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Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Problems: 1. The distance between the two parallel plates is 0.00914m and the lower plate is being pulled at a relative velocity of 0.366 m/s greater than the top plate. The fluid used is soy bean oil with a viscosity of 4 x 10-2 Pa.s at 303K. (a) Calculate the shear stress and the shear rate in fps and SI units. (b) If glycerol at 293 K having a viscosity of 1.069 kg/m-s is used instead of soybean oil. What relative viscosity in m/s is needed using the same distance between the plates so that the same shear stress is obtained as in part (a). Also, what is the new shear rate? Answers: (a) shear stress = 3.34 x 10-2 lbf/ft2, shear rate = 40.0 s-1 (b) relative velocity = 0.01369 m/s, shear rate = 1.5 s -1

GIVEN:

dy=0.00914 m du=0.366 m/s = 4X10-2 Pa.s

REQ’D: a. du/dy,  in Pa.s b. du c. du/dy SOL’N: a.

du 0.366m / s = = 40.044 / s dy 0.00914m

 =

du 4 x10 -2 Pa.s = dy 40.044 / s

 = 0.03347 lbf/ft2 b.  = 1.602 Pa

� du � � �0.014m / s �

1.602 Pa = 1.069 Pa.s � du= 0.014 m/s c. = 

du du  1.602 Pa = ; = dy dy  1.609 Pa.s

du = 1.499/s dy

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

2. A 15-cm long cylindrical metal rod slides inside a tube filled with oil. The inner diameter of the tube is 5 cm and the clearance is 0.05 mm. The mass of the bar is 0.5 kg when immersed in the oil. What is the viscosity of the oil? If the steady-state velocity of the rod is 0.1 m.s? GIVEN:

clearance = 0.05 mm

L=15cm

oil m=0.5 kg v= 0.1 m/s 5cm

REQUIRED:

viscosity of the oil

SOL’N: Oil: Newtonian Fluid = 

=

 =

du ; du= 0.1 m/s ; dy=0.05 mm dy

F mg (0.5)(9.8) = = = 207.962 Pa A p dhgc p (0.15)(0.05)(1)

 207.962 = = 0.10398 Pa.s = 103.98 cp du / dy (0.1/ 0.05) /1000

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

3. Water at 20ºC flows at a velocity of 1 m/s over a plane surface 50 cm wide and I cm long. (Take critical NRex = 3.2 x 105. Calculate: (a) the at a distance of 5cm from the leading edge (b) the at which transaction from streamline to turbulent flow occurs on the boundary layer (c) the at a distance of 50 cm from the leading edge GIVEN:

H2O @ 20C V= 1 m/s NRe,x = 3.2x105

 5cm

1m

x REQUIRED: a)thickness of the boundary layer b)distance from the leading edge c)thickness of the boundary layer as well as of the laminar sublayer SOL’N: a) Since the boundary layer is laminar:

1 = 4.64 NRe,x –0.5 x

1 = 0.0410 cm b) NRe,x =

x x r x(1m / s)(1000kg / m)(998.204kg / m3 ) ;3.2 x105 =  1.04cp (0.001)kg / m.s

x= 0.3334m c)

1 = 0.376 NRe,x –0.2 x

1 = 0.376 NRe,x –0.2 50cm = 1.4898cm

1 = 71.5 NRe,x –0.9 x 1 = 71.5 (3.2x105) –0.9 x 1= 0.0397cm

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

4. Problem 3.1 (McCabe and Smith, 6th ed.) p.64: Calculate the Re, (a) water at 10ºC flowing at an average velocity of 2m/s in a 100-mm pipe; (b) air at 2atm pressure and 180ºF flowing at 50 ft/s in a 12-in.duct; (c) oil with a specific gravity of 0.78 and a viscosity of 20 cP flowing at 5ft/s in a 2in. pipe; (d) polymer melt with a density of 900 kg/m3 and a viscosity of 1 Pa.s flowing at 0.2 m/s in a 15-mm tube. SOL’N: a)

ID=100mm H2O T = 10C V = 2m/s r = 1000.32 kg/m3

Re =

Dv r (0.1m)(2m / s)(1000.32kg / m3 ) = = 152,720.61 (turbulent)  1.31x10-3 cp

b) Air P=2atm T=180F v=50 ft/s

ID=12in

@ 180F =  = 0.02cp

Assume ideal gas behavior:

r=

Re=

(12 /12)(50)(0.123425) (0.02)(6.72 x10-4 ) Re = 459,170.387 (turbulent)

PM (2atm)(28.84) = RT (0.7302)(355.222)

r = 0.123425 ib/ft3

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

c)

Oil =20cp v=5ft/s

ID=2in

Dv r (2 /12) ft (5 ft / s)(0.78 x62.4lbm / ft 3 ) Re = =  �6.72 x10-4 lbm / ft.s � 20cp � � 1cp � � Re = 3017.8571 (turbulent)

d) 15mm tube r=900 kg/m3 =1 Pa.s v=0.2m/s (15/1000)m(0.2m/s)(900kg/m3) Re = 1 N m2.s Re = 2.7 (laminar) ANS. a.Re =152,720.61 (turbulent) b.Re = 459,170.387 (turbulent) c.Re = 3017.8571 (turbulent) d. Re = 2.7 (laminar)

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

5. Problem 3.3 (McCabe and Smith, 6th ed.) p.64 Use the nomograph in Appendix 8 to determine the value of n in the equation of gas viscosity for carbon monoxide and for helium over the ranges 0 to 300ºC to 600ºC. SOL’N: 0 - 300C carbon monoxide

300C - 600C carbon monoxide

/o = (T/273)n CO @ 273.15 = 0.017cp CO @ 300C = 0.028cp

CO @ 300C = 0.017cp CO @ 600C = 0.038cp n

�0.017cp � �573.15 K � � �= � � �0.028cp � �273.15 K � n = -0.6733

n

�0.017cp � �873.15 K � � �= � � �0.038cp � �573.15 K � n = -1.9108

Helium:

Helium:

He @ 0C = 0.0182 cp He @ 300C = 0.032cp He @ 300C = 0.032cpHe @ 600C = 0.043cp n

�0.0182cp � �573.15 K � � �= � � �0.032cp � �273.15 K � n= -0.7614

n

�0.032cp � �873.15 K � � �= � � �0.043cp � �573.15 K � n= -0.7019

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

6. Problem 3.4 (McCabe and Smith, 6th ed.) p.65 a) Estimate the transition length at the entrance to a 15-mm tube through which 100 percent glycerol at 60ºC is flowing at a velocity of 0.3 m/s. The density of glycerol is 1,240 kg/m3. (b) Repeat part (a) for 100 percent n-propyl alcohol entering a 3-in pipe at 30ºC and a velocity of 7 ft/s. The density of n-propyl alcohol is 50 lb/ft 3. SOL’N:

a)

Re =

Dv r (15/1000)m(0.3m /s )(1240kg /m 3 ) =  �0.001 � 85cP � kg /m.s � �1cP �

Re=65.6471(laminar)

XT = 0.05(65.6471) = 49.2353mm 15mm

b)

Re =

Dv r (3/12) ft (7 ft /s )(50lbm / ft 3 ) =  �6.72x10-4 � 1.78cP � lbm / ft .s � � 1cP �

Re=73150.75 (turbulent) XT=(45)(3 in) = 135 in.

