 76

1 y = Vy t  gt 2 2 1 8.0 = V sin t  gt 2 2

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: 359 : 02.

Chapter- 6

Engineering Mechanics Ans: (a)

Sol:

ay = 4 m/sec2

ar

Kinematics of Rigid Bodies Fixed Axis Rotation and General Plane Motion

 Vx = 2 m/sec

01.

Ans: (a)

r =10m

Sol:

V2y

V1y

V0

tan =

aN=ay



3

4



V2x

aN

V

3 4

 = Tan-1 3/4 = 36.60

V1x

ay = aT cos – aN sin

V1x = 100–t3/2

Note: Velocity will always act in the

V2y = 0 100+10t–2t2=0

tangential direction

(t–10)(t+5) = 0

Vx = Vsin

t = 10sec 3/2

V2x at t = 10  V2x = 100–10

= 68.37m/sec Radius of curvature r =

V2 aN

 dVy   Where aN = ay =  dt  at t 10 sec  = (10–4t)t=10 aN = –30m/sec2 r= =

ACE Engineering Academy

2 2x

V aN

68.37 2 = 155.8 m 30

V=

2 sin 36.6

V = 3.33m/sec  aN =

V 2 3.332  r 10

aN = 1.111 m/sec2 ay = aT cos–aN sin 4 = aT cos36.6 – 1.111sin36.6  aT = 5.83 m/sec2 aT = r =

aT 5.83 = = 0.583 rad/sec2 r 10

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: 360 : 03.

Ans: (c)

ME – GATE_Postal Coaching Solutions

05. Ans: (a) Sol:

Sol: Given  = 4 t

V2y

 = 2 radians at t = 1sec

V

 = ?  = ? at t = 3sec

V1y

d   d   dt = dt



V

 g

600

 = 4 t dt =

V1x

8 3/ 2 t  c …(1) 3

Given , v = 100m/sec

From given condition, at t = 1,  = 2rad 2 8 3/ 2 (1)  2 = 1  c1  c1  3 3 8 2   = t 3/ 2  3 3 At t = 3sec ,  =

8 3/ 2 2 (3)  3 3

04.

v1y = vsin60 = 100

 = 3rad/sec , 2

vat t=1 =

a = 30cm/s2 2

aN = r = 2(3) = 18cm/sec since total acceleration a =

a T2  a 2N

v 22 x  v 22 y 50 2  76.8 2

= 91.6m/sec.  vy   76.8     = tan-1  v  = Tan-1  50   x  = 56.9

 a2 = a T2  a 2N

aN = gcos = 9.81cos56.9

30 2  a T2  18 2

= 5.35m/sec2

2

aT = 24cm/sec aT = r = 24

24 = 12rad/sec2 2

ACE Engineering Academy

(use V = u+at)

v2x = v1x = 50 m/sec

=

2

3 2

v2y = 76.8 m/sec

Ans: (b)

=

v1x = 50 m/sec

= 86.6 – 9.8(1)

2  1.15rad / sec 2 3

Sol: r = 2cm,

= 1001/2

v2y = v1y –gt

d d(4 t ) 2   dt dt t

t = 3 =

v1x = vcos600

v1y = 86.6 m/sec

t = 3 = 13.18rad =

V2x

aN

V 2 91.6 2 r=  = 1568.62 m aN 5.35

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: 361 : 06.

Ans: (d)

42  2t1  22

V2y

Sol:

Engineering Mechanics

 2t1  2 2 t1 =  2 V2x

V=50 m/sec V1y

09.

aN = g

Sol: Given retardation

 = –3t2

300

d  3t 2 dt

V1x

v1x =vcos30 = 43.3m/sec

 d =   3t

aN = g = a r= 07.

V12x 43.32  =191.13m aN 9.81

2

dt

 = –t3+c1 From given condition at t = 0,  = 27 rad/sec

Ans: (c)

27 = –03+c1

Sol: Angular distance

 c1 = 27

 = 2t3–3t2

  = –t3 + 27

d Angular velocity () =  6t 2  6t dt Angular acceleration () =

08.

Ans: (c)

Wheel stops at  = 0,  0 = –t3+27 t = 3sec

d  12 t  6 dt

at t = 1 = 12(1) – 6 = 6 rad/sec2

10.

Ans: (d)

Sol: angular speed,  = 5 rev/sec = 52 rad/sec  = 10 rad/sec

Sol: Given angular acceleration,  = rad/sec2

Angular displacement in time t1 and t2 =  rad = 2–1

Ans: (c)

Radius, r = 0.1m If  is constant, d = 0   = 0  aT = 0 (since aT = r)

t2 = 2 rad/sec t1 = ?

Since aT = 0

 2t1  02  21

a=

 2t 2  02  2 2

r = r2 v2 = a = aN = r r 2 = 0.1 10 = 102 m/sec2 2

 2t 2   2t1  2 2  1  ACE Engineering Academy

a 2N  a T2

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: 362 :

ME – GATE_Postal Coaching Solutions

Ans: a = 40m/s2

11. Sol:

 = 73.84

P

VB = rAB = 253 = 75 cm/sec

2m aN Q

VB = rABAB = r0BBC

aT

a

From BC,

=12 rad/s2

r r BC  0 B  0C sin  sin  sin 

=4 rad/s2

r0B = 22.9cm

Tangential acceleration

VB = r0B. BC

aT = r  = 2 12 = 24m/s2

BC =

Normal acceleration, aN = r 2 = 2  42 = 32 m/s2 The resultant acceleration a  a T2  a 2N  24 2  32 2  40m / s 2

13.

VB 75  = 3.274 rad/sec r0 B 22.9

Ans: (b)

Sol: Vb = ?

12.

Ans: (a)

B

Ic

Sol:

ro1B

180–(–)–(+)= r0C r0B

VB

 –



BC

20cm

A

20cm

 20cm 15cm

tan =

C 3m



D

22.5cm

A

VA = 12 m/sec

VA = ro1A   12 = ro1A 6 ro1A = 2m

20  = 53.13 15

10 tan =   = 26.56 20 tan =

1m

VC 10cm



B

ro1A

 = +

 = 6 rad/sec

30  53.13 22.5

4 = 2+ ro1B ro1B = 2m  VB = ro1B  = 26 VB = 12 m/sec

 = 180  (53.13–26.56) – (53.13+26.56) ACE Engineering Academy

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: 363 : 14.

Engineering Mechanics

Ans: (a)

Sol: I

A

45

I

1m/s

VQ=1m/sec

1m

Q

600 B

O

70 VP

Va = 1 m/s

20

P

Va = along vertical

‘I’ is the instantaneous centre.

