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1. 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

2. PART I: MOMENTUM TRANSFER

3. 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 A= 8.219x10-3 m2
MW N2 = 28 g/mol
From Appendix A.3
µ= 1.77x10-5 Pa-s L=3000m
P1 = 200kPa= 2x105 Pa Total mass flowrate= 7.40x10-2 kg/s
Solution:
G=
�
=
. x − kg/s
. x − m
G= 9.0035 kg/ m-s2
Ԑ= 4.6x10-5 m commercial steel Relative Roughness=
Ԑ
=
. x − m
. m
Relative Roughness= 4.497x10-4
NRE =
µ
=
. m . kg/ m−s
. x − Pa−s
= 52,037.18, Thus, Turbulent flow
f= 0.0053
Assume pressure drop is less than 10% of P1
P1
2 – P2
2 =
��
=
. m . kg/ m−s .
�
�−�
. m
g
o
4.000x1010 - P2
2 = 4.457x1010
P2 = 188,528.5 Pa or 188.53 Kpa
%Pressure drop=
. − .
.
x100 = 5.75%
2. Reynolds number for milk flow. Whole milk at 293K having a density of 1030 kg/m3
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= =
.
=3.167x10-3 m2
mdot = 0.605 kg/s
Volumetric flowrate= (0.605kg/s)/(1030kg/m3 )= 5.874x10-4 m3 /s
Velocity=
�
=
. x − m /s
. x − m
= 0.1855m/s

4. a. NRE =
µ
=
. m
.
s
g
. x − � −
= 5,722.94 , Turbulent flow
b.
.
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 D=0.0525m
V=1.22m/s L= 76.2m
From Appendix A.4
Olive Oil:
ρ=919.00kg/m3
µ= 0.084 Pa-s
Solution:
NRE =
µ
=
. m
.
s
g
. � −
= 700.74 , Laminar flow
f= 16/ NRE =16/700.74= 0.02283
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. Z1= 0m P1= 1793kPa
ρoil= 769kg/m3 Z2= 17.4m P2= 862kPa
Solution:
Z1 g + +
�
- Ws = Z2 g +
�
+ ƩF
= Ws =0
0 + - 0= (17.4m)(9.81m/s2 ) +
�
/
+ ƩF
ƩF= 1039.97 J/kg

5. 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 g = 9.81 m/s2
ρseawater= 1020 kg/m3
Lsea lab= 5m
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

6. 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-m-
diameter 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/ m3 ).
Calculate the kW input of the pump.
Given:
m= (0.1133m3/s)(1000 kg/m3 )= 113.3 kg/s P1 =-20.7 kPa below atm
A1 = ( ) = . = .
V1 =
. /
.
= 1.550 m/s
V2 =
. /
.
= 2.236 m/s
A2 = . = .
If µ= 1x10-3 kg/m-s , NRE =
µ
=
.
.
−
−
= 4.73x105
Thus, flow is turbulent and ɑ= 1.0
− + − +
−
+ =
. − . + . . − +
. + .
+ =
Ws = -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 Ws 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 + =
.
+ (
. �
�
) . � +
.
+ =

7. 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 ρA = 1000kg/m3
R= 0.415 m ρB = 1.30kg/m3
Pb= 754 mm Hg
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 4x10-
2 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

8. 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 B= Glycerol
μB= 1.069kg/m-s μA= 4.0x10-2kg/m-s
= 1,069cP
τ1=
� −
−
=
� −
−
=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
.
= 0.5344m
h0 = 0.47 ft. x
.
= 0.1433m
q= 0.415(L-0.2 h0) h0
1.5
√
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

9. 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/ft3. Compare
with the value from the curve.
b.Do the same for a nonviscous liquid having a density of 0.85 g/cm3.
Given:
q= 60 gal/min ρ= 62.4 lbm/ft3
For 60gal/min, brake hp= 0.80 hp
ή=0.58 H= 31ft
mflowrate=
�
�
. . = 8.34 lbm/s
Ws= -Hg/gc = -31ft-lbf/lbm
Brake hp=
−
ή
=
− − − /
.
.
= 0.810 hp
0.810hp x
. ℎ
= 0.604kW broke hp= 0.81hp or 0.604kW
b. ρ= 62.4 lbm/ft3
ρ= 62.4 lbm/ft3(0.85)= 53.1lbm/ft3
brake hp and ρ 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