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

7. Problem 3.8 (McCabe and Smith, 6th ed.) p.65 Air at 30ºC and 5 bars, is flowing inside a ½ -in. Schedule 0 steel pipes. If the flow is at 4.0 ft3/min at standard temperature and pressure (0C and 1atm), is the flow likely to be laminar or turbulent? GIVEN: ½ sched 40 Air 30C & 5 bar q=4 ft3/min REQ’D: type of flow SOL’N: Assume ideal gas behavior

r=

PM = RT

1atm � � 1.01325bar � �

( 5 bars ) ( 28,84 lb/lbmol ) � �

� atm - ft 3 � 0.7302 ( 86 + 460 ) °F � � lbmol - °F � �

= 0.357 lb/ ft3 @30C =0.018 cP; D (1/2 in sched 40) = 0.622in. v = q/s; s=p/4 D2 = (p/4)(0.662/12)2; S= 2.11x103 ft2 m(30C, 5 bars) = m(0C, 1 atm) where m =vrS ; v1r1S1 = v2r2S2 V P ( 1atm ) ( 28.84) = 0.08 lb/ft3 (0°C ,1atm ) V1(30°C ,5bars ) = 2 2 = r1 ( 0.7302) ( 492) V (30°C ,5bars ) =

Re =

( 31.6) ( 0.08 ) ( 0.357 )

( 0.662/12) ( 7.07 ) ( 0.357 )

( 0.018x 6.72x10 ) -4

Re= 10832.89 (turbulent) 8. Problem 3.9 (McCabe and Smith, 6th ed.) p.65

=

= 7.08 ft/s

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Crude oil is pumped at 1.5 m/s through a pipeline 1 m in diameter above. What value of the oil viscosity would laminar flow exist? GIVEN: v = 1.5 m/s D = 1m roil = 0.86x1000 kg/m3 = 860 kg/m3 REQ’D:  when laminar flow exists SOL’N: Re = Dvr 3 Dv r ( 1m ) ( 1.5m /s ) ( 860kg /m ) = = Re 2100

 = 0.6142 kg/m-s

9. Problem 3.10 (McCabe and Smith, 6th ed.) p.65 The apparent viscosity of a non-newtonian liquid at a given shear rate is the value indicated by a viscometer operating on the liquid at that shear rate. It is the viscosity that would be indicated by the viscometer if the liquid were Newtonian. (a) Calculate the apparent viscosity of a 4 percent suspension of paper pulp in water at shear rates du/dy of 10s -1 and 1000s-1. (b) Repeat for a 25 percent suspension of clay in water. Given: a) 4% suspension of paper pulp b) 25% clay suspension du/dy=10/s Req’d:

 du/dy=1000/s

Sol’n: n'

�du � �du � τ = k' � �; τ = μ � � �dy � �dy � a) k’ = 20.02 n’ = 0.575

b) k’ = 1.59 n’=0.185

if du/dy = 10/s

τ = 20.02 [ 10]

if du/dy = 10/s 0.575

 = 1.59 [ 10]

0.185

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

 = 75.2426 Pa

μ=

 = 2.4344 Pa

τ 75.2426 Pa = du 10 dy

μ=

τ 2.4344 Pa = du 10 dy

= 7.524 Pa-s

= 0.24344 Pa-s

if du/dy = 1000/s

 = 20.02 [ 1000]

if du/dy = 1000/s

 = 1.59 [ 1000]

0.575

 = 1062.83 Pa

μ=

0.185

 = 5.707 Pa

τ 1062.83 Pa = du 1000 dy

=

= 1.06283 Pa-s

 5.707 Pa = du 1000 dy = 5.7068x10-3 Pa-s

FLUID FLOW Problems: 1. A 20 wt. % sucrose solution having a density of 1074 kg/m3 is flowing through the piping system shown: 3 – in. 1 – in. 1 ½ - in The steel pipes are schedule 40 pipes. The flowrate entering pipe 1 is 1.892 m 3/hr. the flow divides equally in each of pipes 3. Calculate the following using SI units: (a) the total mass flowrate in pipe 1 and pipes 3. (b) The average velocity in pipes 2 and 3. (c) The mass velocity in pipes 2 and 3. Given: Inside diameter: 1 in = 1.049 in in = 1.610 in 3 in = 3.068 in Req’d:

r= 1074 kg/m3

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

a) m in 1 in. & 3in. pipes b) v in 2in and 3in pipes c) G in 2in and 3in pipes Sol’n: Law of Conservation of mass: m1=m2; m3=m1/2 a) m=qr m1 = 1.892 m3/hr(1074 kg/m3) = 2032.008 kg/hr m3= m1/2= 2032.008/2 kg/hr= 1060.004 kg/hr b) v= q/s

v2 =

v3 =

1.892m3 /hr 2 π 3.068 ( .0254 ) ) ( 4

1.892m3 / hr

p 2 ( 1.61x0.0254 ) 4

= 0.11 m/s

= 0.40 m/s

c) G = m/s � 1hr � 2.032.008 kg/hr � � �3600 s �= 118.36 kg G2 = 2 p m2 -s 3.068 ( 0.0254 ) ) ( 4 2. Water is being pumped from an open water reservoir at a rate of 2.0 kg/s at 10ºC to an open storage tank 1500-m away. The pipe is 3 ½ - in. Schedule 40 pipe and the frictional losses in the system are 625 J/kg. The surface of the water reservoir is 20-m above the level of the storage tank. The pump has an efficiency of 75%. What is the kW power required for the pump? Given: a 

3 ½” sched 40 20m

H2O @ 10C

=0.75 m= 2 kg/s; hf= 625 J/kg

Req’d: Power requirement

b datum 

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Sol’n:

 Wp =

Pb -Pa g a Vb 2 -Va 2 + ( Z b -Za ) + +h f r gc 2gc

Pa=Pb=1 atm Va = Vb=0 (negligible due to large cross-sectional area) Za=20 m Zb=0

