Solution for Transport Process and Separation Process

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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 1 Student View 
Jason M. Keith 
Chapter 1 
Introduction to Engineering Principles and Units 
This chapter presents a series of problems that introduce the student to the use of units and methods 
for expressing different variables such as temperature and composition. Likewise, implementation of 
material balance and energy in fuel cells are illustrated in the following set of problem modules. 
1.3-1 Determination of a Solution Density 
1.4-1 Gas-Law Constant 
1.4-2 Composition of a Gas Mixture 
1.5-3 Combustion of Fuel Gas 
1.6-1 Pre-heating of Methane and Steam 
1.6-2 Heating of an Ethanol Solution 
1.6-3 Calculation of Heat Transfer Rate using Steam Tables 
1.6-4 Incomplete Combustion of Methane 
1.6-5 Standard Enthalpy of Reaction 
1.7-1 Cooling of a Fuel Cell 
1.7-2 Simultaneous Material and Energy Balances 
1.7-3 Oxidation of Woody Biomass 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 2 Student View 
Jason M. Keith 
Example 1.3-1: Composition of a Methanol Solution 
A Direct-Methanol Fuel Cell is being fed with a liquid mixture of 80 mol % methanol and 20 mol % 
water at a temperature of 20 °C. Calculate the mass fractions and the mass of each component of this 
mixture in g and lbm. 
Strategy 
We will select a basis of 100 moles of mixture and use the molecular weights of methanol and water 
to determine the mass. 
Solution 
In the basis of 100 moles of mixture we selected, there will be 80 moles of methanol and 20 moles of 
water. 
We need to use the molecular weight of each component to determine the mass of methanol and 
water in the mixture, as shown in the following equations: 
 
H O H O H O
2 2 2
m n M= 
 
CH OH CH OH CH OH
3 3 3
m n M= 
Substituting the corresponding quantities into these equations yields: 
 ( ) 2H O 2
2
2
18 g H O 
m 20 moles H O
1 mol H O
 
=   
 
 
 
H O 2
2
m __________ g H O= 
 ( ) 3CH OH 3
3
3
32 g CH OH 
m _____ moles CH OH
1 mol CH OH
 
=   
 
 
 
CH OH 3
3
m ____________ g CH OH= 
These values can be converted into lbm by multiplying the results by the conversion factor between 
lbm and g: 
 
m
H O 2
2
1 lb
m 360 g H O
___________ g
 
=  
 
 
 mH O
2
m _________ lb= 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 3 Student View 
Jason M. Keith 
 
m
CH OH 3
3
1 lb
m ____________ g CH OH
453.59 g
 
=  
 
 
 mCH OH
3
m ___________ lb= 
To determine the mass fractions of methanol and water, first we need to calculate the mass of 
mixture by adding the individual weights of its components: 
 
H O CH OHmixture
2 3
m m m= + 
 
m mmixture
m __________ lb __________ lb= + 
 
mmixture
m 6.438 lb= 
Now we can calculate the mass fractions by dividing the mass of each component by the mass of the 
mixture: 
 
H O
m
H O
mmixture
2
2
m ________ lb
x
m 6.438 lb
= = 
 
H O
2
x 0.123= 
 
CH OH
m
CH OH
mmixture
3
3
m _________ lb
x
m 6.438 lb
= = 
 
CH OH
3
x ____________= 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 4 Student View 
Jason M. Keith 
Example 1.4-1: Gas-Law Constant 
Calculate the value of the gas-constant R if the pressure is in mm Hg, the moles are measured in 
g mol units, the volume is in liters and the temperature is in K. Convert this value to
cal
mol K⋅
. 
Strategy 
We can calculate the constant R by solving it from the ideal-gas law at standard conditions. 
Solution 
The ideal-gas equation of state can be solved for R to yield: 
 
pV
R
nT
= 
At standard conditions, P = 760 mm Hg, V = 22.414 L, n = 1 mol and T = 273.15 K. Substituting 
these values into the ideal gas equation of state, we have: 
 
( ) ( )
( ) ( )
760 mm Hg 22.414 L
R
1 mol 273.15 K
= 
 
L mm Hg
R _____________
mol K
⋅
=
⋅
 
To obtain the value of R in 
cal
mol K⋅
, we can start by converting it to SI units as shown below: 
 
3L mm Hg ________________ Pa 1 m
R 62.364
mol K 760 mm Hg 1000 L
 ⋅  
=   
⋅   
 
 
3Pa m
R _____________
mol K
⋅
=
⋅
 
A Pascal is defined as: 
 
2
N
Pa
m
= 
If we multiply the Pa by m
3
, we get: 
 3 3
2
N
Pa m m N m Joule(J)
m
⋅ = = ⋅ = 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 5 Student View 
Jason M. Keith 
Now we can use a conversion factor between Joules and calories to determine the value of R in the 
desired units: 
 
J 1 cal
R ____________
mol K ____________ J
 
=  
⋅  
 
 
cal
R 1.987
mol K
=
⋅
 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 6 Student View 
Jason M. Keith 
Example 1.4-2: Composition of a Gas Mixture 
Hydrogen can be produced by an ethanol reforming process. The gas produced by the reformer is 
exiting at a pressure of 2144 kPa and has the following composition. 
Component Molar Fraction 
H2 0.392 
H2O 0.438 
CO 0.081 
CO2 0.080 
CH4 0.009 
 
Determine the partial pressure of each component. 
Strategy 
The definition of partial pressure will allow us to solve this problem. 
Solution 
To calculate the partial pressure of each gas in the stream exiting the reformer, we can use the 
following equation: 
 
i i
P y P= 
where: 
 yi = molar fraction of the species i present in the gas mixture. 
 Pi = Partial pressure of the species i present in the gas mixture. 
 P = absolute pressure of the system. 
Substituting the corresponding molar fraction and the absolute pressure of the system into the 
definition of partial pressure, we have: 
 ( )
H H
2 2
P y P 0.392 _____________ kPa= = 
 
H
2
P 840.45 kPa= 
( )
H O H O
2 2
P y P 0.438 _____________ kPa= = 
H O
2
P ____________ kPa= 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 7 Student View 
Jason M. Keith 
 
( )
CO CO
P y P _________ _____________ kPa= = 
CO
P 173.66 kPa= 
( )
CO CO
2 2
P y P _________ _____________ kPa= = 
CO
2
P __________ kPa= 
( )
CH CH
4 4
P y P __________ _____________ kPa= = 
CH
4
P __________ kPa= 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 8 Student View 
Jason M. Keith 
Example 1.5-3: Combustion of Fuel Gas 
In the steam-methane reforming process for producing hydrogen, part of the reformer exit gas is 
being burned with 25 % excess oxygen from air in order to supply heat for the reforming reaction to 
occur. The fuel being burned has the following composition: 
Component Mol % 
H2 41.9 
H2O 5.1 
CO 1.7 
CO2 41.9 
CH4 9.4 
 
The combustion reactions occurring inside the firebox are shown below. The combustion of methane 
is only 87 % complete. 
1) CH4 + 2O2 CO2 + 2H2O 
2) CO + 
1
2
O2 CO2 
3) H2 + 
1
2
O2 H2O 
Determine the composition of the gas produced by the combustion reaction assuming an air 
composition of 21 mol % oxygen and 79 mol % nitrogen. 
Strategy 
We can perform molecular material balances around the combustion chamber to determine the 
amounts of each species in the exhaust gases.A basis of 100 moles of fuel will be selected for 
simplicity. 
Solution 
We can start by performing a methane balance around the combustion chamber: 
CH4 balance 
Input = Output + Consumption 
4
Input 9.4 moles CH= 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 9 Student View 
Jason M. Keith 
The amount of methane exiting in the flue gas can be determined using the definition of fractional 
conversion: 
CH ,outCH ,in
CH
CH ,in
44
4
4
n n
x
n
−
= 
Solving for the 
CH ,out
4
n and substituting the corresponding quantities yields: 
 
CH ,out CHCH ,in CH ,in
4 44 4
n n x n= − 
 ( )
CH ,out
4
n 9.4 moles ______ 9.4 moles= − 
 
CH ,out
4
n ______ moles= 
Since no information is given about the fractional conversion for reactions 2) and 3), complete 
combustion will be assumed. Thus, 
 
H ,outCO,out
2
n n 0= = 
We can proceed to perform a material balance on carbon dioxide as follows: 
CO2 balance 
Input + Generation = Output 
Input = __________ moles 
By looking at the chemical reactions, it can be seen that both reactions 1) and 2) are generating 
carbon dioxide. From the stoichiometric coefficients of these reactions, we can see that one mole of 
fuel is producing one mole of CO2. Thus, 
 
CO ,generated CH ,reacted CO,reacted
2 4
n n n= + 
Substituting numeric values into this equation, we get: 
 ( )
CO ,generated
2
n ________ 9.4 moles ________ moles= + 
 
2CO ,generated
2
n 9.88 moles CO= 
We can substitute this result into the material balance equation for CO2 to yield: 
CO ,out
2
n ________ moles 9.88 moles= + 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 10 Student View 
Jason M. Keith 
CO ,out
2
n __________ moles= 
In a similar way, we can perform material balances for water, considering the generation of water by 
reactions 1) and 3). In these reactions, it can be seen that 1 mole of fuel is producing 1 mole of 
water: 
H2O balance 
Input + Generation = Output 
Input = 5.1 moles 
The amount of water generated by the chemical reactions will be given by: 
 
H O,generated H ,reacted CH ,reacted
2 2 4
n n 2n= + 
Substituting numeric values into this equation, we have: 
 ( ) ( )
H O,generated
2
n _________ moles 2 ________ 9.4 moles= + 
 
H O,generated
2
n ____________ moles= 
We can substitute this result into the material balance equation for H2O to get: 
H O,out
2
n 5.1 moles ___________ moles= + 
H O,out
2
n ___________ moles= 
To determine the oxygen exiting in the product stream, we will perform a material balance for 
molecular oxygen: 
O2 balance 
Input = Output + Consumption 
To determine the amount of oxygen that must be fed into the reactor, first we need to determine the 
theoretical amount of oxygen required by each reaction. This value can be calculated by multiplying 
the amount of fuel in the feed (carbon monoxide, methane or hydrogen) by the stoichiometric ratio 
of fuel to oxygen. Thus, 
 2
O ,r1 4 2
4
2
2 moles O
n 41.9 moles CH ___________ moles O
mol CH
 
= =  
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 11 Student View 
Jason M. Keith 
 2
O ,r 2 2
2
________ moles O
n _______ moles CO 0.85 moles O
mol CO
 
= = 
 
 
 2
O ,r3 2 2
2
2
________ moles O
n 41.9 moles H ___________ moles O
mol H
 
= =  
 
 
Knowing the values of oxygen consumed by each reaction, we can calculate the total amount of 
oxygen required by this process: 
O ,r1 O ,r 2 O ,r3
2 2 2
Consumption n n n= + + 
2 2 2
Consumption _________ moles O 0.85 moles O _________ moles O= + + 
2
Consumption 105.6 moles O= 
Since there is 25 % excess oxygen, the number of moles of oxygen entering the reactor will be given 
by: 
( ) ( )2Input ________ Consumption ________ 105.6 moles O= = 
2
Input _________ moles O= 
Solving for the amount of oxygen exiting the reactor in the balance equation for molecular oxygen 
and substituting the amount of oxygen fed and reacted, we get: 
Output = Input – Consumption 
O ,out
2
n ________ moles 105.6 moles= − 
O ,out
2
n ________ moles= 
Finally, since the oxygen is entering the process as air, we need to take into account that nitrogen is 
entering into the firebox in this process. However, all the nitrogen will exit in the product stream 
since it is not being consumed nor generated by the chemical reactions. Thus, 
N2 balance 
 
N ,outN ,in
22
n n= 
The amount of nitrogen fed into the system can be determined by multiplying the molar fraction of 
nitrogen in the air by the amount of air fed: 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 12 Student View 
Jason M. Keith 
 ( )2N ,out 2N ,in
2
22
_______ moles N
n n ________ moles O
0.21 moles O
 
= =   
 
 
 
N ,out 2N ,in
22
n n 496.57 moles N= = 
Now that we know the amount of moles of all the species, we can calculate the total number of 
moles exiting the combustion chamber. 
out N ,outCH ,out CO ,out H O,out O ,out
24 2 2 2
n n n n n n= + + + + 
out
n __________ moles __________ moles 63.36 moles 496.57 moles __________ moles= + + + + 
out
n __________ moles= 
The molar fraction of each component of the exhaust gases can be obtained by dividing the number 
of moles of each component by the total number of moles. Thus, 
CH ,out
CH
out
4
4
n __________ moles
y
n __________ moles
= =
CH
4
y 0.002= 
CO ,out
CO
out
2
2
n __________ moles
y
n __________ moles
= =
CO
2
y 0.081= 
H O,out
H O
out
2
2
n 63.36 moles
y
n __________ moles
= =
H O
2
y __________= 
N ,out
N
out
2
2
n 496.57 moles
y
n __________ moles
= =
N
2
y __________= 
O ,out
O
out
2
2
n __________ moles
y
n __________ moles
= =
N
2
y __________= 
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 13 Student View 
Jason M. Keith 
Example 1.6-1: Pre-heating of Methane and Steam 
a) The steam used for producing hydrogen by steam-methane reforming process is leaving the boiler 
at a temperature of 400 °C, a pressure of 1 atm, and a flow rate of 
kmol
38.31
hr
. However, before 
entering the reformer, it must be heated to a temperature of 450 °C at constant pressure. Determine 
the power required in kW to heat the steam to this temperature. 
Strategy 
Since there is no phase change involved in this heating process, the definition of sensible heat can be 
used to determine the power required. 
Solution 
The following equation is used to determine the heat required for increasing the temperature of a 
substance: 
 
2 1p,i
ˆQ nC (T T )= −� � 
where: 
 Q� = Power required for changing the temperature of a mass of substance from T1 to T2. 
 n� = Molar flow rate of substance i. 
p,i
C = Heat capacity of substance i. 
 T1 = Initial temperature 
 T2 = Final temperature 
The value of the heat capacity of steam at the final temperature of 450°C can be obtained using 
linear interpolation between 400°C and 500°C with the data found in Table 1.6-1 of Geankoplis as 
shown in the following steps: 
 
 
 
p,H O p,H O@ T @ T
mid low
p,H O p,H Ohigh low @ T @ T
2 2mid low
2 2high low
C CT T
T T C C
−−
=
− − 
Inserting the numbers from Table 1.6-1, we have: 
 p,H O @ 450 C 
2
C 35.21450 C 400 C
500 C 400 C __________ 35.21
°
−° − °
=
° − ° − 
 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola14 Student View 
Jason M. Keith 
Solving for the heat capacity at the process temperature of 450 °C, we get: 
 
p,H O
2
J
C __________
mol K
=
⋅
 
Substituting all known values into the equation for the power yields: 
kmol 1 hr 1000 moles J 1 kJ
Q __________ __________ (450 C 400 C)
hr 3600 s 1 kmol mol K 1000 J
      
= ° − °       ⋅      
� 
Q __________ kW=� 
b) Determine the heat transfer rate required to bring methane from the gas lines at room temperature 
to the operating temperature of the steam-methane reforming process from part a). Methane is being 
consumed at a rate of 
kmol
12.8
hr
 
Strategy 
In a similar way to part a) of this problem, we can use the definition of sensible heat to calculate the 
heat required. 
Solution 
We can substitute the corresponding values for the initial and final conditions of methane into the 
equation used to calculate sensible heat in part a). 
kmol 1 hr 1000 moles J 1 kJ
Q 12.8 __________ (450 C 25 C)
hr 3600 s 1 kmol mol K 1000 J
      
= ° − °       ⋅      
� 
Q __________ kW=� 
The heat capacity value for methane was obtained through linear interpolation using data from Table 
1.6-1 of Geankoplis, at a temperature of 450°C. 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 15 Student View 
Jason M. Keith 
Example 1.6-2: Heating of an Ethanol Solution 
A gas mixture containing 87.5 mole % water and 12.5 mole % ethanol is being heated from 
210.4 °C to 350 °C before entering a pre-reformer unit to produce hydrogen for use in Proton-
Exchange Membrane Fuel Cells. This mixture will enter the pre-reformer at a flow rate of 
kmol
61.4
hr
 
Calculate the heat required to bring the ethanol/water mixture to the operating conditions of the 
ethanol-reforming process. The heat capacity of ethanol vapor at the average temperature of 
280.2 °C is 
J
98.9
mol K⋅
. This value was calculated using the equation for heat capacity of ethanol 
vapor as a function of temperature, with parameters obtained from Table B.2 of Elementary 
Principles of Chemical Processes by Felder & Rousseau. 
Strategy 
This problem can be solved by calculating and adding the sensible heat of each component of the 
mixture. 
Solution 
The sensible heat of a mixture can be obtained from the following equation: 
 ( ) 2 1mixture i p,iˆQ n C (T T )= −∑� � 
Applying this equation to the number of components of the mixture in this problem, we get: 
 ( )H O p,H O C H OH p,C H OH 2 1mixture
5 52 2 2 2
ˆ ˆQ n C n C (T T )= + −� � � 
The flow rate of water and ethanol can be calculated by multiplying their corresponding molar 
fractions by the overall flow rate of the mixture: 
 
H O H O
2 2
n y n=� � 
C H OH C H OH
5 52 2
n y n=� � 
 
H O
2
kmol
n 0.875 61.4
hr
 
=  
 
� 
C H OH
52
kmol
n __________ 61.4
hr
 
=  
 
� 
 
H O
2
kmol
n __________
hr
=� 
C H OH
52
kmol
n __________
hr
=� 
The specific heat of water can be obtained using linear interpolation between the temperatures of 
300 °C and 400 °C with data from Table 1.6-1 of Geankoplis. Thus, 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 16 Student View 
Jason M. Keith 
 
p,H O @ 350 C
2
J
C __________
mol K°
=
⋅
 
Now we can substitute all the values we found into the equation for the sensible heat of the mixture 
to yield: 
mixture
kmol 1 hr 1000 moles J 1 kJ
Q (350 C ________ C) ________ ________
hr 3600 s 1 kmol mol C 1000 J
kmol 1 hr 1000 moles
 _________ 98.9
hr 3600 s 1 kmol
     
= ° − °       ⋅°     
   
+    
   
�
J 1 kJ
mol C 1000 J
  
  ⋅°  
 
mixture
Q 102.0 kW=� 
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 17 Student View 
Jason M. Keith 
Example 1.6-3: Calculation of Heat Transfer Rate using Steam Tables 
Liquid water at 30 °C is fed into a steam-methane reforming plant for producing syngas which can 
be used as fuel for solid oxide fuel cells. Before entering the reformer, the water must be boiled and 
heated to a temperature of 450 °C and at a pressure of 2180 kPa. Use steam tables to determine the 
following: 
a) The amount of heat required for heating 1 mol of water from 30 °C to the boiling point at 2180 
kPa. 
b) The amount of heat required to vaporize 1 mol of water. 
c) The amount of heat needed for heating 1 mol of saturated steam to a temperature of 450 °C. 
Strategy 
To determine the amount of heat required for parts a) to c) in this problem we will make use of 
thermodynamic properties of water and steam. 
Solution 
To solve parts a), b) and c) of this problem, we need to look for 4 different enthalpy values of water 
and steam, which are described below: 
 Ha,l = Enthalpy of liquid water at 30 °C 
 Hb,l = Enthalpy of saturated liquid water at the boiling temperature at a pressure of 2180 kPa 
 Hb,v = Enthalpy of saturated steam at the boiling temperature at a pressure of 2180 kPa 
 Hc,v = Enthalpy of superheated steam at the process temperature of 450 °C 
After finding these 4 enthalpies in the steam tables, the enthalpies we need to calculate to solve this 
problem will be given by the following equations: 
 Ha = Hb,l – Ha,l 
 Hb = Hb,v – Hb,l 
Hc = Hc,v – Hb,v 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 18 Student View 
Jason M. Keith 
The enthalpies of liquid water can be found in Table 2-305 of Perry’s Chemical Engineers’ 
Handbook 8
th
 Edition. However, we will have to use linear interpolation for determining the 
enthalpies at the temperature and pressure given in the problem statement. The data given in this 
table is shown below: 
 Pressure (kPa) 
 1000 2180 5000 
T
em
p
er
a
tu
re
 
 (
K
) 
300 
kJ
2.0444
mol 
@ 2180 kPa,300K
Ĥ
 
kJ
____________
mol 
303.15 
@ 1000 kPa,303.15K
Ĥ
 
@ 2180 kPa,303.15K
Ĥ
 
@ 5000 kPa,303.15K
Ĥ
 
400 
kJ
9.6106
mol 
@ 2180 kPa,400K
Ĥ
 
kJ
9.6601
mol 
 
Where the caret (^) indicates the enthalpy is per mole. 
From the table shown above, we need to calculate first the values of the enthalpy at a constant 
temperature of 303.5 K at a pressure of 1000 kPa. We setup the linear interpolation as: 
 
 
 
@ T @ T
mid low
high low @ T @ T
mid low
high low
ˆ ˆH HT T
ˆ ˆT T H H
−−
=
− − 
Inserting the numbers from the table, we have: 
@ 1000 kPa ,303.15K
Ĥ 2.0444_________ 300 K
400 K 300 K _________ 2.0444
−−
=
− − 
Solving for H @ 1000 kPa, 303.15, K: 
 
( )
@ 1000 kPa ,303.15K
_________ K 300 K kJ
Ĥ _________ 2.0444 2.0444 2.2827
400 K 300 K mol
− 
= − + = −  
Repeating this procedure for the enthalpies at a pressure of 5000 kPa we have: 
 
( )
@ 5000 kPa,303.15K
_________ K 300 K kJ
Ĥ _________ 2.1106 2.1106 _________
400 K 300 K mol
− 
= − + = −  
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 19 Student View 
Jason M. Keith 
After substituting the calculated values for the enthalpies at T = 303.15 K in the table, we get: 
 Pressure (kPa) 
 1000 2180 5000 
T
em
p
er
a
tu
re
 
 
(K
) 
300 
kJ
2.0444
mol 
@ 2180 kPa,300K
Ĥ
 
kJ
_________
mol 
303.15 kJ2.2827
mol 
@ 2180 kPa,303.15K
Ĥ
 
kJ
_________
mol 
400 
kJ
9.6106
mol 
@ 2180 kPa,400K
Ĥ
 
kJ
9.6601
mol 
 
At this point, we can solve for the enthalpy at the operation conditions in the processdescribed in 
this problem. 
Interpolating across at a constant temperature and solving for 
@ 2180 kPa,303.15K
Ĥ yields: 
 
( )
@ 2180 kPa ,303.15 K
_________ kPa 1000 kPa kJ
Ĥ _________ 2.2827 2.2827
_________ kPa 1000 kPa mol
 −
= − + 
−  
Thus, the enthalpy of liquid water at 30°C is estimated to be: 
 
a,l @ 2180 kPa,303.15 K
kJˆ ˆH H __________
mol
= ≈ 
Now we can proceed to look in Table A.2-9 of Geankoplis for the enthalpy values of saturated water 
and saturated steam. Again, we will have to use linear interpolation to calculate the enthalpies of 
saturated water and steam at a pressure of 2180 kPa using the values from the following table. 
 
 
 
 
 
 
 
 
 l
kJ
H
kg
 
 
  
v
kJ
H
kg
 
 
  
P
re
ss
u
re
 
 
 
 
(k
P
a
) 
1553.8 ____________ 2793.2 
2180 b,lH 924.46=
�
 
b,v
H ____________=�
 
2548 966.78 ____________ 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 20 Student View 
Jason M. Keith 
The unknown enthalpies were calculated by interpolation as shown in the following equations: 
( )
b,l
2180 kPa 1553.8 kPa kJ
H 966.78 ___________ ___________ 924.46
2548 kPa 1553.8 kPa kg
− 
= − + = − 
� 
( )
b,v
2180 kPa 1553.8 kPa kJ
H __________ 2793.2 2793.2 __________
2548 kPa 1553.8 kPa kg
− 
= − + = − 
�
 
To calculate the remaining unknown enthalpy, we need to look for data for superheated steam as we 
did for calculating Ha,l. With the values obtained from Table A.2-10 of Geankoplis, the table for the 
process conditions will be given by: 
 Pressure (kPa) 
 2000 2180 2500 
T
em
p
er
a
tu
re
 
 
(°
C
) 
420 
kJ
3291.6
kg 
 
kJ
3284.0
kg 
450 
kJ
3357.6
kg 
�
c,v
kJ
H = ______________
kg 
kJ
_________________
kg 
500 
kJ
3467.6
kg 
 
kJ
3462.1
kg 
 
where the values in bold were calculated by interpolation as shown in the steps below: 
( )
@ 2000 kPa,450 C
450 C 420 C kJ
H 3467.6 3291.6 3291.6 3357.6
500 C 420 C kg°
° − ° 
= − + = ° − ° 
�
 
 
( )
@ 2500 kPa,450 C
________ C ________ C kJ
H 3462.1 3284.0 3284.0 _____________
500 C ________ C kg°
 ° − °
= − + = 
° − ° 
�
 
( )c,v @ 2180 kPa,450 C
_________ kPa 2000 kPa kJ
H H ___________ 3357.6 3357.6 ___________
__________ 2000 kPa kg°
 −
= = − + = 
− 
� �
 
Finally, we can apply the equations shown above for the enthalpies Ha, Hb and Hc to yield: 
 
a b,l a ,l
kJ 18 kg 1 kmol kJˆ ˆ ˆH H – H 1 mol 924.46 ______________
kg 1 kmol 1000 mol mol
   
= = −   
   
 
 
 
 
a
Ĥ 14.34 kJ= 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 21 Student View 
Jason M. Keith 
b b,v b,l
kJ kJ 18 kg 1 kmolˆ ˆ ˆH H – H 1 mol ______________ 924.46
kg kg 1 kmol 1000 mol
    
= = −    
   
 
 
 
b
Ĥ ______________ kJ= 
 
c c,v b,v
kJ kJ 18 kg 1 kmolˆ ˆ ˆH H – H 1 mol ______________ ______________
kg kg 1 kmol 1000 mol
    
= = −    
   
 
 
 
 
 cH ______________ kJ= 
 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 22 Student View 
Jason M. Keith 
Example 1.6-4: Incomplete Combustion of Methane 
A mixture of air and methane is being fed to the firebox in a steam-methane reforming process. The 
methane in this mixture is being burned in order to provide heat for the steam-methane reforming 
reaction to occur. 
The fuel mixture is entering at a rate of 
kmol
213.5
hr
and has a molar fraction of 0.0111 of methane. 
90 % of the methane is burning to produce CO2, and the rest is undergoing incomplete combustion to 
produce CO. These combustion reactions are shown below: 
 1) CH4 + 2O2 CO2 + 2H2O(g) 
o
r,1
kJ
Ĥ 802600
kmol
∆ = − 
 2) CH4 + 
3
2
O2 CO + 2H2O(g) 
o
r,2
kJ
Ĥ 519670
kmol
∆ = − 
Calculate the amount of heat transferred by these two reactions in
kcal
hr
. 
Note: The standard heats of reaction may be obtained from Table A.3-2 of Geankoplis or using 
standard heats of formation of the molecules involved in the chemical reaction, shown in Table A.3-
1. 
Strategy 
This problem can be solved by using tabulated data for standard heats of reaction. We need to take 
into account the selectivity of CO2 to CO. 
Solution 
The amount of heat transferred by the two reactions will be given by the following equation: 
 ( ) ( )o oCO r,1 CO r,2
2
ˆ ˆQ H n H n H= ∆ = ∆ + ∆� � � � 
where: 
 
CO
2
n� = Molar flow rate of carbon monoxide being produced by reaction 2 
 ( )or,1Ĥ∆ = Standard enthalpy of reaction 1 
 
CO
n� = Molar flow rate of carbon monoxide being produced by reaction 2 
 ( )or,2Ĥ∆ = Standard enthalpy of reaction 2 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 23 Student View 
Jason M. Keith 
 
The flow rates of carbon dioxide and carbon monoxide can be calculated by multiplying the overall 
flow rate of fuel/air mixture by the molar fraction of methane. Then, we multiply this amount by the 
amount of methane being converted to carbon dioxide and carbon monoxide respectively. 
If we look at the stoichiometric coefficient of both chemical reactions, it can be seen that the ratio of 
carbon dioxide to methane in reaction 1 and the ratio carbon monoxide to methane in reaction 2 are 
both equal to one. Thus, the amount of methane consumed by reaction 1 will be equal to the amount 
of carbon dioxide produced. Similarly, the amount of carbon monoxide produced by reaction 2 will 
be equal to the amount of methane consumed by reaction 2. 
 4
CO CH , r1
2 4
kmol CH kmol
n n ______ ____________ 213.5
kmol hr
  
= =   
  
� �
 
 
 
 2
CO
2
kmol CO
n 2.13
hr
 
=� 
 
For carbon monoxide: 
4
CO CH , r 2
4
kmol CH kmol
n n ______ ___________ 213.5
kmol hr
  
= =   
  
� �
 
 
 
 
CO
kmol CO
n __________
hr
=�
 
 
 
Now we can substitute the molar flow rates and the enthalpies of reaction into the equation for the 
heat transfer rate Q� to get: 
 
2
kmol CO kJ kmol CO kJ 1 kcal
Q H 2.13 802600 ________ ___________
hr kmol hr kmol _______ kJ
     
= ∆ = − +      
      
� �
 
kcal
Q ______________
hr
=�
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 24 Student View 
Jason M. Keith 
Example 1.6-5: Standard Enthalpy of Reaction 
The water-gas shift reaction produces hydrogen from steam and carbon monoxide and is described 
by the following equation: 
 CO + H2O(g) CO2 + H2 
Determine the standard heat of this reaction. 
Strategy 
The heat of a given reaction can be obtained based on the stoichiometric coefficient of the species 
involved in the chemical reaction and their individual heats of formation. 
Solution 
The heat of reaction can be calculated using the following equation: 
 
 
o o o
r i f ,i i f ,i
reactantsproducts
ˆ ˆ ˆH | | H | | H∆ ν ∆ ν ∆= −∑ ∑
 
Applying this equation to the number of products and reactants for the reaction of methanol in the 
fuel cell yields: 
 
 o o o o or HCO H O COf ,CO f ,H f ,H O f ,CO
22 22 2 2
ˆ ˆ ˆ ˆ ˆH | | H | | H | | H | | H∆ ν ∆ + ν ∆ − ν ∆ − ν ∆= 
 
Since an element is a pure chemical substance, there is no energy transfer involved in its formation. 
Thus, the heat of reaction equation will be reduced to: 
 
 o o o or CO H O COf ,CO f ,H O f ,CO
2 22 2
ˆ ˆ ˆ ˆH | | H | | H | | H∆ ν ∆ − ν ∆ − ν ∆= 
 
The individual heats of formation for each one of these molecules can be found inTable A.3-1 of 
Geankoplis: 
 o
f ,CO
2
kJ
Ĥ 393.51
mol
∆ = − 
o
f ,H O
2
kJ
Ĥ ______________
mol
∆ = 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 25 Student View 
Jason M. Keith 
o
f ,CO
kJ
Ĥ ______________
mol
∆ = 
Substituting these values and the stoichiometric coefficients into the heat of reaction equation, we 
have: 
 
o
r
kJ kJ kJ
Ĥ (1) 393.51 (1) ______________ (____) ______________
mol mol mol
     
∆ − − − − −     
     
=
 
o
r
kJ
Ĥ 41.16
mol
∆ −= 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 26 Student View 
Jason M. Keith 
Example 1.7-1: Cooling of a Fuel Cell 
The bipolar plates are the component of a fuel cell responsible of transferring the heat out of the fuel 
cell. A modified design of the bipolar plate is shown in the following figure. This design has 
additional channels which allow cooling of the fuel cell by circulating a fluid such as air or water. 
 
 
 
 
 
 
 
 
A fuel cell stack in a vehicle initially at a temperature of 80 °C is being cooled by water at room 
temperature. The fuel cell is being cooled down to a temperature of 45 °C. Determine the water flow 
rate required in 
L
min
 if the fuel cell reaches the temperature of 45 °C in 10 seconds. 
The following data is available for the bipolar plate. 
 
3
V 168.75 cm= 
 
3
g
1.3413 
cm
ρ = 
 
p
kJ
C 1.62 
kg C
=
⋅°
� 
The cooling water is exiting the cooling channels at a temperature of 33 °C. 
Strategy 
We can set the energy balance by equaling the amount of heat lost by the fuel cell to the amount of 
heat gained by the water. Since only liquid and solid phases are present in this problem, the heat 
capacities of water and the bipolar plates may be assumed constant. 
Solution 
The energy balance for this problem may be written as follows: 
Reactant air 
channels 
Hydrogen 
channels 
Cooling air/water 
channels 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 27 Student View 
Jason M. Keith 
 Heat gained by water = Heat lost by fuel cell 
 
H O fuel cell
2
H H∆ = ∆ 
Both amounts of heat in the energy balance can be obtained using the equation for sensible heat. 
 ( )H O H O p,H O H O
2 2 2 2
H m C T t∆ = ∆�� 
 ( )
fuel cell fuel cell
H _______ _________ T∆ = ∆ 
where: 
 
H O
2
m mass flow rate of water=� 
 
p,H O
2
C mass specific heat of liquid water=� 
 t time required for cooling the bipolar plate to the final temperature of 45 C= ° 
 
bp
m mass of bipolar plate= 
 
p,H O
2
C mass specific heat of bipolar plates=� 
Note that the equation for the change in enthalpy of water is multiplied by a time period t. This is 
done so we can obtain the total amount of energy transferred as heat to the water instead of the heat 
transfer rate. 
We can start by calculating the change in enthalpy of water as shown in the following steps: 
 ( )H O H O p,H O
2 2 2
H m C T t∆ = ∆�� 
The specific heat of water can be obtained from Table A.2-5 of Geankoplis to be: 
 
p,H O
2
kJ
C ______________
kg C
=
⋅°
� 
Substituting this value, the time period t and the corresponding temperatures into the equation for the 
change in enthalpy of water yields: 
 ( ) ( )
H O H O
2 2
kJ
H m ______________ 33 C 25 C ______ s
kg C
 
∆ = ° − ° 
⋅° 
� 
 
 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 28 Student View 
Jason M. Keith 
Now we proceed to calculate the change in the enthalpy of the fuel cell: 
 
fuel cell bp p,bp
H m C T∆ = ∆� 
The mass of the bipolar plate can be calculated using the volume and density data given in the 
problem statement: 
 ( )33bp bp bp
g _________
m V 1.3413 _________ cm 0.226kg
cm _________
  
= ρ = =  
   
 
Substituting this result, the heat capacity value of the bipolar plates and the initial and final 
temperatures of the fuel cell into the equation for 
fuel cell
H∆ we get: 
 ( ) ( )
fuel cell
kJ
H _______ kg 1.62 _____ C _____ C
kg C
 
∆ = ° − ° 
⋅° 
 
 
fuel cell
H __________ kJ∆ = 
Now we can substitute the equations for 
fuel cell
H∆ and 
H O
2
H∆ into the energy balance equation: 
 ( ) ( )
H O
2
kJ
m ___________ 33 C 25 C 10 s _______ kJ
kg C
 
° − ° = 
⋅° 
� 
Solving for the mass flow rate of water we have: 
 
( ) ( )
H O
2
_______ kJ
m
kJ
________ 33 C 25 C 10 s
kg C
=
 
° − ° 
⋅° 
� 
 
H O
2
kg
m 0.038
s
=� 
We need to convert the mass flow rate to volumetric flow rate as follows: 
 
H O
H O
H O
2
2
2
kg ________
0.038m s ________
V
kg
_______
L
 
 
 = =
ρ
�
� 
 
H O
2
L
V _______
min
=� 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 29 Student View 
Jason M. Keith 
Example 1.7-2: Simultaneous Material and Energy Balances 
The steam-methane reforming process is carried out in a plant operating on the scale of a large 
central station, which is capable of producing 6 2
kg H
1 10
day
× (equivalent to 1 million gallons of 
gasoline). Since the reaction for producing hydrogen from methane and steam is endothermic, heat is 
being added to the reactor. The following diagram is describing this process. 
 