10. 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/m3. 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��
������ �� ����� c�b�� =
��
(
��
�
)
= . �
E�. . −
ε =
������ �� ����
���a� ������ �� b��
=
[���a� ��� − ����� ���]
���a� ���
=
. − .
.
� = .
����ac� a��a �� �a���c��
Sp = D
������ �� �a���c��
Vp = D

11. E�. . −
aV =
Sp
Vp
=
D
D
=
D
E�. . −
Dp =
av
=
D
= D �� = . �
E�. . −
a =
− ε
Dp
=
a − .
.
� = . �−
b. ε = . � = . D
Sp =
πD
+ πD . D = . πD
Vp =
πD
. D =
.
πD
E�. . −
aV =
Sp
Vp
=
. πD
.
πD
=
. D
E�. . −
Dp =
av
=
. D
= D = . �� = . �
E�. . −
a =
− 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 = = . =
.

12. = . � = . −
∙
= . = . −
. , =
. . . −
∆ = ∆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:
= .
= = = . −
= .

13. .
= . � = . −
/ ∙
. . −
� =
�
=
[ . −
. ]
. −
= .
. . − � = .
. . −
= = . . −
= .
= = . . . = .
�� = .
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 = . −
� =
. . −
− . −
+
.
[ . −
− . − ]
. − .
� = . −
/ ∙

14. 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/m3 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. Calculate the pressure drop at the minimum fluidizing conditions.
c. Calculate the minimum velocity for fluidization.
d. Using 4.0 times the minimum velocity estimate the porosity of the bed.
Given:
=
.
= . ∅ = . =
= . = .
. = . � = . −
∙ = .
.
= . = . ℎ
=
/
= .
= =
.
.
= .
= = = .
=
−
−
,
.
=
− .
= .
.
∆ = ( − )( − )
= . − . − . .
∆ = . �
.
. � ,
∅ ∙
+
( − )
∅ ∙
� ,
−
( − )
�
=
. � ,
. .
+
− .
. ∙ . � ,
−
. . − . .
. −
=
� ,
= .

15. � ,
= . =
′
�
=
. ( ′
) .
. −
′
= . /
. ′
′
= . = . /
′
=
−
′
= . =
.
− ,
= . /
′
= .
. =
−
= .
−
= .
= .
′ = .
.
− .
= .
= .
′
= .
.
− .
= .
= .
′
= .
.
− .
= .
� = .
19. Minimum Fluidization Velocity Using a Liquid. A tower having a diameter of 0.1524
m is being fluidized with the water at 20.2oC. The uniform spherical beads in the tower
bed have a diameter of 4.42 mm and a density of 1603 kg/m3. Estimate the minimum
fluidizing velocity and compare with the experimental value of 0.02307 of Wilhelm and
Kwauk(W5).
Given:
=
.
= . ∅ = . ℎ
= .
.
= . � = . −
∙

16. Solution:
� ,
= [ . +
. ( − )
�
]
� ,
= [ . +
. . . − . .
. −
]
� = .
� =
′
�
. =
. ( ′
) .
. −
′
= . /
The value predicted is . while the experimental value is .
Thus, the predicted value is slightly higher than the experimental value.
20. Pressure with Two Liquids, Hg and Water. An open test tube at 293 K is filled at
the bottom with 12.1 cm of Hg and 5.6 cm of water is placed above the Hg. Calculate the
pressure at the bottom of the test tube if the atmospheric pressure is 756 mm Hg. Use a
density of 13.55 g/cm3 for Hg and 0.998 g/cm3 for water. Give the answer in terms of
dyn/cm2, psia, and KN/m2. See Appendix A.1 for Conversion factors.
Solution:
=
=
= + = ℎ + ℎ +
= [ . . + . . + .
.
]
= .
∙
∙
= . /
. −
= .
=
.
.
= .