Vb = along horizontal

From sine rule PQ IQ IP   sin 45 sin 70 sin 65

So instantaneous center of Va and Vb will be

IP sin 65  IQ sin 70

perpendicular to A and B respectively IA  OB  l  cos   1  cos 60 0  IB  OA  l  sin   1  sin 60 0 

VQ  I Q    1

1 m 2

 

3 m 2



VQ IQ

VP  IP   

Va    IA

15.

45 65 20

Va  2 rad / sec IA

= 0.9645 16.

Ans: (d)

Sol: Refer the figure shown below, by knowing

the velocity directions instantaneous centre can be located as shown. By knowing

sin 65 IP  VQ  1 IQ sin 70

Ans: (a)

Sol: Instantaneous centre will have zero velocity

because the instantaneous centre is the point of contact between the object and the floor.

velocity (magnitude) of Q we can get the angular velocity of the link, from this we can get the velocity of ‘P using sine rule.

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: 364 : 02.

ME – GATE_Postal Coaching Solutions Ans: (b)

Sol: u = 0, v = 1.828 m/sec

Chapter- 7 Kinetics of Particle and Rigid Bodies

S = 1.825m, v2 – u2 = 2as

01.

1.8282 – 0 = 2a  1.828

Ans: (a)

Sol:

a=

1.828 2

a = 0.914 m/sec2 T

Direction of motion

WQ    g a  

T Q

W

W 



W  W a  g 

T Directon of Inertial force

Direction of motion

W  a  g   

W+Q

W W

For the left cord,

For equilibrium, Fy = 0

Fy = 0

W T = W+  a  g 

W T   a  W ………..(1)  g 

= 4448+

For the right cord

4448  0.194 9.81

T = 4862.42N

Fy = 0 W Q a  W  Q  …(2) T    g  From (1) & (2)

03.

Ans: (a)

Sol:

Py

W

W Q W  a  a  W = W+Q–   g   g 

P 

W W Q  a  W = W +Q–  a   a  g   g  g Qa 2Wa = Q– g g

 g  a  2Wa 2Wa   Q Q = g ga  g  ACE Engineering Academy

Directon motion

4

3

Px

F

3 tan = 4

N

 = tan–1 3 / 4   36.86

Fnet  x = ma

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: 365 :

Engineering Mechanics

W Px–F =  a  g 

From static equilibrium condition

W Pcos36.86–F =  a  g 

N–W = 0 N = W = 44.48N

 2224  0.2g  0.8P – F =  g  

Fx = 0

Fy = 0

From dynamic equilibrium condition F = ma

0.8P – F = 444.8

N =

0.8P – F = 444.8 + F P = 556+1.25F ……(1)

=

Fy = 0 N+Py–W = 0

a g

a = g ….(1)

F N = W–Py (since  = ) N

Since v2–u2 = 2as

F = N

0–(9.126)2 = 2(–a)  13.689

F =  (W–Py)

a = 3.042 ….(2) From (1) & (2)

= 0.2(2224–P sin 36.86)

3.042 = (9.81)

F = 444.8–0.12P …..(2)

 = 0.31

From (1) & (2) P = 556+1.25(444.8 – 0.12P) 1.15P = 1112

ma P

P = 967 N

F N Wcos

Ans: (d)

Sol:

05. Ans: (a) Sol:

P = 966.95

04.

W a g

Wsin W



W u = 9.126 m/s

Y

Q ma

V=0

N

ma F N

mg.sin

s mg cos

W

 X

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: 366 :

Fy = 0 (static equilibrium)

TA = 2TB ….(1)

N – Wcos = 0

Work done by A & B equal

N = Wcos = mgcos

TASA = TBSB

Since F = N =  mgcos….(1)

2TBSA = TB SB

Fx = 0 (Dynamic equilibrium)

2SA = SB

F+ma – Wsin = 0

TB = mBaB+mBg….(3)

F = mgsin–ma…(2)

For ‘A’ body

From (1) & (2)

TA = mAg–mAaA….(4)

mg cos = mgsin – ma

(2), (3) & (4) sub in (1)

a = gsin – gcos

mAg–mAaA = 2(mB(2aA)+mBg)

 a = gcos(tan – )

mAg–mAaA = 4mBaA+2mBg

Given PQ = s

mAaA+4mBaA = mAg–2mBg

1 s = ut+ at2 2 1 s = 0(t)+ at2 2 =

2aA = aB….(2) For ‘B’ body

F = –ma+mgsin

06.

ME – GATE_Postal Coaching Solutions

aA = t=

2s a

=

2s g cos tan    

=

m A g  2m B g m A  4m B

150  2(50) 150  50   4  10  10 

50 50 = = 1.42 15  20 35

Ans: (a) 07. Ans: 4.905 m/s

Sol:

Sol: S = 0.4; K = 0.2 x

TB

TB

TB B 50N

mBaB

FBD of the block a

W = 200 N

mBg

TA A 150N mAg

ACE Engineering Academy

P = 10t

mAaA

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: 367 : W.r.t free body diagram of the block:

Engineering Mechanics 10

V

200 dv 9.81 0

FS = SN;

 (10t  40)dt 

FK = KN

5t

Fy = 0

Velocity (V) = 4.905 m/s

8

2



 40t 8  20.387  V  180  80  20.387  V 10

N–W = 0 N = W= 200 N

08. Ans: 1.198 m/s2 Sol:

Limiting friction or static friction

FBD of the crate

(FS) = 0.4200= 80 N

N

Kinetic Friction

P

(FK) = 0.2  200 = 40 N The block starts moving only when the

10

WX

F

force, P exceeds static friction, FS

WY 100

100

W=1009.81=981N

Thus, under static equilibrium W.r.t. FBD of the crate:

 Fx = 0

WX = Wsin 100 = 981sin 100

 P–FS = 0  10t = 80

= 170.34N

80 t  8 sec 10

WY = W cos100 = 981  cos100 = 966.09 N

 The block starts moving only

FY = 0  N – WY = 0 N = WY = 966.09N;

when t >8seconds

F = N = 0.3 966.09 =289.828 N During 8 seconds to 10 seconds of time:

FX = 0  P+ WX –F = 0

According to Newton’s second law of

 P + 289.828 –170.34 =0

motion

P = 119.488 N Force = mass  acceleration

P  FK   m  dv  (10t  40)  dt

ACE Engineering Academy

200 dv  9.81 dt

P = ma = 119.488 N a

119.488  1.198 m/s 2 100

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: 368 :

ME – GATE_Postal Coaching Solutions

10. Ans: 2.053 rad/s2

09. Ans: 57.67 m Sol:

ma

N

Sol:

Wx=Wsin 45

7m

1m

F

3m

0

W=39.81=29.43N 450

Wy = cos 450

M = I

W=mg = 98.1N

450

M = 29.43  3 = 88.29N-m 3  82 m 2 2 I  I 0  Ad   md   3  32 12 12  16  27  43kg  m 2 2

Wx = W sin 45= 98.1 sin 45 = 69.367 N Wy = W cos 45 = 69.367 N FY = 0



N –WY = 0 N = WY = 69.367 N

11.