 CH4 + H2O(g) CO (g) + 3H2(g) 
o
r,298 K
kJ
H 206.16
mol
∆ = 
 
 
 
 
 
 
Determine the heat added to the reactor if all of the methane is converted into products. 
Strategy 
We can determine the heat transfer rate Q� by performing material and energy balances around the 
reaction chamber. 
Solution 
The flow rates of hydrogen and carbon monoxide out of the reactor can be determined by the ratio of 
their stoichiometric coefficients to the stoichiometric coefficient of methane (limiting reactant). 
Thus, 
 CO
CH
4
n
1
n
=
�
�
 
H
CH
2
4
n
______
n
=
�
�
 
Solving these equations for the unknown molar flow rates 
H CO
2
n , n , we get: 
 
CO CH
4
n n=� �
 
H CH
2 4
n ____ n=� � 
The molar flow rate of methane entering the reactor can be obtained by multiplying the overall flow 
rate by the molar fraction of methane to yield: 
Q� 
mol
439.5
min
 
4
mol CH
0.237
mol
 
2
mol H O
0.763
mol
 
T = 431 °C 
H ,out
2
n� 
CO,out
n� 
T = 810°C 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 30 Student View 
Jason M. Keith 
 4
CH CH
4 4
mol CH mol
n y n 0.237 ________
mol min
 
= =  
 
� � 
 4
CH
4
mol CH
n ________
min
=� 
Substituting this result into the equations for the molar flow rates 
H CO
2
n , n we can obtain the 
individual flow rates of each gas in the product stream: 
 4
CO
4
mol CH1 mol CO
n __________
mol CH min
 
=  
 
�
 
2 4
H
4
2
____ moles H mol CH
n __________
mol CH min
 
=  
 
� 
 
CO
mol CO
n __________
min
=�
 
2
H
2
mol H
n __________
min
=�
 
Now we can proceed to perform energy balances to determine the heat transfer rate. For this we will 
select a reference state of 25 °C (298 K). 
The general balance equation is given by: 
 ( )oR P298H H q H+ −∆ + =∑ ∑ 
Input items 
The heat entering the reactor is equal to the sum of the changes in the enthalpies of the reactants 
between the reference state and the process conditions. Thus, 
( )24
CH p,CH
4 4
mol CH kJ 1 minˆH nC T __________ 4.680 10 ______ C 25 C
min mol C 60 s
−    = ∆ = × ° − °    
⋅°    
� �CH
4
H 32.99 kW=� 
( )2
H O p,H O
2 2
mol H O mol kJ 1 minˆH nC T 0.763 439.5 ______________ ______ C 25 C
mol min mol C 60 s
    
= ∆ = ° − °    
⋅ °    
� �
H O
2
H __________ kW=� 
The heat capacities were obtained using linear interpolation with the data from Table 1.6-1 of 
Geankoplis at the temperature of 431°C (704 K). 
The standard enthalpy of reaction must be included in the input items as shown in the general energy 
balance equation. 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 31 Student View 
Jason M. Keith 
 o
298 K
kJ mol 1 min
H __________ __________
mol min 60 s
   
−∆ =    
   
� 
 
o
298 K
H ____________ kW−∆ =� 
Output items 
In a similar way we can obtain the enthalpies of the gases leaving in the product stream as shown in 
the following steps. The heat capacities of carbon monoxide and hydrogen were obtained by linear 
interpolation of the data in Table 1.6-1 of Geankoplis at a temperature of 810 °C (1083 K). 
( )
CO p,CO
mol CO kJ 1 minˆH nC T __________ _____________ 810 C 25 C
min mol C 60 s
    
= ∆ = ° − °    
⋅°    
� � 
CO
H 42.50 kW=� 
( )2
H p,H
2 2
mol H kJ 1 minˆH nC T __________ _______________ 810 C 25 C
min mol C 60 s
    
= ∆ = ° − °    
⋅°    
� � 
H
2
H __________ kW=� 
Now we can substitute the enthalpies we calculated into the general balance equation and solve it for 
the heat added to the system q to get: 
 ( )oR P298H H q H+ −∆ + =∑ ∑ 
 ( )oP R 298q H H H= − − −∆∑ ∑� 
 ( )oHCO CH H O 298
2 4 2
q H H H H H= + − − − −∆� � � �� 
 q 42.50 kW __________ kW 32.99 kW __________ kW __________ kW= + − − +� 
 q __________ kW=� 
The positive sign of q�
 
indicates that heat is being added to the system. 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 32 Student View 
Jason M. Keith 
Example 1.7-3: Oxidation of Woody Biomass 
Biomass gasification is a process used for producing hydrogen in large scale for use in fuel cells. 
The elemental analysis of dry woody biomass used in the gasification process yielded the following 
results: 
 
 
 
 The following equation represents the combustion reaction of woody biomass. 
 
 C0.333H0.467O0.200 + 0.3498O2 0.333CO2 + 0.2335H2O(g) 
o
r,25 C
kJ
H 116.62
mol°
∆ = − 
 
Calculate the molar heat of combustion at the temperature of 600 °C. 
 
The following equation may be used for estimating the heat capacity of dry wood. 
 
 ( )3p
J
C 0.1031 3.867 10 T
g C
−  = + × 
⋅ ° 
� 
where: 
 T = Temperature in degrees Kelvin. 
 
This equation was obtained from the book Wood as an Engineering Material published by the U.S. 
Department of Agriculture. 
A diagram of the biomass oxidation process is shown in the figure below: 
 
 
 
 
 
Element Mol % 
C 33.3 
H 46.7 
O 20.0 
1
n , mol biomass 
T = 40 °C 
2 2
n , mol O 
T = 40 °C 
o
r,600 C
q H
°
= −∆ 
3 2
n , mol CO 
4 2
n , mol H O 
T = 600 °C 
 
Combustion 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 33 Student View 
Jason M. Keith 
Strategy 
This problem can be solved by using energy balances around the combustion chamber. The 
reference temperature will be selected to be 25 °C and a basis of 1 mol of biomass will be selected. 
Solution 
First we need to determine the amount of steam and carbon dioxide in the exhaust gases as well as 
the amount of oxygen required by this process. These amounts can be calculated based on the 
stoichiometric coefficients of each molecule involved in the chemical reaction. Thus, 
 
2
1
n
0.3498
n
= 3
1
n
__________
n
= 4
1
n
__________
n
= 
Since we selected a basis of n1 = 1 mol, the amount of moles of oxygen 2n , carbon dioxide 3n and 
steam 
4
n are given by: 
 
2 2
n 0.3498 moles O= 
3 2
n __________ moles CO=
 
4 2
n __________ moles H O= 
Now we can proceed to calculate the amount of heat entering through the reactants. Since there are 
no phase changes occurring for any of the species, the amount of heat can be calculated using the 
definition of sensible heat. 
We can start from the general energy balance equation for a reactive process shown below: 
 ( )oR Pr,25 CH H q H°+ −∆ + =∑ ∑ 
Input items 
 ( )
biomass p,biomass
H biomass m C T= ∆� 
The mass heat capacity of biomass can be obtained by substituting the temperature value of 40 °C 
(313.15 K) into the expression given in the problem statement: 
 ( )( )p
J
C __________ _______________ 313.15 K
g C
 
= + 
⋅ ° 
� 
 
p
J
C __________
g C
=
⋅°
� 
 
Substituting this value into the equation for the enthalpy of biomass entering in the reactants stream 
we get: 
 
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Introduction to Engineering Principles and Units 
Daniel López Gaxiola 34 Student View 
Jason M. Keith 
( ) ( )
__________ g J
H biomass 1 mol biomass __________ _____ C 25 C
1 mol biomass g C
  
= ° − °  
⋅ °  
 
 ( )H biomass __________ J= 
Similarly, we can calculate the heat entering through the oxygen as shown in the following steps: 
( ) ( )2 2
J
H O ____________ mol O ___________ _____ C 25 C
mol C
 
= ° − ° 
⋅° 
 
 ( )2H O ____________ J= 
Note: The heat capacity of oxygen was obtained from the data in Table 1.6-1 of Geankoplis. 
The standard enthalpy of reaction at the temperature of 25 °C is given in the problem statement to 
be: 
 ( )o
r,25 C
kJ
H 116.62 __________ mol
mol°
∆ = − 
 
o
r,25 C
H __________ kJ
°
∆ = 
Output items 
The enthalpies of steam and carbon dioxide can be obtained in a similar way to how we calculated 
the enthalpy of oxygen entering the reactor. For carbon dioxide and steam, the specific heat values 
were obtained through linear interpolation of the data from Table 1.6-1 of Geankoplis. 
( ) ( )2 2
J
H CO __________ mol CO ____________ ______ C 25 C
mol C
 
= ° − ° 
⋅° 
 
 ( )2H CO 8861.5 J= 
( ) ( )2 2
J
H H O __________ mol H O __________ _______ C 25 C
mol C
 
= ° − ° 
⋅° 
 
 ( )2H H O ___________ J= 
Substituting the calculated enthalpies and the heat of reaction at 25 °C into the general energy 
balance equation, we get: 
 ( )oR Pr,25 CH H q H°+ −∆ + =∑ ∑ 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 35 Student View 
Jason M. Keith 
 ( )_________ J __________ J _____________ J q 8861.5 J ________ J+ + + = + 
The diagram of the combustion of biomass process is showing that the amount of heat released by 
the reaction is equal to the enthalpy of reaction at 600 °C. Thus, solving for q yields: 
 
( )o
r,600 C
1 kJ
q H 8861.5 J _________ J _________ J _________ J __________ J
1000 J°
 
= ∆ = + − − −    
 
 
o
r,600 C
H ___________ kJ
°
∆ = 
To obtain the molar heat of combustion at 600 °C, we need to divide this value by the number of 
moles of biomass. Since we selected a basis of 1 mole of biomass reacting, the molar enthalpy will 
be given by: 
 
o
r,600 Co
r,600 C
biomass
H ___________ kJ
Ĥ
n ___________ mol
°
°
∆
∆ = = 
 
o
r,600 C
kJ
Ĥ ___________
mol°
∆ = 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 1 Student View 
Jason M. Keith 
Chapter 2 
Principles of Momentum Transfer and Overall Balances 
In fuel cells, the fuelis usually in gas or liquid phase. Thus, the student must be familiar with the 
principles of fluid mechanics or momentum transfer, which will be covered in the following problem 
modules. In later sections of this chapter, situations combining momentum transfer and heat transfer 
will be illustrated. 
2.2-3 Conversion of Pressure to Head of a Fluid 
2.3-1 Diffusivity of Hydrogen inside a Fuel Cell 
2.5-1 Reynolds Number of Hydrogen flowing into a Fuel Cell 
2.6-1 Flow of Hydrogen into Fuel Cells 
2.6-3 Velocity of Hydrogen in Bipolar Plate Channel 
2.7-1 Energy Balance on Ethanol boiler 
2.10-1 Methanol Flow in Fuel Cell 
2.10-2 Use of Friction Factor in Laminar Flow 
2.10-3 Use of Friction Factor in Turbulent Flow 
2.10-4 Trial-and-Error Solution to Calculate Pipe Diameter 
2.10-5 Flow of Gas in Line and Pressure Drop 
2.10-8 Entry Length for a Fluid in a Rectangular Channel 
2.11-1 Compressible Flow of a Gas in a Pipe Line 
2.11-2 Maximum Flow for Compressible Flow of a Gas 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 2 Student View 
Jason M. Keith 
Example 2.2-3: Conversion of Pressure to Head of a Fluid 
Hydrogen is stored in a compressed gas tank with a volume of 50 L at a pressure of 140 atm and a 
temperature of 25 °C. What is the pressure in the tank in mm Hg and inches of water? 
Strategy 
To determine the pressure of hydrogen as head of a fluid we need to use the equation for calculating 
pressure of a fluid as a function of height. 
Solution 
The equation for pressure as a function of height is shown below. 
 P = ρgh 
where: 
 ρ = density of fluid 
 g = acceleration due to gravitational force 
 h = head or height of fluid 
We can solve this equation for the head h to yield: 
 
P
h 
g
=
ρ
 
To determine the pressure in mm of Hg we need to use the density of mercury, which can be 
obtained from Table 2-31 Perry’s Chemical Engineers’ Handbook, 8
th
 Edition to be: 
 
Hg 3
kg
_____________
m
ρ = 
Since the units in the expression for the head of a fluid need to match, the value of g must be in 
2
m
s
 
 
 
and the pressure must be converted to Pa 
2
N
m
 
 
 
. After entering the values into the equation for 
h we get: 
 
22
3 2
kg mN
_________ __________ 
sm140 atm
1 atm 1 N
h 
kg m
_____________ 9.81
m s
⋅  
  
  
  
  =
  
  
  
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 3 Student View 
Jason M. Keith 
 h ___________ m Hg = 
This value can be converted to mm of Hg by multiplying by the conversion factor from m to mm: 
 
_______________
h ___________ m Hg 
_______________
 
=  
 
 
 h ___________ mm Hg= 
To calculate the head in inches of water, we need to follow a similar procedure. The difference is 
that we need to use the density of water instead of the density of mercury, and use a conversion 
factor to convert meters to inches. Thus, 
 
22
3 2
kg mN
_________ _________ 
sm140 atm
1 atm 1 N
h 
kg m
_________ 9.81
m s
⋅  
  
  
  
  =
  
  
  
 
 
2
h 1446 m H O= 
The density of water was obtained from Table A.2-3 of Geankoplis at a temperature of 25 °C. 
Converting this value to inches we get: 
 2
2
2
1 in H O
h 1446 m H O
_____________ m H O
 
=   
 
 
 
2
h ___________ in H O= 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 4 Student View 
Jason M. Keith 
Example 2.3-1: Diffusivity of Hydrogen inside a Fuel Cell 
Hydrogen in a bipolar plate is diffusing in the anode side of a fuel cell. The current density of a stack 
of 440 fuel cells is 
2
mA
600
cm
. 
The hydrogen is entering the fuel cell at a pressure of 2 atm and a temperature of 25 °C. Determine 
the diffusivity of hydrogen through the gas-diffusion layer with a thickness of 100 µm in 
2mm
s
if the 
fuel cell performance is limited by mass transfer. 
The amount of hydrogen reacted as a function of current is described by the following equation: 
 
H reacted
2
IN
n
2F
=�
,
 
where: 
 I = current in amperes (A) 
 N = number of cells in the fuel cell stack 
 F = Faraday’s constant = 
C
96485
mol e−⋅
 
Strategy 
The diffusivity of hydrogen can be determined using Fick’s Law of diffusion and the definition of 
the consumption rate of hydrogen in terms of the current. 
Solution 
Fick’s first law relates the diffusive flux to the difference in concentration of a substance. For this 
problem, Fick’s first law is given by: 
 
H
HH reacted
2
22
dC
n D A
dx
= −�
,
 
where: 
H
2
D diffusivity of hydrogen through the gas diffusion layer= − 
A cross sectional area of the gas diffusion layer = − − 
H
2
dC
change in the concentration of hydrogen along the thickness of the gas diffusion layer
dx
= −
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 5 Student View 
Jason M. Keith 
The left hand side of Fick’s first law can be re-written in terms of the current. Hence, 
H
H
2
2
dC_________
D A
_________ dx
= − 
The problem statement is giving the value of the current density defined as the current of the fuel 
cell divided by the area. Thus, this equation can be rewritten in terms of the current density I
�
as 
follows: 
H
H
2
2
dC_________
D
_________ dx
= − 
Assuming the diffusivity remains constant along the gas-diffusion layer, this equation can be 
separated and integrated as follows: 
 
 
 x m
H Hx 0 m
H x 100 m
2
2 2H x 0
2
C
C
_________
dx D dC
_________
= µ
= µ
= µ
=
= −∫ ∫
_________
@
@
 
Since all the hydrogen entering the fuel cell is reacting when mass transfer is limiting the fuel cell 
performance, the concentration of hydrogen in the anode will be given by: 
 
H x 100 m
2
C 0
= µ
=
@
 
The concentration of a substance is defined as the number of moles of substance in a volume of 
solution. Thus, 
H
H
2
2
n
C
V
= 
The concentration of hydrogen entering the channels in the bipolar plate can be obtained using ideal 
gas law as shown in the following steps: 
PV nRT= 
Solving for the concentration 
H
2
n
V
from the ideal gas law, we have: 
 
H
H
2
2
n ____
C
V _______
= = 
 
 
 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 6 Student View 
Jason M. Keith 
Substituting the corresponding quantities in the right side of this equation yields: 
 
( )
H 3
2
101325 Pa
2 atm
1 atm
C
Pa m
__________ 298.15K
mol K
 
 
 
=
⋅ 
 ⋅ 
 
 
H 3
2
mol
C __________
m
= 
Substituting the concentration values into the integral equation we get: 
 
 
 
3
 1 10 m 0
molH H0 m _________ 
m
4
2 2
dx D dC
× −
= −∫ ∫
_____
_____
 
We can integrate this equation and substitute the values for the current density, number of fuel cells 
and Faraday’s constant to give: 
( )
( )
2
4
H 3
2
mA
 cells
molcm
1 10 m 0 m D
C m
2
mol e
−
−
 
 
    × − = −      
 
⋅ 
______ ______
___________
___________
 
Finally we can solve for the diffusivity 
H
2
D to obtain: 
( )
( )
2
2 2
4
H
3
2
mA 1 A C/s cm
cells
cm 1000 mA 1 A 1 m
D 1 10 m
C mol
2
mol e m
−
−
     
     
       = ×   
  
⋅  
_____ ________
______ ______
__________ _________
 
2
5
H
2
m
D 1 67 10
s
−= ×. 
This diffusivity value can be converted to 
2mm
s
 by multiplying it by a conversion factor from m
2
 to 
mm
2
. 
2
5
H
2
m
D 1 67 10
s
−  = × 
 
____________
.
____________
 
2
H
2
mm
D
s
= _______ 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 7 Student View 
Jason M. Keith 
Example 2.5-1: Reynolds Number of Hydrogen flowing into a Fuel Cell 
A compressed gas tank contains hydrogen at room temperature and a pressure of 140 atm. The valve 
on this tank is opened and the hydrogen enters a fuel cell at a rate of 0.75 
kg
hr
through a steel pipe 
with an inner diameter of 5.46 mm. Determine the type of pipe connecting the fuel cell to the 
hydrogen tank and determine if the flow of hydrogen is laminar or turbulent. 
Strategy 
To determine the type of pipe used to connect the hydrogen tank to the fuel cell we need to find the 
pipe corresponding to the inner diameter given in the problem statement. The flow regime may be 
determined depending on the value of the Reynolds number. 
Solution 
Appendix A.5 of Geankoplis is showing the properties of different types of standard steel pipes. For 
the inner diameter of 5.46 mm, the nominal size of the pipe is 
1
8
in. with a Schedule Number of 80. 
For flow inside a pipe, the Reynolds number is given by: 
D
Re
υρ
=
µ
 
where: 
D = inner diameter of the pipe in meters (m) 
υ = velocity of the fluid inside the pipe in 
m
s
 
ρ = density of fluid in 
3
kg
m
 
µ = viscosity of fluid in 
kg
m s⋅
 
The velocity of hydrogen circulating in the pipes can be obtained by dividing the volumetric flow 
rate of hydrogen by the inner cross-sectional area of the pipe. However, since we are given the mass 
flow rate, we will have to calculate the volumetric flow rate using the ideal gas law. 
 
mRT
PV
M
=
�
� 
Solving for the volumetric flow rate V� yields: 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 8 Student View 
Jason M. Keith 
 
mRT
V
MP
=
�
� 
Substituting the corresponding quantities into this equation we get: 
 
( )
( )
3kg Pa m
0.75 ________ 298 K
hr mol K
V
101325 Pa kg 1 kmol
140 atm ______
1 atm kmol 1000 moles
⋅  
  
⋅  =
   
   
   
� 
Note that conversion factors for the pressure and number of moles were used in order to get the 
correct units for the volumetric flow rate. 
 
3m
V __________
hr
=� 
To calculate the velocity of hydrogen in the pipe, we also need to determine the cross-sectional area 
of the pipe. This value can be obtained directly from Table A.5 of Geankoplis. 
2A _____________ m= 
Now we can obtain the velocity of hydrogen as shown in the following equations. Note that we are 
multiplying the velocity equation by the conversion factor from hours to seconds, hence to obtain the 
velocity in 
m
s
. 
3
2
m 1 hr
________
hr ________ sV
A ____________ m
 
 
 υ = =
�
 
m
0.772
s
υ = 
The density of hydrogen at the pressure of 140 atm and the temperature of 25 °C (298 K) can be 
calculated using the ideal gas equation of state: 
mRT
PV
M
=
�
� 
Since the density r is equal to dividing the mass flow rate by the volumetric flow rate, we can solve 
for the density as shown in the following step. 
m _______
V _______
ρ = =
�
�
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 9 Student View 
Jason M. Keith 
We can enter the numeric values into the right side of this equation to get: 
( )
( )
33
__________ 2 kg 1 kmol
_______ atm
__________ 1 kmol 1000 moles kg
__________
mPa m
8.314 298 K
mol K
   
   
   
ρ = =
⋅ 
 ⋅ 
 
The only parameter left to be determined before calculating Reynolds number is the viscosity, which 
can be obtained from Figure A.3-2 for gases at a pressure of 1 atm. However, for hydrogen at the 
pressure and temperature conditions in this problem, the viscosity does not depend on pressure. 
Thus, it is valid to use Figure A.3-2. 
To obtain the viscosity of hydrogen we need to locate the coordinates for hydrogen in Table A.3-8 
and draw a line that passes through these coordinates and the temperature value of 25 °C. Hence, the 
viscosity will be estimated to be: 
kg
____________
m s
µ ≈
⋅
 
Now we can enter the quantities we found for velocity, density and viscosity into the equation for the 
Reynolds number to yield: 
( )3 3
m kg
5.46 10 m 0.772 _________
s m
Re
kg
____________
m s
−   ×   
  =
⋅
 
Re __________= 
Conclusion: 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 10 Student View 
Jason M. Keith 
Example 2.6-1: Flow of Hydrogen into Fuel Cells 
Hydrogen is exiting the fuel tank in a vehicle through a stainless steel pipe (schedule number 80) 
with a nominal diameter of ¼”. The hydrogen is leaving at a flow rate of 
L
446.4
hr
at room 
temperature and at a pressure of 2 atm. This flow rate is equally distributed between the 325 fuel 
cells required to power this vehicle. 
The hydrogen is being distributed to the fuel cells by a pipe of 1/16” inner diameter. The hydrogen 
flow rate entering the channels of the bipolar plate of a fuel cell is equally distributed between the 3 
channels of the bipolar plate. (Note: The channels in the bipolar plate will be assumed to be 
semicircular with an inner diameter of 1/16”) 
a) Determine the mass flow rate of hydrogen entering each cell and each one of the channels in the 
bipolar plate. 
 
 
 
 
 
 
 
 
Strategy 
To determine the mass flow rate from the volumetric flow rate we need to use ideal gas law. 
Solution 
We can solve the ideal gas law for the mass flow rate as shown in the following steps: 
 overall
overall
PV M
m
RT
=
�
� 
Substituting the corresponding quantities into the right hand side of this equation yields: 
n = 1 n = 325 
… Hydrogen Tank 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 11 Student View 
Jason M. Keith 
 
( )
( )
overall
L g
2 atm 446.4 2
hr mol
m
L atm
____________ 298.15 K
mol K
  
  
  =
⋅ 
 
⋅ 
� 
 
overall
g
m ________
hr
=� 
To determine the flow rate of hydrogen entering each fuel cell, we divide the mass flow rate of 
hydrogen leaving the tank by the total number of fuel cells in the stack. Thus, 
 
fuel cell
fuel cells
g
________
m hrm
n ______ cells
= =
�
� 
 
fuel cell
g
m 0.225
hr
=� 
This value can by divided by 3 (number of channels on each bipolar plate) to determine the flow of 
hydrogen to each channel: 
 fuel cell
channel
channels
g
0.225m hrm
n 3 channels
= =
�
� 
 
channel
g
m __________
hr
=� 
b) What is the average velocity of the hydrogen leaving the tank? 
Strategy 
The velocity of hydrogen can be determined by applying the definition of velocity in terms of the 
flow rate. 
Solution 
The velocity of a fluid can be determined with the following equation: 
 
V
A
υ =
�
 
First we need to determine the volumetric flow rate of hydrogen to each fuel cell and to each channel 
in the bipolar plates. This can be done in a similar way to part a) of this problem. 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 12 Student View 
Jason M. Keith 
 
fuel cell
fuel cells
L
446.4
V hrV
n 325 cells
= =
�
� 
 
fuel cell
L
V ____________
hr
=� 
 fuel cell
channel
channels
L
_______V hrV
n 3 channels
= =
�
� 
 
channel
L
V _________
hr
=� 
Now we need to calculate the cross-sectional areas of different pipes through which hydrogenis 
circulating. First, the area of the ¼” pipe connected to the fuel tank can be found in Table A.5-1 of 
Geankoplis to be: 
 2
1/4" pipe
A ______________ m= 
The cross-sectional area of the pipe entering the fuel cells can be calculated using the equation for 
the area of a circle, where the diameter will be 1/16”. Hence, 
 
2
2
1/16" pipe
1
in
D 16
A
4 4
 
π π   = = 
 
 
 2
1/16" pipe
A _____________ in= 
Converting this value to m
2
, we have: 
 
2
2
21/16" pipe
____________ m
A ____________ in
1 in
 
=  
 
 
 2
1/16" pipe
A ____________ m= 
Since the channels on the bipolar plate are assumed to be semicircular, the cross-sectional area of the 
channel can be calculated as follows: 
 
2
channel
D
4
A
2
π 
 
 = 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 13 Student View 
Jason M. Keith 
Substituting the corresponding quantities into the right hand side of this equation yields: 
 
2 2
2
channel
1 _______________ m
in
16 1 in
A
8
  
π   
   = 
 2
channel
A _______________ m= 
Now we can obtain the velocities in the three different sections, as shown in the following equations: 
 
 
3
2H in 1/4" pipe
2
L 1 m _________
446.4
hr 1000 L _________
_______________ m
   
    
   υ = 
 
 H in 1/4" pipe
2
m
2.68
s
υ = 
 
 
 
3
2H in fuel cell
2
L 1 m 1 hr
___________
hr 1000 L 3600 s
______________ m
    
    
    υ = 
 
 H in fuel cell
2
m
____________
s
υ = 
 
 
 
3
2H in channel
2
L 1 m 1 hr
_________
hr 1000 L 3600 s
_______________ m
    
    
    υ = 
 
 H in channel
2
m
_______
s
υ = 
c) Calculate the mass flux of hydrogen circulating through each channel in 
2
kg
m hr⋅
. 
Strategy 
The flux of hydrogen refers to the amount of hydrogen flowing through an area during a period of 
time. 
 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 14 Student View 
Jason M. Keith 
Solution 
The flux of hydrogen can be obtained by dividing the mass flow rate of hydrogen entering each 
channel, by the cross sectional area of the channel. Thus, 
 channel
channel
m
G
A
=
�
 
Entering numeric values into this equation we get: 
 
2
g _________
_________
hr _________
G
______________ m
 
 
 = 
 
2
kg
G ___________
m hr
=
⋅
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 15 Student View 
Jason M. Keith 
Example 2.6-3: Velocity of Hydrogen in Bipolar Plate Channel 
The following figure is showing the flow of hydrogen along the horizontal channel of length L in a 
bipolar plate. 
 
 
 
 
 
 
 
 
The velocity of hydrogen along a square channel in a bipolar plate is given by the following 
equation: 
 
2 2
z max
x y
1 1
B B
      
υ = υ − −      
         
 
Determine the average velocity of the hydrogen flowing through the channel. 
Strategy 
The average velocity can be calculated from the expression of the velocity as a function of the 
position in the x and y dimensions. 
Solution 
The average velocity can be calculated using Equation 2.6-17 of Geankoplis, shown below: 
 
 
av zA
1
A
υ = υ∫ ∫ dA 
In Cartesian coordinates dA may be written as dxdy. The cross-sectional area of the channel is 
obtained by multiplying the dimensions of the channel. Hence, 
( )
 
 
2 2
B B
av max2 B B
1 x y
1 1 
B B2B − −
      
υ = υ − −      
         
∫ ∫ dxdy 
Reactant air 
channels 
Hydrogen 
channels 
z 
x 
y 
H2 
L 
x = − B x = B 
y = − B 
y = B 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 16 Student View 
Jason M. Keith 
Integrating this equation in the x-direction, we get: 
 
 
 
 B
2
B
max
av 2 B
 B
y
x 1
4B B−
−
   υ  
υ = − −     
      
∫
______
dy
______
 
Evaluating the integrated expression for x yields: 
 
 
 
2
B
max
av 2 B
y
B 1
4B B−
      υ    
υ = − − − − −        
         
∫
______ _____
____ dy
______ _____
 
We can reduce this equation to get: 
 
 
 
2
B
max
av 2 B
y
1
4B B−
  υ  
υ = − −    
     
∫
_____
_____ dy
_____
 
 
 
 
2
B
max
av 2 B
y
1
4B 3 B−
 υ −   
υ = −    
     
∫
_____ _____
dy 
 
 
 
2
B
max
av B
y
1
B−
 υ  
υ = −  
   
∫ dy_______
 
Now we can proceed to integrate and evaluate the equation for the average velocity with respect to y 
as shown in the following steps: 
 
B
max
av
B−
  υ
υ = −  
  
______
____
______ ______
 
 max
av
       υ  
υ = − − −       
        
_____ ______
____ ____
_____ _____ ______
 
Reducing this equation we determine the expression for the average velocity of hydrogen in the 
channels: 
 
3
max
av
2B  υ
υ = −  
  
_____
_____ ______
 
 max max
av
3
υ υ 
υ = = 
 
_________
_____ ____
 
 max
av
4
9
υ
υ = 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 17 Student View 
Jason M. Keith 
Example 2.7-1: Energy Balance on Ethanol Boiler 
A liquid solution of ethanol and water is entering a boiler before undergoing a reforming reaction for 
producing hydrogen for proton-exchange membrane fuel cells. A diagram of the process is shown 
below. 
 
 
 
 
What is the power required to produce 
kg
1323
hr
of vapor? 
Strategy 
The overall energy balance for a steady-state flow system can be used to determine the amount of 
heat required. 
Solution 
The energy balance equation is given by Equation 2.7-10 of Geankoplis and is shown below: 
 ( ) ( )2 2 s2 1 2 1 2 1
1
H H g z z Q W
2
− + υ − υ + − = −
α
 
where: 
 H2 = Enthalpy of the substance exiting the system 
 H1 = Enthalpy of the substance entering the system 
 α = Kinetic-energy velocity correction factor 
 
2
υ = Velocity of the substance exiting the system 
 
1
υ = Velocity of the substance entering the system 
 g = Acceleration due to gravity 
 
2 1
z z− = Difference in height between the outlet and inlet points in the system 
 Q = Heat added (+) or removed (–) to the system 
 Ws = External work applied by (+) or to (–) the system 
Liquid Ethanol/Water Solution 
l
m
0.085
s
υ = 
l
kJ
H 13216
kg
= −� 
Ethanol/Water gas mixture 
g
m
13.71
s
υ = 
g
kJ
H 10878
kg
= −� 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 18 Student View 
Jason M. Keith 
For this problem we will assume turbulent flow, so the value of the kinetic-energy velocity 
correction factor α is close to 1. The external work term Ws will be neglected in the balance 
equations as there are no mechanical parts performing work to the system. Since no information is 
given about the height of the inlet and outlet points in the system, we will assume the values are the 
same. Thus, the difference 
2 1
z z− is equal to zero. 
After applying these assumptions, the energy balance equation will be given by: 
 ( )
2 1
1
Q H H ______ ______
2
= − + −� � � 
Where the tilde represents the amount per unit mass. 
The values given for the mass enthalpies of the ethanol/water mixtures and the velocities of the 
liquid and gas can be now substituted into this equation to obtainthe amount of heat required. Thus, 
 ( )
2 1
1
Q H H ______ ______
2
= − + −� � � 
 
2 2
kJ kJ 1 m m
Q ____________ 13216 13.71 ________
kg kg 2 s s
      
= − − + −      
       
� 
 
kJ
Q 2432
kg
=� 
To determine the power required, we need to multiply the heat per kilogram of vapor by the flow 
rate: 
 
kJ kg ______
Q 2432 ___________
kg hr ______
  
=   
  
� 
 Q ________ kW=� 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 19 Student View 
Jason M. Keith 
Example 2.10-1: Methanol Flow in Fuel Cell 
A solution of 40 wt. % methanol and 60 wt. % water is entering the channels of a bipolar plate in a 
direct methanol fuel cell. The viscosity and density of this solution are shown below: 
31.85 10 Pa s−µ = × ⋅ 
3
kg
931.5
m
ρ = 
The pressure drop along a rectangular channel is shown by Bahrami et al. [1]
 
to be: 
 
2
5
L
P
1 64
c tanh
3 2
µ υ
∆ =
 ε π  
−   π ε  
 
where: 
 P∆ = Pressure drop, Pa 
 µ = Viscosity of the fluid, Pa s⋅ 
 L = Length of the channel, m 
 υ = Velocity of the fluid in the channel, 
m
s
 
The parameters c and ε are related to the dimensions of the channel as shown in the following 
figure: 
 
 
c
b
ε = 
 
 
What is the pressure drop in atm and mm Hg along a single channel if the methanol flowing in the 
fuel cell has a Reynolds number of 300? 
Strategy 
The pressure drop in the channel can be calculated using the equation for the pressure drop given in 
the problem statement. 
1. Bahrami, M., Yovanovich, M. M., Culham, J. R., “Pressure Drop of Fully-Developed, Laminar Flow in 
Microchannels of Arbitrary Cross Section, Journal of Fluids Engineering, 128, 1036-1044 (2006) 
2b = 1 mm 
2c = 1 mm L = 30 mm 
Gas-Diffusion Layer 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 20 Student View 
Jason M. Keith 
Solution 
The equation given in the problem statement for the pressure drop depends on the velocity of the 
fluid in the channel. Since we are not given this value directly, we will have to calculate it from the 
definition of the Reynolds number. For an open channel, the Reynolds number is defined as follows: 
 H
D
Re
υρ
=
µ
 
where DH is the hydraulic diameter, given by: 
 
H
4A
D
P
= 
In this equation, A is the cross-sectional area of the channel, and P is the perimeter of the channel in 
contact with the fluid. Since the fluid in the channel is in contact with the gas diffusion layer, we 
need to consider all the dimensions of the channel when calculating the wetted perimeter. Hence, the 
hydraulic diameter can be calculated using the dimensions of the channel as shown in the following 
step: 
 
( )
H
4 2b 2c
D
____________
⋅
= 
Substituting the values of b and c into this equation yields: 
 
( )
2
H
4 1 mm 1 m
D
_____________ 1000 mm
 
=  
 
 
 3
H
D 1 10 m−= × 
Now we can enter this value for the hydraulic diameter into the equation for Reynolds number and 
solve for the velocity to get: 
 
( )
3
3H
3
kg
300 1.85 10
Re m s
kgD
1 10 m 931.5
m
−
−
 
× µ ⋅ υ = =
ρ  
×  
 
 
 
m
__________
s
υ = 
The only remaining values that need to be calculated before being able to determine the pressure 
drop are the parameters b, c and ε : 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 21 Student View 
Jason M. Keith 
 2b = 2c = ____________ 
 
__________ 1 m
b c ______________ m
2 1000 mm
 
= = = 
 
 
 
4
c ________________ m
b 5 10 m−
ε = =
×
 
 ____ε = 
Finally, the pressure drop along the length of the channel of 30 mm can be calculated as follows: 
 
( )
( )
( )
( )
3
2
5
kg m
1.85 10 0.03 m ___________
1 atmm s s
P
101325 Pa64 11
______________ tanh
3 2 1
−   ×   
⋅     ∆ =  
    π
−  
π   
 
 P ____________ atm∆ = 
Converting this value to mm Hg, we have: 
 
_____________
P ___________ atm
_____________
 
∆ =  
 
 
P __________ mm Hg∆ = 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 22 Student View 
Jason M. Keith 
Example 2.10-2: Use of Friction Factor in Laminar Flow 
An aqueous solution of 40 mol % methanol is flowing along the channels in a bipolar plate of a 
direct-methanol fuel cell at a velocity of 
m
0.601
s
and a temperature of 27 °C. 
Bahrami et al.[1] are showing the following equation to determine the friction factor in a rectangular 
channel: 
 
( )5
A
12
f Re
192
1 tanh 1
2
⋅
 ε π  
− + ε ε  π ε  
= 
where: 
 
A
A
Re
υρ
=
µ
 
The parameter ε is a function of the dimensions of the channel, as shown in the following figure. 
 
 
c
b
ε = 
 
 
Use the definition of friction factor to calculate the pressure drop in atm along a single channel. 
Strategy 
The pressure drop in the channel can be calculated using the equation for the friction factor given in 
Geankoplis. 
Solution 
Equation 2.10-4 of Geankoplis gives the definition of Fanning friction factor: 
w
2
A P
A
f
2
 ∆
 
 =
ρυ
 
1.
 