17. =
( . )
=
.
= . /

18. PART II: HEAT TRANSFER

19. 21. Chilling Frozen Meat. Cold air at -28.90C and 1 atm is recirculated at a velocity of
0.61 m/s over the exposed top flat surface of a piece of frozen meat. The sides and bottom
of this rectangular slab of meat are insulated and the top surface is 254mm by 254mm
square. If the surface of the metal is at -6.70C, predict the average heat-transfer coefficient
that either Eq.(4.6.2) or (4.6-3) can be used, depending on the NRE L.
Given:
V= 0.61m/s Appendix A.3 for air at -17.80C
P=1 atm ρ= 1.379kg/m3 k= 0.0225W/m-k
L=0.254m Npr= 0.720 μ= 1.62x10-5 kg/m-s
Solution:
=
+
=
− . + − .
= − .
� , =
�
=
.
. .
. −
−
= .
Using Eq. 4.6-2
ℎ = . � ,
.
��
=
.
−
.
. . .
.
ℎ = .
−
22. Natural Convection Cooling of an Orange. An orange 102mm in diameter having
a surface temperature of 21.10C is placed on an open shelf in a refrigerator at 4.40C.
Calculate the heat loss by natural convection, neglecting radiation. As an approximation,
the simplified equation for vertical planes can be used with L replaced by the radius of the
sphere (M1). For a more accurate correlation, see (S2).
Given:
Tw= 21.10C Tb= 4.40C
D=0.102 m R= 0.051m
Plate correlation
∆ = − = . − . = . = .
∆ = . . = . −
Thus, use Table 4.7-2
ℎ = . (
∆
)
= . (
.
.
) =
.
−

20. = = . = .
= ℎ ∆ = (
.
−
) . .
= .
23. Effect of Convective Coefficients on Heat Loss in Double window. Repeat
Problem 5 for heat loss in double window. However, include a convective coefficient of
h=11.35 W/m2•K on the outside surface of one side of the window and an h of 11.35 on
the other side surface. Also calculate the overall U.
Given:
= . . = .
ℎ� = ℎ = .
∙
∆ = ∆ = ∆ = .
= = .
∙
= .
∙
∆ = .
Solution:
� =
ℎ� �
=
. .
= . =
=
∆
=
.
. .
= . =
=
∆
=
.
. .
= .
∑ = .
. . −
=
∆
∑
=
.
.
= . �
Using . . −
= ∆ =
∆
∑
=
∑
=
. .
= .
�
∙
= .
∙ ∙ �
24. Heat-Transfer Area and Use of Log Mean Temperature Difference. A reaction
mixture having cpm= 2.85 kJ/kg•K is flowing at a rate of 7260 kg/h and is to be cooled from
377.6 K to 344.3 K. Cooling water at 288.8 K is available and the flow rate is 4536 kg/h.
the overall Uo is 623 W/m2•K.
a. For counterflow, calculate the outlet water temperature and the area Ao of the
exchanger.
b. Repeat for concurrent flow

21. Given:
′
= .
∙
= .
∙
=
∙
′
=
ℎ
= /ℎ
′
= . ′
= . = . =?
Solution:
a. Countercurrent flow
′
= ′ ′ ′
− ′
=
ℎ
(
.
−
) . − . =
ℎ
= ′
= − =
ℎ
(
.
−
) − . = ,
= .
. . −
∆ = . − . = .
∆ = . − . = .
∆ =
∆ − ∆
�n
∆
∆
=
. − .
�n
.
.
= .
. . −
= ∆ =
∙
. =
,
= .
b. Concurrent flow
′
= . ′
= . = . = .
∆ = . − . = .
∆ = . − . = .
∆ =
. − .
�n
.
.
= .
= . =
= .
25. Loses by Natural Convection from a Cylinder. A vertical 76.2 mm in diameter and
121.9 mm high is maintained at 397.1 K at its surface. It loses heat by natural convection
to air at 294.3 K. Heat loss neglecting radiation losses. Use the simplified equations of
Table 4.7-2 and those equations for the lowest range of NGr, Npr. The equivalent L to use
for the top flat surface is 0.9 times the diameter.