Wp = h f -

g J 9.8 Z a = 625 ( 20m ) gc kg 1 = 429 J/kg

wp =

429 J = 572 0.75 kg

P=572

J � kg � �1KW � 2 � � � � kg � s � 1000W � � P= 1.44 KW

3. A pipeline laid cross country carries oil at a rate of 795 m 3/d. The pressure of the oil is 1793 kPa gage leaving pumping station 1. The pressure is 862 kPa gages at the inlet to the next pumping station 2. The second station is 17.4 m higher than the first station. Calculate the lost work (friction losses) in J/kg mass oil. The oil density is 769 kg/m 3. Given:

17.4 m

Req’d:

hf

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Sol’n:

Pb -Pa g a Vb 2 -Va 2  Wp = + ( Z b -Za ) + +h f r gc 2g c Pa = 1793 +10.325 = 1894.3325 KPa Pb= 862 + 101.325 = 963.325 KPa Za = 0; Zb= 17.4 m Wp= 0 (no pump bet pts. A & B)

m = rvs va=vb=q/s

� P -P � g h f = �b a �- Z b � ρ � gc � � � 1894.325x103 Pa - 963.325x103 Pa � �9.8 � hf = � ( 17.4 ) �- �1 � kg 769 3 � �� � � m � hf = 1040.143 J/kg 4. A pump delivers water from a holding tank at atmospheric pressure (100 kPa) to a process equipment at 450 kpa at a flowrate of 6.2 L/s. The process equipment is located 100m higher than the holding tank. Calculate the power requirement for the pump if its efficiency is 70% and if the fluid and the changes in kinetic energy are negligible. Take the density of water to be 995 kg/m3. Given:

b 

Datum a 

100 m

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Req’d: Power Sol’n: Pa= 100 Kpa Pb= 400 KPa Va = Vb=0 (negligible due to large cross-sectional area) Zb= 100m; Za=0 q = 6.2 L/s (1m3/1000L) = 0.0062 m3/s =70% P = Wpm m = qp

m3 kg kg m= 0.0062 x 995 3 = 6.619 3 s m m

 Wp =

Pb -Pa g a Vb 2 -Va 2 3.55 + 0 + ( Z b -Za ) + +h f r gc 2gc 2 .70W p =

100 - 450 + 9.8(20) 995

Wp = 897.487 J/kg P = 6.169 kg/s x 897.487 J/kg = 5,536.6 m P = 5536.6 m 5. Water at 20ºC is pumped at a constant rate 9 m3/hr from a large reservoir resting on the floor to the open top of an experimental absorption tower. The point of discharge if 5m above the floor, and friction losses in the 50-mm pipe from the reservoir to the tower amount to 2.5 J/kg. At what height in the reservoir must the water level be kept if the pump can deliver only 0.1 kW? Given: b a Za

50mm(diameter) Absorption Tower

H2O@ 20C

5m datum

Power= 0.1KW q=9m3/hr

hf=2.5kg/hr

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Req’d: Za Sol’n:

 Wp =

Pb -Pa g a Vb 2 -Va 2 + ( Z b -Z a ) + +h f r gc 2g c Pa=Pb=1 atm Va =0

1hr � 9m3 /hr � � � q �3600s �= 1.27 m/s Vb = = π 2 s ( 0.05m ) 4 Za=? P = Wpm;

m=

Zb=5m but m = qr

9m 3 �1 hr � kg � � 998.204 3 �= 2.5 kg/m3 � � � hr �3600 s � m � � 1000W � � 0.1KW � � �1KW �= 2.5 kg/s Wp = 2.5 kg/s

 Wp =

40 =

Pb -Pa g a Vb 2 -Va 2 + ( Zb -Za ) + +h f r gc 2gc

9.8 1(1.27)2 + 2.5 ( 5 - Za ) + 1 2(1)

Za= 1.26m

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

6. A pump takes water at 60ºF from a large reservoir and delivers it to the bottom of an open elevated tank 25 ft above the reservoir surface through a 3-in. ID pipe. The inlet to the pump is located 10 ft below the water surface, and the water level in the tank is constant at 160 ft above the reservoir surface. The pump delivers 150 gal/min. if the total loss of energy due to friction in the piping system is 35 ft-lbf/lb, calculate the horsepower required to do the pumping. The pump motor set has an overall efficiency of 55%. Given:

b 

160 ft 25 ft Datum

10ft

a  H2O@ 60F

q = 15 gal/min hf= 35 ft-lbf/lbm =55% Req’d: P in hp Sol’n: Basis: 1 min operation

ηWp =

Pb - Pa gαV - V + ( Z b - Za ) + ρ gc

2 b

2gc

2 a

+ hf

Pa=Pb=1 atm Va = Vb=0 (negligible due to large cross-sectional area) Za=0 Zb=160ft

0.55W p =

lb ft.lb f 32.174 ( 160 ft ) f + 35 32.174 lbm lbm

Wp= 354.545 ft-lbf/lbm

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

r@60F= 999.013 kg/m3 = 62.37 lb/ft3

Power= Wpm; where m= qr

m = 150

� 1 ft 3 � gal � 1min � 62.73 lb/ft 3 ) ( � � � � min �60s � �7.481gal �

m= 20.85 lb/s

� � 1 hp � ft-lbf � lb � � \ P= � 354.55 20.85 � � � � lb m � s� � �550 ft-lbf � � lb m �

� � �; � � �

P = 13.4 hp

7. A pump pumps 0.200 ft 3/s or brine solution having a density of 1.15 g/cm 3 from an open feed tank having a large cross-sectional area. The suction line has an inside diameter of 3.458 inches and the discharge line from the pump a diameter of 2.067 inches. The discharge flow goes to an open overhead tank and the open end of this line is 75 ft above the liquid level in the feed tank. If the friction losses in the piping system are 18.0 ft-lb f/lbm, what pressure must the pump develop and what is the horsepower of the pump if the efficiency is 70%? The flow is turbulent.