Bahrami, M., Yovanovich, M. M., Culham, J. R., “Pressure Drop of Fully-Developed, Laminar Flow in Microchannels 
of Arbitrary Cross Section, Journal of Fluids Engineering, 128, 1036-1044 (2006) 
2b = 1 mm 
2c = 1 mm L = 30 mm 
Gas-Diffusion Layer 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 23 Student View 
Jason M. Keith 
where: 
 A = Cross – sectional area of the channel. 
 Aw = Surface area of the channel being wetted by the fluid. 
 P∆ = Pressure – drop along the channel of length L 
 ρ = Density of the fluid circulating through the channel 
 υ = Velocity of the fluid in the channel 
We can start by solving for the pressure drop P∆ from the definition of friction factor, given in 
Equation 2.10-4 of Geankoplis. 
___________ ___________
P
___________ 2
 
∆ =  
 
 
For the rectangular channel in the bipolar plate, the cross sectional area and the wetted surface area 
are given by: 
 ( ) ( )A 2b 2c= 
 ( )wA 2 ____________ L=    
Note that for the wetted perimeter we are considering the area of the channel in contact with the 
gas-diffusion layer. 
Since b = c, we can rewrite the area and the wetted perimeter as follows and substitute the 
dimensions of the channel in these equations to get: 
 ( )
2
2 2
2
1 m
A ___________ 4 __________ mm ___________ m
___________ mm
 
= = = 
 
 
 ( )( )
2
2
w 6 2
1 m
A _______ 8 ________ mm _______ mm ____________ m
1 10 mm
 
= = = 
× 
 
the value of ε and enter the obtained value into the equation for 
A
f Re⋅ . Thus, 
 
c 0.5 mm
______
b _______ mm
ε = = = 
 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 24 Student View 
Jason M. Keith 
( )5
A
12
f Re
192
1 tanh 1 ________ _____
2
⋅
 π  
− +  π   
= 
A
f Re _________⋅ = 
 
The Fanning friction factor f can be obtained from this equation but first we need to calculate the 
Reynolds number 
A
Re using the equation given in the problem statement as shown in the following 
steps: 
( )
A
A
Re
υρ
=
µ
 
The density of methanol at the temperature of 27 °C can be obtained from Table 2-234 of Perry’s 
Chemical Engineers’ Handbook, 8
th
 Edition to be: 
 
CH OH
3
mol
24.486
L
ρ = 
Converting this value to 
3
kg
m, we have: 
 3
3CH OH
3
3
_______ g CH OHmol 1 kg _________ L
24.486
L 1 mol CH OH 1000 g 1 m
   
ρ =         
 
 
3CH OH
3
kg
____________
m
ρ = 
Since the methanol entering the fuel cell is diluted in water, the density will depend on the 
concentration of the methanol solution. Thus, the density can be calculated as shown below: 
 
CH OH H O
3 2
_________ _________ρ = ρ + ρ 
The density of water is given in Table A.2-3 of Geankoplis. Since the value of the density at the 
temperature of 27 °C, we can use the value in the Table at the temperature closest to 27 °C, which is 
25 °C: 
 
3H O
2
kg
_________
m
ρ = 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 25 Student View 
Jason M. Keith 
Substituting the individual densities of water and methanol into the equation for the density of the 
solution ρ yields: 
 
3 3
kg kg
0.4 ____________ 0.6 ____________
m m
   
ρ = +   
   
 
 
3
kg
____________
m
ρ = 
The viscosity of methanol can be determined using Figure A.3-4 of Geankoplis and the coordinates 
in Table A.3-12. Thus, 
 
kg
____________
m s
µ =
⋅
 
Now we can obtain the value of the Reynolds number 
A
Re as follows: 
 
( ) ( )
2
3
A
m kg
0.601 ____________ ____________ mA s m
Re
kg
____________
m s
  
  υρ   = =
µ
⋅
 
 
A
Re ____________= 
From the value calculated for 
A
f Re⋅ we can solve for the Fanning friction factor to yield: 
 
A
14.13 14.13
f
Re ____________
= = 
 f ____________= 
Finally we can determine the value of the pressure drop along the channel in the direct-methanol fuel 
cell by entering all the corresponding values into the equation for ∆P 
( )
2
4 2 3
2
kg m
____________ 0.601
____________ 1.2 10 m m s
P
____________ m 2
−
 
 ×  ∆ = ⋅ 
1 atm
P 942.45 Pa
____________ Pa
 
∆ =  
 
 
P ______________ atm∆ = 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 26 Student View 
Jason M. Keith 
Example 2.10-3: Use of Friction Factor in Turbulent Flow 
Pure hydrogen is exiting a pressure swing adsorption unit at a rate of 
kg
63.12
hr
and a pressure of 
2061 kPa. The hydrogen is entering the storage tank through a pipe with an inner diameter of 100 
mm and a length of 10 m. If the maximum friction loss permitted along the pipe is 
J
3.60
kg
, what 
material would you propose for the pipe transporting the hydrogen from the pressure swing 
adsorption unit to the storage tank? 
The gas leaving the pressure swing adsorption unit has the following properties: 
 
69.769 10 Pa s−µ = × ⋅ 
 
3
kg
1.425 
m
ρ = 
Strategy 
To determine what type of pipe to use in this process, we need to determine the parameter ε for the 
flow conditions in this process. 
Solution 
The parameter 
D
ε
may be obtained from Figure 2.10-3 of Geankoplis. To do this, we need to know 
the values of the Reynolds number and the Fanning friction factor. 
The friction factor f can be calculated from the definition of the friction loss, as shown in the 
following equation: 
 
2
f
L
F 4f
D 2
∆ υ
= 
Solving for the Fanning friction factor f, we get: 
 
_________
f
_________
= 
The value of the velocity of hydrogen inside the pipe is not given. However, its value can be 
calculated by dividing the volumetric flow rate of hydrogen by the cross-sectional area of the pipe. 
Hence, 
 
V
A
υ =
�
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 27 Student View 
Jason M. Keith 
The volumetric flow rate is calculated by dividing the mass flow rate by the density. The calculation 
of the volumetric flow rate and the cross-sectional area of the pipe are shown in the following steps: 
 
( )
2
2
2
0.1 mD
A ______________ m
4 4
ππ
= = = 
( )23
kg
63.12
m 1 hrhr
kgA 3600 s
1.425 _____________ m
m
 
υ = =  
ρ    
 
 
�
 
m
___________
s
υ = 
Substituting this velocity value into the equation for the friction factor we obtained, we have: 
 f __________= 
The other parameter required to locate the parameter 
D
ε
 in Figure 2.10-3 is the Reynolds number, 
which is obtained as follows: 
 
D
Re
υρ
=
µ
 
Substituting the corresponding values into this equation yields: 
 
( ) 3
6
m kg
0.1 m ______ 1.425 
s m
Re
kg
9.769 10
m s
−
  
  
  =
×
⋅
 
 Re ____________= 
Now we can locate the value of 
D
ε
in the graph for the Reynolds number and the friction factor. 
From Figure 2.10-3, this parameter is estimated to be: 
 __________
D
ε
≈ 
 
 
 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 28 Student View 
Jason M. Keith 
Solving for the pipe roughness ε , we get: 
 ( )_________ D _________ 0.1 mε ≈ ≈ 
 _____________ mε ≈ 
Conclusion: 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 29 Student View 
Jason M. Keith 
Example 2.10-4: Trial-and-Error Solution to Calculate Pipe Diameter 
Hydrogen in a fuel cell vehicle is flowing from the tank to the fuel cell stack through a commercial 
steel pipe with a length of 3 m. The hydrogen is being consumed at a rate of 
g
1.17
s
and is entering 
the fuel cell stack at a temperature of 25 °C and a pressure of 2.5 atm. 
Determine the diameter of the pipe connecting the fuel tank to the fuel cell stack if a head of 240 in 
Hg is available to compensate for the friction loss. 
Strategy 
From the value of Ff given as head of fluid we can determine the required diameter for feeding 
hydrogen to the fuel cells. 
Solution 
The friction loss can be calculated by multiplying the head of fluid in m of H2O by the acceleration 
due to gravity. Thus, 
 2
2f
_________ m H Om
F 240 in Hg ____________
s 1 in Hg
  
=   
  
 
 
f
J
F 812.69
kg
= 
To determine the pipe diameter we have to obtain the diameter from the definition of friction force, 
given by the following equation: 
 
2
f
L
F 4f
D 2
∆ υ
= 
Solving for the diameter D, we have: 
 
2
f
L
D 4f
F 2
∆ υ
= 
As it can be seen in this equation, we need to obtain the Fanning friction factor f and the velocity 
υbefore being able to calculate the diameter. 
The friction factor can be obtained from Figure 2.10-3 of Geankoplis. To do this, we need to 
calculate Reynolds number first and obtain the relative roughness 
D
ε
 of commercial steel. Hence, 
 
_____________ m
D D
ε
= 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 30 Student View 
Jason M. Keith 
Before obtaining the velocity of the fluid first we need to convert the mass flow rate to volumetric 
flow rate. To do this, we need to calculate the density of the fluid using the ideal gas equation of 
state. The velocity can then be determined by dividing the volumetric flow rate of hydrogen by the 
cross-sectional area of the pipe as shown in the following steps: 
 
( )
( )
33
2 g 1 kg
2.5 atm
mol 1000 gPM kg
_________
RT mL atm 1 m
_____________ 298.15 K
mol K 1000 L
  
  
  
ρ = = =
⋅  
 ⋅  
 
 
3
3
g 1 kg
1.17
s 1000 gm m
V ___________
kg s
_________
m
 
 
 = = =
ρ
�
� 
 
2D
A
4
π
= 
 
3
2
m
4 ___________
sm V
s A D
 
 
   υ = = 
π 
�
 
 
2
___________
D
υ = 
The other parameter needed to obtain the friction factor from Figure 2.10-3 is the Reynolds number, 
calculated as follows: 
 
( ) 2 3
__________m kg
D m ___________
D D s m
Re
kg
_____________
m s
  
  υρ   = =
µ
⋅
 
 
169.23
Re
D
= 
The viscosity was obtained from Appendix A.3 of Geankoplis. 
As we can see, the diameter of the pipe D appears in all the expressions required for determining the 
friction factor. Therefore, we will select a diameter value and compare the result obtained to the 
calculated friction force. The initial guess for the diameter will be: 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 31 Student View 
Jason M. Keith 
Trial 1 
 D 0.010 m= 
Substituting this value in the equations for Reynolds number, velocity and relative roughness, we 
get: 
 
169.23 169.23
Re _____________
D 0.010
= = = 
 
2 2
__________ __________ m
__________
D (0.010) s
υ = = = 
 
____________ m ____________ m
__________
D D (0.010 m)
ε
= = = 
Locating the values of the relative roughness and the Reynolds number in Figure 2.10-3 we find: 
 f = 0.0085 
Substituting this result as well as the velocity and dimensions of the pipe into the equation for the 
friction force we get: 
 
2
f
L
F 4f
D 2
∆ υ
= 
 ( )
2
f
m
______
_____ m s
F 4 0.0085
0.010 m 2
 
 
 = 
 
f
J
F ______________
kg
= 
It can be seen that this value does not match the calculated friction force. For a second trial, we will 
select a diameter of 0.020 m. Selecting a higher value for the diameter, will result in an decrease in 
the velocity of the fluid, thus reducing the value of the friction force. 
Trial 2 
 D 0.020 m= 
Substituting this value in the equations for Reynolds number, velocity and relative roughness, we 
get: 
 
169.23 169.23
Re _________
D 0.020
= = = 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 32 Student View 
Jason M. Keith 
 
2 2
__________ __________ m
__________
D (0.020) s
υ = = = 
 
__________ m __________ m
__________
D D (0.020 m)
ε
= = = 
 f = _____________ 
 ( )
2
f
m
__________
__________ m s
F 4 __________
0.020 m 2
 
 
 = 
 
f
J
F _____________
kg
= 
 
As we can see, the value obtained for a diameter of 0.020 m yields a friction force of 
J
____________
kg
. The error between this value and the friction force calculated using the head of 
fluid available is given by: 
 
J J
__________ __________
kg kg
Error % (100%)
J
__________
kg
−
= 
 Error __________ %= 
Conclusion: 
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 33 Student View 
Jason M. Keith 
Example 2.10-5: Flow of Gas in Line and Pressure Drop 
Hydrogen at room temperature and standard pressure is entering a fuel cell stack through a smooth 
pipe with an inner diameter of 1 cm and a length of 3 m. Calculate the pressure of hydrogen in the 
fuel tank. The hydrogen consumption rate is 
g
2.053
s
. 
Strategy 
The initial pressure of the gas can be determined using the equation for the pressure drop of a gas 
inside a tube. 
Solution 
We will start by assuming turbulent flow of hydrogen in the pipe. The initial pressure of hydrogen 
can be calculated using Equation 2.10-10 of Geankoplis shown below: 
 
2
2 2
1 2
L G RT
P P 4f
DM
∆ ⋅
− = 
where: 
 P1 = Initial pressure of the fluid in the pipe, Pa 
 P2 = Final pressure of the fluid in the pipe, Pa 
 f = Fanning friction factor 
 ∆L = Length of the tube, m 
 G = Mass flux of the fluid in the pipe, 
2
kg
m s⋅
 
 R = Ideal gas constant, 
3Pa m
mol K
⋅
⋅
 
 T = Temperature of the fluid, K 
 D = Inside diameter of the pipe, m 
 M = Molecular weight of the fluid, 
g
mol
 
The flux of hydrogen may be calculated by dividing the mass flow rate of hydrogen by the cross-
sectional area of the pipe as shown in the following steps. 
 
m
G
A
=
�
 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 34 Student View 
Jason M. Keith 
 
( )
2
2
2
__________ mD
A _____________ m
4 4
ππ
= = = 
Substituting this value and the mass flow rate into the equation for the mass flux of hydrogen we get: 
 
2
g 1 kg
2.053
s 1000 g
G
_____________ m
 
 
 = 
 
2
kg
G __________
m s
=
⋅
 
The friction factor can be obtained from Figure 2.10-3 of Geankoplis. To do this, first we need to 
calculate Reynolds number and the relative roughness of the pipe 
D
ε
. Thus, 
 
DG
Re =
µ
 
Substituting the corresponding values for the parameters on the right side of this equation yields: 
 
2
6
kg
0.01 m __________
m s
Re ____________
kg
8.8 10
m s
−
 
 
⋅ = =
 
× 
⋅ 
 
The viscosity of hydrogen was obtained from Appendix A.3 of Geankoplis. We can see that this 
value for the Reynolds number indicates that the flow of hydrogen to the fuel cell is turbulent. 
Therefore, the equation we selected for calculating the pressure P1 is valid for this problem. 
Now with the Reynolds number value we can obtain the friction factor for a smooth pipe to be given 
by: 
 f = __________ 
Solving Equation 2.10-10 of Geankoplis for the pressure P1 and substituting all the known quantities 
into this equation, we can obtain the pressure of hydrogen leaving the fuel tank as shown in the 
following steps. 
 
2
2
1 2
L G RT
P 4f P
DM
∆ ⋅
= + 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 35 Student View 
Jason M. Keith 
 
( )
( )
( )
2 3
2
2
1
kg Pa m
3 m _________ _________ 298.15 K
m s mol K 101325 Pa
P 4 _________ 1 atm
1 atmg 1 kg
0.01 m 2
mol 1000 g
⋅  
⋅   ⋅ ⋅      = +         
  
  
 
1
1 atm
P _____________ Pa
101325 Pa
 
=  
 
 
1
P __________ atm= 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 36 Student View 
Jason M. Keith 
Example 2.10-8: Entry Length for a Fluid in a Rectangular Channel 
Determine if the velocity profile for the gas flowing through the channels of a fuel cell bipolar plate 
is fully developed for the following cases: 
a) Hydrogen in a Proton-Exchange Membrane Fuel Cell at a temperature of 25 °C with laminar 
flow. The fuel consumption rate is of 2
g H
1.642
s
. 
b) Carbon monoxide in a Solid - Oxide Fuel Cell at a temperature of 800 °C with laminar flow. The 
fuel cell is converting carbon monoxide into products at a rate of 
kg CO
0.51 
hr
. 
The viscosity of carbon monoxide was found in Appendix A.3-2 to be: 
 5
kg
3.8 10
m s
−µ = ×
⋅
 
c) Same gases from parts a) and b) with turbulent flow. 
The dimensions of the channels in the bipolar plate are shown in the following figure: 
 
 
 
 
 
Strategy 
The entry length is the distance required for the establishment of fully-developed velocity profile. 
Solution 
a) To determine if the velocity profile is fully established in the channel of a bipolar plate, we need 
to calculate the entry length Le, defined by the following equation: 
 e
L
0.0575Re
D
= 
For this problem, the diameter D of the channel will be equal to the hydraulic diameter, i.e. the 
perimeter of the channel 'wetted' by the fluid flowing through it. Thus, 
5 mm 
3 mm 20 cm 
Fuel Gas 
Gas-Diffusion Layer 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 37 Student View 
Jason M. Keith 
e
H
L
0.0575Re
D
= (1) 
To determine the entry length, we need to calculatethe Reynolds number, as shown in the following 
steps: 
 H
D G
Re =
µ
 
The mass flux of hydrogen G is obtained by dividing the mass flow rate of hydrogen by the cross-
sectional area of the channel. Hence, 
 
( ) ( )
22
2
2g H 1 kg1.642
s 1000 gm kg
G __________
A m s1 m
____ mm ____ mm
__________ mm
 
 
 = = =
⋅ 
 
 
�
 
The hydraulic diameter is calculated using the following equation 
 
( )
( )
2
H
4 ____________ m4A
D _____________ m
P 2 0.005 m 0.003 m
= = =
+
 
where: 
 A = Cross-sectional area of the channel 
 P = Perimeter of the channel ‘wetted’ by the fluid. 
Substituting this value and the mass flux in the equation for Reynolds number we get: 
 
( ) 2
6
kg
________________ m ___________
m s
Re ____________
kg
8.8 10
m s
−
 
 
⋅ = =
 
× 
⋅ 
 
The viscosity of hydrogen was obtained from Appendix A.3 of Geankoplis. Now we can solve 
equation (1) for the entry length Le and substitute the calculated values to yield: 
 ( )( )e HL 0.0575Re D 0.0575 ____________ _____________ m= = 
 eL __________ m= 
Conclusion: 
 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 38 Student View 
Jason M. Keith 
b) The entry length for the carbon monoxide flowing in the solid-oxide fuel cell will be determined 
in a similar way to part a) of this problem. 
 e
H
L
0.0575Re
D
= 
The calculations of the mass flux and Reynolds number of carbon monoxide are shown in the 
following steps: 
 
( )( )
22
2
kg CO 1 hr
__________ 
m kghr 3600 s
G ____________
A m s1 m
3 mm _____ mm
__________ mm
 
 
 = = =
⋅ 
 
 
�
 
The hydraulic diameter from part a) is given by: 
 
H
D _____________ m= 
Substituting the hydraulic and the mass flux of CO in the equation for Reynolds number yields: 
 
( ) 2
kg
_____________ m __________
m s
Re __________
kg
________________
m s
 
 
⋅ = =
⋅
 
Solving for the entry length, we get: 
 ( )( )e HL 0.0575Re D 0.0575 __________ ____________ m= = 
 eL 0.201 m= 
Conclusion: 
 
 
c) For turbulent flow, the entry length is relatively independent from Reynolds number and 
estimated to be: 
 e HL 50D= 
Substituting the hydraulic diameter of the channel into this equation yields: 
 ( )eL 50 ______________ m= 
 eL __________ m= 
Conclusion:
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 39 Student View 
Jason M. Keith 
Example 2.11-1: Compressible Flow of a Gas in a Pipe Line 
A gas mixture of 12.5 mol % ethanol and 87.5 % water is leaving a boiler at a temperature of 400 °C 
and a pressure of 21.604 atm. 
The ethanol/water mixture enters a reformer in a large-scale ethanol reforming plant at a rate of 
kg
425.96
s
 through a commercial steel pipe with a length of 120 m and an inner diameter of 75 cm. 
Determine the pressure of the ethanol/water vapor mixture entering the reformer if the viscosity of 
the gas is 51.705 10 Pa s−× ⋅ [2]. 
Strategy 
The pressure of the gas entering the reformer can be calculated using the equation for the pressure 
drop for isothermal compressible flow. 
Solution 
Equation 2.11-9 of Geankoplis can be solved for the pressure P2 at the end of the pipe to yield: 
 2 2 1
2 1
2
P____________ _________
P P ln
DM ____ P
 
= − −   
 
 
where: 
 f = Fanning friction factor 
 ∆L = Length of the pipe, m 
 G = Mass flux of gas in the pipe, 
2
kg
m s⋅
 
 R = Gas constant, 
3Pa m
mol K
⋅
⋅
 
 D = Inner diameter of the pipe 
 M = Molecular weight of the gas flowing through the pipe, 
kg
mol
 
 P1 = Pressure at the beginning of the pipe segment, Pa 
 P2 = Pressure at the end of the pipe segment, Pa 
 
2. DOE Hydrogen Program: DOE H2A Analysis Production Case Studies, http://www.hydrogen.energy.gov/ 
h2a_prod_studies.html. Accessed: July 2010. 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 40 Student View 
Jason M. Keith 
To use this equation, the friction factor and the mass flux of the ethanol mixture must be calculated. 
Before determining the friction factor we need to calculate the relative roughness of the pipe and the 
Reynolds number. Thus, 
 
D
Re
υρ
=
µ
 
The velocity of the gas can be calculated with the following equation: 
 
V
A
υ =
�
 
However, to determine the volumetric flow rate, we need to divide the mass flow rate by the density 
of the fluid. The density of this mixture is calculated using ideal gas law equation of state. 
 
________
________
ρ = 
Substituting the corresponding quantities into this equation, after calculating the molecular weight of 
the gas mixture, we get: 
 
H O H OEthanol Ethanol
2 2
M x M x M= + 
 
2
2
mol H Omol ethanol g g
M 0.125 ____ 0.875 ____
mol mol ethanol mol mol H O
  
= +        
 
 
g
M _________
mol
= 
 
( )
( )
3
g 1 kg
21.604 atm 21.5
mol 1000 g
L atm 1 m
0.08206 673.15 K
mol K 1000 L
  
  
  
ρ =
⋅   
   ⋅   
 
 
3
kg
__________
m
ρ = 
Now the volumetric flow rate can be determined as follows: 
 
3
3
kg
425.96
m msV _________
kg s
_________
m
= = =
ρ
�
� 
Substituting this value into the equation for the velocity of the fluid yields: 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 41 Student View 
Jason M. Keith 
 
( )
3
2
m
_________
V ms _________
A s_________ m
4
υ = = =
 π
 
  
�
 
The value of the Reynolds number required to determine the friction factor can now be calculated as 
shown below: 
 
( ) 3
m kg
0.75 m __________ _________
s m
Re
kg
______________
m s
  
  
  =
 
 
⋅ 
 
 Re _______________= 
The relative roughness 
D
ε
 of a commercial steel pipe is given by: 
 5
___________ m
6 10
D ___________ m
−ε = = × 
The friction factor can be obtained from Figure 6-9 of Perry’s Chemical Engineers’ Handbook, 8
th
 
Edition to be: 
 f = __________ 
The only value left to be calculated before being able to solve for the pressure of the gas entering the 
reformer is the mass flux, defined as: 
 
m
G
A
=
�
 
Substituting the mass flow rate and the cross-sectional area of the pipe into this equation, gives: 
 
( )
2 22
kg kg
425.96 425.96
kgs sG ___________
_________ m m s_________ m
4
= = =
⋅ π
 
  
 
 
 
 
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Principles of Momentum Transfer and Overall Balances 
Daniel López Gaxiola 42 Student View 
Jason M. Keith 
Finally we can substitute all the corresponding values into the equation for 2
2
P to get: 
( ) ( ) ( )
( )
2 3
2
2
2
2
2 3
2
kg Pa m
4 _________ 120 m _________ 8.314 673.15
m s mol K101325 Pa
P 21.604 atm
1 atm g 1 kg
0.75 m _________
mol 1000 g
kg Pa m
2 _________ 8.314 673.
m s mol K
 
⋅  
   ⋅ ⋅      = −        
  
  
⋅  
   ⋅ ⋅   −
( )
2
101325 Pa
15 21.604 atm
1 atm
ln
Pg 1 kg
_________
mol 1000 g
  
  
  
    
       
 
2 2 11 2 2
2
2
______________ Pa
P ______________ Pa 4.178 10 Pa ______________ Pa ln
P
 
= − × −   
 
 
This equation can be solved using computer software or trial and error to yield: 
 
2
P ________________ Pa≈ 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 43 Student View 
Jason M. Keith 
Example 2.11-2: Maximum Flow for Compressible Flow of a Gas 
Determine the maximum velocity that can be obtained for the ethanol/watermixture from Example 
2.11-1 and compare to the actual velocity of the fluid fed to the ethanol reformer. 
Strategy 
The maximum velocity of the fluid is obtained using the definition of the velocity of sound in an 
isothermal fluid. 
Solution 
The maximum velocity of the ethanol/water mixture can be determined using Equation 2.11-12 of 
Geankoplis, as shown below: 
 max
RT
M
υ = 
Substituting the ideal gas constant, as well as the temperature of the gas in the pipe lines and the 
molecular weight into this equation yields: 
 
( )
3
max
Pa m
_________ 673.15 K
mol K
g 1 kg
_________
mol 1000 g
⋅ 
 ⋅ υ =
  
  
  
 
 
max
m
___________
s
υ = 
The velocity of this gas in the process at the entrance to the ethanol reformer is given by Equation 
2.11-13 of Geankoplis: 
 
2
2
RTG
P M
υ = 
Substituting the pressure found in Example 2.11-1 and the mass flux of ethanol vapor, we find that 
the velocity is given by: 
 
( )
( )
3
2
2
Pa m kg
_________ 673.15 K ___________
mol K m s
g 1 kg
_______________ Pa _________
mol 1000 g
⋅   
  
⋅ ⋅  
υ =
  
  
  
 
 
2
m
___________
s
υ = 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 1 Student View 
Jason M. Keith 
Chapter 3 
Principles of Momentum Transfer and Applications 
Chapter 3 introduces students to the principles for calculation of power and efficiency of equipment 
such as compressors, pumps and fans. The following problem modules illustrate the application of 
this type of equipment to processes for producing hydrogen for fuel cells as well as the derivation of 
equations for different flow conditions from the general transport equations. 
3.1-3 Surface Area in Packed Bed of Cylinders 
3.1-4 Pressure Drop and Flow of Gases in Packed Bed 
3.2-1 Flow Measurement using a Pitot Tube 
3.3-1 NPSH Available for Pump 
3.3-2 Calculation of Brake Horsepower of a Pump 
3.3-3 Brake-kW Power of a Centrifugal Fan 
3.3-4 Compression of Methane 
3.8-3 Laminar Flow in a Circular Tube 
3.11-1 Dimensionless Groups 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 2 Student View 
Jason M. Keith 
Example 3.1-3: Surface Area in Packed Bed of Cylinders 
Natural gas has been proposed as a source of hydrogen for fuel cell vehicle applications because of 
the existing infrastructure. In a process known as steam reforming, natural gas and steam are reacted 
into mostly carbon monoxide and hydrogen with some carbon dioxide also produced. There is also 
excess water in the reformate stream. 
A water gas shift reactor can be used to convert some of the remaining carbon monoxide into 
hydrogen according to the reaction: 
CO + H2O H2 + CO2 
The following figure shows an axisymmetric view of an annular water gas shift reactor which is 8 
cm high. In the outer (annular) region, an iron chromium oxide catalyst is present to carry out the 
water gas shift reaction. A 20 µm thick palladium membrane separates the reaction (outer) zone 
from the separation (inner) zone. 
 
Cylinders of iron chromium oxide catalyst with diameter and length of 0.1 cm are forming a packed 
bed in the reaction zones with a bulk density of 
3
lb
39.75
ft
. Determine the void fraction ε, the 
effective diameter of the particles Dp and the hydraulic radius rH for the flow through the packed bed 
if the density of the catalyst is 
3
lb
76
ft
. 
Strategy 
The equations defining the parameters ε, Dp and rH can be used for solving this problem. 
Separation Zone 
Reaction Zones 
Gas Flows In 
Annular Membrane Reactor : Top View (Left) and Side View (Right) w
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 3 Student View 
Jason M. Keith 
Solution 
First we will calculate the void fraction ε of the packed bed, defined by the following equation: 
 
volume of voids in bed
total volume of bed
ε = 
For simplicity, we will select a basis of 1 ft
3
 of packed bed. Thus, we can calculate the mass of the 
bed to be: 
 3
3bed
lb
m 1 ft _________ ___________ lb
ft
 
= = 
 
 
This mass of packed bed can be used to calculate the volume of the solid cylinders of catalyst, as 
shown in the following calculation: 
 3
catalyst
3
_________ lb
V _____________ ft
lb
76
ft
= = 
Substituting this volume and the basis of 1 ft
3
 of packed bed into the equation for ε yields: 
 
3 3
3
1 ft ___________ ft
______ ft
−
ε = 
 _________ε = 
For non-spherical particles, the effective diameter is given by the following equation: 
 p
V
6
D
a
= 
where 
V
a is the specific area, given by the ratio of the surface area of the catalyst particle to the 
volume of the particle. Thus, 
V
a is given by: 
 
p
2V
p
_______
2 ___________
S _____
a
DV
______
4
 π
+ π 
 = =
π
 
In this equation, the first term in the numerator represents the area of the ends of the cylinder, while 
the second term is accounting for the area of the walls of the cylinder. We can substitute the values 
of D and L into this equation to determine the specific area 
V
a as follows: 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 4 Student View 
Jason M. Keith 
( )
( )
( )
2
p
2V
p
__________
2 ________
S ______ 1 _________ cm
a 60
V cm 1 ft0.1 cm
__________
4
 π
+ π 
  = = =  
 π
 
1
V
a 1828.8 ft−= 
Now we can calculate the effective diameter to be: 
 
p 1
6
D
1828.8 ft−
= 
 pD ____________ ft= 
The hydraulic radius of an object is defined as: 
 
H
r
a
ε
= 
where a is the ratio of the wetted surface of the particles to the volume of the packed bed. The 
following equation can be used for calculation of a: 
 ( )
V
a a 1= − ε 
The values of the void fraction and the specific area of the catalyst particles can be entered into this 
equation to give: 
 ( )1a 1828.8 ft 1 __________−= − 
 1a ___________ ft−= 
Therefore, the hydraulic radius can now be calculated to yield: 
H 1
__________
r
__________ ft−
= 
4
H
r 4.98 10 ft−= × 
 
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 5 Student View 
Jason M. Keith 
Example 3.1-4: Pressure Drop and Flow of Gases in Packed Bed 
A water-gas shift reactor in a distributed-scale hydrogen plant is producing hydrogen at a rate of 
lb
65.2
h
. The reactor consists of a tubular packed bed of 5.25 cm diameter with 31.4 kg of a catalyst 
with a density of 
3
lb
76
ft
. 
The void fraction of the bed is 0.57 and the spherical catalyst pellets have a diameter of 0.1 cm. 
Determine the pressure drop of the reacting synthesis gas in the packed bed. The synthesis gas is 
entering the reactor at a pressure of 2066 kPa and has the following properties: 
 
lb
0.048
ft h
µ =
⋅
 
 
30
lb
0.356
ft
ρ = 
The differential form of the pressure drop in a packed bed reactor is given by the Ergun equation: 
3
c p p
dP G 1 150(1 )
1.75G
dz g D D
  − φ − φ µ
= − +   ρ φ  
 
The solution to this differential equation is given by: 
 
1/2
o
P
(1 W)
P
= − α 
where: 
o
c c o
2
A (1 ) P
β
α =
− φ ρ
 
o 3
o c p p
G(1 ) 150(1 )
1.75G
g D D
 − φ − φ µ
β = + 
ρ φ   
 
The first term in the brackets in the equation for βo is dominant for laminar flow and the second term 
is dominant for turbulent flow. In these equations, the following notation and units are used: 
P = Pressure 
2
lb
ft
 
 
 
 
φ =Void fraction (dimensionless) 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 6 Student View 
Jason M. Keith 
gc = Gravitational constant = 
8
2
f
lb ft
4.17 10
h lb
⋅
×
⋅
 
Dp = Diameter of particle in the bed (ft) 
µ = Viscosity of gas 
lb
ft h
 
 
⋅ 
 
ρ0 = Gas density 3
lb
ft
 
 
 
 
ρc = Density of the catalyst 3
lb
ft
 
 
 
 
m
G
A
=
�
= Mass flux of synthesis gas 
2
lb
ft h
 
 
⋅ 
 
m� = Mass flow rate of synthesis gas 
lb
h
 
 
 
 
υ= Velocity of the gas in the reactor
lb
h
 
 
 
 
z = Distance down packed bed (ft) 
A = Cross – sectional area of the reactor (ft
2
) 
W = Mass of catalyst in the reactor (lb) 
Strategy 
The Ergun equation can be used for calculating the pressure at the outlet of the packed bed. 
Solution 
First we determine the value of βo. All of the terms in the problem statement are in the appropriate 
units except for the particle diameter. We have: 
p
1 ft
D 0.1cm ______________ ft
30.48cm
 
= = 
 
 
Another value we need to calculate to calculate βo is the mass flux of synthesis gas G, given by: 
m
G
A
=
�
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 7 Student View 
Jason M. Keith 
Where the cross sectional area is obtained as follows: 
 
( )
2
2
2
2
________ cm 1 ft
A ___________ ft
4 ____________ cm
π  
= = 
 
 
Entering this value and the mass flow rate into the equation for the mass flux G we get: 
 
2 2
lb
65.2
lbhG ______________
____________ ft ft h
= =
⋅
 
Substituting this and the other values into the equation for 
o
β : 
o 3
o c p p
G(1 ) 150(1 )
1.75G
g D D
 − φ − φ µ
β = + 
ρ φ   
( ) ( )
o
2
3
3 2
f
2
lblb 150(1 0.57) 0.048_______________ (1 0.57)
ft hft h
______________ ftlb lb ft
0.356 _______________ ____________ ft 0.57
ft h lb
lb
1.75 _______________
ft h
  
−−   ⋅ ⋅ β =
  ⋅
    ⋅ 
 
+  ⋅ 
 
f
o 3
lb
77.91
ft
β = 
Note that this term has units of pressure f
2
lb
ft
 
 
 
per unit length (ft). It is also noted that the second 
term in the brackets is dominant, suggesting turbulent flow in this industrial reactor. 
Now we need to determine the value of αW which is needed in the formula for the pressure drop. 
We first need to obtain the feed pressure, and catalyst weight in the appropriate units. 
The feed pressure in f
2
lb
ft
is: 
f
2
f
o 2
lb
_________ lb___________inP 2066 kPa _____________
___________ kPa ___________ ft
 
  
= =  
   
 
 
 
 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 8 Student View 
Jason M. Keith 
The catalyst weight W in lbm is: 
1 lb
W 31.4 kg ___________ lb
0.454kg
 
= = 
 
 
Thus, 
( )f3
o
2c c o f
3 2
lb
2 77.91 _________ lb
ft2 W
W ________
lbA (1 ) P lb
_____________ ft (1 0.57) 76 _______________
ft ft
 
 
β  α = = =
− φ ρ   
−   
  
 
The exit pressure can be determined by entering this value into the equation for the pressure ratio
o
P
P
: 
1/2
o
P
(1 W)
P
= − α 
Substituting the values of Wα and the pressure at the entrance of the packed bed P0 yields: 
1/2
f
2
P
(________________)
lb
____________
ft
=
 
 
 
 
Solving for the pressure at the outlet P, we get: 
 1/2 f f
2 2
lb lb
P (________________) _____________ 35373
ft ft
 
= = 
 
 
Hence, the pressure drop is given by: 
 f f
2 20
lb lb ___________
P P P 35373 ___________
ft ft ___________
  
∆ = − = −   
  
 
 P ___________ psi∆ = 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 9 Student View 
Jason M. Keith 
Example 3.2-1: Flow Measurement using a Pitot Tube 
A Pitot Tube is used for measuring the flow in a circular pipe. To determine the flow of hydrogen at 
room temperature to a proton-exchange membrane fuel cell, a Pitot tube with a coefficient of 0.84 is 
used in a pipe with a diameter of 1”. The static-pressure of hydrogen is measured to be 12.34 mm of 
Hg above atmospheric pressure. 
Determine the maximum and average velocities and flow rates of hydrogen in the pipe if the reading 
on the manometer is 0.022 in of Hg. The following figure shows the diagram of Pitot Tube for this 
problem. 
 