22. Given:
= . = . = . = .
∆ = . − . = .
Solution:
� = = . = .
= = . = .
∆ = . . = .
= . . = .
ℎ = . (
∆
) = . (
.
.
) = .
∙
ℎ � = . (
∆
) = . (
.
.
) = .
∙
= ℎ ∆ = ( .
∙
) . . = .
� = ℎ ∆ = ( .
∙
) . . = .
= + � = . + .
= . �
26. Insulation in a Cold Room. Calculate the heat loss per m2 of surface area for a
temporary insulating wall of a food cold storage room where the outside temperature is
299.9 K and the inside temperature 276.5 K. The wall is composed of 25.4 mm of cork
board having k of 0.0433 W/m•K. 4.1.1
Given:
= . = .
− =
.
= .
= . ∙
Solution:
. . −
=
−
−

23. =
.
−
.
. − .
= .
�
27. Natural Convection in Enclosed Horizontal Space. Repeat Example 4.7-3 for the
case where the two plates are horizontal and the bottom plate is hotter than the upper
plate. Compare the results.
Given:
= . = .
= . = . = .
−
From example 4.7-3,
�,� �� = .
Using Eq. 4.7-13
ℎ =
�
= . �,� ��
1/3
ℎ =
.
−
.
. .
ℎ = .
−
= . . = .
= ℎ ∆ = .
−
. . − .
= .
28. Removal of Heat from Bath. Repeat problem 2 but for a cooling coil is made of 308
stainless steel having an average thermal conductivity of 15.23 W/m•K.
Solution:
= ( .
∙
)
.
∙
(
ℎ ∙ ∙
)
= .
ℎ ∙ ∙
=
−
−
=
. . −
. − .
= . = , �

24. 29. Radiation to a Tube from a Large Enclosure. Repeat Example 4.10-1 but use the
slightly more accurate Eq. (4.10-5) with two different emissivities. .
Given:
= = . . = .
= = ,
= = . = .
Using 4.9-5
= � −
. . −
. − . ,
= − , − ,
ℎ
30. Hardening of a Steel Sphere. To harden a steel sphere having a diameter of 50.8
mm, it is heated to 1,033K and then dunked into a large water bath at 300K. Determine
the time for the center of the sphere to reach 366.5K. The surface coefficient can be
assumed as 710 W/m2-K. k= 45 W/m-K, and α= 0.0325 m2/h.
Given:
=
.
= . = , = = .
=
−
=
.
ℎ
= =
Solution:
=
ℎ
= −
.
= .
−
−
=
− .
− ,
= .
= . = =
.
ℎ
.
= . ℎ
31. Chilling Slab of Beef. Repeat Example 5.5-1, where the slab of beef is cooled to
100C at the center but use air of 00C at a lower value h= 22.7 W/m2-k.
Given:
ℎ = .
−
=
. −
=
.
−
= .
Solution:
=
ℎ
=
.
−
.
−
.
= .
=
� −�
� −�
=
−
− .
= 0.625

25. = . = =
(
. −
)
.
= . . ℎ
32. Time to Freeze a Slab of Meat. Repeat Example 5.5-2 using the same conditions
except that a plate or contact freezer is used where the surface coefficient can be
assumed as h= 142 W/m2-k.
Given:
ℎ =
−
= . =
.
−
= . =
,
=
.
Solution:
Using Eq. 5.5-11
=
− ℎ
+
=
. � � ,
. − .
.
−
+
.
.
−
= , . ℎ
33. Heat Transfer with a Liquid Metal. The liquid metal bismuth at a flow rate of 2.00
kg/s enters a tube having an inside diameter of 35 mm at 425oC and is heated to 430oC
in the tube. The tube wall is maintained at temperature of 25oC above the liquid bulk
temperature. Calculate the tube length required. The physical properties are as follows
(H1): k = 15.6 W/m•K, cp = 149 J/kg•K, μ = 1.34 x 10-3 Pa•s.
Given:
= = . =
.
= =
= .
∙
=
∙
� = . −
∙
Solution:
= = . = . −
= =
.
. −
=
.
∙
� =
�
=
. .
. −
= .
� =
�
=
. −
.
= .
ℎ = . �
.
� = � � = . .
ℎ =
.
.
. ( . . )
.
= ∆ = . − =