Given:  75ft

datum

brine r=1.15g/cm3 D=2.067” 3.548”(D)

hf=18 ft-lbf/lbm =0.70 q=0.2 ft3/s

Req’d: a) P in hp b)Pressure that the pump must develop Sol’n:

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Pb -Pa g a Vb 2 -Va 2  Wp = + ( Z b -Za ) + +h f r gc 2gc Pa=Pb=1 atm Za=0 Zb=75ft Va=0;

Vb =

q 9 ft 3 / s = = 8.58 ft/s 2 s p �2.067 � � � 4 � 12 �

� lbf � ( 1) ( 8.58) + 18 ft - lbf 0.70Wp = � 1 ( 75) + � 2 ( 32.174 ) lb m � lb m � 2

Wp= 134.49 ft-lbf/lbm Power= mWp

where m= qr

rbrine= 71.77 lb/ft3 m= 0.2 ft3/s (71.77 lb/ft3)=14.354 lb/s

� � 1hp Power = 134.49 ft-lbf/lbm (14.354 lb/s) � ft-lbf �550 lbm �

� � �= P = 3.51 hp � �

Consider the pump: 

a

b



 Wp =

Pb -Pa g a Vb 2 -Va 2 + ( Z b -Za ) + +h f r gc 2g c

Assume= Zb-Za

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Va= q/s =

Va= q/s

0.2 ft 3 / s 2

p �3.548 � � � 4 � 12 �

0.2 ft 3 / s 2

p �2.067 � � � 4 � 12 �

= 2.91 ft/s

= 8.58 ft/s

2 2 � ( 8.58) - ( 2.91) � Pb - Pa 1 � � + ( 0.7 ) ( 134.49 ) = 71.77 2 ( 32.174 )

Pb-Pa = 6683.98 lbf/ft2

PROBLEMS: 1. Calculate the frictional pressure drop in pascal for olive oil at 293 K through a commercial pipe having an inside diameter of 0.0525 m and a length of 76.2 m. the velocity of the fluid is 1.22 m/s. is the flow laminar or turbulent? Use the friction factor method. For olive oil, density= 919 kg/m3; viscosity = 84 x 10-3 Pa.s Given: ID = 0.0525m

oil v=1.22m/s roil=919kg/m3  = 84x10-3 Pa.s T=293K Req’d: Ps, Re, f

L=76.2m

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Sol’n:

Ps = r hf s hf s = Re =

4 fLV 2 2 gcD .0525m ( 1.22 m/s ) ( 919kg / m 3 ) 84 x10 -3 Pa.s

= 700.7375 Flow is laminar

f = 16/Re f = 16/700.7375 = 0.02283 4 ( 0.02283) ( 76.2 ) ( 1.22 ) hf s = = 98.65 J/kg 2 ( 1) ( 0.0525) 2

Ps = 919(98.65) = 90659.35 Pa by Hagen Eqn Ps =

32 ( 84 x10 -3 ) ( 1.22 ) ( 76.2 )

( 0.0525)

2

= 90662.03 Pa

2. A liquid having a density of 801 kg/m3 and a viscosity of 1.49 cP is flowing through a horizontal straight pipe at a velocity of 4.57 m/s. The commercial steel is 1 ½ -in. Schedule 40. For a length of 61 m, calculate (a) friction loss (b) For a smooth tube of the same diameter, calculate the friction loss (c) what is the % reduction? Given: ID = 1.61 in- .04089 m

liquid v=4.57m/s roil=801kg/m3  = 1.49 cP T=293K

L=61m

Req’d: a) friction loss; also for smooth tubes; and % reduction

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Sol’n:

4 fLV 2 hf s = 2 gcD e= .0457mm e 0.0000457 = = 0.0011176 D 0.04089 ( 0.04089 ) ( 4.57 ) ( 801) = 100, 456.85 Re = 1.49 x10-3 Re = 1.0 x105 f= .0055 4 ( .0055) ( 61) ( 4.57 ) hf s = = 367.55 J/kg 2 ( 1) ( .04089 ) 2

b) e/D = 0; f= .0045 4 ( .0045) ( 61) ( 4.57 ) hf s = = 280.34 J/kg 2 ( 1) ( .04089 ) 2

c) % reduction =

367.55 - 280.301 x100 = 23.73 % 367.55

3. Water at 60ºF is pumped from a reservoir to the top of a mountain through a 6-in schedule 120 pipe at an average velocity of 12 ft/s. the pipe discharges into the atmosphere at a level of 3000 ft above the level in the reservoir. The pipeline itself is 4500 ft long. If the overall efficiency of the pump and the motor driving it is 70% and the cost of electric energy to the motor is 4 cents per kW-hr, what is the hourly energy cost of pumping this water? 4. centrifugal pump takes brine from the bottom of a supply tank and delivers it into the bottom of another tank. The line between the tanks is 600 ft of 4 -in schedule 40 pipe. The flowrate is 400 gal/min. In the line are two-gate valves, four standard tees, and four ells. What is the energy cost for running this pump for one 24-hr day? The specific gravity of brine is 1.18, the viscosity of brine is 1.2 cP and the energy cost is $400 per hp-year on a basis of 300 days per year. The overall efficiency of the pump and motor is 60%.

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Given: b 150ft datum a

4” sched 40

Brine SG=1.18  = 1.2 cP

q=400 gal/min

L=600ft

= 0.6 * 2 gate valves Energy cost = $400/hp-yr * 4 standard tees * 4 ells Req’d: cost of pumping per day Sol’n:

 Wp =

Pb -Pa g a V 2 -V 2 + ( Z b -Za ) + b a +hf s + hf e + hf c + hf f r gc 2gc Pb=Pa=Patm Za=0 Zb= 150ft Va=Vb= 0

Skin friction: hf s =

4 fLV 2 ; L= 600ft 2 gcD

�1 � �1 � 400 � � � � q �7.481 � �60 �= 10.08 ft/s v= = 2 s p �4.026 � � � 4 � 12 � f= f (e , Re)

D(4" sched 40) = 4.026 in.

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

�4.026 � ( 10.08) ( 1.18 x62.4 ) � Dv r � 12 � � Re = = = 308794 = 3.1x105  1.2 ( 6.72 x10-4 )

e 0.0457 = = 0.00044 from fig 6.9(HB) f = 0.0044 D 4.026 ( 25.4 ) 4 ( 0.0044 ) ( 600 ) ( 10.08) ft-lbf hf s = = 49.7 2 ( 32.174 ) ( 4.006 /12 ) lbm 2

contraction: Kc=0.4 0.4 ( 10.08) hf c = = 0.6316 2 ( 32.174 ) 2

Expansion: Ke=1 1 ( 10.08) hf e = = 1.579 2 ( 32.174 ) 2

fittings: Kf(gate valves) = 2(0.17) Kf(tees) = 4(0.4) Kf(ells) = 4(0.75) Kf= 4.94 4.94 ( 10.08) ft-lbf hf f = = 7.8003 2 ( 32.174 ) lbm 2

0.6(Wp) = (1)(150 ft)+49.7+0.6316+1.579+7.8 Wp= 349.52 ft-lbf/lbm lbm �1 � �1 � m = r q = ( 1.62 x10-4 ) ( 400 ) � � � �= 65.63 s �7.481 � �60 � P = Wp(m) �1 � P = 349.52 ( 65.63) � �= 41.70 hp �550 � � 400 � � 1 yr � cost = 41.7 hp � � �300 days � 1 hp-yr � � � � cost = $ 55.61/day

