CACHE Modules on Energy in the Curriculum
Fuel Cells
Module 3 (Final Draft): Energy Balance in a Solid Oxide Fuel Cell
Module Author: Donald J. Chmielewski
Module Affiliation: Center for Electrochemical Science and Engineering
Department of Chemical and Biological Engineering
Illinois Institute of Technology, Chicago, IL 60616
Course:
Material and Energy Balances
Text Reference:
Felder and Rousseau (2000), Section 9.5
Concept Illustrated: Energy balances on a reactive process with complex geometry;
Application of shaft work to a reactive process.
Background/Introduction
Fuel cells are a promising alternative energy conversion technology. One type of fuel
cell, the Solid Oxide Fuel Cell (SOFC) uses hydrogen as a fuel. The fuel reacts with
oxygen to produce electricity. Fundamental to SOFC design is an understanding of the
heat generated by the reaction and its impact on efficiency.
The SOFC reactions are:
H2 + O-2  H2O + 2 e1/2O2 + 2 e-  O-2
H2 + 1/2O2  H2O
Electron
Flow
(Current)
e-
e-
N2
H2
2-
O2
N2
H2
H2
O2
O2-
H2O
H2O
H2 H2O
O2O
2-
Cell Voltage
Air
In
Anode
Gas
Chamber
Cathode
Gas
Chamber
Fuel Cell
N2
H2
Electric Load
H2
In
O2
O
H2O
Anode:
Cathode:
Overall:
O2
H2 &
H2O
Out
O2
Anode
Cathode
Electrolyte
Figure 1: Reactions within SOFC
Air
Out
Figure 2: Flow Diagram for SOFC
For each mole of hydrogen consumed, two moles of electrons are passed through the
electric load. To convert electron flow, Faraday’s constant should be used
( F  96,485 coulombs/mole of electrons). The objective of a fuel cell is to deliver power
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to the load: Power = Current · Voltage. ( coulomb  volt  joule and joule / s  watt ). The
fuel cell obtains this power from the enthalpy released during the overall reaction H2 +
1/2O2  H2O; however, only a portion of this enthalpy can be converted to electric
power, the remainder will appear as heat released by the reaction. This heat must be
removed using the flowing gas streams. The performance of a fuel cell is typically
communicated in terms of efficiency, defined as energy delivered to the load divided by
the energy available from reaction.
Problem Information
Example Problem Statement: An adiabatic SOFC is operated at atmospheric pressure
with an inlet flow of pure hydrogen at 20 g/s and a hydrogen utilization of 75%. The term
utilization is synonymous with the percent conversion, as defined in Section 4.6 of Felder
and Rousseau (2000). At the cathode chamber inlet, 2.67x105 standard liters per minute
(slpm) of air is fed at 500oC, and the gas exiting the cathode chamber is at 625oC. If the
exit stream of the anode chamber is 675oC and the cell voltage is 0.7 volts, then
determine the following:
1) The electric current and power delivered to the load.
2) The molar flow from the cathode chamber.
3) The temperature of the gas inlet to the anode.
4) The fuel to power efficiency of the fuel cell.
Example Problem Solution:
1) To determine power to the load we must first determine current to the load and
combine with cell voltage (given to be 0.7 volts). Since current is proportional to the rate
of electrons generated by the half reaction (here we will use the anode reaction), we must
first determine the conversion rate for the anode reaction. If 75% of the hydrogen is
utilized, then the conversion rate,  , is found as:
 =
20 g H 2 fed mole of H 2 0.75 mole H 2 reacted 7.5 mole H 2 reacted



s
2 g H2
mole of H 2 fed
s
This rate along with the anode stoichiometry (2 mole of electrons per mole of H2)
indicate that 15 moles/s of electrons must be delivered to the load. If we now use
Faraday’s constant as a unit conversion, we find the current to be (15 mole of electrons/s)
x (96485C/mole of electrons) = 1.45x106 amps. This current multiplied by the cell
voltage (0.7 volts), gives the load power as 1MW. In the notation of chapter 7 of Felder
and Rousseau (2000), this power should be considered shaft work being removed from
(or being done by) the system.
2) The molar flow out of the cathode is equal to the molar flow into the cathode minus
the molar flow of oxygen through the electrolyte and into the anode. We start by
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calculating the molar flow into the cathode. Assuming air is an ideal gas at standard
conditions (1 atm and 0oC), we find that the number of moles per standard liter is:
1 atm
n
P


 0.045moles / L
V RT 0.08206 L  atm  mol 1  K 1 273K 
and thus the molar flow for 2.67x105 slpm is:
2.67  10 5 L 0.045mole min


 200.25moles / s
min
L
60s
Using the reaction stoichiometry and our calculation that 7.5 moles/s of H2 are reacted in
the anode, we conclude that 3.75 moles/s of O2 are reacted in the cathode. Thus, the exit
flow from the cathode must be 200.25 - 3.75 = 196.5 moles/s.
3) Now that we have performed some initial calculations concerning material and energy
flows, we turn to the energy balance of the entire system. After identifying the
appropriate balance equation, we find the need to calculate the reaction generated
enthalpy, which will leave us with one equation (the energy balance) and one unknown
(the temperature inlet to the anode).
Since the SOFC is a continuous (or open) process, we should apply Equation 7.4-15 of
Felder and Rousseau (2000):
(7.4-15)
H  E k  E p  Q in  W shaft,out
Neglecting E k , E p and Q in , the last being due to assumption of adiabatic operation,
leaves:
(E-1)
H  W shaft,out
To determine H we turn to Equation 9.5-1a of Felder and Rousseau (2000):
H  Hˆ ro 
 n Hˆ   n Hˆ
i
outlet
i
i
(9.5-1a)
i
inlet
From part (1) we have that W shaft,out =1MW. Then using Table B.1 of Felder and Rousseau
(2000) for the reaction H2 + 1/2O2  H2O(g), we find the enthalpy of reaction, ( Ĥ ro ), to
be -241.8 kJ/mole of H2 converted. Combining this with the conversion rate  we find:
 7.5 mole H 2 reacted   241800 J 
  -1800kJ/s = -1.8MW

