Psme Quiz

ENGINEERING MATHEMATICS 1. At a distance of 34 feet from the base of a flag pole, the angle of elevation to the top of a flag that is 5.1 feet tall is 48.6o. The angle of elevation to the bottom of the flag is 44.6o . The pole extends 1 foot above the flag. Find the height of the pole, in feet. Solution: Height of the pole = 1 + 34 x tan 48.6 = 39.6 ft (B)

2. 25 trees are planted in a straight line 5 metres apart from each other. To water them the gardener must bring water for each tree separately from a well 10 metres from the first tree in line with the trees. How far will he move in order to water all the trees beginning with the first if he starts from the well? Solution: a1 = 10 m d=5m n = 25 a25 = a1 + (n-1)d a25 = 10 + (25-1)(5) = 130 m Sum = (n/2)(a1 + a25) = (25/2)(10 + 130) = 1750 m Distance moved = 2 x 1750 – 130 = 3370 m 3. A bag contains 26 tiles, each marked with a different letter of the alphabet. What is the probability of being able to spell the word math with four randomly selected tiles that are taken from the bag all at the same time? Solution: Probability = 1 / 26P4 = 1/14950

4. Serial numbers for a product are to be made using four letters (using any letter of the alphabet) followed by three single-digit numbers. For example, JDGR209 is one such serial number. How many such serial numbers are possible if neither letters nor numbers can be repeated? Solution: 26*25*24*23*10*9*8=258,336,000

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5. What is the differential equation of parabolas having their foci at the origin and axes along the xaxis? Solution: y 2 = 4a (x + a )

2 yy ′ = 4 a a=

yy ′ 2

y2 =

4 yy ′  yy ′  x +  2  2 

y2 =

4 yy ′  yy ′  x +  2  2 

y = y ′(2 x + yy ′)

6. Find the volume of a prism whose base is an equilateral triangle with a side of 4 and whose length is 6. Solution: Area of triangle =

1 (4 ) 4 2 − 2 2 = 2 12 = 4 3 2

( )

Volume = 6 4 3 = 24 3

g(x ) f (a ) − g(a ) f (x ) is ___. x →a x −a

7. If f(a) = 2, f’(a) = 1, g(a) = -1, g’(a) = 2, then the value of lim

Solution: g(x ) f (a ) − g(a ) f (x ) x →a x −a lim

lim

x →a

g′(x ) f (a ) − g(a ) f ′(x ) 1

g′(a ) f (a ) − g(a ) f ′(a)

(2)(2) − (− 1)(1) = 5 2

MACHINE DESIGN 1. A shaft revolving at 1740 rpm is supported by a bearing with a length of 105 mm and a diameter of 64 mm. If the load is light and SAE Oil No. 20 is used with an average temperature of 65 C and diametral clearance Is 0.136 mm., what would be the loss of power due to friction? Express your answer in kW? µ = 2.4 x 10-6 reyns. Solution: Petroff’s equation Tf =

r=

4µπ 2 r 3Lns in − lb cr

D 64 = = 32 mm = 1.25984 in 2 2

L = 105 mm = 4.13386 in ns =

1740 = 29 rps 60

cr =

cd 0.136 = = 0.068 mm = 0.002677 in 2 2

Tf =

4 2.4 ×10 −6 π 2 (1.25984)3 (4.13386)(29 ) in − lb 0.002677

(

)

Tf = 8.48443in − lb Pf =

Tf n 63,000

=

(8.48443)(1740) 63,000

= 0.23433 hp

Pf = (0.23433 hp)(0.746 kW hp) = 0.1748 kW

2. A steam boiler, 1.2 metre in diameter, generates steam at a gauge pressure of 0.7 N/mm2. Assuming the efficiency of the riveted joints as 75% , find the thickness of the shell. Given that ultimate tensile stress = 385 MPa and factor of safety = 5. Solution: D = 1.2 m = 1200 mm p = 0.7 N/mm2 η = 0.75 N=5 σu = 385 MPa = 385 N/mm2