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Problem: Determine the pressure in overcoming the friction in a coil through which water flows with a velocity of 1.2m/s. The coil is made of steel pipe with an inside diameter of 30mm. The diameter of the coil is 1m. The no. of turns is 10. The average temp. of water is 30C. Req’d: Pcoil Sol’n:

hf s =

4 fLV 2 Dvr ( 0.03) ( 1.2 ) ( 995.647 ) ; Re = = = 42,165.58 = 4.2 x10 -5 -3 2 gcD  ( 0.85x10 )

e 0.0457 mm = = 1.52 x10-3 D 30 mm from fig 6.9 (HB) = f = 0.0065 L= Dc(no. of turns) L = (1m)(10) = 31.42 m 4 ( 0.0065) ( 31.42 ) ( 1.2 ) = 19.60 J/kg 2 ( 1) ( 0.030m ) 2

hf s =

Ps = (995.647)(19.6)= 19518.20 N/m 2 � 3.54 ( 0.03) � � 3.54d � Pcoil = Ps � 1+ = 19518.20 � 1+ � � D � 1m � � � Pcoil = 21591.04 Pa

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Assuming the nozzle is to be used supersonically, what is the maximum Mach number at the discharge of the divergent section? Mass velocity at the discharge: G=

3476 = 1738 kg/m2.s 3

17.38 =

1 1 ( 2)(1.4)(12.72)( 20)(1.01325x10 5 )   Pr  1.4 1-    0.4  Po  

1  Pr  1.4      Po  

Pr = 0.0939 Po  2  1 Ma =  1 11.4 - 1   0.0939 1.4

   -1  

Pr = Ma = 2.20 Po

When LB = Lmax, P, T, ρ  negligible (The gas leaves at asterisk condition, Mab = 1) For acoustical velocity: Aa

=

  (1.4)(1000 R)(0.7302atm. ft 3 / lbmol R)(14.7lbf .N / s 2 .m 2 / 1atm)1 ft 2 (32.174) 1 / 144 in   29lbm / lbmol 2

Aa = 1549.4529 ft/s Velocity at the entrance of the pipe is: a = (0.05) (1549.4529 ft/s) = 77.5 ft/s

Ta 1 + [ (  - 1) / 2] Ma, b 2 = TB  T * 1 + [ (  - 1) / 2] Ma, a 2

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

1000 24 = T * 2 1 + [ (1.4 - 1) / 2]0.05 2





T*= 834R

r a Ma, b 1 + [ (  - 1) / 2] Ma, a 2 = r * B Ma, a 1 + [ (  - 1) / 2] Ma, b 2



0.795 1 2 1 + [ (1.4 - 1) / 2]0.05 2 = r * 0.05 2.4 ρ*=0.0435 lbm/ft3

r a Ma, b 1 + [ (  - 1) / 2] Ma, b 2 = r * Ma, a 1 + [ (  - 1) / 2] Ma, a 2 20 1 = 1.2 r * 0.05 ρ*=0.913atm Exit velocity is sonic: T=T*=834R a =  * = 1549.4529

*=1416 ft/s G through the entire pipe: G = (0.795)(77.5 ft / s )

G = 61.61 lb/ft2.s

834 1000



Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Example Problem 6.2: Air flows from a reservoir through an isentropic nozzle into a long, straight pipe. The pressure and temperature in the reservoir are 20 atm and 1000R, respectively, and the Mach number at the entrance of the pipe is 0.05. (a) What is the value of ƒL max/rH? (b) What are the pressure, temperature, density, linear velocity, and mass velocity when L b=Lmax? (c) What is the mass velocity when ƒLmax/rH = 400? Required: a)

fL max RH

b) P*, Aa, a, Ga, T*, ρ* c) G* Solution: a)



fLmax 1  1 1 + [ (  - 1) / 2] Ma, a 2  +1 = -1 ln 2 RH   Ma a 2 2 Ma, a 2 ( + 1)



   

fLmax 1 + [ (1.4 - 1) / 2]0.05 2 1  1 1.4 + 1  = -1 ln 2 RH 1.4  0.05 2 2 0.05 2 (1.4 + 1) fL max = 280.0203 RH

Pa 1 = b) Po 1 +  (  - 1)  Ma 2  1       2   1 -1 /  Pa =

20  1.4 - 1  1 0.05 2  1 +     2   1 - 1 / 1.4

Pa= 19.9650  20 atm

   

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics





fLmax 1  1  + 1 2 1 + [ (  - 1) / 2] Ma, a 2  =  2 -1 ln c)  RH   Ma , a 2 Ma 2 , a(1.4 + 1)  400 =





1  1 1.4 + 1 2 1 + [ (1.4 - 1) / 2] Ma, a 2  1 ln   1.4  Ma 2 , a 2 Ma 2 , a (1.4 + 1) 

Ma, a = 0.04194 a =

0.04194 ( 77.5) 0.05

a = 65 ft/s G = (65)(0.795) G = 51.7 lb/ft2.s Problem 6.4: A standard 1-in schedule 40 horizontal steel pipe is used to conduct chlorine gas. The gas enters the pipe through a rounded entrance at a pressure of 6 atm abs, a temperature of 120C, and a velocity of 35m/s. (a) What is the maximum possible length of the pipe? (b) What are the pressure and stagnation temperature of the gas at the end of the pipe at maximum length? Assume adiabatic flow. For Chlorine,  = 1.36 and M = 70.91. Given:

Gas : Pa = 6 atm Ta = 120C  = 35 m/s  = 1.36 M= 70.91 g/mol Required: a) Lmax b) Pb and T Solution:

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

     -1 2  2 1 +   Ma a    2 fL max  +1 1 1     =  -1ln  RH   Ma 2 a 2 Ma 2 a ( + 1)          2  2M ( 35m / s ) ( 70.91kg / mol )  Ma 2 a = a =  Ta R  m 3 .atm  101325Pa    (1.36)(120 + 273)    0.08205 kmol.k   1atm      

Ma 2 a = 0.019548656

At Point a: ra =

( 6atm )( 70.91) PM = ( 0.08205)(120 + 273) RT

r a = 13.196 kg/m3  @ 120C = 0.018 Cp

 1.049  1    ( 35)(13.196)  12  328  Re = 0.018 x10 - 3

Re = 683, 845.9  0.0457mm = = 0.0017 D (1.049 )( 2.54 )