F = KN = 0.5 × 69.367 = 34.683N

Sol:

M 88.29   2.053 rad / s 2 I 43

Ans: (a) L

Fx = 0 (Dynamic Equilibrium

ma

D Alembert principle)

b

Wx – F– ma = 0

VA

69.367 – 34.683 – 10×a = 0

Fy = 0

a  3.468m / s

VA+ma = mg

2

W = mg

1 S = ut + at 2 2

VA = m(g–a)…(1)

 t is unknown we can not use this equation

Since, M = I

Where a = aT = b

So use V2–u2 = 2as V = 20m/s2; u = 0; a = 3.468m/s2 V 2 2as

S

20 2

V2   57.67m 2  a 2  3.468

ACE Engineering Academy

 mL2  M =   mb 2   12 

M=





m 2 L  12b 2  12

mgb =





m 2 L  12b 2  12

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: 369 :

=

from (1) & (2)

12gb L  12b 2 2

 12gb 2  m g  2 L  12b 2 

  

3  mg  VA = m g  g  = 4 4  

[∵m g  b  ]

 gL2  12gb 2  12gb 2    m L2  12b 2  

VA =

13.

Ans: (d)

R = 0.25m F = 8N

WL2 VA = 2 L  12b 2

Mass moment of inertia, Ix = Iy =

Ans: (d)

Iz =

Sol:

L ma W

VA

mr 2 4

mr 2 2

M = I 80.25 = 5  = 0.4 2– 02 = 2

L/2

2–02 = 2(0.4) (since for half

Fy = 0

revolution  = )

VA+ma = W VA = m(g–a)…(1) Where, a =

W 4

Sol: I = 5kg.m2

mgL2  2 L  12b 2

12.

Engineering Mechanics

L  2

14. Ans: 4.6 seconds Sol: M = 60 N – m

Since, M = I 2 L  mL2  L   W =  m   2  12  2  

L 4mL2 2a  mg = 2 12 L 3 a = g …(2) 4 ACE Engineering Academy

 = 1.58 rad/sec

0 = 0,

L = 2m,  = 200 rpm =  = 20.94

200  2 60

rad sec

Moment, M = I 60 =

mL2  12

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: 370 :

 60 =

ME – GATE_Postal Coaching Solutions

40  2 2  12

Chapter- 8

 = 4.5rad/sec2

Work-Energy Principle and Impulse Momentum Equation

 = 0+t 20.94 = 4.5t t = 4.65 sec 15.

01. Ans: (a)

Ans: (a)

Sol:

Sol:

Reel

30o

L

Lcos30o

L= 3.048m

r

L–Lcos30o W2 = 262.132N

Moment (M) = W r = m g  r Applying D-Alembert’s principle M – I = 0 mgr – (I0 + mk2)  = 0 

mgr mgr gr   2 2 2 2 I 0  mk mr  mk mr  k 2

Linear acceleration of the reel = tangential acceleration to the drum a = aT = r 

rgr gr 2  r2  k2 r2  k2

W1 = 4.448N, u1 = ?

The loss of KE of shell converted to do the work in lifting the sand box and shell to a height of “L – Lcos30o” i.e., Wd =

1 mV 2 2

Where d = L – Lcos30o = 3.048 – 3.048cos30 = 0.41 m 266.580.41=

1  266.58  2  V  2  9.81 

V = 2.83 m/sec Where V is the velocity of block & shell By momentum equation m1u1 + m2u2 = m1v1 + m2v2 ACE Engineering Academy

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: 371 :

Where v1 = v2 = V & u1 = ?, u2 = 0

03.

4.448 4.448  262.132  u1   2.83 9.81 9.81

Ans: (a)

Sol: Given, m = 2kg

Position at any time is given as x = t + 5t2 + 2t3

 u1 = 169.6 m/sec u1 & u2 = Initial velocity of shell and block

At t = 0, x = 0,

respectively

At t = 3sec, x = 3 + 5(32) + 2(33) = 102m

V1 & V2 = Final velocity of block & shell 02.

Engineering Mechanics

Velocity, V =

Ans: (b)

dx  1  10t  6t 2 dt t = 0, is vi = 1m/s

Initial velocity i.e.,

Sol:

Final velocity i.e., at t = 3sec, is vf = 1 + 10(3) + 6(3)2 = 85m/s

F

Work done = change in KE

W S

FS

S

=

1 1 mv f2  mv i2 2 2

=

1  2 85 2  12 = 7224 J 2





F = KS

04.

Strain energy in spring = Area under the

Sol: Given force F = e-2x x2

force displacement curve. =

Ans: (a)

Work done =

1 1 1 F  s = (ks)  s = ks 2 2 2 2

x1

1.5

=

1 2 ks  Gain of KE 2

 e 2 x  dx      2  0.2

= 0.31J.

ks 2 ks 2 = g v = m w 2

v

e

1.5

2 x

0.2

1 2 1 ks  mv 2 2 2

kg .s w

 Fdx

 w  m   g 

05.

Ans: (b)

Sol: F = 4x–3x2

Potential Energy at x = 1.7 = work required to move object from 0 to 1.7m 1.7

PE =

 Fdx 0

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: 372 :

ME – GATE_Postal Coaching Solutions

1.7

=

 4x  3x dx

wLb 

2

0

  x2 = 4   2

b  1 Lv  b L    2 2 g 

1.7

  x 3    3    3  0



= 2x 2  x 3

wb 2 1 wLv 2  2 2 g 2

b   v2 = 2gb 1    2L 



1.7 0

2

= 2(1.7) – (1.7)3 v=

= 0.867 J 06.

Ans: (c)

Sol:

x

07.

dW = wdx

b  gb 2   L 

Ans: (d)

Sol:

b

W1 = 10N

W2 = 20N

V1 = 40m/s

V2 = 10m/s

L–b

m1 = 1kg , m2 = 2kg ,(since g = 10m/sec2) Velocities before impact Where w = weight per unit meter dw = a small work done in moving small elemental “dx” of chain through a d/s “x” Work done = change in KE  b 1  wL  2   dw  x   w L  b  b    v   2  g   0 b

1 wLv g

 wdx.x  w (L  b)b  2 0

wb 2 1 wLv 2  w L  b b  2 2 g wb 2 1 wLv 2  wLb  wb 2  2 2 g ACE Engineering Academy

2

v1 = 40 m/sec, v2 = –10m/s Velocities after impact u1 = ? u 2 = ? Coefficient of restitution e = 0.6 From momentum equation m1v1+m2 v2 = m1u1+m2u2  1(40) + 2(-10) = 1(u1) + 2(u2)  u1 + 2u2 = 20…………………..(1) e

u 2  u 1 relative velocity of Seperation  v1  v 2 relative velocity of approach

0.6 =

u 2  u1 40  (10)

u2 – u1 = 30………………………(2)

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: 373 :

From 1 & 2

Engineering Mechanics

10.

u1 = –13.33 m/sec

Ans: (c)

Sol:

u2 = 16.66 m/sec 

R

08.

