50 SOLVED PROBLEMS TRANSPORT PROCESSES AND UNIT OPERATIONS GEANKOPLIS PDF

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BICOL UNIVERSITY COLLEGE OF ENGINNERING DEPARTMENT OF CHEMICAL ENGINEERING A.Y. 2016-2017 50 SOLVED PROBLEMS TRANSPORT PROCESSES AND UNIT OPERATIONS GEANKOPLIS Prepared by: Neil Dominic D. Careo BS Chemical Engineering IV Professor: Engr. Junel Bon Borbo, MSChE PART I: MOMENTUM TRANSFER 1.Pressure D...

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50 SOLVED PROBLEMS TRANSPORT PROCESSES AND UNIT OPERATIONS GEANKOPLIS Neil Careo

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BICOL UNIVERSITY COLLEGE OF ENGINNERING DEPARTMENT OF CHEMICAL ENGINEERING A.Y. 2016-2017

50 SOLVED PROBLEMS TRANSPORT PROCESSES AND UNIT OPERATIONS GEANKOPLIS

Prepared by: Neil Dominic D. Careo BS Chemical Engineering IV Professor: Engr. Junel Bon Borbo, MSChE

PART I: MOMENTUM TRANSFER

1.Pressure Drop of a Flowing Gas. Nitrogen gas is flowing through a 4-in schedule 40 commercial steel pipe at 298K. The total flow rate is 7.40x10 -2 kg/s and the flow can be assumed as isothermal. The pipe is 3000 m long and the inlet pressure is 200 kPa. Calculate the outlet pressure. Given: From Appendix A.5 Ds40 = 102.3m/1000m= 0.1023 m MW N2 = 28 g/mol From Appendix A.3 µ= 1.77x10-5 Pa-s P1 = 200kPa= 2x105 Pa Solution: � G= =

.

x

A= 8.219x10-3 m2

L=3000m Total mass flowrate= 7.40x10-2 kg/s

− kg/s

. x − m G= 9.0035 kg/ m-s2

Ԑ= 4.6x10-5 m commercial steel

Relative Roughness= =

NRE =

µ

=

.

m

.

x

.

. x

− m

. m Relative Roughness= 4.497x10-4

kg/ m−s

= 52,037.18, Thus, Turbulent flow

− Pa−s

f= 0.0053 Assume pressure drop is less than 10% of P1 P12 – P22 =

��

=

.

4.000x1010 - P22 = 4.457x1010 P2 = 188,528.5 Pa or 188.53 Kpa . − . %Pressure drop= .

Ԑ

m

.

.

kg/ m−s g o

m

.

� �−�

x100 = 5.75%

2. Reynolds number for milk flow. Whole milk at 293K having a density of 1030 kg/m 3 and viscosity of 2.12 cp is flowing at the rate of 0.605 kg/s in a glass pipe having a diameter of 63.5 mm. a. Calculate the Reynolds number. Is this turbulent flow? b. Calculate the flow rate needed in m3 /s for a Reynolds number of 2100 and the velocity in m/s. Given: ρ= 1030 kg/m3

D= 0.0635m

µ= 2.12x10-3 Pa-s

A=

=

.

mdot = 0.605 kg/s Volumetric flowrate= (0.605kg/s)/(1030kg/m3 )= 5.874x10-4 m3 /s � . x − m /s Velocity= = = 0.1855m/s . x − m

=3.167x10-3 m2

a. NRE = b.

.

µ

.

=

m

.

x

.

s −

g

� −

= 5,722.94 , Turbulent flow

5.874x10-4 m3 /s) = 2.155x10-4 m3/s = Vdot 0.1855m/s)= 0.0681 m/s= V

.