From Figure 6.9 (HB): ƒa = 0.0056 At Point b: Ta 1 + [ (  - 1) / 2] Ma 2 b = Tb 1 + [ (  - 1) / 2] Ma 2 a

Ma2b=1

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

1.36 - 1 1+  1 120 + 273  2  = Tb 1.36 - 1 1+   0.01954  2  Tb = 334.22K  61.22C

Pa Ma b = Pb Ma a

1 + [ (  - 1) / 2] Ma 2 b 1 + [ (  - 1) / 2] Ma 2 a

Pb = 0.7735 atm r a Pa Tb = r b Pb Ta 6atm( 334.22 ) 13.19 = rb 0.7735( 393.15) r b = 2.0002 kg/m3  @ 334.22 K = 0.015cP

b =

Tb R = M

(1.36)( 334.22)( 0.08205)(101325) 70.91

 b = 230.85 m/s

 1.049   1   ( 230.85)  ( 2 ) DVr  12   3.28  Re = =  0.015 x10 - 3

Re = 820, 330. 79  8.2 x 105  0.0457 = = 0.0017 D (1.049)(25.4)

; ƒb = 0.0057

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

f =

0.0056 + 0.0057 = 0.00565 2

   1.36 - 1    21 +  0.01954    0.00565( Lmax ) 1  1   1.36 + 1   2  = -1-  ln    0.01954(1.36 + 1)  1.049   1 ft  1m  1.36  0.01954  2     in     4   12in  3.28    L max = 39.65 m

b) Stagnation Condition: (Ma = 0)

Ta Tc ( s )

  - 1  2  1 +   Ma b  2   =   - 1  2  1 +   Ma a   2  

 1.36 - 1   1 +  0  393.15  2   = Tc ( s )  1.36 - 1   1 +  0.01954  2   Tc( s ) = 394.53 K

Or Use data at end of line: T = 334.35  1.36 - 1  1+  0 334.35 2   = Tc  1.36 - 1  1+  1 2   Tc = 394.53 K

FLOW PAST IMMERSED BODIES:

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

(Example Problem 3.1-1, Transport Processes and Unit Operations, 3 rd ed. By Geonkoplis, Christi) Air at 37.8C and 101.3KPa absolute pressure flows past a sphere having a diameter of 42 mm and a velocity of 23 m/s. What is the drag coefficient and the force on the sphere? Given:

V= 23 m/s.

42mm

Air : t=37.8C P=101.3KPa Required: a) CD b) FD Solution: FD CD =

Re p =

ρμ o

AP

2

2gc

Dp vr



 air @ 37.8C = 0.0185cP

Re p =

(421000)( 23m / s )(1.13)

Re p = 59,

0.0185 x10 -3 004

From Figure 7-3: CD = 0.5 FD =

C D A p r 2 2 gc

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

( 0.5)  p ( 0.042) 2 ( 4.13)( 23) 2 FD =

4

2(1)



FD = 0.207 N

FLOW PAST A PACKED BED: 1. Air at 311 K ia flowing through a packed bed of spheres having a diameter of 12.7mm. The void fraction of the bed is 0.38 and the bed has a diameter of 0.61 m and a height of 2.44 m. the air enters the bed at 1.10 atm abs at the rate of 0.358 kg/s. Calculate the pressure drop of the air in the packed bed. Given:

0.61m Dp=12.7mm ε= 0.38

2.44 m

Air: T=311K P=1.10 atm abs m=0.358 kg/s

Required: ∆P Solution:

P 150Vo  (1- ) 2 1.75rVo 2 (1- ) = 2 2 3 + L s Dp  sDp 3 L=2.44m Dp=12.7m = 0.38

 =1

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Vo =

m = rs

r air =

0.358 = 0.88m / s p 2 (1.24) (0.61) 4

(1.1atm)(28.84) = 1.24kg / m 3 (0.08205)(311K )

air@311K=0.018cP 150(0.99)(0.018 x10 -3 )(1 - 0.38) 2 1.75(1.24)(0.99) 2 (1 - 0.38) P = + 2.44m (1) 2 (0.0127) 2 (0.38) 3 (1)(0.0127)(0.38) 3

∆P = 4900.25 Pa 2. A partial oxidation is carried out by passing air with 1.2 mole percent hydrochloride through 40mm tubes packed with 2m of 3-mm by 3-mm cylindrical catalyst pellets. The air enters at 350C and 2 atm with a superficial velocity of 1 m/s. What is the pressure drop through the packed tubes? How much would the pressure drop be reduced by using 4-mm pellets? Assume = 0.40. Given: Catalyst pellets Dp=3mm

2m

40mm

Air with 1.2% mole HCl T=350C P=2 atm V=1 m/s = 0.40.

Required: a) ∆P b) ∆P if dp=4mm Solution:

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

P 150(Vo )( )(1- ) 2 1.75rVo 2 (1- ) a) = 2 2 3 + L (s) Dp  sDp 3 P 150(1)(3.1x10 -4 )(1 - 0.4) 2 1.75(1.1281)(1)(1 - 0.4) = + 2 2m  p  3  3 3  2  p  3 (1)  (0.4) (1)    (0.4)  4  1000   4  1000 

∆P = 89, 717. 0340 Pa b)∆P if dp = 4 mm (150)(1)(3.1x10 -4 )(1 - 0.4) 2 (1.75)(1.1281)(1)(1 - 0.4) P = + 2 2  p  4  3  p  4  3 (1)  (0.4) (1) 2    (0.4) 4 1000     4 1000   

∆P = 53, 411. 4539 P

FLOW IN A FLUIDIZED BED 1. Catalyst pellets 5 mm in diameter are to be fluidized with 45, 000 kg/h of air at 1 atm and 80C in a vertical cylindrical vessel. The density of the catalyst particles is 960 kg/m3; their sphericity is 0.86. If the given quantity is just sufficient to fluidize the solids, what is the vessel diameter? Given: Catalyst pellets Dp=5 mm ρp=960 kg/m3 s=0.86 Air P=1 atm T=80C m=45, 000 kg/h Required: D

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Solution: 2 2  g ( r - r ) = 150Vo  (1- ) + 1.75rVom  gc  p s 2 Dp 2 3 sDp  m 3

Assume:  m = 0.45 r air =

(1atm)(28.84) = 0.996kg / m 3 (0.08205)(80 + 273)

150Vom (0.02 x10 -3 )(1 - 0.45) (1.75)(0.995)(Vom ) 2  9.8  +  ( 960 - 0.996 ) = (0.86) 2 (0.005) 2 (0.45) 3 (0.86)(0.005)(0.45) 3  1 