Ans: (b) 2R

Sol: Given , m1 = 3kg, m2 = 6kg

Velocities before impact KE =

u1 = 4 m/s u2 = –1 m/s Velocities after impact

1 1 mV2+ I2 2 2

From momentum equation

I=

m1u1 + m2u2 = m1v1 + m2v2 3(4) + 6(–1) = 3(0) + 6(v2)



v2 = 1m/s

09.

2

15 1  V  mV 2   mR 2   22 2  2R  V2 5 1 = mV 2  mR 2  4 2 4R 2

2

KE =

v 2  v1 u1  u 2

1 1 0 = 4  (1) 5

e=



1 5 2 m 2R   R 2 = mR 2 2 2

15 1  V   KE = mV 2   mR 2   22 2  2R 

 6 = 6v2

Coefficient of restitution, e =

V 2R

Where,  =

v1 = 0m/s v2 =?

=

Ans: (a)

5 1 mV 2  mV 2 16 2

13mV 2 KE = 16

Sol: Given m1 = 4kg, m2 = 8kg

Velocities before impact u1 = 12m/s u2 = 0 Velocities after impact, v1 = v2 = v From momentum equation m1u1 + m2u2 = m1v1 + m2 v2 4(12) + 8(0) = 4v + 8v 12v = 48 v = 4m/sec

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11. Ans: (b) Sol: mV

m1V r  I C  

3 m r2  2

3 C  20  1   2 (neglecting mass of the clay) 10 1    rad / s 30 3  10  1  1 

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: 374 : 12. Ans: (a)

13.

Sol:

(m+M) g

ME – GATE_Postal Coaching Solutions Ans (a)

1 I(f2  i2 ) 2

Sol: Work done (W) =

=

1 I f2 2

=

1 I(2) 2

(m+M) a N

Fd

W = I

m1 = m  mass of bullet

Where I, , are constant

m2 = M  mass of block

 Hence, work done is same

u1 = V  bullet initial velocity u2 = 0  block initial velocity v1 = v2 = v velocity of bullet and block after impact.

14. Ans: (a) Sol:

K = 10.6kN/m

B =133.44N

A= 222.4N

Fd = N

0.3m

(M+m)a = (M+m)g uA = 0 ,

 a = g

uB = 0

From momentum equation

From momentum equation

m1u1 + m2u2 = m1v1 + m2v2

mAuA+mBuB = mAvA+mBvB

mV + m(0) = (m + M)V

0 = 222.4VA+133.44VB…………..(1)

v=

1 1 2 1 2 2 ks  m A v A  m B v B 2 2 2

mV mM

Now from v2–u2 = 2as

10.61030.152 =

2

 mV  0 –   2gs mM mM 2gs V= m

222.4 2 133.44 2 vA + vB 9.81 9.81 ………….(2)

From 1 & 2 vA = –1.98m/s vB = 3.3m/s

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: 375 : 15.

Engineering Mechanics

Ans: (b)

Sol: Given, t = 20mm = 0.02m

Chapter- 9

R = 200mm = 0.2m

Virtual Work

m = 20kg N = 600rpm KE =

01.

1 2 I 2

mR Where I = 2 =  KE =

Sol: 2

20  0.2 = 2

= 0.4

2N 2  600 = = 20 40 60

1 2  0.4  20 2

= 789.56  790J.

25kN

25kN

2

A

2m

C

2m

3m

D

B

Let RA & RB be the reactions at support A & B respectively. Let y displacement be given to the beam at B without giving displacement at ‘A’

2/7y

y

4/7y

A

B

The corresponding displacement at C & D are

4 2  y and  y 7 7

By virtual work principle, 4 2 RA0–25  y  25   y  R B   y  0 7 7    150   R B  y  0   7 Since y 0, RB– RB = ACE Engineering Academy

150 =0 7

150 kN 7

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: 376 :

ME – GATE_Postal Coaching Solutions

Now let us give virtual displacement at A as y, Therefore corresponding displacement at C

4/8y

3 5 & D are  'y &  'y 7 7

C

A

y

6/8y D

B

By virtual work principle, RA0–25

5/7y 3/7y

A 2m

RA

C

B

3m

D

2m



RB

5 3  y – 25  y + RB0 = 0 7 7

RB =

y

200 0 7

y

200 RA = kN 7 02.

A

35kN

C

RA



B

are D

shown

displacement at A). ACE Engineering Academy

below

C

B

D 2 6

y

2 2 y and y 6 6

2 2 R A  y  25  y  R B  0  35  y  0 6 6

RB

corresponding displacement at different as

2m 2m

Now by virtual work principle,

Let the virtual displacement at D as y, then point

4m

y

The corresponding displacement at C & D

2m

2m

4m



2 6

A

25kN

Ans:

190 3

Now, Let the virgual displacement at A as

125 75  '    y  0 RA  7 7    y  0 , RA –

3R B 25 y   y  35 y  0 2 4

25 3 (since  y0) R B  35  2 4

 By virtual work principle, RA   y – 25

4 6 y+RB y–35y = 0 8 8

(Assume

no

Since y 0, RA–

25 35 =0  3 3

RA =

 10 kN 3

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Strength of Materials

Solutions for Vol – I _ Classroom Practice Questions Chapter- 1 Simple Stresses and Strains 1.1 Fundamental, Mechanical Properties of Materials, Stress Strain Diagram 01. Ans: (b)

04. Ans: (b)

Sol: Ductility: A material which undergoes

Sol:

considerable deformation without rupture (large plastic zone)

4



3

1 2

Brittleness: Failure without warning (No plastic zone) i.e. no plastic deformation A B

Tenacity: High tensile strength

C

D 

Creep: Material continues to deform with 05. Ans: (b)

time under sustain loading Plasticity: Material continues to deform

06. Ans: (a)

without any further increase in stress. high

Sol: Strain hardening: - increase in strength

probability of not failing under Reversal of

after plastic zone by rearrangement of

stress of magnitude below this level.

molecules in material Visco-elastic material

Fatigue: Decreased Resistance of material to

exhibits a mixture of creep + elastic after

repeated reversal of stresses

effects at room temperature. Thus their

Endurance

limit:

Material

has

behavior is time dependant 02. Ans: (a) Sol: For rigid plastic material

07. Ans: (a) 08. Ans: (a)



09. Ans: (a) 

Sol: Addition of carbon will increase strength, thereby ductility will decrease.