 
 
 
 
Strategy 
To solve this problem we need to obtain the velocity value from Bernoulli Equation applied to a 
Pitot Tube. 
Solution 
The Bernoulli equation applied to a Pitot Tube is given by: 
 
( )
p1
2 ________________
C
___________
υ = 
where Cp is the value of the Pitot tube coefficient. 
The viscosity of hydrogen can be obtained from Appendix A.3 of Geankoplis to be. 
 
kg
_________________
m s
µ =
⋅
 
The density can be calculated using Ideal Gas equation of state as shown in the following steps: 
 
( )
( )
H @ P = 1 atm
5
2
kg
_____ atm _________
________ mol
RT _____ _____
8.206 10 298.15 K
______ _____
−
 
 
 ρ = =
 ⋅
× 
⋅ 
 
H2 to Fuel Cell 
∆h = 0.022 in Hg 
1 2 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 10 Student View 
Jason M. Keith 
 
3H @ P = 1 atm
2
kg
_________
m
ρ = 
First we need to convert the manometric static pressure to absolute pressure. To do this, we will use 
the density of mercury as 
3
kg
13533.61
m
, obtained from Table 2-31 of Perry’s Chemical Engineers’ 
Handbook, 8
th
 Edition. Thus, 
 ( )Hgstatic H @ P = 1 atm
2
P gh∆ = ρ − ρ 
 ( )
3 3 2static
kg kg m
P 13533.61 _________ ___________ ____________ m Hg
m m s
  
∆ = −  
  
 
 
static
P 1638 Pa∆ = 
We can add this value to the atmospheric pressure of 101325 Pa to get the absolute pressure to be 
______________ Pa. Since the density of hydrogen was calculated at atmospheric pressure, we need 
to correct the density value for the actual pressure of hydrogen in the pipe. This can be done by 
multiplying the density at 1 atm of pressure by the pressure ratio. Therefore, 
 
H 3H @ P = 1 atm
atm
2 2
P kg __________ Pa
_________
P m 101325 Pa
   
ρ = ρ =   
  
 
 
H 3
2
kg
__________
m
ρ = 
Now we can calculate the pressure difference using the change in the height of mercury in the 
manometer as shown below: 
( )Hg H
2
P g h∆ = ρ − ρ ∆ 
( )
3 3 2
kg kg m 0.0254 m Hg
P 13533.61 __________ ____________ __________ in Hg
m m s 1 in Hg
   
∆ = −    
    
 
P __________ Pa∆ = 
Substituting this pressure drop into the equation for the velocity of hydrogen in the tube yields: 
 
( )
1
3
2 __________ Pa
0.84
kg
__________
m
υ = 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 11 Student View 
Jason M. Keith 
 
1
m
__________
s
υ = 
Since the point 1 in the tube is at the center of the tube, where the velocity reaches its maximum 
value, the velocity we calculated is the maximum velocity. 
 
max
m
__________
s
υ = 
The average velocity of a fluid in a pipe can be estimated using Figure 2.10-2 of Geankoplis as a 
function of Reynolds number. The value of Reynolds number using the maximum velocity for this 
problem is given by: 
 
( ) 3
max H
2
0.0254 m m kg
1 in ___________ ___________D 1 in s m
Re
kg
_______________
m s
   
   υ ρ
   = =
µ
⋅
 
 Re 8507= 
Locating this value in Figure 2.10-2 we can estimate the ratio of the average velocity to themaximum velocity to be: 
 av
max
__________
υ
≈
υ
 
Solving for the average velocity and entering the value of the maximum velocity into this equation 
we get: 
 ( )av max
m
___________ ___________ ___________
s
 
υ ≈ υ ≈  
 
 
 
av
m
27
s
υ ≈ 
To calculate the average and maximum flow rates of hydrogen we need to multiply the 
corresponding velocity value by the cross-sectional area of the pipe. The area of the pipe is 
calculated as shown below: 
 
( )
2
2 2
2
2
1 inD _______________ m
A _______________ m
4 4 1 in
ππ  
= = = 
 
 
Multiplying the cross-sectional area by both velocity values we get: 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 12 Student View 
Jason M. Keith 
 2
av
m
V _______________ m 27
s
 
=  
 
� 
 
3
av
m
V _______________
s
=� 
 2
max
m
V _______________ m ________
s
 
=  
 
� 
 
3
max
m
V _______________
s
=� 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 13 Student View 
Jason M. Keith 
Example 3.3-1: NPSH Available for Pump 
Water at a temperature of 25°C is being pumped to a boiler in a distributed-scale steam-methane 
reforming plant. The water is entering the boiler through a commercial steel pipe with a diameter of 
0.1 m and a length of 25 m. The estimated friction factor for the flow conditions in this process is 
estimated to be 0.0045. 
Determine the available net positive suction head (NPSH) of the pump if the velocity of the water in 
the pipe is of 
m
2.3
s
 
Strategy 
This problem can be solved using the equation for the NPSH as a function of the pressure and 
thermodynamic properties of the fluid in the pipe. 
Solution 
The following equation shows the relation between the NPSH available and the conditions in the 
system: 
 ( )
2
vp1
A 1
P P
g NPSH gz F
2
− υ
= + − −
ρ
∑ 
where: 
 g = Acceleration due to gravitational force = 
2
m
9.80665
s
 
 P1 = Pressure of the fluid before entering the pump (Pa) 
 Pvp = Saturation pressure of fluid at the process temperature (Pa) 
 ρ = Density of fluid 
3
kg
m
 
 
 
 
z1 = Difference in height between the pump and the point at pressure P1 (m) 
υ = Velocity of the fluid 
m
s
 
 
 
 
F∑ = Friction loss in suction line to pump 
J
kg
 
 
 
 
Since there is no difference in the height between the pipe and the pump, the term 
1
gz in the equation 
for g(NSPH)A is neglected in this problem. Thus, 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 14 Student View 
Jason M. Keith 
 ( )
2
vp1
A
P P
g NPSH F
2
− υ
= − −
ρ
∑ 
The density and saturation pressure of water at 25 °C can be found in Geankoplis in Tables A.2-3 
and A.2-9 respectively: 
3
kg
_____________
m
ρ = 
 
vp
P __________ Pa= 
The only remaining unknown value in the equation for NSPHA is the friction loss, which can be 
calculated using the following equation: 
 
2L
F 4f
D 2
∆ υ
= 
Substituting the corresponding values into this equation, we get: 
 ( )
2
m
2.3
25 m Js
F 4 ___________ ________
0.1 m 2 kg
 
 
 = =∑ 
Now we can enter all the known quantities into the equation for NSPH to yield: 
 ( )
( ) ( )
2
2 A
3
m
2.3
101325 Pa __________ Pam Js
9.80665 NPSH 11.9
kgs 2 kg
__________
m
 
 −   = − − 
 
 
Solving for the NPSHA: 
 ( )
( ) ( )
2
A
3 2
m
2.3
101325 Pa __________ Pa J 1s
NPSH 11.9
kg m2 kg
__________ 9.80665
m s
  
  −   = − −
   
     
 
 ( )
A
NPSH _________ m= 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 15 Student View 
Jason M. Keith 
Example 3.3-2: Calculation of Brake Horsepower of a Pump 
Determine the brake horsepower of the pump from Example 3.3-1 operating at a flow rate of 
kg
683.1
hr
 for feeding water to a boiler in a Steam-Methane reforming plant. Assume the 
characteristic curves of this pump are described by Figure 3.3-3 of Geankoplis. 
Strategy 
The brake horsepower of the pump described in this example can be calculated using the mass flow 
rate of fluid and the work performed by the pump. 
Solution 
The brake hp of a pump can be calculated using Equation 3.3-2 of Geankoplis: 
 S
W m
brake hp
550
−
=
η
�
 
where: 
 WS = Work performed by the pump 
f
ft lb
lb
⋅ 
 
 
 
 m� = Mass flow rate of fluid 
lb
s
 
 
 
 
 η= Efficiency of the pump 
First we need to calculate the work performed by the pump, defined as follows: 
 
S
C
g
W H
g
= − 
In this equation, the value of the head of fluid H is unknown. However, it can be read from the 
characteristic curves of the pump, shown in Figure 3.3-3. The volumetric flow rate of water required 
for using Figure 3.3-3 is calculated by multiplying the mass flow rate by the density of water at the 
process conditions as shown below: 
 
3
3 3
kg ________
683.1
hr ________m
V
kg ________________ m
997.08
m ft
 
 
 = =
ρ  
 
 
�
� 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 16 Student View 
Jason M. Keith 
 
3
3
ft ___________ gal gal
V ___________ ___________
min 1 ft min
 
= = 
 
� 
With this volumetric flow rate, we can estimate the head of fluid and the efficiency of the pump 
from Figure 3.3-3 to be: 
 H ___________ ft≈ 
 ___________η ≈ 
We can substitute the head of fluid value into the equation for the work WS to yield: 
 
2
S
2
f
ft
_____________
hrW ________ ft
lb ft
_____________
hr lb
 
 
 = −
⋅ 
 ⋅ 
 
 f
S
lb ft
W ___________ 
lb
⋅
= 
The negative value of the work WS indicates that the fluid is performing work over the pump. Now 
we can enter the values we calculated into the equation for the brake horsepower of the pump to get: 
 
( )
f
lb ft kg _________ _________
___________ 683.1
lb hr _________ _________
brake hp
550 _________
⋅     
−     
     = 
 brake hp ________ hp= 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 17 Student View 
Jason M. Keith 
Example 3.3-3: Brake-kW Power of a Centrifugal Fan 
Air at a flow rate of 
3m
2.90
min
(measured at 1 atm and 298.15 K) and a velocity of 
m
67.1
s
enters a 
proton-exchange membrane fuel cell stack through a centrifugal fan. The air is entering the fan at a 
pressure of 1.009 bar and a temperature of 100°F. What is the discharge pressure of the air if the fan 
has a brake power of 2.12 kW and an efficiency of 70%? 
Strategy 
This problem can be solved by performing a mechanical energy balance on the system, and using the 
definition of brake power of a centrifugal fan. 
Solution 
The mechanical-energy balance for this problem (
1 2
z z= ,
in
0υ = , F 0=∑ ) is given by: 
 
( )
2
out1 2
S
P P
W
2
υ−
= −
ρ
 
 WS = Work performed by the centrifugal fan 
J
kg
 
 
 
 
 P1 = Pressure of air at the suction point (Pa) 
 P2 = Pressure of air at the discharge point (Pa) 
 ρ = Average density of the air 
3
kg
m
 
 
 
 
 
out
υ = Velocity of air at the discharge point 
m
s
 
 
 
 
Before being able to calculate the discharge pressure from this equation, we need to calculate the 
velocity of air in the outlet, the average density and the work performed by the fan on the fluid. 
The density of the air entering the fan is calculated using the ideal gas law and the properties of an 
idealgas at standard conditions. In the following calculation, the sub-index 1 represents the 
conditions of the air at the inlet of the fan. 
 air std 1
31
1 stdstd
kg air
___________M T P ___________ K 1.009 barkmol
ˆ mT P 311.15 K ___________ barV
___________
kmol
     
ρ = =            
 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 18 Student View 
Jason M. Keith 
 
31
kg
___________
m
ρ = 
The density of the air leaving through the fan can be determined using the pressure ratio between the 
suction and discharge points. After substituting the known pressure and density we get: 
 2 2
32 1
1
P Pkg 1 bar
___________
P m 1.009 bar ___________ Pa
    
ρ = ρ =         
 
 ( )
2 2
________________ Pρ = 
Now we can obtain the average density of the air as a function of the discharge pressure P2: 
 
( )3 2
1 2
kg
_________ ________________ P
m
2 2
+ρ + ρ
ρ = = 
 ( )
3 2
kg
_______ ____________ P
m
 
ρ = + 
 
 
Substituting this expression into the mechanical-energy balance equation: 
 
( )
( )
2
out1 2
S
2
P P
W
_______ ____________ P 2
υ−
= −
+
 
Now we need to calculate the left hand side of this equation. Equation 3.3-2 of Geankoplis defines 
the brake power of a centrifugal fan as follows: 
 S
W m
brake kW
1000
−
=
η
�
 
where: 
 m� = Mass flow rate of fluid 
kg
s
 
 
 
 
 η= Efficiency of the fan 
We can solve for the work 
S
W performed by the fan: 
S
________
W _____________
________
 
=  
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 19 Student View 
Jason M. Keith 
The value of the mass flow rate m� can be obtained from the properties of an ideal gas at standard 
conditions. Thus, 
 std
TV ________
m
ˆ ________V
 
=  
 
�
� 
The sub-index 0 of the temperature indicates the temperature at which the air flow rate was 
measured. Substituting the corresponding numeric values into this equation yields: 
 
3
3
m 1 min kg
________ _________
273.15 Kmin 60 s kmol
m
__________ Km
__________
kmol
  
      =  
   
 
 
� 
 
 kg
m _____________
s
=� 
Entering this value and the brake power of the fan into the equation for the work WS: 
 
S
1000(0.70) J
W 2.12 kW _____________
kg kg
_____________
s
 
 
= − = − 
 
 
 
Now, after substituting the calculated values, the energy balance equation is given by: 
 
( )
25
2
2
1 10 Pa m
1.009 bar P 67.1
J 1 bar s
_____________
kg _______ ____________ P 2
×   −   
   − = −
+
 
 
( )
5
2
2
1.009 10 Pa PJ
_______________
kg _______ ____________ P
× −
=
+
 
This equation can be solved for P2 as shown in the following steps: 
 ( ) 5
2 2
J
____________ _______ ____________ P 1.009 10 Pa P
kg
+ = × −   
 ( ) 5
2 2
_______________ ____________ P 1.009 10 Pa P+ = × − 
 ( )52 2____________ P P 1.009 10 13386.4+ = × + 
 
2
____________
P
0.8673
= 
 
2
__________
P ________________ Pa 1.318 bar
__________
 
= = 
 
 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 20 Student View 
Jason M. Keith 
Example 3.3-4: Compression of Methane 
A compressor in a steam-methane reforming process for hydrogen production is compressing 
natural gas at room temperature from a pressure of 1 atm to 21.8 atm before entering the reforming 
reactor. What percent of power is saved when operating the compressor at isothermal conditions if 
compared to adiabatic compression? 
Strategy 
The equation for calculating the brake power of a compressor allows us to compare both adiabatic 
and isothermal compression processes. 
Solution 
The equation for calculating the power required by the compressor is shown below: 
 S
W m
brake kW
−
=
η
�
 
The value of the work WS in this equation is in kJ. The values of the mass flow rate m� and the 
efficiency η are not given in the problem statement. However, for calculating the amount of power 
saved, the values of m� and η are not required, as it will be shown in the following steps. We need 
to calculate the work performed by the compressor at both adiabatic and isothermal conditions. 
For adiabatic compression, the work WS is given by: 
 1 2
S,adiabatic
1
1
RT P
W 1
1 M P
−γ
γ  γ  − = −   γ −    
 
The parameter γ represents the heat capacity ratio. For methane, a value of 1.31γ = is given in 
Geankoplis. Hence, the work for adiabatic compression is calculated as follows: 
 
( )
S,adiabatic
_____________
_____________
kJ
_____________ 298.15 K
_____________ 21.8 atmkmol KW 1
kg_____________ 1 atm
______
kmol
 
     ⋅  − = −    
      
 
 
S,adiabatic
kJ
W 702.9
kg
− = 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 21 Student View 
Jason M. Keith 
 
For isothermal operation, the work WS is calculated using the following equation: 
2
S, isothermal
1
P_______________
W log
M P
− = 
Substituting the pressure ratio into this equation and the temperature of methane before entering the 
compressor yields: 
 
S, isothermal
kJ
W _____________
kg
− = 
The percent of power saved can be calculated dividing the difference in power between adiabatic 
and isothermal compression by the power required for adiabatic compression. Thus, 
 
( ) ( )
( )
adiabatic isothermal
adiabatic
brake kW brake kW
Power saved (%) 100
brake kW
−
= × 
Substituting the equations for the power required by the compressor and the work values in this 
equation, we get: 
 
m
Power saved (%)
η
=
�
( )S,adiabatic S,isothermalW W
m
 − − −
 
η
�
( )S,adiabatic
100
W
×
−
 
 
kJ kJ
_____________ _____________
kg kg
Power saved (%) 100
kJ
_____________
kg
 
− 
 = ×
 
 
 
 
 Power saved _____________= 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 22 Student View 
Jason M. Keith 
Example 3.8-3: Laminar Flow in a Circular Tube 
An aqueous solution of 40 % methanol is flowing from the fuel reservoir to a stack of direct-
methanol fuel cells in a mobile phone through a pipe with an inner diameter of 3 mm and a length of 
2 cm. Use the equation derived in Geankoplis for laminar flow in a circular tube to determine the 
pressure drop along the pipe if the maximum Reynolds number for methanol in the pipe is 1850. 
Strategy 
We can use the Hagen-Poiseuille equation to calculate the pressure drop between the methanol 
reservoir and the fuel cell stack. 
Solution 
The Hagen-Poiseuille equation is derived in Geankoplis for laminar flow in a circular tube, shown 
below: 
 av
21 2
32 L
P P
D
µυ
− = 
The viscosity of the methanol solution at 25 ºC can be obtained from Figure A.3-4 to be: 
 
kg
_________________
m s
µ =
⋅
 
The only parameter we need to calculate before being able to determine the pressure drop is the 
average velocity. We can obtain the ratio of the average velocity to the maximum velocity from 
Figure 2.10-2 of Geankoplis at the Reynolds number of 1850. From this figure, it can be seen that 
for laminar flow the ratio of the velocities remains constant and equal to __________. Thus, 
 
av max
________υ = υ 
From the definition of Reynolds number we can solve for the maximum velocity to yield: 
 
max
_____________
__________
υ = 
The density of an aqueous solution of 40 % methanol can be obtained from Table 2-109 of Perry’s 
Chemical Engineers’ Handbook, 7
th
 Editionto be: 
 
3@20ºC
kg
934.5
m
ρ = 
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 23 Student View 
Jason M. Keith 
Substituting the corresponding quantities into the definition of Reynolds number, we get: 
 
( )
max
3
3
kg
1850 _______________
mm s
_____________
kg s
3 10 m _____________
m
−
 
 
⋅ υ = =
 
×  
 
 
With this value, we can determine the average velocity from the data read from Figure 2.10-2: 
 
av
m m
_____________ _____________ 0.6
s s
 
υ = = 
 
 
Finally, after entering this velocity and the rest of the values into the Hagen-Poiseuille equation we 
can determine the pressure drop as shown in the following steps: 
 
( )
( )
21 2 3
kg m
32 _____________ 0.6 _____________ m
m s s
P P
3 10 m−
  
  
⋅  − =
×
 
 
1 2
P P ______ Pa− = 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 24 Student View 
Jason M. Keith 
Example 3.11-1: Dimensionless Groups 
Using dimensionless numbers determine which of the following forces is the most dominant for the 
flow conditions in a polymer-electrolyte membrane fuel cell and direct-methanol fuel cell: 
 -Inertia Force 
 -Gravity Force 
 -Pressure Force 
 -Viscous Force 
 
The flow conditions in both fuel cells are given in the following set of data. The direct-methanol fuel 
cell is using an aqueous solution of 40 % methanol as fuel. 
Polymer-Electrolyte Membrane Fuel Cell 
 Pressure: 2.5 atm 
 Temperature: 80 ºC 
 Velocity of hydrogen: 
m
15
s
 
 Hydraulic diameter 
 of the channel: 31 10 m−× 
 
 
Direct Methanol Fuel Cell 
 Pressure: 1 atm 
 Temperature: 25 ºC 
 Velocity of methanol: 
m
0.49
s
 
Density:
3
kg
931.5
m
 
Hydraulic diameter 
 of the channel: 48.57 10 m−×
Strategy 
The Froude, Euler and Reynolds numbers establish a relation between the forces given in the 
problem statement and hence can be used for determining the most dominant force for a specified 
flow conditions. 
Solution 
The definitions of the dimensionless groups that will be used for solving this problem are given in 
the following equations. 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 25 Student View 
Jason M. Keith 
2
Fr
inertia force
N
gravity force gL
υ
= = 
 
Eu
pressure force ______
N
inertia force ______
= = 
Re
inertia force L
N
viscous force
υρ
= =
µ
 
In these equations the parameter L represents the characteristic length. For this problem, the 
characteristic length will be given by the hydraulic diameter of the channel DH. Thus, 
2
Fr
H
inertia force
N
gravity force gD
υ
= = 
 
Eu
pressure force ______
N
inertia force ______
= = 
Re
inertia force _________
N
viscous force _________
= = 
For comparison of the forces involved in these dimensionless numbers, we will use the reciprocal of 
the Euler number. By doing this, we can compare the three numbers using the inertia force as 
reference. Therefore, 
 
Eu
1 inertia force _______
N pressure force _____
= = 
The viscosity of the methanol solution at room temperature is obtained from Appendix A.3 to be: 
 
kg
_______________
m s
µ =
⋅
 
For hydrogen at a temperature of 80 ºC, the density and viscosity obtained from Table 2-223 of 
Perry’s Chemical Engineers’ Handbook, 8
th
 Edition and are shown below: 
 
6 kg9.9982 10
m s
−µ = ×
⋅
 
 
3
kg
0.178
m
ρ = 
Now we can proceed to calculate the dimensionless numbers for the flow conditions in both types of 
fuel cell. For the polymer-electrolyte membrane, we can substitute the given parameters into the 
equations for Froude, Euler and Reynolds as shown in the following steps: 
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Principles of Momentum Transfer and Applications 
Daniel López Gaxiola 26 Student View 
Jason M. Keith 
( )
2
2
Fr
H
2
m
_______
s
N
mgD
9.80665 ___________ m
s
 
 υ  = =
 
 
 
 
Fr
N ___________= 
 
2
3
Eu
kg m
________ ________
1 m s
101325 PaN
2.5 atm
1 atm
 
 
 =
 
 
 
 
Eu
1
________________
N
= 
( ) 3
Re
m kg
___________ m ____ 0.178
s m
N
kg
_______________
m s
  
  
  =
⋅
 
Re
N ____________= 
Conclusion: 
 
 
 
Repeating a similar procedure for the direct-methanol fuel cell, we find the following values for the 
dimensionless numbers: 
( )
2
2
Fr
H
2
m
________
s
N
mgD
________ ______________ m
s
 
 υ  = =
 
 
 
 
Fr
N ________= 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 27 Student View 
Jason M. Keith 
 
2
3
Eu
kg m
________ ________
1 m s
N _____________
 
 
 = 
Eu
1
________________
N
= 
( ) 3
Re
m kg
______________ m ________ ________
s m
N
kg
______________
m s
  
  
  =
⋅
 
Re
N ________= 
Conclusion: 
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 1 Student View 
Jason M. Keith 
Chapter 4 
Principles of Steady - State Heat Transfer 
Heat transfer is occurring in many chemical and separation processes as a consequence of a 
temperature difference. In Chapter 4, the following problem modules explain the heat transfer 
processes involved in fuel cell vehicles and in the processes for producing fuel for fuel cells. 
4.1-1 Heat Loss through a Stainless Steel Bipolar Plate 
4.3-1 Cooling of a Fuel Cell 
4.3-2 Heat Loss from an Insulated Pipe 
4.3-3 Heat Loss by Convection and Conduction and Overall U 
4.3-4 Heat Generation in a Solid-Oxide Fuel Cell 
4.5-1 Heating of Natural Gas in Steam-Methane Reforming Process 
4.5-2 Trial-and-Error Solution for Heating of Steam 
4.5-3 Heating of Ethanol in Reforming Process 
4.5-4 Heat-Transfer Area and Log Mean Temperature Difference 
4.5-5 Laminar Heat Transfer and Trial and Error 
4.6-3 Heating of Steam by a Bank of Tubes in High-Temperature Electrolysis 
4.7-3 Natural Convection in Bipolar Plate Vertical Channel 
4.8-2 Steam Condensation in a Fuel Cell 
4.9-1 Temperature Correction Factor for a Heat Exchanger 
4.9-2 Effectiveness of Heat Exchanger 
4.11-1 Radiation in Cylindrical Solid-Oxide Fuel Cell 
4.15-1 Cooling Channels in Fuel Cell Bipolar Plates 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 2 Student View 
Jason M. Keith 
Example 4.1-1: Heat Loss through a Stainless Steel Bipolar Plate 
Calculate the heat flux through a stainless steel bipolar plate in a polymer-electrolyte membrane fuel 
cell with a thickness of 4.5 mm. The fuel cell is operating at a temperature of 80 ºC during the 
summer season in Houghton, Michigan where the temperature is 70 ºF. 
Strategy 
The equation for the heat flux obtained from Fourier’s Law can be used to obtain the solution to this 
problem. 
Solution 
Equation 4.1-10 of Geankoplis is defining the heat transfer per unit area as follows: 
 ( )1 2
2 1
q k
T T
A x x
= −
−
 
We can substitute the values given in the problem statement into this equation, but first we need to 
convert the temperature outside the fuel cell to ºC: 
 ( )
( )T º F 32 70º F 32
T º C
1.8 1.8
− −
= = 
 T _________º C= 
Entering the temperatures inside and outside the fuel cell stack into the heat transfer equation, as 
well as the thickness of the bipolar plate represented by 
2 1
x x− , we get: 
 ( )
W
________
q m K ________ K 294.25 K
A __________m
⋅= − 
 
2
q W
_____________
A m
= 
The thermal conductivity of steel was obtained from Table 4.1-1 of Geankoplis. 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 3 Student View 
Jason M. Keith 
Example 4.3-1: Cooling of a Fuel Cell 
Air at a temperature of 25 °C is being used for cooling a single cell fuel cell. The convective heat 
transfer coefficient of the air is 
2
W
61.2
m °C⋅
 and is capable of removing heat at a rate of 183.6 W. 
What would be the dimensions of the square surface of the fuel cell if its temperature must not 
exceed 50°C? 
 
 
 
 
 
 
 
 
Strategy 
The heat transfer rate by convection can be obtained using Newton's Law of Cooling. 
Solution 
The heat flux q� when heat is being transferred by forced convection is defined as follows: 
 ( )Sq hA T T∞= −� 
where: 
 
2
W
q heat transfer rate, 
m
=� 
 
2
W
h convective heat transfer coefficient, 
m K
=
⋅
 
 ST temperature on the surface of the object, °C= 
 T temperature of the air, °C∞ = 
 A = surface area of the fuel cell, m
2
 
Air @ 25 °C Air @ 25 °C 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 4 Student View 
Jason M. Keith 
To determine the dimensions of the surface of the fuel cell, we can substitute the given temperatures 
and heat transfer rate and solve for the area A to yield: 
 
( )2
____ ________ W
A
W________________
_________ 50 C 25 C
m °C
= =
° − °
⋅
 
 2A 0.12 m= 
Since the heat is being removed from the fuel cell through both the left and right faces of the fuel 
cell, this value of A must be divided by 2. Thus, 
 
2 2
fuel cell 2
0.12 m ___________ cm
A
2 1 m
 
=  
 
 
 2fuel cellA ________ cm= 
The dimensions of a fuel cell with a square surface could be obtained as follows: 
 2fuel cellL A ________ cm= = 
 L ________ cm= 
Therefore, for a heat transfer rate of 183.6 W, air at 25 °C can be used to keep the surface area of a 
________ cm x ________ cm fuel cell at a temperature of ________ °C. 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 5 Student View 
Jason M. Keith 
Example 4.3-2: Heat Loss in Fuel Reforming Applications 
A pipe made of 308 stainless steel (schedule number 80) with a nominal diameter of 1.5" is carrying 
methane at a temperature of 400°C in a steam-methane reforming process for producing hydrogen. 
The pipe is insulated with a layer of glass-fiber with a thickness of 1". Determine the temperature at 
the interface between the pipe and the glass fiber and the heat loss through the insulated pipe with a 
length of 15 m. The surface of the insulating material is at a temperature of 25°C. 
A schematic of the pipe is shown below: 
 
 
 
 
 
 
 
 
Strategy 
The equation for the heat loss through a pipe can be applied to the different layers in the pipe. 
Solution 
The heat loss through the walls of a cylinder is given by: 
 in out
T T
q
R
−
= 
where: 
 Tin = temperature at the inner wall of the pipe, K 
 Tout = temperature at the outer wall of the pipe, K 
 R = resistance of the pipe to the heat transfer through its walls, 
K
W
 
 
Natural Gas 
@ 400°C 
T1 
T2 
T3 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 6 Student View 
Jason M. Keith 
In this problem, we need to apply this equation for both the steel pipe and the insulated pipe. The 
overall heat loss will be obtained using this equation for the insulated steel pipe. The resistance to 
heat transfer in cylindrical coordinates is calculated with the following equation: 
 out in
lm
r r
R
kA
−
= 
in this equation: 
 rout = outer radius of the cylinder, m 
 rin = inner radius of the cylinder, m 
 k = thermal conductivity of the material, 
W
m K⋅
 
 Alm = log mean area, m
2
 
The log mean area of the pipe is defined as: 
 out inlm
A A
A
______
ln
 ______
−
=
 
 
 
 
where Aout and Ain are the outer and inner surface areas of the cylinder, respectively. 
Applying the equations for resistance and the log mean areas to the steel and the overall pipe we 
have:
Steel Pipe 
 1 21 2
1 2
T T
q
R
→
→
−
=� 
 2 11 2
steel lm,1 2
r r
R
k A
→
→
−
= 
 2 1lm,1 2
2
1
A A
A
A
ln
A
→
−
=
 
 
 
 
 
Overall 
 1 31 3
1 2 2 3
T T
q
R R
→
→ →
−
=
+
� 
 
 3 1lm,1 3
3
1
A A
A
A
ln
A
→
−
=
 
 
 
 
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 7 Student View 
Jason M. Keith 
To determine the log mean areas we need to look for the radius of the steel pipe in Appendix A.5 of 
Geankoplis. For the 1.5" pipe: 
 rin = r1 = __________ m 
 rout = r2 = __________m 
The radius of the pipe including the insulation is obtained by adding the thickness of 1" to the outer 
radius of the steel pipe. Hence, 
 r3 = __________m + 0.0254 m = __________m 
With these diameter values and the length of the pipe, the areas A1, A2 and A3 can be calculated as 
shown in the following steps. 
 A1 = 2πr1L = 2π(0.01905 m)( __________ m) = 1.795 m
2
 
 A2 = 2πr2L = 2π(__________m)( __________ m) = __________m
2
 
 A3 = 2πr3L = 2π(__________m)( __________ m) = __________m
2
 
The thermal conductivities for the glass fiber and the steel can be found in Appendix A.3 of 
Geankoplis and are shown below. The conductivity of the glass fiber was selected at the highest 
temperature available in Table A.3-15. The thermal conductivity of steel was obtained from Table 
A.3-16. 
 steel
W
k ____________________
m C
=
⋅°
 
 glass fiber 
W
k __________
m C
=
⋅°
 
Substituting the values we obtained into the equations for the individual layers yields: 
Steel Pipe 
1 2 2
1 2
1 2
T T __________ °C T
q
CR
_____________
W
→
→
− −
= =
°
� 
( )
4
1 2
2
____________ m ____________ m C
R 1.17 10
W W
21.6 ____________ m
m C
−
→
− °
= = ×
⋅°
 
2 2
2
lm,1 2 2
2
___________ m ___________ m
A 2.025 m
2.274 m
ln
1.795 m
→
−
= =
 
 
 
 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 8 Student View 
Jason M. Keith 
Overall 
1 3
___________ °C ___________ °C
q 2695.6 W
C C
___________ 0.139
W W
→
−
= =
° °
+
� 
( )
2 3
2
___________ m 0.02415 m C
R 0.139
W W
___________ ___________ m
m C
→
− °
= =
⋅°
 
2 2
2
lm,2 3 2
2
___________ m ___________ m
A ___________ m
4.670 m
ln
2.274 m
→
−
= =
 
 
 
 
Hence, the heat loss through the insulated pipe is 2695.6 W. 
Since this answer represents the amount of heat lost per unit time, if we assume that the system is at 
steady state. The heat loss per unit time will be the same in the individual layers. Thus, we can use 
the equation for the heat loss through the steel pipe to determine the temperature at the steel-glass 
fiber interface. 
 21 2
______ °C T
q 2695.6 W
C
_____________
W
→
−
= =
°
� 
Solving for the temperature T2, we get: 
 2
C
T ___________ °C 2695.6 W _______________
W
° 
= −  
 
 
 2T ___________ °C= 
As it can be seen, the temperature at the pipe - insulator interface is almost the same as the 
temperature of the inner wall of the steel pipe. This is because most of the heat is lost through the 
metal pipe due to the high thermal conductivity of steel in comparison to the thermal conductivity of 
the insulating material. 
 