26. = = ℎ − =
= . = = .
= .
34. Temperature Rise in Heating Wire. A current of 250 A is passing through a stainless
steel wire having a diameter of 5.08 mm. the wire is 2.44 m long and has a resistance of
0.0843 Ω. The outer surface is held constant at 427.6K. The thermal conductivity is k=
22.5 W/m•K. Calculate the center-line temperature at steady state.
Given:
= . = . = .
∙
= .
= =
.
= . ℎ =
Solution :
= = .
= ̇ = ̇
.
.
̇
. = ̇
.
.
̇ = .
=
̇
+ =
. .
.
+ .
= .

27. PART III: MASS TRANSER
]

28. 35. Equimolar Counterdiffusion of a Binary Gas Mixture. Helium and nitrogen gas are
contained in a conduit 5 mm in diameter and 0.1 m long at 298 K and a uniform constant
pressure of 1.0 atm abs. The partial pressure of He at one end of the tube is 0.060 atm
and 0.020 atm at the other end. The diffusivity can be obtained from Table 6.2-1.
Calculate the following for the steady-state equimolar counterdiffusion.
a. Flux of He in kgmol/s•cm2
b. Flux of N2
c. Partial Pressure of He at a point 0.05 m from either end.
Given:
=
. −
. − − = .
Solution:
. . . −
∗
=
−
−
=
. −
. − .
. − .
∗
= . −
∙
.
∗
=− ∗
∗
= − . −
∙
. = .
∗
= . −
=
. −
. −
. − .
= .
36. Mass Transfer from a Napthalene Sphere to Air. Mass transfer is occurring from a
sphere of naphthalene having a radius of 10 mm. the sphere is in a large volume of still
air at 52.6oC and 1 atm abs pressure. The vapor pressure of naphthalene at 52.6oC is 1.0
mmHg. The diffusivity of naphthalene in air at 0oC is 5.16 x 10-6m2/s. Calculate the rate
of evaporation of naphthalene from the surface in kgmol/s•m2. [Note: the diffusivity can
be corrected for temperature using the temperature-correction factor of the Fuller et al.
Eq (6.2-25).]
Given:
= = . −
= = . = .
= . −
.
Solution:
=
−
�n
= − . −
= .
=
=
− .
�n
.
= .
.
= . −
(
.
)
.
= . −

29. =
−
=
. −
. . −
− .
. . . .
= . −
∙
37. Diffusion of Methane Through Helium. A gas of CH4 and He is contained in a tube
at 101.32 kPa pressure and 298 K. At one point the partial pressure of methane is
pA1=60.79 kPa and at a point 0.02 m distance away, pA2=20.26 kPa. If the total pressure
is constant throughout the tube, calculate the flux of CH4 (methane) at steady-state for
equimolar counterdiffusion.
Given:
= . = = . = .
− = .
Solution:
= . −
. −
=
∙
∙
. −
. . −
∗
=
−
−
=
. −
. − .
.
∗
= . −
(
∙
)
38. Diffusion Through Membranes in Series. Nitrogen gas at 2.0 atm and 30oC is
diffusing through a membrane of nylon 1.0 mm thick and polyethylene 8.0 mm thick in
series. The partial pressure at the other side of the two films is 0 atm. Assuming no other
resistances, calculate the flux NA at steady state.
Given:
= = = = . = .
. −
= . −
ℎ = . −
Solution:
=
−
. +
=
.
.
. − +
.
. −
= . −
∙
39. Loss from a Tube of Neoprene. Hydrogen gas at 2.0 atm and 27oC is flowing in a
neoprene tube 3.0 mm inside diameter and 11, outside diameter. Calculate the leakage
of H2 through a tube 1.0 m long kgmol H2/s at steady state.
Given:
=
.
= . =
.
= .
= = = . =
. −
= . −
= .