03. Ans: (a) ACE Engineering Academy

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: 380 :

ME – GATE _ Postal Coaching Solutions

1.2 Elastic Constants and Their Relationships 01. Ans (c)

D lateral strain D  Sol: Poisson’s ratio  linear strain L L

D





PL AE

02. Ans: (c)

P

8 B

L

1

 2.5105 =

A,L,E

200  20 V

 V = 0.016 m3

C

D AE 8 P

P V V

Sol: Bulk modulus =

P

 2 (8)  10 6 D 4 0.25  8 50000 D = 1.98 × 103 = 0.00198  0.002 cm 1.3 Linear and Volumetric Charges of Bodies

01. Ans: (d)

0

Pz

Sol:

 P     P    Px 

Py

P Px

Px

E

E

E

 . Px 1   

02. Ans: (a) Py

Pz

Let Py = Pz = P

Punching force = shear resistance of plate  c / s area   ( surface Area)

y = 0 , z = 0 y 

y E



Sol: c = 4 ---- (given)

z   . x E E

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4 

. D 2  (  . D . t ) 4

D = t = 10 mm

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: 381 :

03. Ans: (d) Sol:

3P

Strength of Materials

04. Ans: (c) Sol:

3P

Steel

K

2K a

s = 140 MPa 

Ps As

A

P

Aluminium

Al = 90 MPa 

B = 100 MPa  PB =

B

From similar triangle 3a 2a  A B 3B = 2A ……. (1)

2P

PB AB

100  200 = 10,000 N 2

Ps = 22,300 N ,

D

P

PA A A

Bronze

C

W

PAl = 90  400 = 36,000 N

2P

a

B

A

140  500  23,300 N 3

PS =

a

PAL = 36,000 N

Stiffness K   KA 

W 

WA W  A  A A 2K

Similarly

B 

WB K

From equation (1) 

3

W WB  2 A K 2K

WA 3 WB

PB = 10,000 N Take minimum value  P = 10,000 N

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: 382 :

ME – GATE _ Postal Coaching Solutions

04. Ans: (b) 1.4 Thermal Stresses

Sol: Stress in steel

Ps 8  103   80 MPa As 100

Common Data for Question Nos. 01 & 02

Stress in Gunmetal = 01. Ans: (b)

  t s    t Al   P  6

6

1110  20  24 10  20

A gm

8 103   40MPa 200

Sol: Free expansion = Expansion prevented  P    AE  s  AE  AL

Pgm

05. Ans: (a) Sol:



P P   3 100 10  200 200 103  70  P = 5.76 kN



02. Ans: (b) P 5.76 10 3 Sol:  s  s   57.65 MPa As 100  Al

5.76 10 3 P    28.82 MPa A al 200





  Strain in X-direction due to temperature,

 t    T  Strain in X-direction due to volumetric stress

Common Data for Question Nos.03 & 04

03. Ans: (d) Sol:  s   gm    s t L   gm tL

PL PL  3 100  200  10 200  100  10 3   6  10 6  200  L  10  10 6  200  L 2 P  8  10  4 3 200  100  10 P = 8 kN

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x 

y x    z E E E

x 

 1  2  E

 

 x E  1  2

 

T E 1  2 

 

 ET  1  2

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: 383 :

Strength of Materials Common Data for Question Nos. 04 & 05

Chapter- 2 Complex Stresses and Strains

04. Ans: (a) Sol: 1 

x  y 2

65   13  65  13  2     20 2  2  2

01. Ans: (b) Sol: Maximum principal stress 1 = 18

Minimum principal stress 2 = 8 Maximum shear stress =

1   2 = 13 2

Normal stress on Maximum shear stress plane =

1   2 18  (8)  5 2 2

02. Ans: (b)   2 Sol: Radius of Mohr’s circle max = 1 2

20 =

1  10 2

1 = 50 N/mm2

=

 1  70 MPa ,  2   18 MPa

05. Ans: (a) Sol: 1 =

1   2 ; E E

1 =

Change in diameter Original diameter

1 =

70  0.3  18 = 3.77  10 4 5 2  10

Change in diameter = 3.77 10 4  300  0.1131 mm Length of major axes of ellipse = 300 + change in diameter = 300.113 mm

03. Ans: (b) Sol: Long dam  plane strain member

z  0 

2

   y     2xy   x  2 

 z  x  y   E E E

 x  150 MPa ,  y   300 MPa ,   0.3  0   z  0.3  150  0.3  300   z  45 MPa

Similarly length of minor axes 2 

 2 1  E E

2 

D  18  0.3  70   1.95  10  4 300 2  10 5

D  1.95  10 4  300  0.0585 mm  Minor axis length, = 300 – 0.0585 = 299.94 mm

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: 384 :

ME – GATE _ Postal Coaching Solutions

Common Data for Question Nos. 06 & 07

06. Ans: (b) Sol:

Chapter- 3 Shear Force and Bending Moment



01. Ans: (b) Sol: Contra flexure is the point where BM is

becoming zero.

 1 = 2 = 175

X

17.5 kN/m A C 4m 4m RA

07. Ans: (d)

X

08. Ans: (c) Sol:  2  0(Given) 2 

x  y 2

x  y 2

 x  y  2 

A

C

2

 x  y   2 

2

2

 2 xy   x . y   xy   x . y

B

D

MA = 0

    2 xy    x  y    2  

P.O.C

3.78

2

   y     2 xy   x  2 

    y    x 2  

 2 xy

RB

2

    2 xy 

D 2m

50  x  y   2 

20 kN

B x

17.54   

2

4  20  10  R B  8  0 2

RB = 42.5 kN Mx = 20x + RB(x  2) For bending moment be zero Mx = 0 20x + 42.5(x  2) = 0 x = 3.78m From right ie. D

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: 385 : Common Data for Question Nos. 02 & 03

Strength of Materials

05 Ans: (a) Sol:

02. Ans: (b) Sol:

A

25 kN/m

100 kN

4 4kN 2m

2m

2 kN

B 2 kN

S

P

2m

Q 1.5m

2m

41.41 kN 25 kN/m

77.34 kN

06. Ans: (c) Sol:

P

Take M P  0 1  1. 5   25  1.5    4   R Q  4   100  2  25  0 2  3 

R Q  77.34 kN

l P 2 RA= 2

V = 0 1 R P  R Q  100   25  1.5 118.75 kN 2 R p  41.41 kN

l 2

RB=

P 2

Pl 4 BMD Diagram

Pl l from left is 4 2 The given beam is statically determinate

(BM) at

S. F. at P = 41.41 kN 03. Ans: (c)

structure. Therefore equilibrium equations

Sol: MS = RP (3)+25 1001 = 49.2 kN-m

are sufficient to analyze the problem.