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Supplemental Material for Transport Process and SeparationProcess Principles 
Daniel López Gaxiola 9 Student View 
Jason M. Keith 
Example 4.3-3: Heat Loss by Convection and Conduction in a Steam-Methane 
Reforming Process 
Natural gas at 400°C is flowing inside a steel pipe with an inner diameter of 1.5 in and an outer 
diameter of 1.9 in. The pipe is insulated with a layer of glass-fiber with a thickness of 1 in. The 
convective coefficient outside the insulated pipe is 
2
btu
1.23
ft hr F⋅ ⋅°
. The temperature on the external 
surface of the pipe is 43.4°C. 
Calculate the convective coefficient of natural gas and the overall heat transfer coefficient U based 
on the inside area Ai, if heat is being lost at a rate of 
btu
7115
hr
in a pipe with a length of 49.2 ft. 
Strategy 
To determine the heat transfer coefficients, we will use the equation for heat loss for a multilayer 
cylinder. 
Solution 
The heat loss through a cylinder with different layers is defined by the following equation: 
 i o i o
i A B o
T T T T
q
R R R R R
− −
= =
+ + +∑
� 
where: 
 Ti = Temperature on the internal surface of the pipe 
 To = Temperature on the external surface of the pipe 
 Ri = Convective resistance inside the pipe 
 RA = Conductive resistance through the steel pipe 
 RB = Conductive resistance through the insulation layer 
 Ro = Convective resistance outside the pipe 
The resistance to heat transfer due to convection is defined as follows: 
 conv
1
R
hA
= 
 
 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 10 Student View 
Jason M. Keith 
where: 
 h = Convective heat transfer coefficient 
 A = Area of heat transfer 
The resistance of a cylinder to heat conduction is calculated as follows: 
 out incond
lm
r r
R
kA
−
= 
where: 
 rout = Outer radius of the cylinder 
 rin = Inner radius of the cylinder 
 k = thermal conductivity of the material 
 Alm = log mean area of the cylinder 
We can enter the definitions of the resistances due to conduction and convection into the equation 
for the heat loss to yield: 
 i o
o 11 i
i i steel A,lm glass fiber B,lm o o
T T
q
r rr r1 1
h A k A k A h A−
−
=
−−
+ + +
� 
In this equation: 
 ri = inner radius of the steel pipe = 
1.5 in
2
= ____________ in 
 r1 = outer radius of the steel pipe = 
1.9 in
2
= ____________ in 
 ro = outer radius of the insulated pipe = ___________ in 
With these values we can calculate the log mean areas AA,lm and AB,lm and the inner and outer areas 
of the insulated pipe. Thus, 
 ( ) ( ) 2i i
1 ft
A 2 r L 2 0.75 in ___________ ft ___________ ft
12 in
 
= π = π = 
 
 
 ( ) ( ) 21 1
1 ft
A 2 r L 2 ___________ in ___________ ft 24.48 ft
12 in
 
= π = π = 
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 11 Student View 
Jason M. Keith 
 ( ) ( ) 2o o
1 ft
A 2 r L 2 ___________ in ___________ ft ___________ ft
12 in
 
= π = π = 
 
 
 
2 2
21 i
A,lm 2
1
2
i
A A 24.48 ft ___________ ft
A 21.80 ft
24.48 ftA
lnln
___________ ftA
− −
= = =
   
   
  
 
 
2 2
2o 1
B,lm 2
o
2
1
A A ___________ ft 24.48 ft
A ___________ ft
___________ ftA
lnln
24.48 ftA
− −
= = =
   
   
  
 
The temperature of the inner and outer surfaces of the pipe are given in °C and therefore, they must 
be converted to °F: 
 ( )iT ( F) 400 C 1.8 32 ___________ F° = ° × + = ° 
 ( )oT ( F) 43.4 C 1.8 32 ___________ °F° = ° × + = 
Now we can substitute all the values into the heat loss equation and solve for the convective 
coefficient hi: 
 
1
i o o 11 i
i
i steel A,lm glass fiber B,lm o o
T T r rr r1 1
h
A q k A k A h A
−
−
 − −−
= − − − 
  �
 
The thermal conductivity values can be obtained from Appendices A.3-15 and A.3-16 of 
Geankoplis. However, since the values are given in the SI system they must be converted to the 
English system. Hence, 
 
steel
J 1 btu ___________ m 3600 s 1 °C btu
k 45 26
m s C ___________ J 1 ft 1 hr ___________ °F ft hr F
     
= =     
⋅ ⋅ ° ⋅ ⋅°     
 
glass fiber
J 1 btu _________ m 3600 s 1 °C btu
k ______ 0.0317
m s C _________ J 1 ft 1 hr _____ °F ft hr F
−
     
= =     
⋅ ⋅° ⋅ ⋅ °     
 
The thermal conductivities of steel and glass-fiber were obtained at the highest temperature available 
in Appendix A.3. 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 12 Student View 
Jason M. Keith 
Now we can calculate the convective heat transfer coefficient hi as shown below: 
( )
( )
( )
i 2
2
1ft
___________ in 0.75 in
___________ in1 ___________ F 110.1 F
h
btu btu___________ ft
7115 26 ___________ ft
hr ft hr F
1 ft
___________ in 0.95 in
___________ in
 
btu
0.0317 _____
ft hr F
  
−  
° − °  = −

 ⋅ ⋅°
 
−  
 −
⋅ ⋅°
( ) ( )
1
2 2
2
1
btu
______ ft ___________ ___________ ft
ft hr F
−


−

⋅ ⋅ ° 
 
1
5
i 2
1 hr F hr F hr F hr F
h __________ 2.95 10 __________ 0.0162
________ ft btu btu btu btu
−
−⋅° ⋅° ⋅° ⋅° = − × − −  
 
i 2
btu
h __________
ft hr F
=
⋅ ⋅°
 
To calculate the overall heat transfer coefficient U we need to use the equation for the heat loss in 
terms of U. Thus, 
 i i i oq U A (T T )= −� 
We can solve this equation for the coefficient U and substitute the corresponding values to get: 
( )i 2i i o
btu
7115
q hrU
A (T T ) ________ ft ________ F 110.1 F
= =
− ° − °
�
 
i 2
btu
U ________
ft hr F
=
⋅ ⋅°
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 13 Student View 
Jason M. Keith 
Example 4.3-4: Heat Generation in a Tubular Solid-Oxide Fuel Cell 
A tubular solid-oxide fuel cell with an outer diameter of 2.2 cm and a length of 150 cm is operating 
at a current density of 
2
mA
202.6
cm
. Determine the heat generation rate in 
3
W
m
 if the voltage of the 
fuel cell is 1 V. Assume that the thickness of the electrodes and electrolyte membrane are small 
compared to the overall diameter of the fuel cell. 
The following figure shows a tubular solid-oxide fuel cell: 
 
 
Strategy 
The heat generation rate of the fuel cell can be obtained from the power of the fuel cell, which 
depends on the current and the voltage. 
Solution 
The heat generated by the fuel cell in terms of the power is given by the equation shown below: 
 
2
P
q
R L
=
π
� 
The power of the fuel cell can be obtained by multiplying the current by the voltage of the fuel cell. 
Hence, 
 P = IV 
 
 
Cathode 
Fuel flow 
Anode Air Flow 
Electrolyte 
Cathode interconnection 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 14 Student View 
Jason M. Keith 
Substituting this equation into the equation for the heat generation rate yields: 
 
2
IV
q
R L
=
π
� 
The problem statement is not giving the value of the current. However, if we calculate the cross-
surface area of the fuel cell we can determine the value of the current in A: 
 
surface
I I
i
A 2 RL
= =
π
 
Solving for the current I and substituting the dimensions of the fuel cell into this equation, we get: 
 ( )( ) 2
mA 1 A
I DLi ________ cm ________ cm 202.6
cm 1000 mA
  
= π = π   
  
 
 I ________ A= 
Entering this value into the heat generation equation we have: 
 
( )
( )
2 3
3
________ A 1 V
q
________ cm 1 m
________ cm
2 __________ cm
=
  
π   
   
� 
 5
3
W
q 3.68 10
m= � 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 15 Student View 
Jason M. Keith 
Example 4.5-1: Heating of Natural Gas- in Steam-Methane Reforming Process 
 
Natural gas at a temperature of 310°C is flowing inside a steel pipe Schedule 80 with an inner 
diameter of 1.5 in at a rate of 
3
3 m7.79 10
s
−× . The natural gas is being heated by the product of the 
reforming process at 850°C. The convective heat transfer coefficient of the reformate is 
2
W
1025
m K⋅
. 
Calculate the heat transfer rate in W through a pipe with a length of 7 m. The properties of natural 
gas are given in the following table. 
ρ 
3
kg
7.859
m
 
Cp 
J
3087
kg K⋅
 
k W0.0803
m K⋅
 
µb 5 kg1.942 10
m s
−×
⋅
 
µw 5 kg2.909 10
m s
−×
⋅
 
 
Strategy 
The equation for heat transfer through a pipe will be used to determine the heat flux. 
Solution 
When heat is being transferred through a fluid, the heat flux is given by: 
 
( )r nT T
q
R
−
=
∑
� 
where: 
 q =� Heat transfer rate in W 
 Tr = Temperature of the heating medium (reformate), °C or K 
 Tn = Temperature of the fluid inside the pipe (natural gas), °C or K 
 R∑ = Sum of resistances to heat transfer through the pipe, 
C K
 or 
W W
°
 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 16 Student View 
Jason M. Keith 
In this problem, the sum of the resistances is given by the sum of two convective resistances (fluid 
inside and outside the pipe) and the resistance to heat flow through the steel pipe. Thus, 
 o i
i i steel lm o o
r r1 1
R
h A k A h A
−
= + +∑ 
The parameters we need to calculate before being able to use the equation for heat transfer in terms 
of the resistances are: Ai, Alm, Ao and hi. The heat transfer areas are calculated using the inner and 
outer diameters of the pipe from Appendix A.5. 
 ( ) ( ) 2i iA D L 0.0381 m 7 m _________ m= π = π = 
 ( )( ) 2o oA D L _________ m 7 m _________ m= π = π = 
 
2 2
2o i
lm 2
o
2
i
A A _________ m _________ m
A 0.945 m
_________ mA
lnln
_________ mA
− −
= = =
   
   
  
 
The following correlation can be used for calculating the heat transfer coefficient for an aspect ratio 
(length/diameter) of the pipe greater than 60. 
 
0.14
1
0.8 b3
L Re Pr
w
k
h 0.027 N N
D
 µ
=  
µ 
 
where: 
 k = Thermal conductivity of the fluid inside the pipe, 
W
m K⋅
 
 NRe = Reynolds number 
 NPr = Prandtl number 
 bµ = Viscosity of the fluid in the pipe at the bulk temperature,
kg
m s⋅
 
 wµ = Viscosity of the fluid in the pipe at the temperature of the inner wall,
kg
m s⋅
 
In this problem, 
L 7 m
_________
D 0.0381 m
= = . Hence, we can use this correlation to calculate hL. 
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 17 Student View 
Jason M. Keith 
The dimensionless quantities NRe and NPr are defined as follows: 
 Re
D
N
υρ
=
µ
 
p
Pr
C
N
k
µ
= 
The velocity of the natural gas is obtained by dividing the volumetric flow rate by the cross-sectional 
area of the pipe. Thus, after substituting the corresponding quantities into the equations for Reynolds 
and Prandtl numbers, we have: 
 
3
3
2
i
Re
m
______________
kgs0.0381 m 7.859
V ______________ mD
r
N __________________
kg
_______________
m s
 
 
  
 ρ   π   = = =
µ
⋅
�
 
 
p
Pr
J kg
3087 ________________
C kg K m s
N _____________
Wk
______________
m K
 
 µ ⋅ ⋅ = = =
⋅
 
Now we can substitute the dimensionless numbers we just calculated and the properties of the fluid 
into the equation for the heat transfer coefficient to yield: 
 ( ) ( )
0.14
10.8
3
L
kgW
______________0.0803
m sm Kh 0.027 ________________ _______
kg0.0381 m
______________
m s
 
 ⋅⋅=  
 
⋅ 
 
 L 2
W
h _____________
m K
=
⋅
 
The convective heat transfer coefficient hL was obtained using the properties of the fluid inside the 
pipe. Therefore, the heat transfer coefficient hL is equal to the heat transfer coefficient hi. 
Substituting the corresponding quantities into the equation for the heat transfer rate q� . The thermal 
conductivity was obtained from Appendix A.3. 
 
( )
( ) ( ) ( )2 2 22 2
850 C 310 C
q
1 0.04826 m 0.0381 m 1
W W W
_________ _________ m 45 0.945 m _________ _________ m
m K m K m K
° − °
=
−
+ +
⋅ ⋅ ⋅
�
 
q ____________ W=� 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 18 Student View 
Jason M. Keith 
Example 4.5-2: Trial-and-Error Solution for Heating of Steam 
Steam at a temperature of 150°C is being heated before entering a steam-methane reforming unit to 
produce hydrogen for fuel cells. The heating medium is the synthesis gas produced by the steam-
methane reforming unit at 850 °C. The convective heat transfer coefficient of the syngas is 
2
W
950
m K⋅
. The steam is flowing in a 1.5-in schedule 40 steel pipe at a velocity of 
m
20.1
s
. 
Determine the overall coefficient Ui for a pipe with a length of 6.2 m. 
Strategy 
We can determine the convective heat transfer coefficient using the properties of steam at the 
temperature of the inner wall of the pipe. This temperature will be determined by trial and error. 
Solution 
The overall heat transfer coefficient can be determined from the equation for the heat transfer rate 
through the pipe: 
 o ii i o i
T T
q U A (T T )
R
−
= − =
∑
� 
The temperature difference To - Ti can be eliminated from this equation to yield: 
 i
1
U
___________
= 
where: 
 o i
i i lm o o
r r1 1
R
h A kA h A
−
= + +∑ 
The convective coefficient hi can be calculated using the following correlation. 
 
0.14
1
0.8 b3
L Re Pr
w
k
h 0.027 N N
D
 µ
=  
µ 
 
The dimensionless quantities in this equation will be determined using the properties of steam at the 
temperature of the inner wall of the pipe. For the first trial, this temperature will be assumed to be 
about one-quarter the difference between the temperatures of the steam and the air. Thus, 
 w,assumed
850 C 150 C
T 150 C __________ C
4
° − °
= + ° = ° 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 19 Student View 
Jason M. Keith 
From Table A.2-12 of Geankoplis we can get the properties of steam at a temperature of 148.9 °C, 
which is relatively close to the bulk temperature of the steam inside the pipe (150°C). 
 
3
kg
__________
m
ρ = 5b
kg
1.488 10
m s
−µ = µ = ×
⋅
 PrN __________= 
 p
J
C 1909
kg K
=
⋅
 
W
k __________
m K
=
⋅
 
The other parameter required to use the correlation for calculating hL is the viscosity of steam at Tw. 
This can be obtained using linear interpolation from the data in Table A.2-12. Hence, 
 
5
w
5
kg
2.113 10
325 C 315.6 C m s
kg kg371.1 C 315.6 C
______________ 2.113 10
m s m s
−
−
µ − ×
° − ° ⋅=
° − ° − ×
⋅ ⋅
 
Solving for the viscosity µw we get: 
 
5
w
325 C 315.6 C kg kg kg
_______________ 2.113 10 _______________
371.1 C 315.6 C m s m s m s
−° − °  µ = − × + 
° − ° ⋅ ⋅ ⋅ 
 
w
kg
_______________
m s
µ =
⋅
 
With the properties of steam, we can now determine the Reynolds number as shown in the following 
steps. The diameter of the pipe was obtained from Table A.5-1 of Geankoplis 
3
Re
5
m kg
___________ m _______ 0.525
D s m
N _______________
kg
1.488 10
m s
−
  
  υρ   = = =
µ ×
⋅
 
Substituting this value into the correlation for hL we get: 
 
( ) ( )
0.14
5
10.8
3
L
kgW
1.488 10____________
m sm Kh 0.027_______________ 0.95
kg0.04089 m
_______________
m s
− × ⋅⋅=  
 
⋅ 
 
L 2
W
h ____________
m K
=
⋅
 
Now we can proceed to calculate the sum of the resistances to heat transfer as follows: 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 20 Student View 
Jason M. Keith 
 o i
i i lm o o
r r1 1
R
h A kA h A
−
= + +∑ 
The heat transfer areas are calculated using the inner and outer diameters of the pipe from Appendix 
A.5. 
 ( ) ( ) 2i iA D L 0.04089 m ______ m ___________ m= π = π = 
 ( )( ) 2o oA D L 0.04826 m ______ m ___________ m= π = π = 
 
2 2
2o i
lm 2
o
2
i
A A _______ m _______ m
A 0.866 m
_______ mA
lnln
_______ mA
− −
= = =
   
   
  
 
These values can be entered into the equation for the sum of resistances to yield: 
 
( ) ( ) ( )2 2 22 2
1 0.04826m 0.04089 m 1
R
W W W
_______ _________ m 45 0.866 m 950 _______ m
m K m K m K
−
= + +
⋅ ⋅ ⋅
∑ 
K
R _________
W
=∑ 
The thermal conductivity value of steel was obtained from Table A.3-16. To determine if the value 
of Tw we selected is correct, we need to solve for the temperature from the equation for the heat 
resistance due to the steam in the pipe: 
 ( )iw,calculated b o b
R
T T T T
R
− = −
∑
 
Solving for Tw and substituting the rest of the values into this equation, we have that: 
 ( )w,calculated b o b
i i
1
T T T T
h A R
= + −
∑
 
 
( )
( )w,calculated
2
2
1
T 150 C 850 C 150 C
W K
_______ _______ m _______
m K W
= ° + ° − °
 
 
⋅  
 
w,calculatedT _________ C= ° 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 21 Student View 
Jason M. Keith 
It can be see that the Tw,assumed does not match Tw,calculated. Hence, we have to repeat the procedure for 
determining the temperature of the inner wall. The Tw value will affect the value of hL since we have 
to look for a new value of µw in Appendix A.3. 
Selecting a higher Tw for the second trial will yield a higher µw, resulting in a lower heat transfer 
coefficient hL and a higher R∑ . For the second trial, we will select: 
 Tw,assumed = 800°C 
Substituting the viscosity µw into the equation for the heat transfer coefficient hL yields: 
 ( ) ( )
0.14
5
10.8
3
L
5
kgW
1.488 10__________
m sm Kh 0.027 ____________ _____
kg0.04089 m
3.95 10
m s
−
−
 
× ⋅⋅=  
 ×
⋅ 
 
 L 2
W
h ________
m K
=
⋅
 
For this value of hL we will get the sum of the resistances as follows: 
 
( ) ( ) ( )2 2 22 2
1 0.04826m 0.04089 m 1
R
W W W
62.3 _________ m _______ 0.866 m 950 _________ m
m K m K m K
−
= + +
⋅ ⋅ ⋅
∑ 
K
R ________
W
=∑ 
Solving for Tw and substituting the rest of the values into this equation, we have that: 
( )
( )
( )w,calculated b o b
2i i
2
1 1
T T T T 150 C 850 C 150 C
W Kh A R
62.3 _______ m _______
m K W
= + − = ° + ° − °
 
 
⋅  
∑
 
w,calculatedT __________ C= ° 
The only property that will change for the third trial is the viscosity µw. By changing the temperature 
again, the effect on the convective coefficient hL will be negligible. Hence, we can use the calculated 
Tw value of ________°C. 
Now we can substitute the values of R∑ at Tw = _________°C and the inner area of the pipe to 
obtain the overall heat transfer coefficient Ui as shown in the following steps: 
 i
i
1
U
A R
=
∑
 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 22 Student View 
Jason M. Keith 
 
( )
i
2
1
U
K
________ m ________
W
=
 
 
 
 
 i 2
W
U 59.8
m K
=
⋅
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 23 Student View 
Jason M. Keith 
Example 4.5-3: Heating of Ethanol in Reforming Process 
A vapor mixture of ethanol and water is used in a reforming process to produce hydrogen for proton-
exchange membrane fuel cells. In a distributed-scale process, the ethanol mixture is flowing at a rate 
of 
kg
0.367
s
and is entering a 1" steel pipe Schedule 40 at a temperature of 210.4°C. Determine the 
length of the pipe required if the vapor is exiting at 350°C and the inner wall of the pipe is at a 
constant temperature of 270°C. The properties of the vapor mixture are summarized in the following 
table. 
µ 51.284 10 Pa s−× ⋅ 
ρ 
3
kg
13.2
m
 
Cp 
J
2211
kg K⋅
 
k W0.03631
m K⋅
 
 
Strategy 
To determine the heat transfer area we can use the equation for heat flux through a fluid. 
Solution 
The heat flux when heat is being transferred by a fluid is given by: 
 ( )L w
q
h T T
A
= −
�
 
where Tw is the temperature of the inner wall of the pipe, and T is the bulk temperature of the 
ethanol and water mixture. Since the problem statement is giving the flow rate and properties of the 
fluid, the heat transfer rate can be calculated with the equation for sensible heat: 
 ( )p out inq mC T T= −� � 
In this equation, Tin and Tout are the temperatures of the ethanol/water mixture at the inlet and outlet 
points, respectively. 
Substituting this equation into the equation for heat flux, we get: 
 
( )p out inmC T T
______________________
______
−
=
�
 
Solving for the heat transfer area we have: 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 24 Student View 
Jason M. Keith 
 
( )p out inmC T T
A
______________
−
=
�
 
To determine the heat convective heat transfer coefficient, the following correlation can be used for 
a pie with constant wall temperature and if NPe>100 and L/D>60: 
 ( )0.8L Pe
k
h 5.0 0.025N
D
= + 
where: 
 NPe = Peclet number 
We can calculate Peclet number by multiplying Reynolds number by Prandtl number and thus 
determine if it is valid to use this correlation. 
 NPe = NReNPr 
 Re
D
N
υρ
=
µ
 
 
( )
p
Pr
J
2211 ______________ Pa s
C kg K
N ______
Wk
0.03631
m K
⋅
µ ⋅
= = =
⋅
 
The velocity of the fluid is calculated by dividing the volumetric flow rate by the cross-sectional area 
of the pipe. The diameter of the pipe is obtained from Appendix A.5 of Geankoplis. 
 
( )
2
cross
3
kg
________
m ms ________
A s________ mkg
13.2
m 4
υ = = =
ρ  π
 
  
�
 
Entering this velocity value into the definition of Reynolds number yields: 
 
3
Re
5
m kg
________ m ________ 13.2
D s m
N _________________
kg
1.284 10
m s
−
  
  υρ   = = =
µ ×
⋅
 
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 25 Student View 
Jason M. Keith 
Now we can determine the Peclet number to be given by: 
 ( )PeN ____________ ______= 
 PeN ______________= 
Therefore, the correlation we selected is valid for this problem and the convective heat transfer 
coefficient is found to be: 
 ( )
0.8
L
W
0.03631
m Kh 5.0 0.025 _______________
__________ m
⋅  = +
 
 
 L 2
W
h ___________
m K
=
⋅
 
This value can be entered into the equation for the heat transfer area to yield: 
 
( )
( )
( )
( )
p out in
L w
2
kg J
0.367 2211 _______ C 210.4 C
mC T T s kg K
A
Wh T T
________ _______ C 210.4 C
m K
 
° − ° − ⋅ = =
− ° − °
⋅
�
 
 2A __________ m= 
The area for heat transfer is given by: 
 A DL= π 
This equation can be solved for the length of the pipe L to give: 
 
( )
2________ m
L
___________ m
=
π
 
 L ________ m= 
 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 26 Student View 
Jason M. Keith 
Example 4.5-4:Heat-Transfer Area and Log Mean Temperature Difference 
An ethanol/water vapor mixture with a heat capacity of 
kJ
2.23
kg C⋅ °
 in a mid-scale ethanol-
reforming plant is heated from 210.4 °C to 350 °C. This mixture is flowing at a rate of 
5 kg5.18 10
day
× . The vapor is being heated by air flowing at a rate of 6
kg
2.752 10
day
× , temperature of 
560.6 °C and a heat capacity of 
kJ
1.166
kg C⋅°
. What type of flow for this heat exchanger will you 
select between countercurrent and parallel flow if the overall heat transfer coefficient is 
2
W
92.4
m C⋅ °
?. 
Strategy 
To determine which type of flow is more efficient for this process we need to determine the heat 
transfer area for both types of flow. 
Solution 
The amount of heat gained by the ethanol mixture in terms of the overall heat transfer coefficient is 
given by the equation shown below: 
 i i lmq U A T= ∆� 
where lmT∆ is the log mean temperature difference defined as: 
 2 1lm
2
1
T T
T
T
ln
T
∆ − ∆
∆ =
 ∆
 
∆ 
 
The q� equation can be solved for the area Ai to yield: 
 i
q
A
_________
=
�
 
Since we know the inlet and outlet temperatures of the ethanol/water mixture we can calculate the 
amount of heat gained as shown in the following steps: 
 ( )ethanol p,ethanol ethanol,out ethanol,inq m C T T= −� � 
 ( )5
kg kJ 1 day
q 5.18 10 2.23 _______ C _______ C
day kg C __________
   
= × ° − °   
⋅°   
� 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 27 Student View 
Jason M. Keith 
 q __________ kW=� 
To determine the log mean temperature difference required to calculate the heat transfer area we 
need to obtain the temperature of the air leaving the heat exchanger. We know that the amount of 
heat gained by the ethanol mixture is being lost by the heating air. Thus, the temperature can be 
determined as follows: 
 ( )air p,air air,in air,outq m C T T= −� � 
 
air,out
6
__________ kW
T _________ ____________ _______ C
kg kJ 1 day
2.752 10 1.166
day kg C _________
= − = ° −
  
×   
⋅ °  
 
air,outT _________ C= ° 
Now that we know the inlet and outlet temperatures of both the air and the ethanol mixture, we can 
calculate the log mean temperature difference. Hence, 
 2 1lm
2
1
T T
T
T
ln
T
∆ − ∆
∆ =
 ∆
 
∆ 
 
where: 
 1,countercurrent air,in ethanol,outT T T _________ C _________ C∆ = − = ° − ° 
 1,countercurrentT _________ C∆ = ° 
 2,countercurrent air,out ethanol,inT T T _________ C _________ C∆ = − = ° − ° 
 2,countercurrentT _________ C∆ = ° 
Substituting the values of ∆Ti into the definition of the log mean temperature difference we get: 
 lm,countercurrent
_________ C _________ C
T
_________ C
ln
_________ C
° − °
∆ =
 °
 
° 
 
 lm,countercurrentT _________ C∆ = ° 
 
 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 28 Student View 
Jason M. Keith 
For countercurrent flow, we can find the heat transfer area to be given by: 
 
( )
i,countercurrent
2
1000 W
_________ kW
1 kW
A
W
92.4 _________ C
m C
 
 
 =
°
⋅°
 2i,countercurrentA _________ m= 
To determine the heat transfer area for parallel flow, we can use Figure 4.5-3 of Geankoplis to 
calculate the log mean temperature difference for parallel flow, which will be given by: 
 2 1lm
2
1
T T
T
T
ln
T
∆ − ∆
∆ =
 ∆
 
∆ 
 
where: 
 1,parallel air,out ethanol,outT T T ________ C ________ C∆ = − = ° − ° 
 1,parallelT _________ C∆ = ° 
 2,parallel air,in ethanol,inT T T ________ C ________ C∆ = − = ° − ° 
 2,parallelT _________ C∆ = ° 
 lm,parallel
_________ C _________ C
T _________ C
_________ C
ln
_________ C
° − °
∆ = = °
 °
 
° 
 
Substituting the log mean temperature difference into the equation for Ai,parallel we have: 
 
( )
i,parallel
2
1000 W
_________ kW
1 kW
A
W
92.4 _________ C
m C
 
 
 =
°
⋅°
 
 2i,parallelA _________ m= 
Conclusion: 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 29 Student View 
Jason M. Keith 
Example 4.6-3: Heating of Steam by a Bank of Tubes in High-Temperature 
Electrolysis 
High-temperature electrolysis is a process for producing hydrogen from water for use in fuel cells. 
Before entering the electrolysis stack, steam at a pressure of 50 bar is being heated from 650°C to 
850°C by a bank of 1" (nominal diameter) commercial steel tubes containing 12 rows normal to the 
flow and 7 staggered rows in the direction of flow. The length f the tubes is 0.4 m. 
The heating medium circulating in the tubes is helium coming from a nuclear source and the outer 
surface of the tubes is at a temperature of 1000°C. Determine the heat-transfer rate to the steam if the 
velocity of steam is 
m
16.7
s
. 
A diagram of this heating process is shown below: 
 
 
 
 
 
 
 
Strategy 
To solve this problem we need to calculate the amount of heat transferred by convection. The heat 
transfer area will depend on the number of tubes. 
Solution 
The amount of heat gained by the steam can be calculated with the following equation: 
 ( )w bq hA T T= −� 
The bulk temperature of steam Tb is obtained taking the average of the inlet and outlet temperatures: 
 b
850 C 650 C
T _________ C
2
° + °
= = ° 
The temperature of the outer surface of the tube is constant and equal to 1000°C. 
Steam 
Sp=0.0418 m 
Sn=0.0418 m 
Helium from nuclear 
source flowing inside the 
tubes 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 30 Student View 
Jason M. Keith 
To calculate the amount of heat transferred, we need to obtain the heat transfer area of the tubes. The 
area of a single tube is calculated as follows: 
 tubeA DL= π 
Substituting the diameter from Appendix A.5 of Geankoplis and the length of 0.4 m into this 
equation, we can determine the area of a single tube to be: 
 ( )( )tubeA _________ m ______ m= π 
 2tubeA ________ m= 
Since there are 7 columns and 12 rows of tubes, this area must be multiplied by the total number of 
tubes in the bank. Thus, 
 ( )( )( )2rows columns tubeA n n A 7 12 _________ m= = 
 2A _________ m= 
Now we need to determine the heat transfer coefficient of steam using the correlation for flow past a 
bank of tubes, shown in Section 4.6 of Geankoplis. 
 m 1/3calculated Re Pr
k
h CN N
D
= 
In this equation, the parameters C and m will depend on the ratio of the distance between the tubes 
and their outer diameter. For this electrolysis process, the ratio is given by: 
 
pn
SS _________ m
1.251
D D _________ m
= = = 
Using Table 4.6-2 of Geankoplis, we can find the values of C and m for staggered tubes to be: 
 C = _________ 
 m = _________ 
The dimensionless parameters NRe and NPr are calculated from the properties of steam at the film 
temperature, obtained as follows: 
 w bf
T T _________ C 750 C
T
2 2
+ ° + °
= = 
 fT _________ C= ° 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 31 Student View 
Jason M. Keith 
The properties of steam were obtained from Table 2-305 of Perry's Chemical Engineers' Handbook, 
8th Edition and are shown below: 
 
W
k 0.1203
m K
=
⋅
 
 p
J
C 2441.3
kg K
=
⋅
 
 54.34 10 Pa s−µ = × ⋅ 
 
3
kg
9.508
m
ρ = 
Substituting these properties into the definition of Prandtl number we get: 
 
( )
Pr
__________________ ____________ Pa s
N
_______________
⋅
= 
 PrN _________= 
The maximum velocity required to calculate Reynolds number isobtained using the outer diameter 
of the pipes and the distance between the pipes normal to the direction of flow as shown in the 
following equation: 
 nmax
n
S
S D
υ
υ =
−
 
We can enter the velocity of steam, the diameter and the distance between the pipes into this 
equation to yield: 
 
( )
max
m
16.7 __________ m
s
__________ m __________ m
υ =
−
 
 max
m
__________
s
υ = 
Now the Reynolds number can be determined as shown in the next steps: 
 maxRe
D
N
υ ρ
=
µ
 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 32 Student View 
Jason M. Keith 
 
3
Re 5
m kg
_________ m ______ 9.508
s m
N
4.34 10 Pa s−
  
  
  =
× ⋅
 
 ReN ____________= 
Substituting the dimensionless parameters we calculated and the corresponding values into the 
equation for hcalculated, we have: 
 m 1/3calculated Re Pr
k
h CN N
D
= 
 ( )( ) ( )
0.556 1/3
calculated
W
0.1203
m Kh _______ _____________ ________
__________ m
⋅= 
 calculated 2
W
h ____________
m K
=
⋅
 
The value of the heat transfer coefficient that has to be entered into the equation for the heat transfer 
rate has to be multiplied by a factor that depends on the amount of rows in the direction of flow. 
For 7 rows and staggered tubes we can find this factor in Table 4.6-3 of Geankoplis to be _______. 
Thus, the actual value of the heat transferred coefficient will be given by: 
 calculated 2
W
h ______ h ______ ____________
m K
 
= =  
⋅ 
 
 
2
W
h ___________
m K
=
⋅
 
Entering this value into the equation for q� , we get: 
 ( )( )22
W
q ___________ ___________ m 1000 C 750 C
m K
= ° − °
⋅
� 
 q ______________ W=� 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 33 Student View 
Jason M. Keith 
Example 4.7-3: Natural Convection in Bipolar Plate Vertical Channel 
Hydrogen at standard pressure is flowing by natural convection in the bipolar plate channels of a fuel 
cell. These channels have a length of 24.94 cm and a thickness and height of 1 mm. Determine the 
heat transfer rate across the channel if the temperature of the walls is constant and equal to 82 °C. 
The surface of the gas diffusion layer adjacent to the channel is at a temperature of 85.17 °C. A 
schematic of the bipolar plate channels is shown in the following figure. 
 
 
 
 
Strategy 
The heat transfer rate by convection can be calculated by using correlations that involve 
dimensionless groups. 
Solution 
When heat is being transfer by convection, the heat transfer rate q� is given by: 
 ( )1 2q hA T T= −� 
The heat transfer area can be calculated from the dimensions of the channel as shown below: 
 ( ) ( )( )A 2L t H 2 0.2494 m 0.001 m 0.001 m= + = + 
 2A ___________ m= 
To determine the heat transfer coefficient, we can use the definition of Nusselt number: 
 Nu
hH
N
k
= 
where H is the height of the channel. 
However, since the Nusselt number is not given, we need to use another correlation in terms of 
dimensionless groups. In section 4.7 of Geankoplis, multiple correlations are shown as function of 
the Grashof and Prandtl numbers. We need to select the adequate correlation depending on the value 
yielded by the product of these dimensionless groups defined as follows: 
 
( )3 1 2
Gr 2
H g T T
N
ρ β −
=
µ
 
1 mm 
1 mm 
249.4 mm 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 34 Student View 
Jason M. Keith 
In this equation: 
 H = height of the bipolar plate channel, m 
 ρ = density of hydrogen, 
3
kg
m
 
 g = acceleration due to gravity = 
2
m
9.80665
s
 
 β = volumetric coefficient of expansion of hydrogen = 
f
1
T
, K
-1
 
 µ = viscosity of hydrogen, 
kg
m s⋅
 
We can find the properties of hydrogen in Appendix A.3 of Geankoplis at the film temperature given 
by: 
 1 2f
T T ________ C ________ C
T
2 2
+ ° + °
= = 
 fT ________ C ___________ K= ° = 
The properties of hydrogen at this temperature (shown below) can be substituted into the equation 
for Grashof number to get: 
3
kg
0.068
m
ρ = 
6 kg9.92 10
m s
−µ = ×
⋅
 
3 11 2.80 10 K
356.74 K
− −β = = × 
 
( ) ( )( )3 13 2
Gr 2
6
kg m
0.001 m 0.068 __________ ___________ K ________ C ________ C
m s
N
kg
9.92 10
m s
−
−
  
° − °  
  =
 
× 
⋅ 
 
GrN ________= 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 35 Student View 
Jason M. Keith 
In Appendix A.3 we can also find the value of Prandtl number for hydrogen to be: 
 PrN ________= 
The product of Grashof and Prandtl numbers can now be obtained as follows: 
 ( )Pr GrN N ________ ________ ________= = 
Looking at the correlations in Section 4.7 of Geankoplis, we find that the Nusselt number 
corresponding to this value of Pr GrN N is 1. Thus, we can solve for the heat transfer coefficient from 
the definition of Nusselt number to yield: 
 Nu
W
________ ________
N k m K
h
H 0.001 m
 
 
⋅ = = 
 
2
W
h ________
m K
=
⋅
 
Now we can enter the corresponding quantities into the equation for the heat transfer rate to obtain: 
 ( )( )22
W
q ________ ______________ m ________ C ________ C
m K
= ° − °
⋅
� 
 q ________ W=� 
 
 
 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 36 Student View 
Jason M. Keith 
Example 4.8-2: Condensation in Bipolar Plate Channels in Fuel Cells 
The reaction occurring in a proton-exchange membrane fuel cell is producing water at a rate of 
5 kg7.65 10
s
−× through each channel on the bipolar plate in the cathode side. This amount of water is 
produced as steam at a temperature of 77°C. Determine if the water is condensing in a single channel 
if the partial pressure of water is 37.91 kPa. The dimensions of the channel are shown in the 
following figure: 
 
 
 
 
 
The Nusselt number for a square tube with constant temperature at the boundaries is 2.98. Frano 
Barbir in Section 6.5.2 of the book PEM Fuel Cells - Theory and Practice published by Prentice 
Hall estimates the average temperature in the bipolar plate channels to be 64.1°C. 
Strategy 
In this problem, condensation will occur if the amount of heat removed by convection is higher than 
the latent heat of condensation of steam. 
Solution 
The amount of heat lost by the steam can be calculated from the equation for convective heat 
transfer as shown below: 
 convq hA T= ∆� 
where the change in temperature ∆T is given by: 
 wT T T∆ = − 
in this equation: 
 T = Temperature of steam 
 Tw = Temperature of the walls of the square channel 
t = 1 mm 
H = 1 mm 
L = 249.4 mm 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 37 Student View 
Jason M. Keith 
The convective heat transfer coefficient is obtained from the definition of Nusselt number: 
 Nu
hL
N
k
= 
Solving for the convective heat transfer coefficient h and substituting the values of the length L and 
the thermal conductivity of steam we get: 
 
( )
Nu
W
________ 2.98
kN m Kh
L ________ m
⋅= = 
 
2
W
h ________
m K
=
⋅
 
The thermal conductivity value used in this equation was obtained from Table 2-305 of Perry's 
Chemical Engineers' Handbook, 8th Edition. 
Substituting this value and the heat transfer area into the equation for convq� yields: 
 ( )( ) 2A 4 ________ m ________ m ______________ m= = 
 ( )( )2conv 2
W
q ________ ______________ m 77 C 64.1 C
m K
= ° − °
⋅
� 
 convq_____________ W=� 
To determine if the steam is condensing in the fuel cell, we need to compare this amount of heat to 
the latent heat of vaporization, defined as follows: 
 vap fgq mh=� � 
The heat of vaporization hfg can be obtained from Table A.2-9 of Geankoplis. When condensation is 
occurring, the saturated vapor pressure is equal to the partial pressure. Hence, we will look for the 
enthalpy of vaporization at a pressure of 37.91 kPa and substitute into the equation for vapq� . 
 5vap
kg J J
q 7.65 10 _____________ _____________
s kg kg
−  = × − 
 