3. Frictional Pressure Drop in Flow of Olive Oil. Calculate the frictional pressure drop in pascal for olive oil at 293 K flowing through a commercial pipe having an inside diameter of 0.0525m and a length of 76.2m. The velocity of the fluid is 1.22m/s. Use the friction factor method. Is the flow laminar or turbulent? Given: T=293K V=1.22m/s From Appendix A.4 Olive Oil: ρ=919.00kg/m3 µ= 0.084 Pa-s

D=0.0525m L= 76.2m

Solution: .

NRE =

.

m

s

= . � − µ f= 16/ NRE =16/700.74= 0.02283

g

= 700.74 , Laminar flow

.

=

Frictional Pressure drop=

.

m

.

.

Frictional Pressure drop= 90,649.78 N/m2 or 90.65 KN/ m2

4. Pipeline Pumping of Oil. A pipeline laid cross country carries oil at the rate of 795 m3/d. The pressure of the oil is 1793 kPg gage leaving pumping station 1. The pressure is 862 kPa gage at the inlet to the next pumping station 1. The pressure is 862 kPa gage at the inlet to the next pumping station, 2. The second station is 17.4m higher than the first station. Calculate the lost work (ƩF friction loss) in J/kg mass oil. The oil density is 769kg/m3 . Given: Flowrate= 795 m3/d. ρoil= 769kg/m3

Z1= 0m Z2= 17.4m

P1= 1793kPa P2= 862kPa

Solution: Z1 g +

0+

=

+

�

ƩF= 1039.97 J/kg

- Ws = Z2 g

+

�

+ ƩF

W s =0 - 0= (17.4m)(9.81m/s2 ) +

�

/

+ ƩF

5. Pressure in a Sea Lab. A sea lab 5.m high is to be designed to withstand submersion to 150m, measured from the sea level to the top of the sea lab. Calculate the pressure on top of the sea lab and also the pressure variation on the side of the container measured as the distance x in m from the top of the sea lab downward. The density of seawater is 1020 kg/m3 . Given: h1 = 150m from sea level ρseawater= 1020 kg/m3 Lsea lab= 5m

g = 9.81 m/s2

Solution: P1= h1 ρseawater g = (150m)(1020 kg/m3 )(9.81 m/s2 ) = 1.500x106 N/m2 or 1500KN/m2 Pvariation= h1 ρseawater g + x seawater g = (150+x)(1020 kg/m3 )( 9.81 m/s2 ) =(150+x)(10,006.2N/m2) = (150+x)(10.0062) KN/m2 6. Pressure in a Spherical Tank. Calculate the pressure in psia and KN/m2 in a spherical tank at the bottom of the tank filled with oil having a diameter of 8.0 ft. The top of the tank is vented to the atmosphere having a pressure of 14.72 psia. The density of the oil is 0.922g/cm3 . Given: P1 = 14.72 psia

h2= 8.0 ft

ρ oil = 0.922g/cm3

P2 English Units ρ oil = 0.922g/cm3 (62.43) = 57.56 lbm/ft3 P2 = (h2 )(ρ oil )(g/gc) + P0 =

.

.

.

= 17.92 psia

.

+

.

SI Units P1 = (h2 )(ρ oil )(g/gc) + P0 P1= (14.72m)(6.89476)= 101.5x 10 N/m2 h2= 8.0ft(1m/3.2808ft) = 2.438m ρ oil = 0.922g/cm3 (1000) = 922 kg/m3 P2 = 2.438m(922kg/m3 )(9.81m/s2 )+ 101.5 x103 N/m2 P2 = 123.5KN/m2

7. Test of Centrifugal Pump and Mechanical –Energy Balance. A centrifugal pump is being tested for performance and during the test the pressure reading in the 0.305-mdiameter sunction line just adjacent to the pump casing is -20.7 kPa (vacuum below atmospheric pressure). In the discharge line with a diameter of 0.254 m at point 2.53 above the sunction line, the pressure is 289.6 kPa gauge. The flow of water from the pump is measured as 0.1133m3 /s. (The density can be assumed as 1000kg/ m 3 ). Calculate the kW input of the pump. Given: m= (0.1133m3/s)(1000 kg/m3 )= 113.3 kg/s . = . A1 = ( )= V1 =

V2 =

.