Vom = 1.35 m/s m = VrS ; S =

S=

(450009kg / hr )(1hr / 3600s ) = 9.3m 2 (1.35m / s )(0.996)

p 2 D ;D = 4

4(9.3) = 3.44m p

2. Solid particles having a size of 0.12 mm, a shape factor of 0.88 and a density of 100 kg/m3 are to be fluidized using air at 20 atm and 25c > the voidage at minimum fluidizing conditions is 0.42. (a) If the cross section of the empty bed is 0.30 m 2 and the bed contains 300 kg of solid, calculate the minimum height of the fluidized bed. (b) Calculate the pressure drop at minimum fluidizing conditions. (c) Calculate the minimum velocity for fluidization. Given: Dp=.12 mm ρp=100 kg/m3 s=0.88

Air P=20 atm T=25C Voidage=0.42. Required: a) Lm

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

b) ∆P c) Vom Solution: r air =

(2atm)(28.84) PM = = 2.36kg / m 3 RT (0.08205)(298.15)

V solid =

300kg = 0.3m 3 1000kg / m 3

V = LS , V solid = V fluidized (1-  m)

V fluidizedbed =

V solid 0.3m 3 = (1-  m) 1 - 0.42

V fluidizedbed = 0.517m 3

0.517 m 3 a) 0.3m 2 L m = 1.72m Lm =

b) P = L m ( g / gc)(1-  m)( r p - r )  9. 8  P = 1.72 (1 - 0.42)(1000 - 2.36)  1  P = 9753.41Pa

c) 150(Vom )(0.018 x10 -3 )(1 - 0.42) 1.75( 236)(V 0m 2 )  9.8  ( 1000 2 . 36 ) = +   (0.88) 2 (.00012) 3 (0.42) 3 (0.88)(0.00012)(0.42) 3  1 

Vom = 5.15x10-3m/s

FLOW IN EXPANDED BED Using data in problem no.2, estimate the voidage of the bed if an operating velocity of 3 times the minimum is used. Given:

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Dp=.12 mm ρp=100 kg/m3 s=0.88

Air P=20 atm T=25C V=3(minimum)

Required:  ; Le Solution:  g  150V (1- ) 1.75 rVom 2  ( r p - r ) = + s 2 Dp 2  m 3 sDp  m 3  gc 

Flow is Laminar: 150Vo  3 = = Km.Vom 2 1-  s Dp 2 g ( r p - r ) 0.42 3 (1 - 0.42) Km = = 24.8 = Ke 5.15 x10 - 3  e3 = ( 24.8)(3)(5.15 x10 - 3 ) 1-  e  e = 0.555

Vsolid @min. condition=Vsolid @exp.condition V bed (1-  m) = V bed (1-  e) L m S (1-  m) = L e S (1-  e) (1.72)(1 - 0.42) = L e (1 - 0.555)

Le = 2.22m

MOTION OF PARTICLES THROUGH FLUIDS

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

1. Oil droplets having a diameter of 20 microns are to be settled from air at an air temperature of 37.8C at 101.3 KPa pressure. The density of the oil 900 kg/m 3. Calculate the terminal setting velocity of the drops. Given: Dp= 20 microns Ρ=900 kg/m3

Air P=101.3 KPa T=37.8C

Required: Ut Solution:  gr ( r p - r )  K = Dp   2  

r air

1/ 3

 101.3   ( 28.84) PM  101.325  = = RT (0.08205)(37.8 + 273)

r air = 1.13kg / m 3  air = 0.018cP

 (9.8)(1.13)(900 - 1.13)  K = 20 x10 - 6 m   (0.0185 x10 - 3 ) 2  

K = 0.62 (Stoke’s Law range) Ut =

(9.8)(20 x10 -6 ) 2 (900 - 1.13) 18(0.0185x10 - 3 ) U t = 0.0106m / s

BAROMETRIC CONDITION

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

2.2  The P.42  

temperature of the Earth’s atmosphere drops about 5C for every 1000 m of

elevation above the Earth’s surface. If the air temperature at ground level is 15C and the pressure is 760mmHg, at what elevation is the pressure 380mmHg? Assume that the air behaves as an ideal gas. Given: P= 380mmHg Z

1000m above ground T drop of 5C

Ground level T=15C P=760mmHg, Required: Z Solution: For non-isothermal case: P2  BZ  gM = 1  P1  To  gcRB -

dT 5C 5K =B= = dz 1000m 1000m

   5   ( Z )   380mmHg 9.8( 28.84)  1000   = 1 760mmHg 288 K   1(0.08205 x101325)(0.005)    

Z = 5582.68m

HYDROSTATIC PRINCIPLE IN A CENTRIFUGAL FIELD

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

2.6  A P.42  

centrifuge bowl with 250 mm ID (Internal Diameter) is turning at 4000 r/min. It

contains a layer of aniline 50 mm thick. If the density of the aniline is 1002 kg/m 3 and the pressure at the liquid surface is atmospheric, what gauge pressure is exerted on the wall of the centrifuge bowl? Given:

ID=250 mm 4000 r/min 50 mm ρ=1002 kg/m3

Required: P2 – P1 (Pressure exerted by the fluid) Solution: P2 - P1 =

(

2r 2 2 r2 - r1 2 gc

)

=

2p ( 4000) = 418.88 / s 60

r2 =

250 = 125mm = 0.125m 2

r1 = r2 - thickness r1 = 0.125m - 0.050m r1 = 0.075m P2 - P1 =

[

( 418.88) 2 (1002kg / min) (0.125) 2 - (0.075) 2 2(1kg.m / N .s 2 )

]

P2 - P1 = 879056.88 Pa 2.5  What P.42  

should be the volume of the separator in Prob. 2.4 to separate 1600 kg/h of

wash liquid? The wash liquid is to be continuous phase; its viscosity is the same as that of the operating temperature of 35C.