� 
 vapq _______ W=� 
Conclusion: 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 38 Student View 
Jason M. Keith 
Example 4.9-1: Temperature Correction Factor for a Heat Exchanger 
The synthesis gas produced in a steam-methane reforming process for hydrogen production is being 
cooled in a heat exchanger before entering the water-gas shift reaction chamber at 846°C to 600°C. 
The cooling medium is air entering a heat exchanger at 255.3°C and leaving at 381.3°C. The syngas 
is flowing at a rate of 
kg
1322
hr
and has a heat capacity of 
J
2584
kg K⋅
. The air flow is being 
distributed through 100 commercial steel pipes with a nominal diameter of 1" Schedule number 40 
and a length of 1.2 m. Calculate the mean temperature difference in the exchanger and the overall 
heat transfer coefficent Uo for the 4 heat exchanger configurations shown in Section 4.9B of 
Geankoplis. 
Strategy 
The heat transfer rate can be obtained using the definition of sensible heat. This can be used to 
determine the overall heat transfer coefficient Uo. The mean temperature difference will depend on 
the type heat exchanger selected. 
Solution 
The heat transfer coefficient can be obtained from the equation shown below: 
 o o mq U A T= ∆� 
Solving for Uo, we get: 
 oU ___________= 
where: 
 Ao = total outer surface area of the pipes distributing the air flow, m
2
 
 ∆Tm = mean temperature difference = T lmF T∆ , K or °C 
The mean temperature difference can be obtained by multiplying a factor FT depending on the heat 
exchanger type by the log mean temperature difference, defined as: 
 
( ) ( )
( )
( )
hi co ho ci
lm
hi co
ho ci
T T T T
T
T T
ln
T T
− − −
∆ =
−
−
 
where: 
 Thi = temperature of the synthesis gas entering the heat exchanger, K or °C 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 39 Student View 
Jason M. Keith 
 Tho = temperature of the synthesis gas exiting the heat exchanger, K or °C 
 Tci = temperature of the air entering the heat exchanger, K or °C 
 Tco = temperature of the air exiting the heat exchanger, K or °C 
The amount of heat removed from the synthesis gas can be calculated as follows: 
 ( )p hi hoq mC T T= −� � 
Substituting the syngas flow rate, specific heat and the inlet and outlet temperatures yields: 
 ( )
kg 1 hr J
q 1322 _______ ______ C ______ C
hr 3600 s kg K
  
= ° − °  
⋅  
� 
 q ____________ W=� 
We can now determine the log mean temperature difference as shown in the following steps: 
 
( ) ( )
( )
( )
lm
______ C ______ C ______ C ______ C
T
______ C ______ C
ln
______ C ______ C
° − ° − ° − °
∆ =
° − °
° − °
 
 lmT ________ C∆ = ° 
To calculate the heat transfer area of the tubes, we need to look for the outer diameter of 1" 
commercial steel pipes in Appendix A.5 of Geankoplis. Hence, the area can be determined as shown 
below: 
 ( )( ) ( ) 2oA DLn _________ m _____ m _____ _________ m= π = π = 
In this equation, n is the number of tubes. 
Now we will proceed to calculate the mean temperature difference for the 4 different types of heat 
exchanger. For all heat exchanger configurations we need to calculate the parameters Y and Z, 
defined as: 
 co ci
hi ci
T T ______ C 255.3 C
Y
T T ______ C 255.3 C
− ° − °
= =
− ° − °
 hi ho
co ci
T T 846 C ______ C
Z
T T ______ C 255.3 C
− ° − °
= =
− ° − °
 
 Y ______= Z ______= 
 
 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 40 Student View 
Jason M. Keith 
For these values of Y and Z, we find the factors FT in Figures 4.9-4 and 4.9-5 and use them to 
determine the mean temperature differences and overall heat transfer coefficients, given by: 
1-2 exchanger 
FT = ______ 
( )m T lmT F T ______ _______ K∆ = ∆ = 
mT 391.68 K∆ = 
 
( )o 2o m
q _________ W
U
A T ________ m 391.68 K
= =
∆
�
 
o 2
W
U ________
m K
=
⋅
 
Cross-flow exchanger with shell fluid mixed 
FT = ______ 
( )m T lmT F T _____ _________ K∆ = ∆ = 
mT _______ K∆ = 
( )o 2o m
q _________ W
U
A T ______ m _________ K
= =
∆
�
 
o 2
W
U ________
m K
=
⋅
 
2-4 exchanger 
FT = ___ 
( )m T lmT F T ____ _________ K∆ = ∆ = 
mT _________ K∆ = 
 
( )o 2o m
q _________ W
U
A T 12.59m ________ K
= =
∆
�
 
o 2
W
U ________
m K
=
⋅
 
Cross-flow exchanger with fluids unmixed 
FT = _____ 
( )m T lmT F T ______ _________ K∆ = ∆ = 
mT _________ K∆ = 
( )o 2o m
q _________ W
U
A T _________ m _________ K
= =
∆
�
 
o 2
W
U 47.10
m K
=
⋅
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 41 Student View 
Jason M. Keith 
Example 4.9-2: Temperature Correction Factor for a Heat Exchanger 
Synthesis gas is being produced in a steam-methane reforming process at a rate of 
kg
1322
hr
and a 
temperature of 846°C. This gas is to be cooled to a temperature of 473.86°C before entering a water-
gas shift reactor to produce additional hydrogen for fuel cells. The cooling medium is air entering at 
255.3°C, a flow rate of 
kg
6307
hr
 and a heat capacity of 
J
1058
kg K⋅
. The overall heat transfer 
coefficient is 
2
W
90
m K⋅
for a heat transfer area of 8.92 m
2
. Determine the type of flow at which the 
heat exchanger is operating and the heat transfer rate if the effectiveness is 0.45. The composition of 
the syngas is shown in the following table: 
 wt. % 
CO 20.61 
H2 7.18 
H2O 72.21 
 
Strategy 
The charts showing the effectiveness of heat exchangers operating at countercurrent flow and 
parallel flow can be used to determine the type of operation. 
Solution 
The heat transfer rate in a heat exchanger can be calculated as a function of the effectiveness ε as 
shown in the following equation: 
 ( )min Hi Ciq C T T= ε −� 
To determine the value of Cmin, we need to calculate the values of CH and CC. These parameters 
depend on the flow rate and the heat capacities of the fluids in the heat exchanger. Thus, 
 H syngas p,syngasC m C= � 
 C air p,airC m C= � 
We can see that the heat capacity of syngas is not given in the problem statement. However, we can 
use Figure A.3-3 of Geankoplis to determine the heat capacity of each individual component. The 
heat capacity of the syngas can then be determined by multiplying the mass fraction of each 
component by its corresponding heat capacity at the film temperature given by: 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 42 Student View 
Jason M. Keith 
 Hi Hof
T T 846 C 473.86 C
T
2 2
+ ° + °
= = 
 fT ________ C= ° 
 Thus, at this temperature value: 
 p,CO
J
C __________
kg K
=
⋅
 
 p,H2
J
C 14853.2
kg K
=
⋅
 
 p,H O2
J
C __________
kg K
=
⋅
 
With these individual heat capacities we can determine the heat capacity of syngas as follows: 
p,syngas CO p,CO H p,H H O p,H O2 2 2 2
C x C x C x C= + + 
 
p,syngas
J J J
C 0.2061 _________ _________ 14853.2 0.7221 _________
kg K kg K kg K
    
= + +     
⋅ ⋅ ⋅     
 
p,syngas
J
C __________
kg K
=
⋅
 
Now we can substitute the heat capacities and flow rates of air and syngas to obtain CH and CC. 
Thus, 
H
kg 1 hr J
C 1322 __________
hr 3600 s kg K
  
=   
⋅  
 
H
W
C __________
K
= 
C
kg 1 hr J
C 6307 __________
hr 3600 s kg K
  
=   
⋅  
 
C
W
C __________
K
= 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 43 Student View 
Jason M. Keith 
We can see from the values of CC and CH that the smaller of these values is CH, thus Cmin = CH. 
Substituting this value into the equation for the heat transfer rate, we get: 
 ( )
W
q __________ __________ 846 C ________ C
K
 
= ° − ° 
 
� 
 q __________ kW=� 
Figure 4.9-7 of Geankoplis is showing the effectiveness of a heat exchanger operating at both 
countercurrent flow and parallel flow as function of the number of transfer units and the ratio min
max
C
C
. 
If we calculate these two values, we can look in this Figure which type of heat exchanger will yield a 
value of ________ε = . Thus, 
 min
max
W
__________
C K 0.59
WC
__________
K
= = 
 
( )22
min
W
90 8.92m
UA m KNTU
WC
__________
K
⋅= = 
 NTU ________= 
Conclusion: 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 44 Student View 
Jason M. Keith 
Example 4.11-1: Radiation in Cylindrical Solid - Oxide Fuel Cell 
The following figure is a schematic of a cylindrical solid-oxide fuel cell. Jiang et al. [1] developed a 
thermoelectrical model to estimate the temperature at different parts of this type of fuel cell. The 
temperatures are estimated to be 1125 K for the air tube and 1200 K for the solid part (membrane 
electrode assembly). Determine the heat flux due to radiation. 
Xue et al. [2] estimated the average emissivity of the membrane electrode assembly to be 0.33. The 
air is being fed to the system through a commercial steel pipe. 
 
 
 
 
 
 
 
 
Strategy 
The heat transferred due to radiation can be estimated using the radiation equation for gray bodies 
given in Section 4.11 of Geankoplis. 
Solution 
The following equation is the definition of heat flux due to radiation: 
 ( )4 41 2
1 2
1
q" T T
1 1 1
 
 = σ −
 + − ε ε 
� 
where: 
 q"� = Heat flux due to radiation, 
2
W
m
 
1. Jiang, W., Fang, R., Dougal, R. A., Khan, J. A., Journal of Energy Resources Technology, 130, 2008. 
2. Xue, X., Tang, J., Sammes, N., Du, Y., Journal of Power Sources, 142, 211 − 222 (2005) 
Air Feed 
Fuel 
Fuel electrode 
Air electrode 
Electrolyte membrane 
Interconnect 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 45 Student View 
Jason M. Keith 
 σ = Stefan – Boltzmann constant = 8
2 4
W
5.676 10
m K
−×
⋅
 
 Ti = Temperature of surface i, K 
 εi = Emissivity of surface i, K 
We can substitute the given quantities into the equation for heat flux to solve this problem. Hence, 
 ( ) ( )
4 4
2 4
W 1
q" _______________ ______ K 1125 K
1 1m K 1
_____ _____
 
  = −
   ⋅ + − 
 
� 
 
2
W
q" __________
m
=� 
The emissivity of steel was obtained from Table 5 − 4 of Perry's Chemical Engineers' Handbook, 8th 
Edition. 
 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 46 Student View 
Jason M. Keith 
4.15-1 Cooling Channels in Fuel Cell Bipolar Plates 
The following figures show the top and isometric views of a fuel cell bipolar plate with 10 cooling 
channels through which air is circulating with a heat transfer coefficient of 
2
W
400
m K⋅
and a 
temperature of 10°C. The outer walls of the bipolar plates are held at a temperature of 60°C. 
Determine the steady – state heat loss in one bipolar plate using finite difference numerical methods, 
with grids 1 mm x 1 mm. The bipolar plates are made of 304 stainless steel. 
 
 
 
 
 
Strategy 
To determine the amount of heat removed we need to determine the temperatures at the different 
nodes, using the Equations in Section 4.15B of Geankoplis. 
Solution 
Since the area surrounding the channel is symmetrical, we can calculate it for one channel and 
multiply it by the number of times this area is repeated in the whole bipolar plate. If we zoom into 
the first channel from the left edge of the bipolar plate: 
 
 
 
 
The shaded areas in this figure indicate the sets of nodes that will be repeated along the bipolar plate. 
Set of Nodes at the Edges of the Bipolar Plate 
We will start by using the finite difference method on the edges of the bipolar plate. For a grid of 1 
mm by 1 mm, the following nodes will be used: 
 
 
 
 
0.15 m 
 
 
6 mm 
108 mm 
8 mm 8 mm 
2 mm x 2 mm 
channels 
 
T2,1 T2,2 T2,3 T2,4 T2,5 T2,6 T2,7 T2,8 T2,9 T2,10 
T3,1 T3,2 T3,3 T3,4 T3,5 T3,6 T3,7 T3,8 T3,9 T3,10 
T4,1 T4,2 T4,3 T4,4 T4,5 T4,6 T4,7 T4,8 T4,9 
T1,1 T1,2 T1,3 T1,4 T1,5 T1,6 T1,7 T1,8 T1,9 T1,10 
2
W
h 400
m K
=
⋅
 
T 10 C∞ = ° 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 47 Student View 
Jason M. Keith 
The outer nodes are kept at a temperature of 60°C. Thus: 
 T1,1 = T1,2 = T1,3 = T1,4 = T1,5 = T1,6 = T1,7 = T1,8 = T1,9 = T1,10 = T2,1 = T3,1 = T4,1 = _____°C 
For the first calculation, we will assume the following temperature values (in °C) for the rest of the 
nodes: 
 
 
 
 
 
 
 
 
We can obtain the first temperature estimation for the interior nodes using Equation 4.15-11 from 
Geankoplis: 
 
 n,m n 1,m n 1,m n,m 1 n,m 1 n,mq T T T T 4T− + − += + + + − 
 
This equation is applicable to the nodes highlighted below. 
 
 
 
 
To start the calculations, we will select node T2,2. We can apply this equation to get: 
 2,2 1,2 2,1 2,2q T ______ T ______ 4T= + + + − 
 ( )2,2q ____ 50 ____ 55 4 ____ ____= + + + − = 
T1,1 T1,2 T1,3 T1,4 T1,5 T1,6 T1,7 T1,8 T1,9 T1,10 
T2,1 T2,2 T2,3 T2,4 T2,5 T2,6 T2,7 T2,8 T2,9 T2,10 
T3,1 T3,2 T3,3 T3,4 T3,5 T3,6 T3,7 T3,8 T3,9 T3,10 
T4,1 T4,2 T4,3 T4,4 T4,5 T4,6 T4,7 T4,8 T4,9 
60 60 60 60 60 60 60 60 60 60 
60 55 55 55 55 55 55 55 55 35 
60 50 50 50 50 50 50 50 50 10 
60 45 40 35 30 25 20 15 10 
T1,1 T1,2 T1,3 T1,4 T1,5 T1,6 T1,7 T1,8 T1,9 T1,10 
T2,1 T2,2 T2,3 T2,4 T2,5 T2,6 T2,7 T2,8 T2,9 T2,10 
T3,1 T3,2 T3,3 T3,4 T3,5 T3,6 T3,7 T3,8 T3,9 T3,10 
T4,1 T4,2 T4,3 T4,4 T4,5 T4,6 T4,7 T4,8 T4,9 
Tn,m Tn,m – 1 
Tn + 1,m 
Tn – 1,m 
Tn,m + 1 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 48 Student View 
Jason M. Keith 
Since the heat q2,2 is not equal to zero, the value of T2,2 we assumed is not the temperature at steady 
state. Setting the value of q2,2 to zero we calculate a new value of T2,2 as shown below: 
1,2 2,1
2,2
T ____ T ____
T
4
+ + +
= 
2,2
____ C 50 C ____ C ____ C
T _________ C
4
° + ° + ° + °
= = ° 
This new value of T2,2 will be used to calculate the temperatures at other nodes. Thus, for q2,3: 
 2,3 3,3 2,4q ____ T ____ T _______ 60 50 ________ 55 _______ _______= + + + − = + + + − = 
Setting q3,2 = 0 and solving for T3,2 yields: 
 2,3
60 50 _______ 55
T _______ C
4
+ + +
= = ° 
We can repeat the same procedure for all the interior nodes. The first iteration will yield the 
following temperature values: 
 
 
 
 
 
Note that we have not done any calculations for the edge nodes. This is because we need additional 
equations for these nodes, describedin the following sections. 
Section 4.15B-3 of Geankoplis and Section 4.5-3 of Incropera and DeWitt [3] gives the following 
equations for different boundary conditions. 
For nodes T2,10, T4,2, T4,3, T4,4, T4,5, T4,6, T4,7, and T4,8, an equation with an adiabatic boundary is 
needed [3]. 
 
 
 
3. Incropera, F. P., DeWitt, D. P., Fundamentals of Heat and Mass Transfer, Fourth Edition, John Wiley & Sons, New 
York (1996). 
 
10 
______ 
60 60 60 60 
60 ______ ______ 
60 
60 45 40 35 
60 60 60 
______ ______ ______ 
30 25 20 
60 60 60 
______ ______ 55 
10 
15 
______ ______ ______ ______ ______ ______ ______ 50 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 49 Student View 
Jason M. Keith 
This equation is applied for nodes with heat conduction from three adjacent nodes with an adjoining 
adiabat, as shown in the following figures: 
 
n,m n,m 1 n 1,m n 1,m n,mq 2T T T 4T− − += + + − 
 
 
 n,m n,m 1 n,m 1q 2 _______ T T 4 ______− += + + − 
 
Setting qn,m = 0 in these equations, the temperatures are given by: 
 
n,m 1 n 1,m
n,m
2T ______ T
T
___
− ++ +
=
−
 
 
n,m 1 n,m 1
n,m
2 ______ T T
T
____
− ++ +
= 
Applying this equation to node T4,2 (with q4,2 = 0), we have: 
 
 
( )4,1
4,2
2 _____ T _____ 2 ______ C 60 C 40 C
T _____ C
___ ___
+ + ° + ° + °
= = = ° 
 
 
 
 
In a similar way, we can find the first estimate of the temperatures at rest of the nodes that follow 
this equation, highlighted in the grid below: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
T1,1 T1,2 T1,3 T1,4 T1,5 T1,6 T1,7 T1,8 T1,9 T1,10 
T2,1 T2,2 T2,3 T2,4 T2,5 T2,6 T2,7 T2,8 T2,9 T2,10 
T3,1 T3,2 T3,3 T3,4 T3,5 T3,6 T3,7 T3,8 T3,9 T3,10 
T4,1 T4,2 T4,3 T4,4 T4,5 T4,6 T4,7 T4,8 T4,9 
Tn,m 
Tn – 1,m 
Tn + 1,m 
Tn,m – 1 
Adiabat 
Tn,m 
Tn – 1,m 
Tn,m + 1 Tn,m – 1 
Adiabat 
Tn,m 
Tn – 1,m 
Tn + 1,m 
Tn,m – 1 
Adiabat 
Tn,m 
Tn – 1,m 
Tn,m + 1 Tn,m – 1 
Adiabat 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 50 Student View 
Jason M. Keith 
 
 
 
 
 
To finish the first temperature estimation, we need to determine 3 more nodes: T3,9, T3,10 and T4,9. In 
nodes T4,9 and T3,10, heat transfer is occurring at an insulated boundary with convection from internal 
flow and conduction from the two adjacent nodes. The following equation (at steady state) can be 
used for this case, also given in page 191 of Incropera [3]. The thermal conductivity of stainless steel 
304 was obtained from Appendix A.3 of Geankoplis. 
 
n,m 1 n 1,m
n,m
h x
T T T
k
T
h x
2
k
− − ∞
∆ 
+ +  
 =
∆ 
+ 
 
 
 
Substituting the corresponding temperatures into this equation yields: 
 
( )
( )
( )
2
3,9
3,10
2
W
400 ______ m
m K____ C 45 C 10 C
h x W
T _____ T ______
k m KT ______ C
h x W
2 400 ______ m
k m K2
W
______
m K
∞
 
 ⋅° + ° + ° ∆ 
+ +   
   ⋅ = = = °
∆   
+   ⋅  + 
 
 ⋅ 
 
Now for T4,9 we have: 
( )
( )
( )
2
4,8 3,9
4,9
2
W
400 ______ m
m K31.3 C ____ C 10 C
h x W
T T T ______
k m KT ______ C
h x W
2 400 ______ m
k m K2
W
______
m K
∞
 
 ⋅° + ° + ° ∆ 
+ +   
   ⋅ = = = °
∆   
+   ⋅  + 
 
 ⋅ 
 
10 
______ 
60 60 60 60 
60 ______ ______ 
60 
60 
60 60 60 
______ ______ ______ 
60 60 60 
______ ______ ______ 
10 ______ ______ ______ ______ ______ ______ ______ 50 
______ ______ ______ ______ ______ ______ ______ 
Tn,m 
Tn – 1,m 
Tn,m – 1 
T∞, h 
Adiabat 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 51 Student View 
Jason M. Keith 
For node T3,9, we can use Equation 4.15-19 of Geankoplis, which corresponds to an interior corner 
with convection at the boundary: 
 
( )n 1,m n,m 1 n,m 1 n 1,m
n,m
1 h x
T T T T T
2 k
T
h x
3
k
+ + − − ∞
∆ 
+ + + +  
 =
∆ 
+ 
 
 
 
( )4,9 3,8
3,9
1 h x
T ______ T ______ T
2 k
T
h x
3
k
∞
∆ 
+ + + +  
 =
∆ 
+ 
 
 
 
( )
( )
( )
( )
2
3,9
2
W
400 ______ m
1 m K______ C ______ C 40.56 C 55 C 10 C
W2
______
m KT
W
400 ______ m
m K3
W
______
m K
 
 ⋅° + ° + ° + ° + ° 
 
 ⋅ =
 
 ⋅+ 
 
 ⋅ 
 
 3,9T ______ C= ° 
The following grid shows the temperatures obtained after the first iteration. The highlighted 
temperatures are the values we just calculated. 
 
 
 
 
 
After completing the first calculation across the grid, we can start a new approximation using the 
new temperature values. Hence, starting with q2,2 and T2,2, we have: 
 2,2 1,2 3,2 2,1 2,3 2,2q T T T T 4T= + + + − 
 ( )2,2q 60 52.81 60 ______ 4 ______ ______= + + + − = 
 
______ 
______ 
60 60 60 
60 ______ ______ 
60 
60 
60 60 60 
______ ______ ______ 
60 60 60 
______ ______ ______ 
______ ______ ______ ______ ______ ______ ______ ______ ______ 
______ ______ ______ ______ ______ ______ ______ 
Tn,m Tn,m – 1 
Tn – 1,m 
T∞, h 
Tn,m+1 
Tn + 1,m 
60 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 52 Student View 
Jason M. Keith 
Setting the value of q2,2 to zero we calculate a new value of T2,2 as follows: 
2,2
60 _______ 60 _______
T _______ C
4
+ + +
= = ° 
The same procedure used for the first approximation is repeated until the assumed and new 
temperatures are similar. In this case we will select a tolerance of 0.01. 
This numerical problem can also be solved using computer software such as Excel or Matlab. The 
final temperature values in the set of nodes for the edges of the bipolar plate are shown below: 
 
 
 
 
 
 
To calculate the total heat lost by the bipolar plate we use Fourier’s Law of Heat Conduction for the 
interior and exterior nodes. 
 ( )
T T
q kA k xL kL T
x x
∆ ∆
= = ∆ = ∆
∆ ∆
 
This amount must be multiplied by 4, since this set of nodes is repeated 4 times in the fuel cell 
bipolar plate (4 external corners), as shown in the shaded areas below: 
 
 
 
 
 
 
 
 
 
______ 
 59.98 
60 60 60 
60 ______ ______ 
60 
60 
60 60 60 
______ 59.88 ______ 
60 60 60 
______ 59.55 ______ 
 58.88 ______ ______ 59.93 ______ ______ 59.65 ______ ______ 
______ 59.98 ______ 59.86 ______ ______ 59.30 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 53 Student View 
Jason M. Keith 
The total heat conducted is the sum of the heat equations applied for all the interior temperature 
differences. The following figure illustrates the nodes used for the heat loss calculation. The nodes 
adjacent to the shaded squares were used for determining qinterior, with a direction for heat flow 
indicated by the arrows. The heat flux for nodes with an adjoining adiabat has to be multiplied by 
1
2
because of symmetry. 
 
 
 
 
Hence, for the interior nodes we have: 
 ( ) ( ) ( ) ( )interior,corner 3,10 2,10 4,9 4,8 q 4kL 0.5 T T 0.5 T T _____ _____ _____ _____ = − + − + − + −  
 
( ) ( ) ( )
( ) ( )
interior,corner
W
q 4 16.3 _____ m 0.5 _____ 59.50 0.5 58.78 ______
m K
 ______ 59.55 59.02 ______ C
 
= − + −  ⋅ 
+ − + − °
 
 interior,cornerq _______ W= 
To determine qexterior, we substituted the temperature differences in the nodes adjacent to the shaded 
squares in the following figure. The direction of heat flow is indicated by the arrows , with a The 
heat flux for nodes withan adjoining adiabat has to be multiplied by 
1
2
because of symmetry. 
 
 
 
 
 
 
 
 
 
Symmetry adiabat 
Symmetry adiabat 
T2,10 T2,8 T2,7 T2,6 T2,5 T2,4 T2,3 T2,2 T2,9 
T2,1 
T1,10 T1,8 T1,7 T1,6 T1,5 T1,4 T1,3 T1,2 T1,9 T1,1 
T3,1 
T4,1 
T3,2 
T4,2 
Symmetry adiabat 
Symmetry adiabat 
T4,9 T4,8 
T3,10 T3,9 
T2,10 
T3,8 
T2,9 T2,8 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 54 Student View 
Jason M. Keith 
Applying Fourier’s law to the exterior nodes, the heat transfer rate will be given by: 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
exterior,corner 1,10 4,2 1,9 2,8
1,7 2,6 1,5 2,4 1,4 2,3 1,3
2,2 1,2 2,2 2,1 3,2 3,1
 q 4kL 0.5 _____ T 0.5 T _____ _____ T T _____
 _____ T T _____ _____ T T T T T
 T T T T T T
= − + − + − + −
+ − + − + − + − + −
+ − + − + − 
 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
exterior,corner
W
 q 4 16.3 ______ m 0.5 ______ 60 0.5 ______ 60 ______ 60
m K
 ______ 60 ______ 60 ______ 60 ______ 60 ______ 60
 ______ 60 ______ 60 ______ 60 ______ 60 C
 
= − + − + −  ⋅ 
+ − + − + − + − + −
+ − + − + − + − °
 
 exterior,corner q ___________ W= 
Finding the same value as qinterior, corner proves that this system is at steady – state. The heat transfer 
rate for this set of nodes will be obtained from the average between the heat transfer for the interior 
and exterior nodes. 
 corners q __________ W= 
Set of nodes between cooling channels 
We need to establish a different nodal network for the spaces between cooling channels. The set of 
nodes for a 1 mm x 1 mm grid is shown below: 
 
 
 
 
 
The same equations used for calculating the temperatures at the exterior corners of the bipolar plates 
will be used again for this set of nodes. The only exception is for node T4,16, which represents the 
case for heat conduction with two adjoining adiabats. Thus, for this node (setting q4,16 = 0) we have: 
 
n,m 1 n 1,m
n,m
T T
T
2
− −+= 
4,15
4,16
______ T
T
2
+
= 
T2,11 T2,12 T2,13 T2,14 T2,15 
T3,11 T3,12 T3,13 T3,14 T3,15 
T1,11 T1,12 T1,13 T1,14 T1,15 
T4,12 T4,13 T4,14 T4,15 
T1,16 
T2,16 
T3,16 
T4,16
Tn,m 
Tn – 1,m 
Tn,m – 1 
Adiabats 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 55 Student View 
Jason M. Keith 
Following the same procedure described for the first set of nodes in this problem (for exterior 
corners), or using computer software, we can find the temperatures for the interior nodes to be: 
 
 
 
 
Now we can proceed to calculate the heat transfer rate using Fourier’s law for the temperature 
difference at the interior and exterior nodes. If we look at the top view of the bipolar plates, we can 
see that this set of nodes is repeated 4 times in the space between two channels as illustrated below. 
 
 
 
 
Since there are 10 channels, there will be 9 spaces between the cooling channels (see figure below). 
Therefore the heat transfer rate in a single set of nodes must be multiplied by 36 to obtain the heat 
transfer in the whole bipolar plate. 
 
 
 
In a similar way as we did for the set of nodes for the corners of the bipolar plate, we can determine 
the heat flow for the set of nodes including the spaces between cooling channels. Thus, 
( ) ( ) ( ) ( )interior,middle 3,11 2,11 3,12 3,13q 36kL 0.5 T T 0.5 _____ _____ _____ _____ T T = − + − + − + −  
( ) ( ) ( )
( ) ( )
interior,middle
W
q 36 16.3 _______ m 0.5 _______ 59.49 0.5 _______ 59.26
m K
 58.99 _______ 58.99 _______ C
 
= − + −  ⋅ 
+ − + − °
 
interior,middleq __________ W= 
Similarly for exterior nodes, we have: 
T1,11 T1,12 T1,13 T1,14 T1,15 T1,16 
T2,11 T2,12 T2,13 T2,14 T2,15 T2,16 
T3,11 T3,12 T3,13 T3,14 T3,15 T3,16 
T4,12 T4,13 T4,14 T4,15 T4,16 
 
 
60 60 60 60 60 60 
 59.49 
______ 
______ 
 58.99 
______ 
______ 
 59.78 
______ 
______ 
 59.71 
______ 
______ 
______ 59.26 ______ ______ 59.71 
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Principles of Steady-State Heat Transfer 
Daniel López Gaxiola 56 Student View 
Jason M. Keith 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( )
exterior,middle 2,12 1,12
2,14 1,14
W
q 36 16.3 _______ m 0.5 _____ _____ 0.5 _____ _____ T T
m K
 _____ _____ T T _____ _____ _____ _____
 _____ _____
  = − + − + −  ⋅ 
+ − + − + − + −
+ − 
 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( )
exterior,middle
W
q 36 16.3 0.15 m 0.5 59.49 60 0.5 _______ 59.71 _______ 60
m K
 _______ 60 59.78 60 _______ 60 _______ _______
 59.71 _______ C
 
= − + − + −  ⋅ 
+ − + − + − + −
+ − °
 
exterior,middleq ________ W= 
 
Taking the average of interior,middleq and exterior,middleq we get: 
 middle
________ W ________ W
q
2
−
= 
 middleq ________ W= 
To obtain the overall heat transfer in the bipolar plate we need to add cornersq and middleq : 
 q ________ W ________ W= − 
 q _________ W= 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 1 Student View 
Jason M. Keith 
Chapter 5 
Principles of Unsteady - State Heat Transfer 
In this chapter, we will study chemical processes where heat transfer is taking place due to a 
temperature difference within the system which is changing with time. The following problem 
modules illustrate different examples where unsteady – state heat transfer processes are occurring in 
fuel cell vehicles and in the processes for producing fuel for fuel cells. 
5.2-1 Cooling of a Cylindrical Solid – Oxide Fuel Cell 
5.2-2 Total Amount of Heat in Cooling of a Solid – Oxide Fuel Cell 
5.3-2 Heat Conduction in a Fuel Cell Stack 
5.3-3 Transient Heat Conduction in a Cylindrical Solid – Oxide Fuel Cell 
5.3-4 Two – Dimensional Conduction in a Cylindrical Solid – Oxide Fuel Cell 
5.4-1 Unsteady – State Conduction and the Schmidt Numerical Method 
5.4-3 Unsteady – State Conduction with Convective Boundary Condition 
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Principles of Unsteady-State Heat Transfer 
Daniel López Gaxiola 2 Student View 
Jason M. Keith 
Example 5.2-1: Cooling of a Cylindrical Solid – Oxide Fuel Cell 
A cylindrical solid – oxide fuel cell has an inner radius of 3.1 mm, outer radius of 3.9 mm and a 
length of 0.2 m. The fuel cell initially at a uniform temperature of 873.15 K enters a medium whose 
temperature is 60°C. Determine how much time in minutes is required for the fuel cell to be cooled 
down to a temperature of 340.23 K if the convective coefficient of the medium is 
2
W
12
m K⋅
. 
The properties of the membrane electrode assembly for this type of fuel cell are estimated by Xue et 
al. [1] to be: 
 
3
kg
6337.3
m
ρ = p
J
C 594.3
kg K
=
⋅
 
W
k 2.53
m K
=
⋅
 
Strategy 
This problem can be solved using the simplified equations for systems with negligible internal 
resistance. 
Solution 
The equation for the dimensionless temperature as a function of time is given by: 
 
p 1
0
h
t
C xT T
e
T T
 
 
 
∞  
∞
−
ρ−
=
−
 
In this equation: 
 T∞ = Temperature of the medium 
 0T = Initial temperature of the fuel cell 
 x1 = Characteristic dimension of the body 
However, this equation is only applicable when the Biot number NBi is less than 0.1. Hence, we need 
to obtain this dimensionless number first, defined as: 
 BiN _________= 
For a cylindrical object, the characteristic length is calculated asfollows: 
 1
r
x
2
= 
1. Xue, X., Tang, J., Sammes, N., Du, Y., Journal of Power Sources, 142, 211 −222 (2005) 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 3 Student View 
Jason M. Keith 
In this case, since the fuel cell is a hollow cylinder, the characteristic dimension will be obtained as 
shown below: 
 
( )
o i
1
1 m
3.9 mm 3.1 mm
r r 1000 mm
x _____________ m
2 2
 
−  −  = = = 
Substituting this parameter into the definition of Biot number, we get: 
( )2
Bi
W
12 _____________ m
m K
N _____________
W
2.53
m K
 
 
⋅ = =
 
 
⋅ 
 
Since this value is less than 0.1, we can proceed to enter the corresponding values into the equation 
for the dimensionless temperature, to yield: 
 
( )
2
3
W
12
m K
kgJ
594.3 6337.3 ______________ m
kg K m
t
340.23 K _________ K
e
_________ K _________ K
 
 
 
 
          
⋅
⋅
−
−
=
−
 
Solving for the time, we have: 
 
( )1
1 340.23 K _________ K
t ln
_________ K _________ K_______________ s
−
 −
= −  
− 
 
_________
t _________ s
_________
 
=  
 
 
t _________ min= w
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Principles of Unsteady-State Heat Transfer 
Daniel López Gaxiola 4 Student View 
Jason M. Keith 
Example 5.2-2: Total Amount of Heat in Cooling of a Solid – Oxide Fuel Cell 
Determine the total amount of heat removed from the fuel cell in Problem 5.2-1 after 5 minutes. The 
characteristic length of the fuel cell is 44 10 m−× . 
Strategy 
The equation for the total amount of heat in J can be used to solve this problem. 
Solution 
The equation for the total amount of heat is given by: 
p 1
p 0
h
t
C x
Q C V(T T ) 1 e
 
 
 
 
∞
−
ρ
 
 = ρ − −
 
  
 
The volume of the cylindrical fuel cell is calculated as follows: 
 ( )V ______ ______ L= π − 
 [ ]V _______________ _______________ 0.2 m= π − 
 6 3V 3.52 10 m−= × 
Substituting the calculated volume and the rest of the values into the equation for the total amount of 
heat transferred, yields: 
( )
( )
2
4
3
3
3
W
12
m K ____ s
kgJ
594.3 6337.3 4 10 m
kg K mJ kg
Q 594.3 6337.3 __________ m (873.15 K 333.15 K) 1 e
kg K m
−
 
 
 
  
     
⋅−
×
⋅
 
 
   
= − −   
⋅     
 
  
Q ____________ J= 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 5 Student View 
Jason M. Keith 
Example 5.3-2: Heat Conduction in a Fuel Cell Stack 
A stack of 220 fuel cells is initially at a temperature of 353.15 K. This stack enters a room at a 
temperature of 266.48 K. Assuming all the sides but the front face of the stack are insulated, 
determine the temperature of the 60
th
 fuel cell after 90 minutes. The thickness of the bipolar plates in 
this fuel cell stack is 2 mm and it is much thicker than the membrane electrode assembly. The heat 
transfer coefficient of the air in this room is 
2
W
91
m K⋅
 
The properties of the bipolar plates in this fuel cell stack are given below [2]: 
 
3
kg
1632
m
ρ = p
J
C 1414
kg K
=
⋅
 
W
k 24.42
m K
=
⋅
 
 A schematic of this fuel cell stack is shown in the following figure: 
 
 
 
 
 
 
 
Strategy 
The temperature at a point inside the fuel cell stack can be obtained from the charts for unsteady – 
state heat conduction of a slab. 
Solution 
To use Figure 5.3-5 of Geankoplis to determine the unsteady – state heat conduction in a flat plate, 
we need to find the parameters X, m and n, defined as: 
 X ________= 
1
k
m
hx
= n __________= 
 