s

 mole H 2 reacted 
Hˆ ro  
Applying this to Equation 9.5-1a and combining with Equation E-1 gives:
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 1MW  1.8MW 
 n Hˆ   n Hˆ
i
i
outlet
or
0.8MW 
 n Hˆ   n Hˆ
i
outlet

i
i
i
inlet
i
i
inlet
 n H 2 ,out Hˆ H 2 ,Ta ,out  n H 2O ,out Hˆ H 2O , Ta ,out  n N 2 ,out Hˆ N 2 ,Tc ,out  n O2 ,out Hˆ O2 , Tc ,out

 n H 2 ,in Hˆ H 2 , Ta ,in  n N 2 ,in Hˆ N 2 ,Tc ,in  n O2 ,in Hˆ O2 ,Tc ,in


(The subscripts of T correspond to ‘a’ for anode ‘c’ for cathode ‘in’ for inlet and ‘out’ for
outlet.) Since we know all of the molar flows into and out of the system, each of
the ni terms is known. Summarizing from parts (1) and (2), we have
n H 2 ,in 
20 g H 2 fed mole of H 2 10 mole H 2 fed


s
2 g H2
s
n N 2 ,in 
200.25 mole air fed 0.79 mole of N 2 158.2 mole N 2 fed


s
mole of air
s
200.25 mole air fed 0.21 mole of O2 42.05 mole O2 fed


s
mole of air
s
10 mole H 2 fed 7.5 mole H 2 reacted 2.5 mole H 2 exiting
n H 2 ,out 


s
s
s
0 mole H 2 O fed 7.5 mole H 2 O produced 7.5 mole H 2 O exiting
n H 2O ,out 


s
s
s
158.2 mole N 2 fed 0 mole N 2 reacted 158.2 mole N 2 exiting
n N 2 ,out 


s
s
s
42.05 mole H 2 fed 3.75 mole O2 reacted 38.3 mole O2 exiting
n O2 ,out 


s
s
s
n O2 ,in 
Additionally, we can utilize Table B.8 of Felder and Rousseau (2000) to determine Ĥ i for
all of the streams except for the inlet to the anode, since temperature of this stream is not
known. (Linear interpolation used whenever needed.)
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Hˆ H 2 , Ta ,o u t  Hˆ H
o
2 , 675 C
 19.0 kJ / mole H 2
Hˆ H 2O , Ta ,o u t  Hˆ H O , 675o C  23.9 kJ / mole H 2 O
2
Hˆ N 2 , Tc ,o u t  Hˆ N
 18.2 kJ / mole N 2
o
2 , 625 C
Hˆ O2 , Tc , o u t  Hˆ O , 625o C  19.3 kJ / mole O2
2
Hˆ H 2 , Ta ,in  to be found kJ / mole H 2
Hˆ N 2 , Tc ,in  Hˆ N
 14.2 kJ / mole N 2
o
2 , 500 C
Hˆ O2 , Tc ,in  Hˆ O
o
2 , 500 C
 15.0 kJ / mole O2
This gives:
800 kJ / s 
 n Hˆ   n Hˆ
i
outlet

i
i
i
inlet
 n H 2 ,out Hˆ H 2 , Ta ,out  n H 2O ,out Hˆ H 2O , Ta ,out  n N 2 ,out Hˆ N 2 , Tc ,out  n O2 ,out Hˆ O2 , Tc ,out

 n H 2 ,in Hˆ H 2 ,Ta ,in  n N 2 ,in Hˆ N 2 , Tc ,in  n O2 ,in Hˆ O2 , Tc ,in


 2.5  19.0  7.5  23.9  158.2  18.2  38.3  19.3
 10  Hˆ , 158.2  14.2  42.05  15.0


H 2 Ta , in
Solving this equation for Hˆ H 2 ,Ta ,in , yields 16.8 kJ/mole H2. From Table B.8 of Felder and
Rousseau (2000), this enthalpy is achieved at 600oC, which would need to be the inlet
temperature to the anode. (It should also be noted that implicit in our application of
Equation 9.5-1a, we have used 1 atm and 25oC as the reference state, which was dictated
by the data received from Tables B.1 and B.8 of Felder and Rousseau (2000).)
4) If we define cell efficiency as the ratio of useful power to chemical energy input, we
find
Power to the Load W shaft,out
1 MW


 0.56
o
Enthalpy Re leased  Hˆ
1.8 MW
r
This assumes the un-utilized hydrogen can be recycled. If not, and we are assuming the
un-utilized hydrogen is lost, then a more appropriate efficiency value would be based on
total conversion of the hydrogen,  = 10 mole of hydrogen converted / s. The result of
this assumption is be to calculate Ĥ o as (10 mole/s of H2) x (-241.8 kJ/mole H2),
r
which gives the new efficiency as:
W shaft,out
Power to the Load
1 MW


 0.42
o

ˆ
Combustion Enthalpy in the H 2 Feed  H r
2.4 MW
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Home Problem Statement:
An atmospheric pressure, adiabatic SOFC is operated with the following inlet and exit
conditions:
Anode In:
Anode Out:
Cathode In:
Cathode Out:
200 slpm of pure H2 at 800oC
200 slpm of H2 and H2O at 900oC
2700 slpm of air at 750oC
2633 slpm at 850oC
Determine the cell voltage.
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