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

Given: m=1600 kg/h T=35C. Required: Volume of Decanter Solution: V mixture = VC 6 HCl + V washliquid V decanter =

V mixture 0.90

q chlorobenzene = q wash liquid = t=

1600kg / hr = 1.44m 3 / hr 3 1109 kg / m

2100kg / hr = 2.06m 3 / hr 3 1020kg / m

100(0.723cP) 100 = r A - r B (1109 - 1020)

t = 0.81hr

H

2O

@ 35C = 0.723cP) Appendix 6

VC6 H 5Cl = 1.44m 3 / hr (0.81hr ) = 1.17 m 3 V washliquid = 2.06m 3 / hr (0.81hr ) = 1.67 m

3

2.84m3 2.84m 3 0 .9 = 3.16m 3

V decanter = V decanter

CENTRIFUGAL DECANTER 2.7  The P.42  

liquids described in Problem 2.4 are to be separated in a tubular centrifuge bowl

with an inside diameter of 150mm, rotating at 8000 r/min. The free liquid surface inside the

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

bowl is 40mm from the axis of rotation. If the centrifuge bowl is to contain equal volumes of the two liquids, what should be the radial distance from the rotational axis to the top of the overflow dam for the heavy liquid? Given:

ID= 150 mm 8000 r/min d=40mm

Required: TA=? Solution:

( rA ) 2 - ( r B / r A ) r 2 B ri = 1- r B / r A VC6 H 5 Cl = V washliquid

Wash liquid

Chlorobenzene

2

pri b - pr 2 B b = V washliquid 2

2

pr2 b - pri b = VC6 H 5Cl

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

2

2

2

r2 - ri = ri - rB 2

2

2

r + rB 150 ri = 2 ; r2 = = 75mm 2 2 2

ri =

(75) 2 + ( 40) 2 2

ri = 60.1mm 2

60.1mm =

(

rA - 1020

)(40) 1109

2

 1020  1-    1109  rA = 41.97mm

2.3  How P.42  

much error would be introduced in the answer to Prob. 2.2 if the equation for

hydrostatic equilibrium (Eq. 2.4) were used, with the density evaluated at 0C and an arithmetic average pressure? Given: Assumption: Density (ρ) is constant Required: % error Solution: Pb Pa = g ( za - zb) gc r r

r=

Where: P =

PM RT

Pa + Pb 380 + 760 = = 570mmHg 2 2

T = 0C + 273 = 273K

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

( 28.84kg / kmol ) (570mmHg )1atm 760 mmHg   r= 3 (0.08205m .atm / Kmol.K )(273K ) r = 0.9656kg / m 3 

( 760 - 380) mmHg  101325N / m 

760mmHg

2

  9.8  N  =  0.9656kg / m 3 ( z )    1  kg

(

z = 5353.81m %error =

5583 - 5354 (100) = 4.1% 5583

4) Given:

Φ=2 in Kerosene SG= 0. 815

Methane T=60F

Φ = ¼ in Required:

H 2O

)

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

a) PA-PB (reservoir levels neglected) b) PA-PB (reservoir levels accounted) % error Solution: 3 *@60F, r CH 4 = 0.042358lbm / ft a) PA - PB = ( R - RO )( r A - r B )

[(

)

]

 1 ft  PA - PB = (5.72in - 0) 0.042358lbm / ft 3 - (0.815)(62.37lbm / ft 3 )   12in  PA - PB = -0.1681inH 2 O

3

PB - PA = 0.1681inH 2 O

 a a  b) PA - PB = ( R - RO )  r A - r B +   r B -   r C   A  A  

� ( 0.042358lbm/ft 3 ) - ( 0.815) ( 62.37lbm/ft 3 ) � 3 � � 2 2 1ft/in � �� �1 � PA - PB = (5.72 - 0)in � �1 � � � �� � �4 � lbm ��12 � 4� � � � + � ( 0.815) ( 62.37 ) - 2 ( 62.37 ) 3 � 2 ft � � 22 PA - PB = -0.1687inH 2 O PB - PA = 0.1687inH 2 O

%error =

0.1687 - 0.1681 x100 ; % error = 0.3557% 0.1687

7. Earth’s atmospheric condition varies somewhat on a certain day. The sea level temperature is 45F and sea level pressure is 28.9 in Hg. An airplane overhead registers a temperature of 23F and a pressure of 12 psi, estimate the plane’s altitude in feet. Given: P=12 psi; t=23F

X

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

sea level t= 45F ;P=28.9 in Hg Required: X (height or altitude of plane)

Solution: PB Bz   = 1 PA Ta  

w=

g / gc

M RB

; Let g/gc

M =w RB

lbf 29lbm x lbm lbmol 1.8 R  K 10.73 ft 3 .lbf x0.0065 x 1 K m in 2 .lbmol  R

2

 12in  1m  x x 3 . 28 ft 1 ft  

w = 5.26

1    0.0065( z ) 3.29  12 psi ln = 5.26. ln 1  14.7  280.37    28 . 9 x psia      29.92  

z=4454ft

8. What is the pressure at the top of an 80 floor building on a day when air can be assumed still and that the temperature of air at the first floor level is 24C? Temperature change is not significant? Each floor in the building measures 10 feet high. Given:

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

PTop?

TA = TTop

800ft

PA=1 atm TA=24C

Required: PTop Formula:

PTop PA

=e

-g

M Z gc RT

Solution:

PTop

 - 9.8KN  2 gkg   1m     x800 ft   1000kg  1Kmol  air  3.28 ft  = 1atm e  KN .m 3 ( 24 + 273N ) K 8.314 2 m .kmol.K

PTop = 0.9723atm

9. Resolve number 8 accounting for temperature change. Given: Pa = 1 atm

; Nair= 29lb/lbmol

; z=800 ft

; T=297.15

Required: PTop Solution: PTop PA

 Bz  = 1  TA  

g

M gc RB

g M

; Let gc RB = w 2

lbm  1.8 R 0.0065K 10.73 ft 3 .lbf  12in  1m  1lbf w= x 29 x 2 x  x / K x 1 lbm lbmol m 3 . 28 ft 3 . 28 ft m . lbmol  R    

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

w = 5.26   1m   0.0065 K / m 800 1     3.28   = 1atm 1 297.15 K      

(

PTop

)

5.26

PTop = 0.9723atm

EQUIVALENT DIAMETER OF DIFFERENT CROSS SECTIONAL AREAS 1. SQUARE a

a

Deq = 4xRH RH =

Area WettedPeri meter

RH =

a2 4a

Deq = 4 x Deq = a

2. RECTANGLE

a 4

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

W L Deq = 4 x

LxW 2 L + 2W

4 LW

Deq = 2( L + W ) 2 LW

Deq = ( L + W ) 3. EQUILATERAL TRIANGLE

h b

1 bh 2 Deq = 4 x 2 = h 3b 3

4. HALF-CIRCLE (OPEN)

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

p 2  1  D  4 2 Deq = 4 x  ( pb ) 1 2

= 4x

D 4

Deq = b 5. HALF-CIRCLE (CLOSE)

Deq = 4 x = 4x

pD 4p + 8

Deq =

6. ANNULUS

pD 2 4pD + 8

pD p +2

Complilation of Notes in Che 411 Unit Operations 1- fluid Mechanics

p p 2 2 D o - Di 4 Deq = 4 x 4 pDo + pDi

=

( Do + Di )( Do - Di ) ( Do + Di ) Deq = Do – Di