.

A2 =

/

.

.

= 1.550 m/s

/

.

P1 =-20.7 kPa below atm

= 2.236 m/s

= .

If µ= 1x10-3 kg/m-s , NRE =

=

µ

.

.

−

Thus, flow is turbulent and ɑ= 1.0 +

− .

− .

+ .

= 4.73x105

−

− .

−

+ +

−

+

. +

W s = -336.4 J/kg Pump input= 336.4x113.3 x1/1000 = 38.11kW

.

= +

=

8. Friction Losses and Pump Horsepower. Hot water in an open storage tank at 82.2°C is being pumped at the rate of 0.379m3 /min from this storage tank. The line from the storage tank to the pump sunction is 6.1m of 2-in. schedule 40 steel pipe and it contains three elbows. The discharge line after the pump is 61 m of 2-in. pipe and contains two elbows. The water discharges to the atmosphere at a height of 6.1 m above the water level in storage tank. a. Calculate all frictional losses ƩF. b. Make a mechanical-energy balance and calculate W s of the pump in J/kg. c. What is the kW power of the pump if its efficiency is 75%? Given: f= 0.0048 Ff=

.

=

.

.

. = 104.47 J/kg Total loss, ƩF= 2.333+ 15.965+ 104.48= 122.77 J/kg a. ƩF= 122.77 J/kg

V

.

+�

+ Ʃ +

+(

.

�

�

)

=

. � +

.

+

=

b. Ws = -186.85 J/kg mdot = (6.317x10-3 m3/s)(970.4 kg/m3 )= 6.130 kg/s Ws= -ɦWp -186.9 J/kg = -0.75 Wp Wp= 249.13

.

= 1.527 kW

9. Measurement of Pressure. An open U-tube manometer is being used to measure the absolute pressure Pa in a vessel containing air. The pressure Pb is atmospheric pressure, which is 754 mm Hg. The liquid in the manometer is water having a density of 1000kg/m3 . Assume that the density ρB is 1.30kg/m3 and that the distance Z is very small. The reading R is 0.415 m. Calculate Pa in psia and kPa. Given: Z= negligible R= 0.415 m Pb= 754 mm Hg

ρ A = 1000kg/m3 ρB = 1.30kg/m3

Solution: Pa – Pb = R(ρA - ρB ) g = (0.415m)( 1000kg/m3 -1.30kg/m3 )(9.81m/s2 ) Pa – Pb = 4065.86 N/m2 = 4.066 KN/m2

.

Pb =

= 100.53 KN/m2

Pa =100.53 KN/m2 + 4.066 KN/m2 =104.596 KN/m2 10. Shear Stress in Soybean Oil. Using Fig. 2.4-1, the distance between the two parallel plates is 0.00914 m 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 soybean oil with viscosity of 4x102 Pa-s at 303K (Appendix A.4). a.Calculate the shear stress τ and the shear stress rate using lb force, ft, and s units. b. Repeat, using SI units c. If glycerol at 293 K having a viscosity of 1.069 kg/m-s is used instead of soybean oil, what relative velocity in m/s is needed using the same distance between plates so that the same shear stress is obtained as in part (a)? Also, what is the new shear rate? Given: V2 = 0

Δy=0.00914m

y=0

V1= 0.366m/s

μ= 4.0 x10-2 Pa-s 303K=40cP y2= 0.00914 m= 0.030 ft V2 = 0.366 m/s = 1.20 ft/s a.English units

τ=

�

−

−

=

�

. .

− (

3.34x10-2 lbf/ft2 Shear stress

�

−

− ) −

.

.

−

−

/

=

− −

Shear rate=

.

=

−

.

/

−

= 40.0 s-1 shear rate b. SI Units τ=

�

−

=

−

−

. �

−

.

.