30. Solution:
=
.
̅ = −
�n
. − . −
̅ =
−
. �n
=
. −
. .
. �n
.
.
̅ = . −
�
40. Time to Completely Evaporate a Sphere. A drop of liquid toluene is kept at a uniform
temperature of 25.9oC and is suspended in air by a fine wire. The initial radius r1=2.00
mm. The vapor pressure of toluene is 866 kg/m3.
a. Derive Eq.(6.2-34) to predict the time tF for the drop to evaporate completely in
a large volume of still air.
b. Calculate the time in seconds for complete evaporation
Solution:
.
̅
= = ∙
−
=
= =
= −
=
̅
= −
̅
=
−
= −
−
∫ = − ∫
=
� �
� � −
.
= . −
. − =
= = . = . = .

31. = . = = . = .
=
+
=
. + .
= .
=
. . .
. . − . .
=
41. Numerical Method for Steady-State Diffusion. Using the results from Example 6.6-
1, calculate the total diffusion rate in the solid using the bottom nodes and paths of C2,2
to C3,2 , C2,3 to C3,3, and so on. Compare with the other diffusion rates in example 6.6-1
Solution:
∆ = ∆
Writing Eq. 6.6-5
=
∆ °
∆
( , − , ) + ( , − , ) + ( , − , )
= . −
. − . + . − . + . − . −
= . −
/
42. Diffusion of A Through Stagnant B and Effect of Type of Boundary on Flux.
Ammonia gas is diffusing through N2 under steady state conditions with N2 nondiffusing
since it is insoluble in one boundary. The total presuure is 1.013 x 105 Pa and at the other
point 20 mm away it is 6.666 x 103 Pa. The DAB for the mixture at 1.013 x 105 Pa and 298
K is 2.30 x 10-5 m2/s.
a. Calculate the flux of NH3 in kgmol/s•m2
b. Do the same as (a) but assume N2 also diffuses; i.e., both boundaries are
permeable to both gases and the flux is equimolar counterdiffusion. In which
case is the flux greater?
Given:
= . = = . = .
− = = . = . −
/
Solution:
. ℎ ℎ −
=
−
�n
= − = . − . = .
= − = . − . = .
=
. − .
�n (
.
.
)
= .
=
−
−
=
. −
. . − .
. .
= . −
∙

32. .
= ∗
=
−
−
=
. −
. − .
.
= . −
∙
The flux is greater than case (a)
43. Diffusion Flux and Effect of Temperature and pressure. Equimolar
counterdiffusion ion is occurring at steady state in a tube 0.11 m long containing N2 and
CO gases at a total pressure of 1.0 atm abs. The partial pressure of N2 is 80 mmHg at
one end 10mm at the other end. Predict the DAB by the method of Fuller et al.
a. Calculate the flux in kgmol/s•m2 at 298 K for N2
b. Repeat at 473 K. Does the flux increase?
c. Repeat at 298 K but for a total pressure of 3.0 atm abs. The partial pressure of
N2 remains at 80 and 10 mmHg, as in part (a). Does the flux change?
Given:
− = . = . =
= = . = = .
Solution:
= . = .
. −
= . = . = .
= . = . = .
=
− .
+
(∑ + ∑ )
=
− .
+
. + .
� = . −
.
= ∗
=
−
−
=
. −
. − .
. − .
= . −
∙
.
= . −
( )
.
=
. −
=
. −
. − .
. − .
= . −
∙

33. . =
= . −
( ) = . −
=
. −
. − .
. − .
= . −
∙
44. Diffusion of A Through Stagnant B in a Liquid. The solution HCl(A) is diffusing
through a thin film of water (B) 2.0 mm thick at 283 K. the concentration of HCl at point
1 at one boundary of the film is 12.0 wt % HCl (density ρ1=1060.7 kg/m3), and at the
other boundary at point 2 it is 6.0wt % HCl (density ρ2=1030.7 kg/m3) thye diffusion
coefficient of |HCl in water is 2.5 x 10-9 m2/s. assuming steady state and one boundary
impermeable to water, calculate the flux HCl in kgmol/s•m2.
Given:
= . −
− = . % = . % % = %
= . = . = . =
Solution:
:
=
.
.
.
.
+
.
= .
= − . = .
=
.
= .
=
.
.
.
+
= .
= − . = .
=
.
= .
. . −
=
−
�n
=
. + .
�n
.
.
= .
. . −
=
+
=
.
.
+
.
. = .
. . −
=
−
−
=
. −
. . − .
. .
= . −
∙