 
2. King, J.A., Lopez Gaxiola, D., Johnson, B.A., Keith, J.M., Journal of Composite Materials, 44 (7), 839 – 855 (2010). 
x 
x1 
ncell = 1 
T = 266.48 K 
The sides and back face 
of the fuel cell stack are 
insulated 
ncell = 220 
ncell = 60 
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Principles of Unsteady-State Heat Transfer 
Daniel López Gaxiola 6 Student View 
Jason M. Keith 
The characteristic length x1 in these equations corresponds to the distance from the front face to the 
center of the fuel cell stack, obtained by multiplying the number of fuel cells by the thickness of a 
bipolar plate, and dividing it by 2. Thus, 
 
( )
1
_______ m 220
x
2
= 
 
1
x __________ m= 
The parameter α is the diffusivity of the fuel cell stack defined as: 
 
k
______
α = 
Substituting the properties of the bipolar plates into this equation, we get: 
 
2
3
W
24.42
mm K _______________
skg J
_________ _________
m kg K
⋅α = =
 
 
⋅ 
 
To determine the value of n, we need to know the distance at which the 60
th
 fuel cell is located from 
the center of the stack. This distance can be calculated by multiplying the thickness of a single 
bipolar plate by (110-60) which is the number of fuel cells from the center. 
 ( )x 110 60 (_________ m) _________ m= − = 
Now we can calculate the three values required to use Figure 5.3-5, as shown in the following steps: 
 
( )
( )
2
2
m 60 s
_______________ 90 min
s 1 min
X _________
_________ m
   
   
  = = 
 
( )2
W
24.42
m Km 1.22
W
91 _________ m
m K
⋅= =
⋅
 
_________ m
n 0.45
_________ m
= = 
From Figure 5.3-5 we can read a value for a dimensionless temperature Y = _______. This value can 
be used to solve for the temperature in the 60
th
 fuel cell as follows: 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 7 Student View 
Jason M. Keith 
 1
1 0
T T
Y _________
T T
−
= =
−
 
where: 
 T1 = Temperature of the cooling medium 
 T0 = Initial temperature of the fuel cell stack. 
We can enter the corresponding temperatures and solve for the temperature T to get: 
 ( )1T T _________ _______ _______= − − 
 ( )T 266.48 K _________ 266.48 K _________ K= − − 
 T _________ K= 
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Principles of Unsteady-State Heat Transfer 
Daniel López Gaxiola 8 Student View 
Jason M. Keith 
Example 5.3-3: Transient Heat Conduction in a Cylindrical Solid – Oxide Fuel Cell 
A cylindrical solid – oxide fuel cell with a diameter of 3.9 mm and a length of 0.2 m is initially at a 
temperature of 1150 K. The fuel cell is shut down in a room where the air is at a temperature of 303 
K. Calculate the temperature at the center of the fuel cell after 5 minutes, assuming it is insulated on 
the flat ends. The heat transfer coefficient of the air is 
2
W
10
m K⋅
. The thermal conductivity and 
diffusivity of the fuel cell are 
W
2.53
m K⋅
and 
2
7 m
6.72 10
s
−× , respectively. 
Strategy 
The solution to this problem can be found by using the charts for unsteady – state heat conduction in 
a cylinder. 
Solution 
Since heat is only being transferred through the walls, the fuel cell can be considered as a long 
cylinder. The parameters n, m and X required to determine the dimensionless temperature Y from 
Figure 5.3-8 are calculated as shown in the following steps: 
At the center of the cylinder, x = 0. Thus, the value of n will also be equal to zero: 
1
x 0 m
n
x ____________ m
= = 
n 0= 
To calculate the value of X, we need to substitute the diffusivity, radius of the fuel cell and the time 
elapsed, as shown below: 
2
1
t
X
x
α
= 
( )
( )
2
2
m 60 s
_____________ ___ min
s 1 min
X _________
_____________ m
   
   
  = = 
Finally, we can determine the parameter m: 
1
k
m
hx
= 
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Supplemental Material for Transport Process and SeparationProcess Principles 
Daniel López Gaxiola 9 Student View 
Jason M. Keith 
 
( )32
W
_______
m Km ________
W
_______ 3.9 10 m
m K
−
⋅= =
×
⋅
 
For these values, the corresponding Y from Figure 5.3-8 will be equal to ______. Now we can solve 
for the temperature at the center of the fuel cell from the definition of the dimensionless temperature 
Y: 
 1
T T
Y ______
_________
−
= = 
 
1
T T __________________= − 
where: 
 T1 = Temperature of cooling medium 
 T0 = Initial temperature of the fuel cell 
Substituting the temperature values in this equation, we get: 
 ( )T 303 K ______ 303 K ________ K= − − 
 T _________ K= 
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Principles of Unsteady-State Heat Transfer 
Daniel López Gaxiola 10 Student View 
Jason M. Keith 
Example 5.3-4: Two – Dimensional Conduction in a Cylindrical Solid – Oxide Fuel 
Cell 
Determine the temperature at the center of the solid – oxide fuel cell from Example 5.3-3 now 
considering heat conduction also occurring through the ends of the cylinder. What does this result 
indicate? 
Strategy 
The procedure to solve this problem consists of using the charts for unsteady – state heat transfer in a 
cylinder for both radial and axial directions. 
Solution 
We need to calculate the required dimensionless quantities X, m and n for both radial and axial 
directions. For the radial direction, these will be given by: 
radial
1
x ______ m
n
x _____________ m
= = 
radial
n ____= 
radial
1
k
m
hx
= 
 
( )
radial
3
2
W
________
m Km _______
W
_____ 3.9 10 m
m K
−
⋅= =
×
⋅
 
radial 2
1
t
X
x
α
= 
( )
( )
2
radial 2
m 60 s
______________ 5 min
s 1 min
X ________
______________ m
   
   
  = = 
These values will yield a dimensionless temperature Yradial = ______ 
For heat conduction in the axial direction, we need to calculate the parameters n, m and X and locate 
them in Figure 5.3-6, applicable for two parallel planes. Thus, 
axial
1
y 0 m
n
y 0.1 m
= = 
axial
n 0= 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 11 Student View 
Jason M. Keith 
axial
1
k
m
hy
= 
 
( )
axial
2
W
______
m Km ______
W
______ ______ m
m K
⋅= =
⋅
 
axial 2
1
t
X
x
α
= 
( ) ( )
( )
2
axial 2
m
_____________ ______ min ______
s
X ______
______ m
 
 
 = = 
Locating these three parameters in Figure 5.3-6 gives a value of Yaxial = ___. 
Now that we have both Y values for both directions, we can obtain a Y for the overall heat transfer 
process as follows: 
 
axial radial
Y Y Y ________= = 
Now we can solve for the temperature at the center of the cylinder to get: 
 1
1 0
T T
Y __________
T T
−
= =
−
 
 
1
T T _________________= − 
 ( )T ______ K ______ ____________________= − 
 T _________ K= 
Conclusion: 
 
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Principles of Unsteady-State Heat Transfer 
Daniel López Gaxiola 12 Student View 
Jason M. Keith 
Example 5.4-1: Unsteady – State Conduction and the Schmidt Numerical Method 
A proton – exchange membrane fuel cell stack has a thickness of 0.3 m and is initially at a uniform 
temperature of 60°C. The front face of the fuel cell stack suddenly exposed to an environmental 
temperature of -6.67°C. The bulk thermal diffusivity of the fuel cell stack is 
2
6 m
8.69 10
s
−× . 
Assuming the convective resistance is negligible and that the back face of the stack is insulated, 
determine the temperature profile after 24 minutes using the Schmidt numerical method with M = 2 
and dividing the fuel cell stack into slices with a thickness of 0.05 m. Follow the special procedure 
for the first time increment. 
The next figure illustrates the conditions in this cooling process: 
 
 
 
 
 
 
 
Strategy 
The equations we need to use to determine the temperature profile will depend on the boundary 
conditions. 
Solution 
The number of time steps to use in this problem will be determined from the definition of the 
parameter M: 
 
2
( x)
M
t
∆
=
α∆
 
Solving for the time increment t∆ and substituting the corresponding values into this equation, 
yields: 
 
2 2
2
( x) (______ m)
t
M m
2 ____________
s
∆
∆ = =
α  
 
 
 
Insulated Face 
T0 = 60 °C 
Ta = -6.67 °C 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 13 Student View 
Jason M. Keith 
 t _________ s∆ = 
The number of time steps needed for this t∆ is given by: 
 
( )
time steps
(______ min) ______t
n 10
t ________ s
= = ≈
∆
 
The front surface of the fuel cell stack corresponds to n =1. In this point, the temperature 
1 a
T used 
for the first time increment is given by: 
 a 0 1
1 a
T T
T
2
+
= 
where: 
 
0 1
T = Initial temperature at point 1. 
 
a
T = Temperature of the environment = -6.67°C 
Since there is no convective heat resistance at the interface, for the remaining time increments: 
 
1 a
T T= 
The general equation for determining the temperature for the slabs n = 2 to 6 is given below: 
 t n 1 t n 1
t t n
T T
T
2
− +
+∆
+
= 
We need an additional equation for the insulated face. This is the point where n = 7 and its 
corresponding equation is given by: 
 
( ) 7 t 6t
t t 7
M 2 T 2 T
T
M
+∆
− +
= 
Now we can proceed to calculate the temperatures for the first time increment. Thus, for n = 1, 
 a 0 1
t+ t 1
T T 6.67 C _____°C
T ______ C
2 2
∆
+ − ° +
= = = ° 
Since the problem indicates we should use the special procedure for the first time increment, this 
temperature value we just obtained is equal to the ambient temperature at the first time increment: 
 
t+ t 1 1 a
T T∆ = 
For n = 2: 
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Principles of Unsteady-State Heat Transfer 
Daniel López Gaxiola 14 Student View 
Jason M. Keith 
 t 1 t 3
t+ t 2
T T
T
2
∆
+
= 
The special procedure used at n = 1 for the first time increment also affects this equation: instead of 
using the temperature of -6.67°C we will use 
1 a
T . Hence, 
 1 a t 3
t+ t 2
T T _________ °C ______ °C
T ________ C
2 2
∆
+ +
= = = ° 
The following four slabs corresponding to n = 3 to 6 can be calculated with the general equation for 
n = 2 to 6, given in previous steps. Substituting the corresponding temperatures into this equation 
yields: 
 t 2 t 4
t t 3
T T 60 C 60 C
T 60 C
2 2
+∆
+ ° + °
= = = ° 
 t 3 t 5
t t 4
T T 60 C 60 C
T 60 C
2 2
+∆
+ ° + °
= = = ° 
 t 4 t 6
t t 5
T T 60 C 60 C
T 60 C
2 2
+∆
+ ° + °
= = = ° 
 t 5 t 7
t t 6
T T 60 C 60 C
T 60 C
2 2
+∆
+ ° + °
= = = ° 
For n = 7, we use the equation for the insulated face: 
 
( ) 7 t 6t
t t 7 t 6
2 2 T 2 T
T T 60 C
2
+∆
− +
= = = ° 
Now we can proceed to calculate the temperatures from the second to the tenth time increments, as 
shown in the following steps: 
For 2∆t: 
t+2 t 1 a
T T C______∆ = = ° 
t+ t 1 t+ t 3
t+2 t 2
T T ______ 60°C
T C
2 2
______∆ ∆∆
+ +
= = = ° 
t+ t 2 t+ t 4
t 2 t 3
T T ______ C 60 C
T C
2 2
______∆ ∆+ ∆
+ ° + °
= = = ° 
4
t+ t 3 t+ t 5
t 2 t
T T ______ C 60 C
T C
2 2
______∆ ∆+ ∆
+ ° + °
= = = ° 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 15 Student View 
Jason M. Keith 
t+ t 4 t+ t 6
t 2 t 5
T T ______ C 60 C
T C
2 2
______∆ ∆+ ∆
+ ° + °
= = = ° 
t+ t 5 t+ t 7
t 2 t 6
T T ______ C 60 C
T C
2 2
______∆ ∆+ ∆
+ ° + °
= = = ° 
t 2 t 7 t t 6
T T ______ C+ ∆ +∆= = ° 
For 3∆t: 
t+3 t 1 a
T T 6.67 C∆ = = − ° 
t+2 t 1 t+2 t 3
t+3 t 2
T T 6.67°C ______°C
T C
2 2
______∆ ∆∆+ − +
= = = ° 
t+2 t 2 t+2 t 4
t 3 t 3
T T _______ C _______ C
T C
2 2
_______∆ ∆+ ∆
+ ° + °
= = = ° 
4
t+2 t 3 t+2 t 5
t 3 t
T T _______ C _______ C
T C
2 2
_______∆ ∆+ ∆
+ ° + °
= = = ° 
t+2 t 4 t+2 t 6
t 3 t 5
T T _______ C _______ C
T C
2 2
_______∆ ∆+ ∆
+ ° + °
= = = ° 
t+2 t 5 t+2 t 7
t 3 t 6
T T _______ C _______ C
T C
2 2
_______∆ ∆+ ∆
+ ° + °
= = = ° 
t 3 t 7 t 2 t 6
T T ____ C+ ∆ + ∆= = ° 
 
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Principles of Unsteady-State Heat Transfer 
Daniel López Gaxiola 16 Student View 
Jason M. Keith 
We can continue to repeat this procedure up to 10∆T. The solutions for the all the temperatures (in 
°C) are given in the following table: 
 n 
 1 2 3 4 5 6 7 
t 60 60 60 ______ 60 ______ 60 
t+∆t ______ ______ 60 60 ______ 60 ______ 
t+2∆t -6.67 ______ ______ 60 ______ ______ ______ 
t+3∆t ______ 22.50 ______ ______ 60 ______ 60 
t+4∆t -6.67 ______ 39.17 ______ ______ 60 ______ 
t+5∆t -6.67 ______ ______ 48.54 ______ ______ ______ 
t+6∆t -6.67 14.16 ______ ______ 53.75 ______ 58.96 
t+7∆t -6.67 ______ 29.79 ______ ______ 56.35 ______ 
t+8∆t -6.67 ______ ______ 40.73 ______ ______ ______ 
t+9∆t -6.67 10.65 ______ ______ 47.76 ______ 54.79 
t+10∆t -6.67 ______ 24.74 ______ ______ 51.28 ______ 
 
We can plot these temperature results to observe the change in temperature within the fuel cell stack: 
 
-10
0
10
20
30
40
50
60
1 2 3 4 5 6 7
T
 (
°C
)
n
t
t+Δt
t+2Δt
t+3Δt
t+4Δt
t+5Δt
t+6Δt
t+7Δt
t+8Δt
t+9Δt
t+10Δt
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 17 Student View 
Jason M. Keith 
Example 5.4-3: Unsteady – State Conduction with Convective Boundary Condition 
Determine the temperature profile for the same fuel cell stack as in Example 5.4-1, now with a 
convective coefficient of 
2
W
13
m K⋅
. The bulk thermal conductivity of the fuel cell stack is 
W
20
m K⋅
. 
Use a value of M = 4. 
 
 
 
 
 
 
 
Strategy 
In this problem we need to use Schmidt numerical method depending on the boundary conditions. 
Solution 
Before we start applying Schmidt method, we need to determine the number of time increments we 
need to use. From the definition of the parameter M, we can solve for the time increment t∆ , as 
shown in the following steps: 
 
2
( x)
M
t
∆
=
α∆
 
 
( )
2 2
2
6
( x) (0.05 m)
t ________ s
M m
8.69 10 4
s
−
∆
∆ = = =
α  
× 
 
 
The number of time steps needed for this t∆ is given by: 
 
time steps
60 s
(______ min)
t 1 min
n ______
t 71.92s
 
 
 = = ≈
∆
 
As stated in Section 5.4B of Geankoplis, when the value of M is greater than 3, the value of the 
environmental temperature Ta will be the same for all time increments. Therefore, 
Insulated Face 
T0 = 60 °C 
Ta = -6.67 °C 
2
W
h 13
m K
=
⋅
 
 
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Principles of Unsteady-State Heat Transfer 
Daniel López Gaxiola 18 Student View 
Jason M. Keith 
 
a
T 6.67 C= − ° 
For the node n = 1, corresponding to the front face of the fuel cell stack, the temperature can be 
calculated using Equation 5.4-7 of Geankoplis: 
 [ ]{ }t t 1 t a 1 t 2t
1
T 2N T M (2N 2) T 2 T
M
+∆ = + − + + 
The value of N is a function of the convective heat transfer coefficient and the thermal conductivity 
of the fuel cell, described by the following equation: 
 
h x
N
k
∆
= 
Entering the corresponding values into this equation, yields: 
 
( )2
W
______ ______ m
m KN _________
W
20
m K
⋅= =
⋅
 
In order to use Equation 5.4-7, the value of M must satisfy the following constraint: 
 M 2N 2≥ + 
Substituting numeric quantities into this constraint, we get: 
 4 2(0.0325) 2≥ + 
 4 ______≥ 
Hence, we can use Equation 5.4-7. From this equation, we can find the temperature 
t t 1
T+∆ to be: 
 ( )( ) [ ]( ){ }t t 1
1
T 2 ________ 6.67 C 4 (________) 60 C 2(60 C)
4
+∆ = − ° + − ° + ° 
t t 1
T ________ C+∆ = ° 
For the points at n = 2, 3, 4, 5, 6, we use Equation 5.4-2: 
 ( )t t n t n 1 n t n 1t
1
T T M 2 T T
M
+∆ + −
 = + − +  
By entering the value of M into this equation, we can get a general equation for 
t t n
T+∆ : 
 
t t n
T ____________________________________+∆ = 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 19 Student View 
Jason M. Keith 
Now we need an equation for the insulated boundary at n = 7, which is Equation 5.4-10 of 
Geankoplis: 
 
( ) 7 t 6t
t t 7
M 2 T 2 T
T
M
+∆
− +
= 
 t 7 t 6
t t 7
T T
T
_____
+∆
+
= 
We can now proceed to determine the temperatures for n = 2 to 7 for the first time increment. Thus, 
 ( )t t 2 t 3 t 1 t 2T 0.25 T T 0.5 T+∆ = + + 
 ( ) ( )t t 2T 0.25 ______ C 60 C 0.5 ______ C ______ C+∆ = ° + ° + ° = ° 
 ( )t t 3 t 4 t 2 t 3T 0.25 T T 0.5 T+∆ = + + 
 ( ) ( )t t 3T 0.25 ______ C 60 C 0.5 ______ C ______ C+∆ = ° + ° + ° = ° 
 ( )t t 4 t 5 t 3 t 4T 0.25 T T 0.5 T+∆ = + + 
 ( ) ( )t t 4T 0.25 ______ C 60 C 0.5 ______ C ______ C+∆ = ° + ° + ° = ° 
 ( )t t 5 t 6 t 4 t 5T 0.25 T T 0.5 T+∆ = + + 
 ( ) ( )t t 5T 0.25 ______ C 60 C 0.5 ______ C ______ C+∆ = ° + ° + ° = ° 
 ( )t t 6 t 7 t 5 t 6T 0.25 T T 0.5 T+∆ = + + 
 ( ) ( )t t 6T 0.25 ______ C 60 C 0.5 ______ C ______ C+∆ = ° + ° + ° = ° 
t 7 t 6
t t 7
T T
T
2
+∆
+
= 
( ) ( )
t t 7
____ C ____ C
T ____ C
2
+∆
° + °
= = ° 
For 2∆t: 
 [ ]t 2 t 1 t t a t t 1 t t 2
1
T 0.065 T _______ T 2 T
4
+ ∆ +∆ +∆ +∆= + + 
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Principles of Unsteady-State Heat Transfer 
Daniel López Gaxiola 20 Student View 
Jason M. Keith 
( ) ( )t 2 t 1 t t t t t t
1
T 0.065 (_______ C) _______ _______ C 2 60 C _______ C
4
+ ∆ +∆ +∆ +∆= ° + ° + ° = °   
 ( )t 2 t 2 t t 3 t t 1 t t 2T 0.25 T T 0.5 T+ ∆ +∆ +∆ +∆= + + 
 ( ) ( )t 2 t 2T 0.25 _____ C 58.92 C 0.5 60 C 59.73 C+ ∆ = ° + ° + ° = ° 
 ( )t 2 t 3 t t 4 t t 2 t t 3T 0.25 T T 0.5 T+ ∆ +∆ +∆ +∆= + + 
 ( ) ( )t 2 t 3T 0.25 ____ C 60 C 0.5 ____ C ____ C+ ∆ = ° + ° + ° = ° 
 ( )t 2 t 4 t t 5 t t 3 t t 4T 0.25 T T 0.5 T+ ∆ +∆ +∆ +∆= + + 
 ( ) ( )t 2 t 4T 0.25 ____ C ____ C 0.5 ____ C ____ C+ ∆ = ° + ° + ° = ° 
 ( )t 2 t 5 t t 6 t t 4 t t 5T 0.25 T T 0.5 T+ ∆ +∆ +∆ +∆= + + 
 ( ) ( )t 2 t 5T 0.25 ____ C ____ C 0.5 ____ C ____ C+ ∆ = ° + ° + ° = ° 
 ( )t 2 t 6 t t 7 t t 5 t t 6T 0.25 T T 0.5 T+ ∆ +∆ +∆ +∆= + + 
 ( ) ( )t 2 t 6T 0.25 ____ C ____ C 0.5 ____ C ____ C+ ∆ = ° + ° + ° = ° 
t t 7 t t 6
t 2 t 7
T T
T
2
+∆ +∆
+ ∆
+
= 
( ) ( )
t 2 t 7
____ C ____ C
T ____ C
2
+ ∆
° + °
= = ° 
For 3∆t: 
 [ ]t 3 t 1 t 2 t a t 2 t 1 t 2 t 2
1
T 0.065 T 1.935 T 2 T
4
+ ∆ + ∆ + ∆ + ∆= + + 
( ) ( )t 3 t 1 t 2 t t 2 t t 2 t
1
T 0.065 (______ C) 1.935 _______ C 2 59.73 C ______ C
4
+ ∆ + ∆ + ∆ + ∆= ° + ° + ° = °   
 ( )t 3 t 2 t 2 t 3 t 2 t 1 t 2 t 2T 0.25 T T 0.5 T+ ∆ + ∆ + ∆ + ∆= + + 
 ( ) ( )t 3 t 2T 0.25 60 C ______ C 0.5 59.73 C ______ C+ ∆ = ° + ° + ° = ° 
 ( )t 3 t 3 t 2 t 4 t 2 t 2 t 2 t 3T 0.25 T T 0.5 T+ ∆ + ∆ + ∆ + ∆= + + 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 21 Student View 
Jason M. Keith 
 ( ) ( )t 3 t 3T 0.25 _______ C _______ C 0.5 _______ C _______ C+ ∆ = ° + ° + ° = ° 
 ( )t 3 t 4 t 2 t 5 t 2 t 3 t 2 t 4T 0.25 T T 0.5 T+ ∆ + ∆ + ∆ + ∆= + + 
 ( ) ( )t 3 t 4T 0.25 _______ C _______ C 0.5 _______ C _______ C+ ∆ = ° + ° + ° = ° 
 ( )t 3 t 5 t 2 t 6 t 2 t 4 t 2 t 5T 0.25 T T 0.5 T+ ∆ + ∆ + ∆ + ∆= + + 
 ( ) ( )t 3 t 5T 0.25 _______ C _______ C 0.5 _______ C _______ C+ ∆ = ° + ° + ° = ° 
 ( )t 3 t 6 t 2 t 7 t 2 t 5 t 2 t 6T 0.25 T T 0.5 T+ ∆ + ∆ + ∆ + ∆=+ + 
 ( ) ( )t 3 t 6T 0.25 _______ C _______ C 0.5 _______ C _______ C+ ∆ = ° + ° + ° = ° 
t 2 t 7 t 2 t 6
t 3 t 7
T T
T
2
+ ∆ + ∆
+ ∆
+
= 
( ) ( )
t 3 t 7
____ C ____ C
T ____ C
2
+ ∆
° + °
= = ° 
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Principles of Unsteady-State Heat Transfer 
Daniel López Gaxiola 22 Student View 
Jason M. Keith 
In a similar way, we can continue to perform this calculation for all the 20 time increments. The 
results for all temperatures (in °C) are shown in the following table: 
 n 
 1 2 3 4 5 6 7 
t 60 60 60 60 60 60 60 
t+∆t _______ _______ _______ 60 _______ _______ 60 
t+2∆t _______ 59.73 _______ _______ 60 _______ _______ 
t+3∆t 58.00 _______ 59.93 _______ _______ 60 _______ 
t+4∆t _______ _______ _______ 59.98 _______ _______ 60 
t+5∆t _______ 58.99 _______ _______ 60 _______ _______ 
t+6∆t 57.15 _______ 59.59 _______ _______ 60 _______ 
t+7∆t _______ _______ _______ 59.85 _______ _______ 60 
t+8∆t _______ 58.38 _______ _______ 59.94 _______ _______ 
t+9∆t 56.52 _______ 59.21 _______ _______ 59.98 _______ 
t+10∆t _______ _______ _______ 59.64 _______ _______ 59.99 
t+11∆t _______ 57.87 _______ _______ 59.84 _______ _______ 
t+12∆t 56.00 _______ 58.84 _______ _______ 59.93 _______ 
t+13∆t _______ _______ _______ 59.40 _______ _______ 59.95 
t+14∆t _______ 57.42 _______ _______ 59.70 _______ _______ 
t+15∆t 55.54 _______ 58.48 _______ _______ 59.85 _______ 
t+16∆t _______ _______ _______ 59.15 _______ _______ 59.87 
t+17∆t _______ 57.02 _______ _______ 59.54 _______ _______ 
t+18∆t 55.14 _______ 58.15 _______ _______ 59.73 _______ 
t+19∆t _______ _______ _______ 58.90 _______ _______ 59.77 
t+20∆t _______ 56.65 _______ _______ 59.36 _______ _______ 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 23 Student View 
Jason M. Keith 
The graphical representation of these temperatures is shown in the following figure: 
 
54
55
56
57
58
59
60
1 2 3 4 5 6 7
T
 (
°C
)
n
t t+Δt
t+2Δt t+3Δt
t+4Δt t+5Δt
t+6Δt t+7Δt
t+8Δt t+9Δt
t+10Δt t+11Δt
t+12Δt t+13Δt
t+14Δt t+15Δt
t+16Δt t+17Δt
t+18Δt t+19Δt
t+20Δt
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 1 Student View 
Jason M. Keith 
Chapter 6 
Principles of Mass Transfer 
Chapter 6 illustrates different situations where the third fundamental transfer process, mass transfer, 
is occurring in fuel cells or in the chemical processes to produce fuel for fuel cells. The following 
examples explain in detail how the flux of different species is affecting the performance of fuel cells. 
6.1-1 Molecular Diffusion of Water in Air at the Cathode of a Fuel Cell 
6.2-1 Equimolar Counterdiffusion at the Cathode Chamber of a Proton – Exchange Membrane 
Fuel Cell 
6.2-5 Diffusivity in Steam-Methane Reforming Process 
6.3-1 Diffusion of Methanol in Water in Direct Methanol Fuel Cells 
6.3-2 Prediction of Diffusivity of Methanol in Water 
6.5-1 Diffusion of Hydrogen through Nafion 
6.5-3 Diffusion of Oxygen through Gas Diffusion Layer 
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Principles of Mass Transfer 
Daniel López Gaxiola 2 Student View 
Jason M. Keith 
Example 6.1-1: Molecular Diffusion of Water in Air at the Cathode of a Fuel Cell 
Proton – Exchange Membrane Fuel Cells consume hydrogen and oxygen from air to generate 
electricity, with heat and steam as byproducts. The steam being produced in the cathode side has a 
partial pressure of 0.22 bar. Some of the steam stays inside the fuel cell to keep the polymer – 
electrolyte membrane at an adequate humidity. The steam in the bipolar plate channels has a partial 
pressure of 0.024 bar. The distance between the gas diffusion layer (GDL) and the bipolar plate is 
0.729 mm. Determine the diffusive flux from the gas diffusion layer to the bipolar plates of water in 
the air. The fuel cell is operating at a pressure and temperature of 2 bar and 40°C, respectively. 
The following schematic illustrates the diffusion process taking place in the fuel cell: 
 
 
 
 
 
 
 
Strategy 
We can use Fick’s law of diffusion to determine the steam flux through the air. 
Solution 
The equation that describes the flux of a chemical species due to diffusion is Fick’s Law, given by: 
 * AAz AB
dC
J D
dz
= − 
This equation can be separated and integrated to yield: 
 
2
1
 z
*
Az
 z
J dz _________________________=∫ 
 
*
AzJ ___________________= 
 
Anode Cathode 
Electrolyte 
Membrane 
GDL 
H2O 
diffusing in 
air 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 3 Student View 
Jason M. Keith 
The point where the steam is being produced will be considered z1 = 0, whereas z2 = 0.729 mm, since 
the outlet point is 0.729 mm away. The concentrations of the steam A2
C and A1
C will be determined 
from the ideal gas law, because the fuel cell is operating at relatively low pressure. Thus, 
 
A
A
1
1
P
C
RT
= 
A
A
2
2
P
C
RT
= 
Substituting the partial pressures, temperature and ideal gas constant into these equations, we have: 
 
( )
A 331
0.22 bar mol
C ______
mbar m
_______________ 313.15 K
mol K
= =
 ⋅
 
⋅ 
 
 
( )
A 332
0.024 bar mol
C ________
mbar m
______________ 313.15 K
mol K
= =
 ⋅
 
⋅ 
 
Now we can enter the concentrations we calculated into Fick’s law to obtain the flux of steam, as 
shown below: 
 
( )
2 3 3
*
Az
mol mol
________ ________
m m mJ ________________
1 ms
0.729 mm 0 mm
1000 mm
 
− 
 =
  −     
 
 *Az 2
mol
J ________
m s
=
⋅
 
The diffusivity value was obtained from Table 6.2-1 at a temperature of 42°C, which is relatively 
close to the operating temperature of 40°C. 
 
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Principles of Mass Transfer 
Daniel López Gaxiola 4 Student View 
Jason M. Keith 
Example 6.2-1: Equimolar Counterdiffusion at the Cathode Chamber of a Proton – 
Exchange Membrane Fuel Cell 
In the diffusion process illustrated in Example 6.1-1, we calculated the flux of water in a fuel cell. For 
the same fuel cell, determine the flux of air diffusing in the steam. 
Strategy 
In a similar way to example 6.1-1 this problem can be solved using Fick’s Law of Diffusion. 
Solution 
Since we need to determine the flux of air in the steam, we can start from Fick’s Law. Thus, 
 
*
BzJ _______________= 
 
We can separate and integrate this equation to get: 
 
 ( ) ( )*Bz BAJ _____________ D ______________= − 
 
 *BzJ ____________________= 
 
Since the pressure of the system is low, we can use Ideal Gas Law equation of state to calculate the 
pressures of air, as shown in the following equations: 
 
B
B
1
1
P
C
RT
= 
B
B
2
2
P
C
RT
= 
We can substitute these equations into Fick’s Law equation to yield: 
 
( )
*
Bz BA
2 1
__________________
J D
RT z z
 
=  
− 
 
Since there is only air and steam in the cathode chamber of the fuel cell, we can calculate the partial 
pressures of air from Dalton’s Law, which states: 
 iP P=∑ 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 5 Student View 
Jason M. Keith 
where: 
 P = Absolute pressure of the system 
 Pi = Partial pressure of component i in the gas mixture 
Applying this equation for the number of gases in this problem and solving for the partial pressure of 
air, we have: 
 A BP P P= + 
 B AP P P= − 
Substituting this equation into Fick’slaw, gives: 
 
( ) ( )
*
Bz BA BA
2 1 2 1
__________ __________ ________________
J D D
RT z z RT z z
   −
= =   
− −   
 
Now we can enter the corresponding quantities into this equation to obtain the solution to this 
problem: 
 
( ) ( )
2
* 5
Bz 3
m ________ bar ________ bar
J 2.88 10
s bar m
_______________ 313.15 K _____________ m
mol K
−  −= × 
 ⋅ 
 
⋅ 
 
*
Bz 2
mol
J __________
m s
=
⋅
 
The negative in this result indicates that the flux is going from point 2 (outlet stream) to point 1 
(steam generation). 
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Principles of Mass Transfer 
Daniel López Gaxiola 6 Student View 
Jason M. Keith 
Example 6.2-5: Diffusion in Steam – Methane Reforming Process 
Hydrogen for Proton – Exchange Membrane Fuel Cells is being produced from natural gas in a 
Steam – Methane Reforming Process. The natural gas reacts with steam to produce a mixture of 
Hydrogen and Carbon Monoxide, known as synthesis gas at a pressure of 20.45 atm and a 
temperature of 850°C. Determine the diffusivity of hydrogen in CO at these conditions using the 
Fuller method. 
Strategy 
We can obtain the diffusivity using Fuller method given in Section 6.2E of Geankoplis. 
Solution 
The diffusivity of a gas mixture can be calculated using the following equation: 
 
( ) ( )
1/2
7 1.75
A B
AB 2
1/3 1/3
A B
1 1
1 10 T
M M
D
P
−  × + 
 =
 υ + υ  ∑ ∑
 
For this problem we will select A = hydrogen, and B = carbon monoxide 
The atomic diffusion volumes of hydrogen and carbon monoxide can be obtained from Table 6.2-2 of 
Geankoplis. Hence, 
 A 7.07υ =∑ 
 B ________υ =∑ 
We can substitute the process conditions, as well as the molecular weights and volumes of the gas 
species into the equation for DAB to get: 
 
( )
( ) ( )
1/2
1.757
AB 2
1/3 1/3
1 1
1 10 1123.15K
________ 28
D
__________ atm 7.07 ________
−  × + 
 =
 +
 
 
2
AB
m
D ________________
s
= 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 7 Student View 
Jason M. Keith 
Example 6.3-1: Diffusion of Methanol in Water in Direct Methanol Fuel Cells 
Direct – methanol fuel cells are a type of fuel cells used for portable devices such as laptops and cell 
phones. The concentration of the methanol solution at the bottom of the reservoir is 32 wt. %. At the 
top of the tank, located 2 inches away, the concentration is 40 wt. %. Determine the diffusive flux of 
methanol in water at steady state at a temperature of 15°C. The densities of these two solutions at this 
temperature are 
3
kg
950.6
m
for 32 wt. % CH3OH at the bottom of the tank and 3
kg
937.2
m
for 40 wt. % 
CH3OH at the top. 
Strategy 
The flux of methanol required in this problem can be calculated using the equation for flux of liquids 
at steady state, given in Section 6.3B of Geankoplis. 
Solution 
The equation for NA for liquid mixtures is given by Equation 6.3-3 of Geankoplis. 
 ( )AB avA A A1 2
D C
N x x
________________
= − 
The diffusivity of methanol in water can be obtained from Table 6.3-1 of Geankoplis to be: 
 
2
AB
m
D ________________
s
= 
The average concentration of methanol can be obtained from the density and molecular weight of the 
methanol solution at both concentrations, as shown below: 
 
1 2
1 2
av
M M
C
2
ρ ρ
+
= 
To calculate the molecular weights at points 1 and 2, we use the following equation: 
1
A B
100kg 100kg kg
M _________
wt.% A @ point 1 wt.% B @ point 1 32 68 kgmol
kgmol
M M 32 18
= = =
 
+ + 
 
 
 
2
A B
100kg 100kg kg
M _________
wt.% A @ point 2 wt.% B @ point 2 kgmol______ ______
kgmol
M M 32 ______
= = =
 
+ + 
 
 
 
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Principles of Mass Transfer 
Daniel López Gaxiola 8 Student View 
Jason M. Keith 
Now we can substitute the molecular weights to obtain the average concentration: 
 
3 3
av 3
kg kg
950.6 937.2
m m
kg kg
________ ________
kgmolkgmol kgmol
C ________
2 m
+
= = 
The remaining unknowns are the molar fractions of methanol at points 1 and 2, and the average molar 
fraction of water in the tank. This can be done as follows: 
 AA
A B
1
wt.% A @ point 1
M
x
wt.% A @ point 1 wt.% B @ point 1
M M
=
+
 
 A
B
2
_____________________
________
x
wt.% A @ point 2 wt.% B @ point 2
________ M
=
+
 
 
B B A
BM
1 2 1
x x (1 x ) (__________)
x
2 2
+ − +
= = 
Notice that the average molar fraction of B was written in terms of A1
x and A2
x , since only methanol 
and water are present in the reservoir. Substituting the weight fractions and molecular weights into 
these equations yields: 
 A1
________
32x ________
________ 68
32 ________
= =
+
 