−

−

/

= 1.602 N/m2 shear stress . − / − Shear rate= = = 40.04 s-1 shear rate − .

c. A= Soybean μB= 1.069kg/m-s = 1,069cP τ1=

�

−

B= Glycerol μA= 4.0x10-2kg/m-s

�

−

=τ2 − − (y2 – y1)A = (y2 – y1)B (4.0x10-2kg/m-s )(0.366m/s-0m/s)A= 1.069kg/m-s(V2 – V1)B (V2 – V1)B = 0.01369 m/s or 0.0449 ft/s

=

−

Shear rate=

−

=

.

.

/

= 1.50 s-1

11. Water Flow Rate in an Irrigation Ditch. Water is flowing in an open channel in an irrigation ditch. A rectangular weir having a crest length L= 1.75 ft is used. The weir head is measured as h0 = 0.47 ft. Calculate the flow rate in ft3 /s and m3/s. Given: L= 1.75ft= 1.75ft x h0 = 0.47 ft. x

.

.

= 0.5344m = 0.1433m

q= 0.415(L-0.2 h0) h01.5√

English units

q= 0.415[1.75ft-0.2 (0.47ft) (0.47ft)1.5 ] √

.

q= 1.776 ft3/s SI units

q= 0.415[0.5334m-0.2 (0.1433m) (0.1433m)1.5 ] √ q= 0.0503 m3/s

.

12. Brake Horsepower of Centrifugal Pump. Using Fig. 3.3-2 and a flowrate of 60 gal/min, do as follows. a.Calculate the brake hp of the pump using water with a density of 62.4 lbm/ft 3. Compare with the value from the curve. b.Do the same for a nonviscous liquid having a density of 0.85 g/cm3. Given: ρ= 62.4 lbm/ft3

q= 60 gal/min For 60gal/min, brake hp= 0.80 hp ή=0.58

H= 31ft

mflowrate=

�

�

.

.

Ws= -Hg/gc = -31ft-lbf/lbm Brake hp= 0.810hp x

−

=

ή .

b. ρ= 62.4 lbm/ft3

ℎ

− −

−

.

/

= 0.604kW

= 8.34 lbm/s

.

= 0.810 hp

broke hp= 0.81hp or 0.604kW

ρ= 62.4 lbm/ft3(0.85)= 53.1lbm/ft3 ρ

brake hp and

brake hp= 0.810hp(53.1lbm/ft3)/62.4lbm/ft3 = 0.689hp

0.689hp

.

ℎ

= 0.513 kW

brake hp= 0.689hp or 0.513 kW

13. Power for Liquid Agitation. It is desired to agitate a liquid having a viscosity of 1.5 x10-3 Pa-s and a density of 969kg/m3 in a tank having a diameter of 0.91m. The agitator will be a six-blade open turbine having a diameter of 0.305m operating at 180 rpm. The tank has four vertical baffles each with a width J of 0.076m. Also, W= 0.0381m. Calculate the required kW. Use curve 2, Fig 3.4-4 Given: μ=1.5x10-3 Pa-s

N= 180/3= 3 s-1

ρ= 969kg/m3

J= 0.076m

Dtank = 0.91m

4 vertical baffles

Dturbine = 0.305m at 180 rpm

Dtank/J= 0.91m/0.076m= 12.0

Using the curve 2, Fig 3.4-4 �

NRE=

.

=

�

.

−

.

−

� −

=180, 300 Turbulent

Curve 2, Np = 2.5 �

Np= 2.5=

P= 172.0 W

�

�

−

.

P= 0.172 kW or 0.231hp 14. Surface Area in a Packed Bed. A packed bed is composed of cubes 0.020 m on a side and the bulk density of packed bed is 980 kg/m 3. The density of the solid cubes is 1500 kg/m3. a. Calculate ε, effective diameter Dp, and a. b. Repeat for the same conditions but for cylinders having diameter of D = 0.02 m and a length h=1.5D. Solution: a. ba��� = . � �� �ac��� b��

B��� ��n���� =

��n���� �� ����� c�b�� =

�a�� ��b�� = (

�

��

)

. �

������ �� ����� c�b�� =

E�.