34. 45. Diffusion of Uric Acid in Protein Solution and Binding. Uric acid (A) at 37oC is
diffusing in an aqueous solution of proteins (P) containing 8.2 percent protein/100 mL
solution. Uric acid binds to the proteins for every 3.0 g mol of total acid present in solution.
The diffusivity DAB of uric acid in water is 1.21 x 10-5- cm2/s and Dp = 0.091 x 10-5 cm2/s.
a. Assuming no binding, predict the ratio DAP/DAB due only to blockage effects.
b. Assuming blockage plus binding effects, predict the ratio DAP/DAB due only to
blockage effects
c. Predict the flus in g uric acid/s•cm2 for a concentration of acid of 0.05 g/L at
point (1) and 0 g/L at point (2) a distance 1.5 μm away.
Given:
= . −
= . = . −
Solution:
.
=
.
= .
. . −
� = ( − . −
) = ( − . −
. ) = .
� �
�
= .
. . . −
� = [ ( − . −
) (
%
) + (
%
)]
= [ . −
. (
.
) + . −
(
.
) ]
� � = . −
�
=
.
.
= . = .
. . . −
=
� −
−
=
. − .
−
. −
= . −
∙
46. Prediction of diffusivity of Enzyme Urease in Solution. Predict the diffusivity of
the enzyme urease in dilute solution water at 298 K using the modified Polson equation
and compare the result with the experimental value in table 6.4-1.
Solution:
. − = , =
D exp
= . � −
�
�
. = . −
∙
. . −
=
. −

35. =
. −
. − ,
= . −
47. Relation Between Diffusivity and Permeability. The gas hydrogen is diffusing
through a sheet of vulcanized rubber 20 mm thick at 25oC. The partial pressure of H2
inside 1.5 atm and 0 outside. Using the data from Table 6.5-1, calculate the following.
a. The diffusivity DAB from the permeability PM and solubility S and compare
with the value in Table 6.5-1.
b. The flux NA of H2 at steady state.
Solution:
. . − = . −
= .
= . −
= .
. . −
= ; =
=
. −
.
= . −
. . . −
=
−
. −
; = . =
=
. −
.
. .
= . −
∙
48. Diffusion of CO2 through Rubber. A flat plug 30 mm thick having an area of 4.0 x
10-4 m2 and made of vulcanized rubber used for closing an opening in a container. The
gas CO2 at 25oC and 2.0 atm pressure is inside the container. Calculate the total leakage
or diffusion of CO2 through the plug to the outside in kgmol/CO2/s at steady state. Assume
that the partial pressure of CO2. The diffusivity is 0.11 x10-9 m2/s.
Given:
= = . = . −
= = .
= . = . −
Solution:
=
.
=
. .
.
= . −
=
̅
̅
̅
̅
=
−
−
̅ =
−
. −
. −
.
= . −

36. 49. Estimation of Diffusivity of Methanol in H2O. The diffusivity of dilute methanol in
water has been determined experimentally to be 1.26 x 10-9m2/s at 288 K.
a. Estimate the diffusivity at 293 K using the Wilke-Chang Equation.
b. Estimate the diffusivity at 293 K by correcting the experimental value at 288 K
to 293 K (Hint: Do this by using the relationship DAB α T/μB).
Given:
= =
. = . −
∙
Solution:
= . + . + . = .
= . −
�
. � = .
.
=
. −
.
. − . .
� = . −
. = . = . −
∙
= . −
( ) (
.
.
) = . −
50. Estimation of Diffusivity of a Binary Gas. For a mixture of ethanol (CH3CH2OH)
vapor and methane (CH4), predict the diffusivity using the methanol of Fuller et al.
a. At 1.0132 x 105 Pa and 298 and 373 K
b. At 2.20265 x 105 Pa and 298 K
Given:
= = = = . = = .
Solution:
∑ = . + . = .
∑ = . + . + . = .
=
− .
+
(∑ + ∑ )
=
− .
.
+
.
. ( . + . )
.
� = . −
= . −
( )
.
= . −
.
= . −
( ) = . −
,