 A2
________
________
x ________
40 60
________ ________
= =
+
 
 BM
(__________) (__________)
x ________
2
+
= = 
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 9 Student View 
Jason M. Keith 
We can now enter the values we calculated into the equation for the flux of A to get: 
 
( ) ( )
( )
2
9
3
A
m kgmol
1.26 10 ________
s m
N ________ ________
0.0254 m
0 2 in ________
1 in
−  ×  
 = −
 
−  
 
 
As we can see in this equation, point 1 is at z = 2 in and point 2 is at z = 0, thus: 
( ) ( )
( )
2
9
3
A
m kgmol
1.26 10 ________
s m
N ________ ________
0.0254 m
2 in ________
1 in
−  ×  
 = −
 
−  
 
 
A 2
kgmol
N __________________
m s
=
⋅
 
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Principles of Mass Transfer 
Daniel López Gaxiola 10 Student View 
Jason M. Keith 
Example 6.3-2: Prediction of Diffusivity of Methanol in Water 
Determine the diffusion coefficient of methanol in water using the Wilke – Chang equation at 15 °C. 
How does this value compare to the experimental diffusion coefficient in Table 6.3-1? Also calculate 
the diffusion coefficient of methanol in water at the operating temperature of Direct – Methanol Fuel 
Cells of 50°C. 
Strategy 
Wilke – Chang equation predicts the diffusivity of liquid mixtures, therefore, it can be used to 
calculate the diffusion coefficient of methanol in water at a given temperature. 
Solution 
Wilke – Chang equation is given in Section 6.3D of Geankoplis: 
 ( )
1/216
AB B 0.6
B A
T
D 1.173 10 M
V
−= × φ
µ
 
where: 
 φ = Association parameter of the solvent = 2.6 (This value is given for water in Section 6.3D) 
 MB = Molecular weight of the solvent 
 µB = Viscosity of the solvent in Pa s⋅ 
VA = Molar volume of the solute at the boiling point (Calculated using values from Table 6.3-
2) 
First we can calculate the molar volume of methanol. Since the molecular formula is CH3OH, the 
molar volume is calculated as follows: 
 
3 3 3
A
m m m
V 1 _________ 4 _________ 1 0.0074
kgmol kgmol kgmol
     
= + +     
     
 
 
3
A
m
V _________
kgmol
= 
Substituting the rest of the parameters into Wilke – Change equation, we have: 
 ( )
( )( )
1/216
AB 0.6
288.15
D 1.173 10 2.6 _____
___________ _________
−
 
= ×     
  
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 11 Student View 
Jason M. Keith 
 
2
AB
m
D _____________
s
= 
The viscosity for water was obtained from Appendix A.2 of Geankoplis at 15°C. We can observe this 
diffusivity value is relatively close to the experimental value in Table 6.3-1 of 
2
9 m1.26 10
s
−× . 
We can repeat this procedure to determine the diffusivity of methanol in water at the operating 
temperature ofdirect – methanol fuel cells. Again, the viscosity of water was obtained from 
Appendix A.2 at the temperature of 50°C. 
 ( )
( )( )
1/216
AB 0.6
323.15
D 1.173 10 2.6 18
_____________ _________
−
 
= ×     
  
 
 
2
AB
m
D _______________
s
= 
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Principles of Mass Transfer 
Daniel López Gaxiola 12 Student View 
Jason M. Keith 
Example 6.5-1: Diffusion of Hydrogen through Nafion 
Nafion is a polymer manufactured by DuPont that is widely used as a selective membrane in polymer 
– electrolyte membrane fuel cells. The protons produced from the electrochemical reaction in the 
anode side of the fuel cell travel through the membrane to the cathode. The Nafion membrane must 
be humidified to facilitate the flow of protons through it. At a water uptake of 22 (molecules 
H2O/molecules of sulfonic groups in the membrane), the concentration of protons in the membrane 
was estimated by Spry and Fayer [1] to be 
mol
0.54
L
. Determine the flux of protons through a 
membrane with a thickness of 0.175 mm. The diffusivity at a temperature of 30.5°C and a water 
uptake of 22 was estimated from data obtained by Ochi et al. to be 
2
9 m1.9 10
s
−× [2]. Assume that the 
reaction in the cathode chamber to form water is occurring instantaneously. 
Strategy 
We can use the diffusive flux equation to calculate the flow of protons through the membrane in the 
fuel cell. 
Solution 
The equation for diffusive flux is given by: 
 
 AN _____________________= 
 
Since the problem is stating that the reaction on the cathode chamber is occurring instantaneously, we 
can assume the concentration A2
C _________= . 
We can now calculate the flux by substituting the concentrations, thickness of the membrane and 
proton diffusivity into the equation shown above, as shown in the following steps: 
( )
2
A
m mol 
_______________ _________ _________
s L 
N
1 m
_________ mm
1000 mm
  
−  
  =
 
 
 
 
A 2
mol
N _______________
m s
=
⋅
 
1. Spry, D.B., Fayer, M.D., Journal of Physical Chemistry B, 113, 10210 – 10221 (2009) 
2. Ochi, S., Kamishima, O., Mizusaki, J., Kawamura, J., Solid State Ionics, 180, 580 – 584 (2009) 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 13 Student View 
Jason M. Keith 
Example 6.5-3: Diffusion of Oxygen through Gas – Diffusion Layer 
The oxygen in the cathode side of a proton – exchange membrane fuel cell diffuses to the catalyst 
layer through a gas – diffusion layer (GDL). The GDL has a thickness of 100 µm, and is typically 
made of a porous material which may be fabricated with either carbon cloth or carbon paper. The 
GDL has a porosity of 0.5 and a tortuosity of 1.63 [3]. Determine the flux of oxygen through the 
GDL if its partial pressure is 98.96 kPa and the operating temperature of the fuel cell is 50 °C. 
Assume there is steam in the pores of the GDL and that the reaction in the catalyst layer in occurring 
instantaneously. 
Strategy 
The equation defining diffusive flux in porous solids will be used to calculate the flux of oxygen 
through the GDL. 
Solution 
Section 6.5C of Geankoplis gives the equation for steady state flux in a porous media: 
 
 AN __________________= 
 
The diffusivity DAB can be obtained using the Fuller method from Section 6.2E of Geankoplis to be: 
 
2
AB
m
D __________________
s
= 
The concentrations of oxygen can be obtained from the ideal gas law, as shown below: 
 
( )
A 31
_________ kPa
C
kPa m
_________ _________ K
kgmol K
=
 ⋅
 
⋅ 
 
 A 31
kgmol
C ___________
m
= 
Since we are assuming the reaction in the cathode is occurring instantaneously, the concentration of 
oxygen at the catalyst layer will be A 2
C 0= 
 
 
3. Hao, L., Cheng, P., Journal of Power Sources, 186, 104–114 (2009) 
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Principles of Mass Transfer 
Daniel López Gaxiola 14 Student View 
Jason M. Keith 
Now we can substitute the corresponding values into the equation for diffusive flux to get: 
 
( )
2
5
3
A
m kgmol
_________ 3.04 10 _________
s m
N
 
_________ _________ m
 
−  ×   
  =
 
µ  
 
 
 A 2
kgmol
N _____________
m s
=
⋅
 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 1 Student View 
Jason M. Keith 
Chapter 7 
Principles of Unsteady - State and Convective Mass Transfer 
This chapter covers different situations where mass transfer is taking place, when the conditions are 
changing with time and where there is a fluid stream that contributes to these mass transfer processes. 
The following examples illustrate convective and transient mass transfer in fuel cell systems. 
7.5-2 Diffusion and Chemical Reaction in the Anode Chamber of a Direct Methanol Fuel Cell 
7.5-4 Diffusion of CO2 and O2 through stagnant Nitrogen in a Solid – Oxide Fuel Cell 
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Principles of Unsteady – State and Convective Mass Transfer 
Daniel López Gaxiola 2 Student View 
Jason M. Keith 
Example 7.5-2 Diffusion and Chemical Reaction in the Anode Chamber of a Direct 
Methanol Fuel Cell 
An aqueous 40 mole % methanol solution is entering the anode of a direct – methanol fuel cell. The 
fuel is diffusing through the gas diffusion layer (GDL) with a thickness of 0.018 cm [1]. The 
diffusion coefficient of the fuel in the GDL is estimated by García et al. [1] to be 
2
9 m1 10
s
−× . The 
governing equation for methanol in the GDL is given by: 
 
2
A
2
d C
0
dz
= 
with the following boundary conditions: 
 At z = 0 : 
A b
C C= 
 At z = δ : 
A 1 A
N k C= 
The rate constant for the chemical reaction occurring at the catalyst layer located at z = δ is 
6
1
m
k 2.8 10
s
−= × . Determine the molar fraction of methanol at z = δ and steady state, if the initial 
concentration Cb of methanol is 3
mol
500
m
 . 
Strategy 
The molar fraction at the catalyst layer can be obtained by solving the governing differential 
equation. 
Solution 
We can start by solving the given differential equation given in the problem statement, as shown in 
the following steps: 
 
2
A
2
d C
0
dz
=∫ ∫ 
 
1
_______ c= (1) 
 
 
A 1
dC c dz=∫ ∫ 
 
A
C _________________= 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 3 Student View 
Jason M. Keith 
1. García, B.L., Sethuraman, V.A., Weidner, J.W., White, R.E., Dougal, R., Journal of Fuel Cell Science and 
Technology, 1, 43 – 48 (2004). 
Applying the first boundary condition at z = 0, we have: 
b
C ________________= 
2
c ______= 
Substituting c2 into Equation 2 yields: 
A
C _______________= (2) 
At z = δ: 
 
A
C _______________= (3) 
The equation for methanol flux through the GDL is described by Fick’s Law, given by: 
 A
A
dC
N D
dz
= − (4) 
At z = δ, the molar flux of methanol is equal to the reaction rate. Therefore: 
 
A
z
N __________
=δ
= (5) 
Since this process is at steady state, we can equal Equations 4 and 5 to get: 
 A
dC
D _________
dz
− = (6) 
From Equation 1, we have that 
1
c _________= . Substituting this into Equation 6 gives: 
 D ____ _________− = (7) 
We can substitute Equation 3 into Equation 7 to solve for c1, as shown in the following steps: 
( )
1 1
Dc k _______________− = 
1 1 1 1 b
Dc k c k C− = δ + 
( )
1
c ______________ ___________− =1
c ________________= (8) 
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Principles of Unsteady – State and Convective Mass Transfer 
Daniel López Gaxiola 4 Student View 
Jason M. Keith 
 
Now we can substitute this equation for c1 into Equation 2 to yield: 
 
 
A b
C ______________ C= + (9) 
 
Equation 9 can be evaluated at the boundary condition for z = δ to obtain the molar fraction at this 
point. Hence, 
 
 
A
C ______________________________= 
 
Reducing this equation and writing CA in terms of the molar fraction of methanol, we have: 
 
( )
______
C ____
D ______
=
+
 (10) 
where C is the overall concentration of the fuel entering the fuel cell. The overall concentration can 
be obtained by dividing the feed concentration of methanol Cb by the feed molar fraction of 0.4. 
Substituting numeric values into Equation 10 gives: 
 
( )
2
3 3
3 3
2
3 4
mol CH OH mol CH OH m
______ ______ _______________
m m s
_____
mol CH OH m m
0.4 _______________ _______________ 1.8 10 m
mol s s
−
   
   
   
=
    
+ ×   
   
 
( )
2
3
A 2
4
mol m
______ _______________
cm s
x
m m
_______________ _______________ 1.8 10 m
s s
−
 
 
 
=
    
+ ×   
   
 
A
x ______=
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 5 Student View 
Jason M. Keith 
Example 7.5-4: Diffusion of CO2 and O2 through stagnant N2 in a Solid – Oxide Fuel 
Cell 
A solid - oxide fuel cell operating at a temperature of 923.15 K and a pressure of 1.9 atm, is 
producing CO2 from an electrochemical reaction of CO with oxygen from air. The partial pressures of 
each gas at the gas diffusion layer the bipolar plate channels, located 0.729 mm away, are given in the 
following table. 
Label Gas Partial Pressure (atm) at 
Gas – Diffusion Layer 
Partial Pressure (atm) 
at Bipolar Plate 
A CO2 0.47 0.01 
B O2 0.31 0.47 
C N2 1.12 1.42 
 
Determine the molar flux of CO2 and O2 in non – diffusing N2. The diffusion coefficients are given 
below: 
 
2
5
AB
m
D 5.84 10
s
−= × 
 
2
5
AC
m
D 5.93 10
s
−= × 
 
2
5
BC
m
D 7.49 10
s
−= × 
Where: A = CO2, B = O2 and C = N2. These diffusion coefficients were estimated using Fuller et al. 
method described in Section 6.2E of Geankoplis. 
The following equations describe multicomponent diffusion for two components diffusing in stagnant 
C [2]: 
( )
CA B
AC BC 2 1 C
2
1
PN N P
ln
D D RT z z P
 
 + =
 −  
 
 
 
 
2. Geankoplis, C.J., Mass Transport Phenomena, Holt, Rinehart and Winston Inc., New York, 1972. 
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Principles of Unsteady – State and Convective Mass Transfer 
Daniel López Gaxiola 6 Student View 
Jason M. Keith 
( )
AB AC AC BCA B A B
B A
B B
AB BC AB BCAB
A B
2 1
AB AC AC BCA B A B
B A
B B
AB BC AB BC
2 2
1 1
1 1 1 1
D D D DN N N N
P P P
1 1 1 1N N
D D D DD P
N N ln
1 1 1 1RT z z
D D D DN N N N
P P P
1 1 1 1N N
D D D D
 
− − + +
 − +
 − −
 
+ =  
−  − −
+ + 
− + 
− − 
 
 
In this problem, the flux will be assumed positive from the bipolar plate to the gas – diffusion layer. 
Strategy 
We can determine the fluxes NA and NB by simultaneously solving the equations given in the 
problem statement. 
Solution 
First, we can substitute the pressures given and the diffusivity coefficients, as well as the operating 
conditions of the fuel cell. Thus, 
( )( )
A B
2 2
5 5
N N 1.9 atm ______ atm
ln
L atmm m ______ atm
_______________ 923.15K _______________ m5.93 10 7.49 10
mol Ks s
− −
+ =
⋅
× ×
⋅
 
 
   
 
 
 
We can solve this equation for NB by following the next steps: 
( ) ( )
A 3
2
B
2
5
1.9 atm ______ atm
N ln
1.12 atmm atm
______________ 923.15K ______________ m
mol K
N m
 _________________
m s
7.49 10
s
−


 =   ⋅  
 
⋅ 


− 
×

 
B 2
mol
N 0.612 ____________
m s
= −
⋅
 (1) 
So far we have obtained one of the simultaneous equations for the diffusion process occurring in this 
fuel cell. However, in order to solve this problem, we need a second equation, obtained from the 
equation for A BN N+ , as shown below: 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 7 Student View 
Jason M. Keith 
( )
( ) ( )
( ) ( )
2
A B 3
4
2 2 2
5 5
A B A B
B B
2 2
5
m
_____________ 1.9atm
sN N
m atm
_____________ 923.15K 7.29 10 m
mol K
1 1 1 1
m m m
_____________ 5.93 10 5.93 10 _____________
N N N Ns s s____ atm 0.01 atm
1 1 N N
m m
5.84 10 _____________
s sln
−
− −
−
+ =
 ⋅
× 
⋅ 
− −
× ×
+ +
− +
−
×
×
( )
( ) ( )
2
2 2
5
2 2 2 2
5 5
A B A B
B B
2 2 2 2
5 5
m
s 1.9 atm
1 1
m m
5.84 10 _____________
s s
1 1 1 1
m m m m
5.84 10 _____________ _____________ 7.49 10
N N N Ns s s s0.31 atm ____ atm 1.9
1 1 1 1N N
m m m m
_____________ 7.49 10 _____________ 7.49 10
s s s s
−
− −
− −
−
×
− −
× ×
+ +
− +
− −
× ×
( ) atm
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
In this equation, it can be seen that some terms have the factor A B
B
N N
N
+
in common. Thus, this 
expression can be simplified as follows: 
 
( )( ) ( ) ( )( )
( )( ) ( ) ( )( )
A B
B
A B 2
A B
B
N N
_________ 0.47 atm _____ atm ______ 1.9 atm
Nmol
N N ______ ln
N Nm s
_________ ______ atm 0.47 atm ______ 1.9 atm
N
+ 
− +    
+ = ×  
+⋅  − +    
 
We can further simplify this equation by removing the molar flow of oxygen in the denominator as 
shown in the following step: 
 
( )
( )
A B B
A B 2
A B B
_______ N N __________ Nmol
N N _______ ln
m s 0.449 N N __________ N
 + +
+ = ×  
⋅ − + + 
 (2) 
Now we can substitute Equation 1 into Equation 2 to get: 
 
( )
( ) ( )
( ) ( )
A A A
A A 2
A A A
________ N 0.612 1.263N ________ 0.612 1.263Nmol
N 0.612 1.263N ________ ln
m s 0.449 N 0.612 1.263N ________ 0.612 1.263N
 + − + −
+ − = ×  
⋅ − + − + − 
 Simplifying similar terms and moving all terms to the left side, we have: 
A
A
A
________ N 1.096
________ N ln ________ 0
________ N ________
 +
+ − = 
+ 
 
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Principles of Unsteady – State and Convective Mass Transfer 
Daniel López Gaxiola 8 Student View 
Jason M. Keith 
This equation can be solved by trial and error or using computer software to obtain the molar flow 
rate of carbon dioxide to be: 
 A 2
mol
N ________________
m s
=
⋅
 
We can enter this value into Equation 1 to determine the flux of oxygen as shown below: 
 B 2 2
mol mol
N 0.612 1.263 ________________
m s m s
 
= −  
⋅ ⋅ 
 
 B 2
mol
N ________
m s
=
⋅
 
The following figure illustrates the diffusion process occurring in the fuel cell cathode. 
 
Bipolar Plate Gas – Diffusion Layer 
NA 
NB 
NC = 0 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 1 Student View 
Jason M. Keith 
Chapter 9 
Drying of Process Materials 
The drying processes discussed in this chapter relate to the removal of water or other organic liquids 
from solids or fluid substances. The following examples explain different situations where drying and 
humidity concepts affect the performance of a fuel cell. 
9.3-1 Humidity from Vapor – Pressure Data 
9.3-2 Use of Humidity Chart 
9.3-3 Adiabatic Saturation of Feed Air for Proton – Exchange Membrane Fuel Cells. 
9.3-4 Wet Bulb Temperature and Humidity 
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Drying of Process Materials 
Daniel López Gaxiola 2 Student View 
Jason M. Keith 
Example 9.3-1: Humidity from Vapor – Pressure Data 
The air in the cathode chamber of a proton – exchange membrane fuel cell is at a temperature of 80°C 
and a pressure of 1 atm. Determine the partial pressure of water, humidity, saturation humidity, 
percentage humidity if the air has a relative humidity of 8.55 %. 
Strategy 
The definitions of the required parameters can be used to solve this problem. 
Solution 
The following definitions can be used to obtain the solution to this problem: 
 
H O
R
2
P
H 100
______
= P
S
H
H 100
H
= 
 
H O H O
S
Air H O
2 2
2
s
s
M P
H
M P P
=
−
 
H O H O
Air
2 2
M P
H
M _________
= 
First we can solve for the partial pressure of water from the equation of relative humidity, as this 
value is given in the problem statement. Thus, 
 
( )( )
H O2
8.55 _______ atm______
P
100 100
= = 
 H O2
P 0.04 atm= 
The saturation pressure of water at the temperature of 80°C was obtained from Appendix A.2 of 
Geankoplis. 
The partial pressure of water can now be substituted into the Equation for the absolute humidity H to 
yield: 
 
( )
2______ kg H O
0.04 atm1 kgmol
H
______ kg Air 1 atm 0.04 atm
1 kgmol
=
−
 
 2
kg H O
H _______
kg Air
= 
 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 3 Student View 
Jason M. Keith 
In a similar way, we can use the saturation pressure of 0.468 atm obtained from Appendix A.2 to 
determine the saturation humidity of the air in the fuel cell, as shown in the following steps: 
 
( )
2
S
_______ kg H O
0.468 atm1 kgmol
H
_______ kg Air 1 atm _______ atm
1 kgmol
=
−
 
 2S
kg H O
H _______
kg Air
= 
The only remaining value to be calculated is the percentage humidity HP from the ratio of the 
absolute humidity to the saturation humidity. Therefore, 
 
2
P
2
kg H O
0.026
kg Air
H 100
kg H O
_______
kg Air
 
 
 =
 
 
 
 
 PH _______ %= 
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Drying of Process Materials 
Daniel López Gaxiola 4 Student View 
Jason M. Keith 
Example 9.3-2: Use of Humidity Chart 
The air in a fuel cell has a dry – bulb temperature of 80°C and a humidity of 2
kg H O
0.035
kg Air
. Use the 
humidity chart to determine the percentage humidity HP, humid volume Hυ , humid heat CS, and dew 
point of this air/steam mixture. 
Strategy 
We can locate the given information in the humidity chart to determine the values required to solve 
this problem. 
Solution 
First we locate the point that corresponds to a humidity of 2
kg H O
H 0.035
kg Air
= at a temperature of 
80°C. From this point we can move horizontally to the left until reaching the saturation line (HP = 
100%). The temperature in this point corresponds to the dew point, found to be: 
 satT _______ C= ° 
At the point we located initially for the dry bulb temperature and absolute humidity, we can read 
directly the percentage humidity. For the air in the fuel cell, we find that: 
 PH _______ %= 
In section 9.3B of Geankoplis, we are given the following equations for the humid heat and volume 
as a function of the absolute humidity: 
 ( ) ( )3 3H 2.83 10 4.56 10 H T
− − υ = × + ×  
 SC ____________________________= 
In these equations, Hυ is in 
3m
kg dry air
, H is in 2
kg H O
kg Air
, T is in K, and CS is in 
kJ
kg dry air K⋅
. 
Entering the humidity and the temperature into these equations, we get: 
 ( ) ( )3 3 2H
kg H O
2.83 10 4.56 10 0.035 __________ K
kg Air
− −
    
υ = × + ×   
    
 
 
3
H
m
__________
kg dry air
υ = 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 5 Student View 
Jason M. Keith 
 2S
kg H O
C 1.005 1.88 0.035
kg Air
 
= +  
 
 
 S
kJ
C _______
kg dry air K
=
⋅
 
 
The following figure illustrates how to determine the saturation temperature and percentage humidity 
from the chart: 
Initial point corresponding to 
T = 80°C and 2
kg H O
H 0.035
kg Air
= 
This point is also located 
approximately at a percentage 
humidity of 6.25% 
100 % 
Percentage Humidity 
Lines 
Adiabatic Saturation 
Curve 
T = ______°C 
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Drying of Process Materials 
Daniel López Gaxiola 6 Student View 
Jason M. Keith 
Example 9.3-3: Adiabatic Saturation of Feed Air for Proton – Exchange Membrane 
Fuel Cells. 
Air to be used as reactant in a PEMFC with an initial humidity of 2
kg H O
0.057
kg Air
enters an adiabatic 
saturator before being fed to the cathode side of the fuel cell. The air enters the saturator at a dry bulb 
temperature of 60 °C and must enter the fuel cell with a humidity of 2
kg H O
0.063
kg Air
. Determine the 
final temperature and percent humidity of the air. 
Strategy 
Both HP and the temperature of the air entering the fuel cell can be obtained using the humidity chart. 
Solution 
First we need to locate the point that corresponds to the temperature of 60°C and humidity of 
2kg H O0.057
kg Air
. Once we located this point in the humidity chart, we move parallel to the adiabatic 
saturation curves, until reaching a point where the absolute humidity is 2
kg H O
0.063
kg Air
. 
Now we are at the point where the air has the conditions required for entering the fuel cell and 
therefore we can read the temperature and percentage humidity to be: 
 T ______ C≈ ° 
PH ______ %≈ 
In the following page we can see a chart indicating the method to determine these values. 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 7 Student View 
Jason M. Keith 
The next chart indicates the procedure followed to determine the temperature and percentage 
humidity: 
Initial point corresponding to 
T = 60°C and 2
kg H O
H 0.057
kg Air
= 
Percentage Humidity 
Lines 
Adiabatic Saturation 
Curve 
T = ______°C 
2kg H OH 0.063
kg Air
=
 
Point corresponding to 
2kg H OH ________
kg Air
= 
Also we can read the percent 
humidity to be approximately ___ % 
70% 
80% 
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Drying of Process Materials 
Daniel López Gaxiola 8 Student View 
Jason M. Keith 
Example 9.3-4: Wet Bulb Temperature and Humidity 
Estimate the humidity of the reactant air in a proton – exchange membrane fuel cell if it has a dry and 
wet bulb temperature of 80 °C and 42.5 °C, respectively. 
Strategy 
The humidity of the air can be determined from the humidity chart using the temperature data given 
in the problem statement. 
Solution 
The wet bulb temperature corresponds to the temperature of the air when it has 100 % humidity. 
Hence, we need to move vertically from the temperature axis in the chart until reaching the curve 
corresponding to 100 % humidity. From this point we move downwards to the right, parallel to the 
adiabatic saturation curves until we reach the vertical line for T = 80°C. Now we can read the 
humidity of the air in the fuel cell to be: 
 2
kg H O
H ______
kg Air
= 
The method to determine the previous humidity value was obtained as follows: 
 
Percentage Humidity 
Lines 
Adiabatic Saturation 
Curve 
Twet = 42.5 °C 
Initial point 
2kg H OH ______
kg Air
= 
100% 
Tdry = 80 °C 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 1 Student View 
Jason M. Keith 
Chapter 11Vapor – Liquid Separation Processes 
Separation processes in Chemical Engineering are used to transform a mixture of substances into two 
or more different products. In this chapter, the following problem modules illustrate the principles of 
chemical processes in which the separation involves the vapor and liquid phases. 
11.1-1 Use of Raoult’s Law for Methanol – Water Equilibrium Data 
11.3-1 Relative Volatility of Methanol – Water Mixture 
 
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Vapor – Liquid Separation Processes 
Daniel López Gaxiola 2 Student View 
Jason M. Keith 
Example 11.1-1: Use of Raoult’s Law for Methanol – Water Equilibrium Data 
A methanol – water mixture is assumed to be in equilibrium with the vapor phase in the reservoir for 
a Direct – Methanol Fuel Cell. Using the following equilibrium data [1,2] and Raoult’s Law 
determine the composition of the vapor and liquid phases at 72°C and 101.325 kPa. The vapor 
pressure data for water was obtained from Appendix A.2 of Geankoplis. 
 
 
 
Strategy 
We need to use Raoult’s Law to obtain the composition of both phases at the given temperature and 
pressure. 
Solution 
The following equation describes Raoult’s Law: 
 A
_______
y
P
= 
1. Wankat, P.C., Separation Process Engineering, Second Edition, Prentice Hall, 2007. 
2. Methanex Corporation. "Technical Information & Safe Handling Guide for Methanol” September 2006. Accessed: 
February 2011. http://www.methanex.com/products/documents/TISH_english.pdf 
T (°C) 
H O2
satP (kPa) CH OH3
satP (kPa) 
64.5 24.521 100.74 
65 25.03 102.66 
66 26.57 106.73 
67.6 28.233 113.68 
69.3 30.328 121.06 
71.2 32.964 129.41 
73.1 35.772 139.55 
75.3 39.109 151.28 
78 43.866 167.02 
81.7 50.94 190.44 
84.4 56.577 209.96 
87.7 64.477 234.53 
89.3 68.417 248.34 
91.2 73.598 264.85 
93.5 80.227 285.04 
96.4 89.254 313.83 
100 101.325 351.38 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 3 Student View 
Jason M. Keith 
In this problem, A = Methanol and B = Water. As we can see, we cannot yet determine the 
composition of the vapor phase because we need to calculate the composition of the liquid phase first. 
This can be done using Equation 11.1-3 of Geankoplis, given by: 
 ( )A A B AP x P 1 x ____+ − = 
We can use the pressures PA and PB, using linear interpolation of the given data, at the temperature of 
72°C. Thus, 
 
A@T 72 C A@T 71.2 C
A@T 73.1 C A@T 71.2 C
P P _____ C _____ C
P P 73.1 C _____ C
= ° = °
= ° = °
− ° − °
=
− ° − °
 
Solving for the pressure A@T 72 CP = ° we determine the pressure to be: 
 A@T 72 CP _______ kPa= ° = 
Using the same procedure we can determine the pressure of water at 72°C to be: 
 B@T 72 CP ________ kPa= ° = 
Now we can substitute the pressure values into the equation for the total pressure, and solve for xA to 
yield: 
 ( )A A_______ x _______ 1 x 101.325+ − = 
 A99.53x _______= 
 Ax _______= 
Hence the molar fraction of water is given by: 
 B Ax 1 x 1 _______= − = − 
 Bx _______= 
Now we can enter the value of xA into Raoult’s Law to determine yA and yB. Thus, 
( )
A
_________ kPa _______
y
101.325 kPa
= 
Ay _______= 
B Ay 1 y 1 _______= − = − 
 By 0.109= 
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Vapor – Liquid Separation Processes 
Daniel López Gaxiola 4 Student View 
Jason M. Keith 
Example 11.3-1: Relative Volatility of Water – Methanol Mixture 
Determine the relative volatility of the methanol – water mixture described in problem 11.1-1 at the 
temperature of 72°C. 
Strategy 
Section 11.3B of Geankoplis gives an Equation for determining the relative volatility 
Solution 
The relative volatility of a mixture is defined by equation 11.3-3 of Geankoplis, shown below: 
 AAB
B
P
P
α = 
We can substitute the saturation pressures at the temperature of 72°C, which are given in Example 
11.1-1. Thus, the relative volatility is found to be: 
 AB
_________ kPa
_______ kPa
α = 
 AB _______α = 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 1 Student View 
Jason M. Keith 
Chapter 12 
Liquid – Liquid and Fluid – Solid Separation Processes 
This chapter includes examples of adsorption processes where one or more components of a gas or 
liquid stream are adsorbed on the surface of an adsorbent material. The following problem module 
illustrates pressure swing adsorption process for purifying hydrogen for proton – exchange membrane 
fuel cells. 
12.3-1 Hydrogen Purification in Pressure Swing Adsorption Process 
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Liquid – Liquid and Fluid – Solid Separation Processes 
Daniel López Gaxiola 2 Student View 
Jason M. Keith 
Example 12.3-1 Hydrogen Purification in Ethanol – Reforming Process 
The dry reformate stream coming from an ethanol reforming process contains 70 mole % H2 and 30 
mole % CO2. Lopes et al. [1] study the adsorption of CO2 on a packed bed of activated carbon with a 
length of 0.267 m. The bed contains 245.6 kg of adsorbent material consisting of activated carbon, 
with a particle diameter of 2.9 mm. The average flow rate of the dry reformate is 
3
5 m5 10
s
−
× and has 
a density of 
3
kg
0.587
m
. 
Use the data in the following table to determine the time required to reach the break – point 
concentration 
0
C
0.01
C
 
= 
 
, the time equivalent to the total capacity of the bed, and the time 
equivalent to the usable capacity of the bed up to the break – point time. Also calculate the length and 
capacity of unused bed after the break-point time. What is the saturation capacity of the bed? 
Breakthrough Concentration of CO2 in the packed bed 
t(s) C/C0 t(s) C/C0 
0 0 1493 0.8814 
27 0.0063 1640 0.9066 
187 0.0125 1827 0.9318 
480 0.0358 1933 0.9444 
600 0.1391 2107 0.9569 
613 0.3487 2240 0.9695 
640 0.4249 2373 0.9757 
667 0.4884 2560 0.9819 
747 0.5600 2707 0.9817 
853 0.6661 2853 0.9879 
973 0.7168 3120 0.9940 
1067 0.7611 3280 0.9938 
1200 0.8055 3440 1.0000 
1360 0.8561 3600 1.0000 
Strategy 
The break-through and capacity of the packed bed can be determined using the design equations 
given in Section 12.3D of Geankoplis. 
Solution 
First, the break – point time can be obtained from the tabulated data, at the point where 
0
C
0.01
C
= to 
be equal to tb = ______ s 
1. Lopes, F.V.S., Grande, C.A., Rodrigues, A.E., Chemical Engineering Science, 66, 303 – 317 (2011) 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 3 Student View 
Jason M. Keith 
The time tt for the total capacity of the bed can be calculated using the integral of the 
0
C
C
curve as a 
function of time: 
 
 
t 1 2
 0
0
C
t 1 dt A A
C
∞ 
= − = + 
 
∫ 
where A1 and A2 are the shaded areas shown in the following figure: 
 
 
 
As it can be seen in this figure, we can determine the break – through time by calculating the areas A1 
and A2. Thus, 
 ( ) ( )1A _____ s 0s 1 ___= − − 
 
1
A _____ s= 
 
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0 500 1000 1500 2000 2500 3000 3500 4000
C
/C
0
t(s)
A2 
A
1
 
C
b
 
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Liquid – Liquid and Fluid – Solid Separation Processes 
Daniel López Gaxiola 4 Student View 
Jason M. Keith 
To determine the area A2, we can use the trapezoidal method to calculate the area under the curve and 
subtract it from the total area A of the rectangle, as shown below: 
 
 
The area of the rectangle to the right of thebreak point can be calculated as follows: 
 ( ) ( )A ______ s _____ s 1 ___= − − 
 A 3413 s=
 
The results for the numerical integration using trapezoidal method will yield: 
 3
A _________ s=
 
Now we can obtain the area A2 to be: 
 2 3
A A A 3413 s _________ s= − = −
 
 2
A _________ s=
 
We can substitute the areas A1 and A2 into the equation for the break-through time tt, to obtain: 
 t
t ______ s _________ s= +
 
 
tt _________ s=
 
 
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0 500 1000 1500 2000 2500 3000 3500 4000
C
/C
0
t(s)
A2 
A3 
A1 
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Supplemental Material for Transport Process and Separation Process Principles 
Daniel López Gaxiola 5 Student View 
Jason M. Keith 
The time equivalent to the available capacity of the bed before the break – point time can be obtained 
as shown in the following steps: 
 
 t
u
 0
0
b C
t 1 dt ____
C
 
= − = 
 
∫
 
 
ut _____ s=
 
The length of the unused bed can be calculated using Equation 12.3-4 of Geankoplis: 
 
u
UNB T
t
t
H 1 H
t
 
= − 
  
where: 
 HT = Total length of the packed bed 
Substituting the times we calculated and the length of the packed bed into this equation, we get: 
 
( )UNB
_____ s
H 1 ______ m
_________ s
 
= − 
  
 
UNBH 0.212 m=
 
Finally, to determine the saturation capacity of the activated carbon in the bed, we need to obtain the 
moles of carbon dioxide adsorbed on the bed. 
The carbon dioxide can be obtained by multiplying the initial concentration of CO2 in the gas by the 
mass of gas in the time tt of _______ s. Thus, 
( )
3
52
3
2
2
2
kmol CO m kg gas
0.30 5 10 ________ s 0.587
kmol gas s m kg CO
Total CO adsorbed ____
kg gas kmol CO
_____
kmol gas
−   
×       
=  
 
 
2 2Total CO adsorbed ________ kg CO=
 
The molecular weight of the gas mixture was obtained by multiplying the molar fraction of each 
component by its corresponding molecular weight and adding the results, as shown below: 
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Liquid – Liquid and Fluid – Solid Separation Processes 
Daniel López Gaxiola 6 Student View 
Jason M. Keith 
 
2 2
gas H H CO CO
2 2
2 2 2 2
kg H kg CO
M y M y M ______ ____ 0.3 _____
kmol H kmol CO
   
= + = +   
    
 
gas
kg
M ________
kmol
=
 
Now we can divide the total CO2 adsorbed by the mass of activated carbon on the bed. Hence, 
2
________ kg CO
 Saturation Capacity
245.6 kg adsorbent
= 
2
kg CO
 Saturation Capacity _____________
kg adsorbent
= 
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	EL SOLUCIONARIO
	Chapter 1 - Student.pdf
	Chapter 2 - Student.pdf
	Chapter 3 - Student.pdf
	Chapter 4 - Student.pdf
	Chapter 5 - Student.pdf
	Chapter 6 - Student.pdf
	Chapter 7 - Student.pdf
	Chapter 9 - Student.pdf
	Chapter 11 - Student.pdf
	Chapter 12 - Student.pdf