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σu

pD N 2tη 385 (0.7)(1200) = 5 2t (0.75) t = 7.27 mm (answer) 3. The lead screw of a lathe has square threads of 24 mm outside diameter and 5 mm pitch. In order to drive the tool carriage, the screw exerts an axial pressure of 2.5 kN. Find the efficiency of the screw, if it is to rotate at 30 r.p.m. Neglect bearing friction. Assume coefficient of friction of screw threads as 0.12. Solution: Do = 24 mm p = 5 mm W = 2.5 kN N = 30 rpm f = 0.12 p 5 Dm = Do − = 24 − = 21.5 mm 2 2 5 p tan λ = = πDm π (21.5)

σd =

=

λ = 4.2336 o tan β = f = 0.12

β = 6.8428o Efficiency tan λ tan(4.2336 ) = = 0.3781 η= tan(λ + β ) tan(4.2336 + 6.8428 ) η = 37.81% (answer) 4. The ram of a hydraulic press 200 mm internal diameter is subjected to an internal pressure of 10 N/mm2. If the maximum stress in the material of the wall is not to exceed 28 MPa, find the external diameter. Solution: D = 200 mm p = 10 N/mm2 σd = 28 MPa = 28 N/mm2  D  σ t + p − 1   2  σt + p   200  28 + 10  t= − 1   2  28 − 10  t = 45.3 mm t=

Do = D + 2t = 200 + 2(45.3) = 290.6 mm (answer)

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5. A flat spiral steel spring is to give a maximum torque of 1500 N-mm for a maximum stress of 1000 MPa. Find the thickness and length of the spring to give three complete turns of motion, when the stress decreases from 1000 to zero. The width of the spring strip is 12 mm. The Young’s modulus for the material of the strip is 200 kN/mm2. Solution: M = Tmax = 1500 N-mm σb = 1000 MPa b = 12 mm E = 200 kN/mm2 = 200,000 N/mm2 θ = 3 turns = 6π radians 12M σb = 2 bt 12(1500) 1000 = 12t 2 t = 1.225 mm (answer)

θ=

12ML Ebt 3

12(1500 )L (200,000 )(12)(1.225)3 L = 4620 mm = 4.62 m (answer) 6π =

6. The bus bar in a power-generating station consists of a ¼- by 2-in. copper bar rigidly anchored at 3-ft intervals. Power is transmitted over this bus bar at 2,300 volts, and the resistance causes the temperature to rise to 160 F when the room temperature is 80 F. The ultimate strength of the copper is 36,000 psi, and the coefficient of thermal expansion is 0.0000109 in./(in.)(F). What change in stress is caused by the operation of the plant? Solution: ∆t = 160 – 80 = 80 F E = 17 x 106 psi ∆F L L δ = αL∆t = = ∆s A E E ∆s = αE∆t = (0.0000109 ) 30 × 10 6 (180 − 60 ) ∆s = 26,160 psi (answer) 7. Determine the diameter of the piston rod of the hydraulic cylinder of 100 mm bore when the maximum hydraulic pressure in the cylinder is limited to 14 N/mm2. The length of the piston rod is 1.2 m. The factor of safety may be taken as 5 and the end fixity coefficient as 2.

(

)

Solution: p = 14 N/mm2 L= 1.2 m = 1200 mm N=5 C=2 E = 210,000 N/mm2

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Wcr =

I=

π 64

Cπ 2EI L2

D4

π  F = pA = (14 ) (100 )2 = 109,956 N 4

Wcr = NF = 5(109,956 ) = 549,780 N

(2)(π )2 (210,000) π 549,780 =

 4 D 64  

(1200)2

D = 44.4 mm (answer)

THERMODYNAMICS 1. There are withdrawn 200 ft3 of air measured at 15 psia and 90 F from a 50-ft3 tank containing air initially at 100 psia and 140 F. What is the pressure of the air remaining in the tank if its temperature is 130 F? Solution: pi = 100 psia , Ti = 140 + 460 = 600 R , Vi = 50 ft 3 pw = 15 psia , Tw = 90 + 460 = 550 R , Vw = 200 ft 3 Tr = 130 + 460 = 590 R , Vr = 50 ft 3 pV p V pV mi = i i , mw = w w , mr = r r RTi RTw RTr mi − mw = mr piVi pwVw pr Vr − = RTi RTw RTr piVi pwVw pr Vr − = Ti Tw Tr (100)(50) (15)(200) pr (50) − = 600 550 590 pr = 34 psia - Answer 2. Air enters a compressor operating at steady state with a pressure of 14.7 lbf/in.2 and a volumetric flow rate of 8 ft3/s. The air velocity in the exit pipe is 225 ft/s and the exit pressure is 150 lbf/in.2 If each unit mass of air passing from inlet to exit undergoes a process described by pV1.3 = constant, determine the diameter of the exit pipe, in inches. Solution: p1 = 14.7 psia V1 = 8 ft3/s v2 = 225 ft/s p2 = 150 psia n = 1.3