04. Ans: (c)

In statically determinate structure the BMD,

Sol:

3 kN VA

A

B

1m 1m 3 kN-m

C 1m

SFD and Axial force are not affected by section (I), material (E), thermal changes.

VB

07. Ans: (a) Sol: As the given support is hinge, for different

–VB  3 + 3 = 0

set of loads in different direction beam will

VC = 1kN  Bending moment at B

experience only axial load.

 VC  1 = 1 kNm ACE Engineering Academy

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: 386 :

ME – GATE _ Postal Coaching Solutions

9 2b3d  = bd 3 2 12 3

M.I about CG = ICG =

Chapter- 4

M.I about X  X | at d

Centre of Gravity & Moment of Inertia

y

E 1 y1  E 2 y 2 E1  E 2

h h  2E 2  h    E 2  2 2  y 2E 2  E 2

E1  2E 2 

Ix =

9 3 5 bd  6bd  d 2 2 4

=

111 3 bd = 13.875bd3 8

  30  45 3 60  120 3  2  30  45  37.5 2  12   12

= 4.38106 mm4

Sol: y  A1E1Y1  A 2E 2 Y2 A1E1  A 2 E 2

05. Ans: Ix = 152146 mm4, Iy = 45801.34 mm4

1.5a  3a 2  E1  1.5a  6a 2  2E1 3a 2 E1  6a 2 (2E1 ) 22.5a 3 E 1  1.5a 15a 2 E 1

Sol:

Ix 

30  403   20 4   152146 mm 4 12 64

Iy 

2  40  30 3    20 4 4  10       2  10 2  15   2  64 12 3       

= 45801.34 mm4

03. Ans: 13.875 bd3 Sol:

=

Sol:

02. Ans: (b)



= IG + Ay2

04. Ans: 4.38106 mm4

y  1.167h from base



dis tan ce

2

01. Ans: (a) Sol:

4

2b

5 y= d 4 CG X

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3d d/4

X

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: 387 :

Strength of Materials

MA = 0

P 100 + 2P  200 + 3P300 = RB 400

Chapter- 5

 RB = 3.5P, RA = 2.5P

Theory of Simple Bending

Take moments about F and moment at F MF = RB 150 3P50 = 375P 01. Ans: (b) Sol:

b/2

b

375P 1.5  10 6  200  10 3   2176 6

b/2

A



M F b  I yF

b

B

 P = 0.29 N

M Z

03. Ans: (b) Sol:

1   M is same Z

2  105 E  b   b  0.5 / 2 250 R y max

b = 200 MPa

b 2  b  2  A ZB 6    =2 2 B ZA b b  2 6

04. Ans: (b) Sol: (max)steel = m  ()timber = 20 7 = 140 MPa

05. Ans: (a) Sol:

02. Ans: (b)

10 kN 1m

Sol: 4mm 10mm

NA

P A 100

100

RA

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2P

 f = 1.510

-6

BMD Diagram

3P

50 F 50 100

1m

B RB

M = 10 kN-m f=

M 10 10 3   60 MPa Z 10  10 2 6

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ME – GATE _ Postal Coaching Solutions

06. Ans: (c)

09. Ans: 80 MPa

Sol:

Sol:

10 mm

75

100 mm

10 mm

100 mm

25 50

200 mm

Force on the hatched area = Avg. Stress  hatched Area

Modular ratio, m = 20

M f  I y 

Maximum stress in timber = 8 MPa Stress in timber in steel level,

f 16  10 6   f = 14.22 MPa 3 25 100  150 12

 Force on hatched area

100  8 50  fw fw = 4MPa Maximum stress developed in steel is = mfw = 204 = 80 MPa

= Average stress  hatched area  0  14.22  =  (25 50) = 8.9 kN 2  

using modular ratio. 10. Ans: 8 kN-m

07. Ans: (c)

Sol: Moment of inertia,

f M Sol: Tensile  y top I  f Tensile =

Convert whole structure as a steel structure by

2  10  100 3 5  200 3  I= = 5  106 mm4 12 12 0.3  3  10 6  70 3  10 6

(maximum bending stress will be at top fibre

fs = 820 = 160 MPa

so y1=70mm)  fTensile = 21 kN/m2

M=

Common Data for Question Nos. 09 & 10 ACE Engineering Academy

10 mm

Bending equation M fs  I y

08. Ans: (c)

5 mm 10 mm

200 mm

5  10 6  160 200 2 5  10 6  160  2 = 8 kN-m = 200

11. Refer GATE - 16 book

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: 389 :

Strength of Materials

03. Ans: 48.7

Chapter- 6

FAy Ib

max =

Shear Stress Distribution in Beams

= 01. Ans: (a)

= 48.70 MPa

3 3 f Sol: max =  avg =  2 b.d 2

04. Ans: 3.5

3 50  10 3 3=  2 100  d

in flange just above web

200  103 =  160  20  150 171.65  106  160

d = 250 mm = 25 cm

= 3.5 MPa

Common Data for Q. 02, 03 and 04: Sol:

B=160

05. Ans: 61.43 MPa 20

D=320

200  10 3  140  15  70  160  20  150 171.65  10 6  15

2.7963

d=280

37.28

Sol:

120

max =48.70

15

(2) 160

CG

02. Ans: 37.3

INA = 13  106 mm4

Bending moment (M) = 100 kN-m, Shear Force (SF) = f = 200 kN 160  320 3 145  280 3  12 12 6

4

= 171.65  10 mm

at interface of flange & web = =

107

20 All dimensions are in mm

All dimensions are in mm

I=

33 (1)

20

20

FAy Ib

yCG = 107 mm from base max =

FAy Ib

A y = (120 2043) + (33 2016.5) = 114090 mm3 max =

140  10 3  114090 = 61.43 MPa 13  10 6  20

200  10 3  160  20  150 171.65  10 6  15

= 37.28 MPa ACE Engineering Academy

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ME – GATE _ Postal Coaching Solutions

04. Ans: (c) Sol: Series

Chapter- 7 Torsion

 T.L   T.L  max = AB + BC =      CJ  AB  CJ  BC

01. Ans: (a) Sol: Twisting moment = 1  0.5



 0.128  

=

1 1 K4

  