. −

ε=

�� �

�� �� ����� c�b��

�� = . �� ( ) �

�

[���a� ��� − ����� ���] . − . ������ �� ���� = = ���a� ��� . ���a� ������ �� b��

����ac� a��a �� �a���c�� ������ �� �a���c��

=

�� �

�= .

Sp = D Vp = D

E�.

. −

E�.

. −

Dp =

av

aV =

Sp D = = D D Vp

=

=D

ε= .

�= . D

πD

Sp =

. −

E�.

. −

πD

E�.

aV =

Dp =

. −

av

=

. D a=

. −

�=

Vp = E�.

E�.

−ε = Dp

a= b.

D

�� = .

�

a− . .

. �−

+ πD . D = . πD . D =

.

πD

Sp . πD = = . Vp . D πD

= D= −ε = Dp �=

.

�� = .

�

a− . .

. �−

15. Flow Measurement Using a Pitot Tube. A pitot tube is used to measure the flow rate of water at 20oC in the center of a pipe having an inside diameter of 102.3 mm. The manometer reading is 78 mm. the manometer reading is 78 mm carbon tetrachloride at 20oC. the pitot tube coefficient is 0.98. a. Calculate the velocity at the center and the average velocity. b. Calculate the volumetric flow rate of the water. Solution:

.

=

.

= .

= .

∆ℎ =

= .

=

=

.

.

.

. −

=

,

=

.

∆ = ∆ℎ

.

. −

= �

. .

=

= −

√

=

�

.

= .

.

.

− = .

.

=

−

−

.

.

.

. −

−

.

−

=

. .

= .

= .

= .

= .

.

√

= .

∙

−

= .

.

= .

−

�= .

.

=

/ −

16.Force on a Cylinder in a Wind Tunnel. Air at 101.3 kPa absolute and 25oC is flowing at a velocity of 10m/s in a wind tunnel. A long cylinder having a diameter of 90 mm is placed in the tunnel and the axis of the cylinder is held perpendicular to the air flow. What is the force on the cylindrical per meter length.

Solution: =

=

= .

=

.

= .

−

. .

.

. −

= . =

�

. −

�

[ .

=

=

=

−

�= . .

−

. . −

=

�

.

=

.

= .

−

.

.

−

.

.

�� = .

/ ]

∙

= .

= .

= .

17. Molecular Transport of a Property with Variable Diffusivity. A property is being transported through a fluid at steady state through a constant cross-sectional area. At point 1 the concentration Γ1 is 2.78 x 10-2 at point 2 at a distance of 2.0 m away. The diffusivity depends on concentration Γ as follows. =

+ Γ= .

+ .

Γ

a. Derive the integrated equation for the flux in terms of Γ1 and Γ2. Then calculate the flux. b. Calculate Γ at z = 1.0 m Solution: � =− � ∫

−

�

= [− Γ −

. .

� = = .

� =

.

= . .

� = .

−

=−

=∫

Γ

+ Γ

Γ

+ Γ

Γ

]

Γ Γ

−

− −

= .

+

.

Γ Γ

Γ −Γ +

=

� −� +

= .

− .

Γ

Γ −Γ

(� − � ) Γ = .

[ .

. − .

−

−

−

/ ∙

.

Γ = . −

−

]

18. Minimum Fluidization and Expansion of Fluid Bed. Particles having size of 0.10 mm, a shape factor of 0.86, and a density of 1200 kg/m 3 are to be fluidized using air at 25oC and 207.65 kPa abs pressures. The void fraction at minimum fluidizing conditions is 0.43. The bed diameter is 0.60 m and the bed contains 350 kg of solids. a. Calculate the minimum height of the fluidized bed. b. Cal...