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p1V1n = p2V2n

(14.7)(8 )1.3 = (150 )V21.3 V2 = 1.34 m 3 s V2 = Av 2

π 2 d (225) 4 d = 0.087 ft = 1.044 in (answer)

1.34 =

3. Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 12 m3/min and exits at 12 bar, 400 K. Heat transfer occurs at a rate of 2 kW from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW. Solution: p1 = 1.05 bar = 105 kPa T1 = 300 K V1 = 12 m3/min p2 = 12 bar T2 = 400 K qloss = 2 kW pV (105)(12) = 0.2439 kg s m= 1 1 = RT1 0.287(300)(60) W = mqloss + mc p (T2 − T1 ) W = 2 + 0.2439(1.0 )(400 − 300 ) W = 26.39 kW (answer) 4. Air enters a compressor operating at steady state with a pressure of 14.7 lbf/in.2, a temperature of 80 oF, and a volumetric flow rate of 18 ft3/s. The air exits the compressor at a pressure of 90 lbf/in.2 Heat transfer from the compressor to its surroundings occurs at a rate of 9.7 Btu per lb of air flowing. The compressor power input is 90 hp. Neglecting kinetic and potential energy effects and modeling air as an ideal gas, determine the exit temperature, in oF. Solution: q = 9.7 Btu/lb p1 = 14.7 psia T1 = 80 + 460 = 540 F V1 = 18 ft3/s p2 = 90 psia W = 90 hp x 2544 Btu/hp-hr = 228,960 Btu/hr W = mqloss + mc p (T2 − T1 )

p1V1 (14.7 )(144 )(18) = = 1.3228 lb s RT1 53.34(540) 228,960 = (1.3228)(9.7) + (1.3228)(0.24)(t 2 − 80) 3600 t 2 = 240 F (answer) m=

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5. Air is compressed at steady state from 1 bar, 300 K, to 6 bar with a mass flow rate of 4 kg/s. Each unit of mass passing from inlet to exit undergoes a process described by pV1.27 = constant. Heat transfer occurs at a rate of 46.95 kJ per kg of air flowing to cooling water circulating in a water jacket enclosing the compressor. If kinetic and potential energy changes of the air from inlet to exit are negligible, determine the compressor power, in kW. Solution: p1 = 1 bar = 100 kPa T1 = 300 K m = 4 kg/s p2 = 6 bar qloss = 46.95 kJ/kg n = 1.27 W = mqloss + mc p (T2 − T1 ) T2  p2  =  T1  p1 

n −1 n

1.27 −1

T2  6  1.27 =  300  1  T2 = 439.09 K W = 4[46.95 + 1.0(439.09 − 300 )] W = 744.16 kW (answer) 6. A refrigeration system operating on the reversed Carnot cycle rejects 5800 kJ/min. The minimum and maximum temperatures are 250 K and 345 K, respectively. Find the available part of the heat rejected for a sink temperature of 5oC in kJ/min. Solution: QR = 5800 kJ/min Th = 345 K Tl = 250 K To = 5 + 273 = 278 K Available Part of Energy Rejected A.E . = QR − To ∆S = QR − To

QR Th

 T   278  A.E . = QR  1 − o  = 5800 1 −  = 1,126.4 kJ min  345   Th  7. Five kmols of air initially at one atmosphere and 299 K are compressed isothermally to 8 atmospheres. How much total heat is removed during the compression? a. 25,846.3 kJ

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b. 28,922.9 kJ c. 39,345.6 kJ d. 44,680.9 kJ Solution: p  Q = nRT ln 1   p2   1 atm  kJ    Q = (5 kmol ) 8.314 (299 K )ln  kmol − K    8 atm 

Q = −25,846.27 kJ (a)

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