2.14  10  2000 8.5  10   32  75 6

4



 2

4



16 15

Sol:  =

T  l1 l 2     C  J 1 J 2 

Common Data for Question Nos. 03 & 04

5000  10 3 10 5

03. Ans: (a)

=

5000  10 6  1 1     5 4 10 10  10 4   25  10

Sol: (P)AB = 30 kW

 100  10 100  10   25  10 4  10  10 4  

= 0.7 radians

(P)BC = 45 kW 2NTAB  TAB = 1.43 kN-m 60

2NTBC PBC =  TBC = 2.14 kN-m 60 T    J R

06. Ans: 43.27 MPa & 37.5 MPa Sol: Given Do = 30 mm ,

TBC 2.14  106 BC = =  ZP  753 16 BC = 25.83 MPa

 Take maximum value of ‘’ i.e, 58.26 MPa

t = 2 mm

 Di = 30 – 4 = 26 mm

We know that

T 1.43  106 AB = AB  = 58.26 MPa  3 ZP  50 16

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  

180 = 7.21  7.14 

=

PAB =

4

05. Ans: (d)

1

1 1

4



= 0.128 radian

02. Ans: (d)

strength solid strength hollow

6

4

= 0.5 kN-m

Sol:

1.43  10  4000 8.5  10   32  50



 q  J R

q 100  10 3  max 4 4  30  26  30    32  2





qmax = 43.279 N/mm2 q 100  10 3  min 4 4  30  26  26    32  2 





qmin = 37.5 N/mm2

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: 391 :

Strength of Materials

03. Ans: (c) Sol:

Chapter- 8

ymax = 18mm

Deflections and Slopes

01. Ans: (c) Sol:

d d

B

A

b

wl 3  0.02 6 EI

ymax =

wL4  0.018 8EI

 WL3  L  6   =  8  6EI 

b



1 ymax  I

0.02  L  6  0.018 8

L = 1.2 m

y I  A  B yB IA

04. Ans: (a) 2

y A  bd / 12 d  yB =   yA 3 db / 12 b 3

yB =

max =

Sol: W y





y

02. Ans: (b) Sol: Total load W= wl

(L-l/2)

l

(L-l/2)

L

y max  y max 

3

W 8EI

Conditions given

W 3 3EI

ynet =  yudl yw 3

Total Net deflection = 

3

WL WL  8E1 3EI  5WL3 24EI

y=

wl 3 48EI

wl 2 = 16EI tan  =

y L  l / 2

 is small  tan  =  =

y L  l / 2

( indicates upward) ACE Engineering Academy

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Ll y =     2 

ME – GATE _ Postal Coaching Solutions

07. Ans : 0.05 Sol:

Ll y=    2 

A

5m

C

5m

B

10 m

Thus y = y  w 3 w 2  L  l      48EI 16EI  2 

 Curvature,

l 5  L 3

d2y  0.004 dx 2

Integrating w.r.t. x, We get,

05. Ans: (c) Sol: By using Maxwell’s law of reciprocals

theorem

W

0.004 x 2 2

y = 0.002x2

C

A

y

dy  0.004 x dx

@ mid span, x = 5 m

B

 y = 0.002 x2

CB = BC Deflection at ‘C’ due to unit load at B

y = 0.05 m

= deflection @ B due to unit load at C As the load becomes half deflection becomes half 06. Ans: (c)

W=1kN

Sol:

W=1kN

20 mm A

15 mm B

40 mm

30 mm

 wL3   wL3      yA = yB    3EI  A  48 EI  B

LB = 400mm ACE Engineering Academy

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Chapter- 9 Thin Cylinders

Strength of Materials

v 

V  1.86  10 4 V

h 

PD P  800   50P 4t 4 4

h  Common Data for Question Nos. 1 & 2



01. Ans: (b) Sol: max = l =

max =

h  0 PD = 2 4t

(R = 0)

 V  3 h  5.25  10 4 P P = 0.355 MPa 04. Ans: (c)

02. Ans: (b)

Sol: Maximum stress h = 50P

V Sol: v =  2 h   l V

= 50 0.355 = 17.77 MPa

   . h   = 2  h       E E E  E  2  10

50P 50P  0.3  5 2  10 2  10 5

 1.75  10 4 P

1.6  900  30 MPa 4  12

= 2  60

h    h  R E E E

5



0.3  30   30 0.3  60    5  5 2  10   2  10 2  10 5 

= 5.7  10–4 900  2000  V = 5.710–4  4 2

Common Data for Question Nos. 05 & 06

05. Ans:1.25 MPa & 2.5 MPa Sol: R = 0.5 m, D = 1m, t = 1mm

P = gh At 0.5 m depth P = h (10103)0.5 = 5000 N/m2

3

 V = 725.23 cm

Common Data for Question Nos. 03 & 04

= (510–3) MPa Hoop stress,  h  1 

5  10 3  1000   2.5MPa 2 1

03. Ans: (b) Sol: D = 800mm, t = 4mm

V= 50cm3 = 50000mm3  = 0.3 4 4 V   R 3     400 3 = 268 106 mm3 3 3 ACE Engineering Academy

PD 2t

Longitudinal stress,     2  

PD 4t

5  10 3  1000  1.25MPa 4 1

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ME – GATE _ Postal Coaching Solutions

06. Ans: (2.125105 & 5  106)

Chapter- 10

Sol: E = 100 GPa ,  = 0.3

σ σ  εh  h  μ    E E 2 .5  1.25    0.3  2.125  10 5 3 3  100  10  100  10 

εl   

σl σ  μ. h E E 1 l  h  E

1 1.25  0.3  2.5  = 510–6 3 100  10

Columns and Struts 01. Ans: (c)  2  EI l e2

Sol: Pe =

For a given system , le =  Pe =

l 2

4 2  EI l2

02. Ans: (b)

P1 l 22e Sol:  P2 l12e P1 l2  P2 (2l ) 2 P1 : P2 = 1 : 4 03. Ans: 4 Sol: P =

2

l2 PI

EI

P I bonded  Po I loose

 b2t 3    12   4   bt 3  2   12 

04. Ans: (b) Sol: Euler’s theory is applicable for axially loaded columns 2 EI l2 2 2 EI There fore, F = l2 Fcos 45 =

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Strength of Materials  P2 P2  A2  L2   2  A 1  L1   A 22 =  A1   P 2  A1  L1 P 2  A 2  L 2     A 12 A 22  A

Chapter- 11 Strain Energy Resilience 01. Ans: (a) Sol: 1000 kg

1000 kg

22

44

5cm

 L1 L 2      U B  A1 A 2  B 7.165 3    U A  L1 L 2  4.77 2     A1 A 2  A

10cm

03. Ans: (c)

U = U1 +U2

Sol: A1 = modulus of resilience

   V1   V2 2E 2E 2 2

2 1

=

A1 + A2 = mod of toughness A1 =

1000 4   4  4  10 2 2

=

2  4.1  10 6

1000 2 

1 A2 =  0.008  50  106   0.008  70  106  2

2



2

2  4.1  10

6

 [ 225]

04. Ans: (d)

02. Ans: (a) 2cm

2cm 10cm

20cm

= 76 104 A1 + A2 = (14 + 76)  104 = 90 104

U = 0.228  0.23 kg-cm

Sol:

1 4  0.004  70  10 6 = 14 10 2

U  P2

20cm

1

1

2

10cm

2

40cm

4cm

A

B

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Due to the application of P1 & P2 one after the other (U1 + U2)  P12 + P22 ……….. (1) Due to the application of P1 and P2 together at the same time.

U B V1  V2 B  U A V1  V2 A

 12  V1   U B  2E   U A  12  V1    2E

P2 Sol: U = .V 2A 2 E

U  (P1 + P2)2 ………...........(2)   22  V2  2E B   22  V2  2E A

It is obvious that (P12 + P22) < (P1 + P2)2  (U1 + U2 ) < U

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ME – GATE _ Postal Coaching Solutions

05. 2 2 Sol: U  U1  U 2  T L  T L

2GJ 1

Chapter- 12

2GJ 2

Propped and Fixed Beams

  4 4 J1  50 ; J 2  26 32 32

L  100 mm

01. Ans: (d)

G  80  10 3 N mm 2 on substitution,

Sol:

w/unit run

A

U = 1.5 N-mm

B K 

RB = ?

K  Stiffness  K 

Load deflection

RB 

 Compatibility condition Deflection @ B =  K 

RB R  B  K

A

y2

B

RB

y1

w 4 y1  8EI

,

R 3 y2  B 3EI

(+) (–)

y1 = y2 =  w 4 R B  3    8EI 3EI w 4 R B  3 R B   K 8EI 3EI w 4 R B R B  3   K 3EI 8EI

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Strength of Materials

By conjugate beam method

w 4 1   1  R B3  3  8EI 3EI   K

2Pa

 3EI  K 3  3 w 4  RB  3  8EI  3EI  K 

2Pa EI

A

L

w R B  3EI  K 3     8EI 3EI  K 3 

 y c  deflection @ C

 3EI  K  3w  RB  3 8  K 

yc 

= B.M.D. @ C by conjugate beam

3

3w  3EI   R B 1  3 8  K  RB 

02. Ans:

=

3w

2Pa L   L  L   EI 2  2Pa 3L 3PaL2  L EI EI 2

Compatibility Condition (yB) = 0

8 3EI 1 K 3

 y1 = yc 8R B L3 3PaL2  3EI EI

9pa 8L

RB 

Sol:

P

L

A

C

B

a

03. Ans: 12.51 kN Sol:

P

40 kN 2m

M=2Pa L

L

A

2m

B

C RB = ?

E = 200 GPa +6

Applying, superposition principle y2 2L 3

R B ( 2L) 8R B L  3EI 3EI

As per compatablity 3EI

3 

3 3 (40  10 )(2000) 3  EI



3 2 40  10  (2000) 2EI

 2000 1mm

Where EI = 41011 N/mm2

(RB )(4000)3 40103 (2000)3 40103 (2000)3    1 341011 341011 241011

M=2Pa L

(RB) 4

I = 2  10 mm (R B )(4000)

RB

3

L A

9Pa () 8L

a

L

y1 

C

L

B

RB = 12.51 kN

C

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ME – GATE _ Postal Coaching Solutions

03. Ans: (b)

Chapter- 13 Theories of Failure

Sol: 1 = 1.5 (T) ,

2 =  (T),

3 = –/2 (c)

2

fy = 2000 kg/cm ,

 = 0.3

In which theory of failure  = 1000 kg/cm2 Check

01. Ans: (d)

f = fy = 2500 kg/cm2

Sol:

1 = 2000 kg/cm2 Maximum shear stress theory



1   3  2

1 = 1333 kg/cm2 (b) Max shear stress theory

fy

≯

2

2000   3 2500  2 2

3 = –500 (comp) 02. Ans: (b) Sol: Diameter of plate, D = 100 cm

Internal pressure, P = 10 kg/cm2 2

f = fy = 2000 kg/cm FOS = 4 ,

 1   3  f y    2  2    1.5   2000 2  2 2      

4  2000 2   = 1000 kg/cm2 04. Ans: (c)

t=? Maximum, Principal stress theory 1   h 

1= fy 1.51 = 2000

3 = ?

 max 

(a) Max principal stress theory

PD ≯ fy 2t

Sol: 1 = 800 kg/cm2

2 = 400 kg/cm2 1 = fy/E fy  1   2  3  E E E E

10  100  2000 2 t  t = 2.5 mm Safe thickness of plate = 2.5  FOS = 2.5  4

400  f y 800  0.25 E E E  fy = 800 – 100 = 700 kg/cm2

= 10 mm

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: 399 :

Strength of Materials

05. Ans: (a)

Chapter - 14

Sol: For springs in series: effective stiffness is

Springs 01. Ans: (a)

1 1 1   K e K1 K 2 Therefore, K e 

02. Ans: (b) Sol: Stiffness of Spring (S)

Gd 4 G ( 2r ) 4 S  64nR 3 64nR 3

K 1K 2 K1  K 2

06. Ans: (d) Sol: Deflection of closely coiled spring

Gr 4 S 4nR 3



64 R 3 wn G d4

n 03.

Ans: (d)

G d4 Sol: k  64 R 3 n

07. Ans: (d) Sol: For springs connected in series

4

kd

Let d1= d If d is doubles i.e d2= 2d k 1 d 14 k d4  4  1  k2 d2 k 2 (2d) 4

2S 1 1 1    Ke  3 K e S 2S For springs connected in parallel (Ke) = K1 + K2 = S + 2S = 3S

k2 = 16k1

 04. Ans: (a) Sol:  

64WR 3 n Gd 4

 R

08. Ans: (d) Sol: When one spring placed in other then those

3

1 R 13 R 13 8    2 R 32  R 1  3    2 

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(K e ) series 2S / 3 2   (K e ) parallel 3S 9

two springs will be in parallel. Hence combined stiffness is given by Ke = KA+ KB

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: 400 :

ME – GATE _ Postal Coaching Solutions

09. Ans: (a) Sol:

12. Ans: (d) Sol:

1 1 1   k e 10 40

20kN/m

8kN/m

Ke = 8

30kN/m n = 10

K

2K

2K

n = 20

Ke = 20+8+30 = 58 kN/m

1000N 2

10. Ans: (a)

W = 1000N

1=10 mm

System - 2

Sol: K1 = K System - 1

K2 = 2K + 2K = 4K

From system

K 2 4K  =4 K1 K

1 

1000 K

11. Ans: (a)

K

1000  100 N/m 10

Sol: Equivalent Load Diagram:

From system

 (1)

 (2)

Keq = 2K + 2K = 4K K1

K2

Keq = 4× 100 = 400 2 

Keq = K1 + K2

1000 = 2.5m 400

F

= 300 + 100 Keq = 400 MN/m  

F 400kN  K eq 400  10 3 kN

m

1 m = 1mm 1000

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