Module Title: 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 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 overall SOFC reaction is H2 + 1/2O2  H2O.
e
e
H2
N2
O2
O
H2O
Electron
Flow
(Current)
-
-
O2
H2
H2
O2
O2-
H2O
H2
H2 H2O
O2O2-
Air
In
Anode
Gas
Chamber
Cathode
Gas
Chamber
Fuel Cell
N2
H2O
Cell Voltage
H2
In
2-
N2
Electric Load
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
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 via reaction, which 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 over energy available.
Draft 2
-1-
July 31, 2007
Problem Information
Example Problem Statement: An atmospheric pressure, adiabatic SOFC is operated
with an inlet flow of pure hydrogen at 20 g/s and a hydrogen utilization of 75%. 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) If 75% of the hydrogen is utilized, then
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 indicates that 15 moles of electrons/s will be delivered to the load, which, via
Faraday’s constant, is equal to 1.45x106 amps. This current multiplied by the cell voltage,
then 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) Assuming air is an ideal gas at standard conditions (1 atm and 0oC), we find:
1 atm
n
P


 0.045moles / L
V RT
0.08206 L  atm  mol 1  K 1 273K 


and
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 of H2 are reacted per
second we conclude that 3.75 moles of O2 are reacted per second. Thus, the exit flow
from the cathodes must be 196.5 moles/s.
3) 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.
This leaves:
H  W shaft,out
Draft 2
(E-1)
-2-
July 31, 2007
From part (1) we have that W shaft,out =1MW. To determine H we turn to equation 9.5-1a
of Felder and Rousseau (2000):
H  Hˆ ro 
 n Hˆ   n Hˆ
i
i
outlet
i
(9.5-1a)
i
inlet
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 equation (E-1) we find:
 1MW  W shaft,out  Hˆ ro 
 n Hˆ   n Hˆ
i
i
outlet
i
i
inlet
 7.5 mole H 2 reacted   241800 J 



s

 mole H 2 reacted 
  n Hˆ   n Hˆ
i
i
outlet
i
i
inlet
which gives:
0.8MW 
 n Hˆ   n Hˆ
i
outlet

i
i
i
inlet
 n H 2 ,out Hˆ H 2 , Ta ,o u t  n H 2O ,out Hˆ H 2O , Ta ,o u t  n N 2 ,out Hˆ N 2 , Tc ,o u t  n O2 ,out Hˆ O2 , Tc ,o u t

 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


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 
Draft 2
-3-
July 31, 2007
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.
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 out 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
1 MW

 56%
Enthalpy Re leased 1.8 MW
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
Power to the Load
1 MW

 42%
Combustion Enthalpy in the H 2 Feed 2.4 MW
Draft 2
-4-
July 31, 2007
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.
Home Problem Solution:
From the cathode stream flow rates, we conclude that 67 slpm of O2 are being reacted.
This amounts to 0.05 moles / sec of O2 and a hydrogen consumption / water production
rate of 0.1 mole / sec, which yields an electron flow of 0.2 moles / sec (or 19,297 amps of
current). Converting all of the flows to molar flow we find:
n H 2 ,in  200 slpm H 2  0.045 / 60  0.15 mole H 2 / s
n N 2 ,in  2700 slpm Air  0.79  0.045 / 60  1.60 mole N 2 / s
n O2 ,in  2700 slpm Air  0.21  0.045 / 60  0.43 mole O2 / s
n H 2 ,out  0.15 mole H 2 fed / s  0.1 mole H 2 reacted / s  0.5 mole H 2 exiting / s
n H 2O ,out  0 mole H 2 O fed / s  0.1 mole H 2 O produced / s  0.1 mole H 2 O exiting / s
n N 2 ,out  1.60 mole N 2 / s
n O2 ,out  0.43 mole O2 fed / s  0.05 mole O2 reacted / s  0.38 mole O2 exiting / s
Combining equations 7.4-15 and 9.5-1a of Felder and Rousseau (2000), we find the
following expression for the shaft work delivered to the system.
 W shaft,out  Hˆ ro 
 n Hˆ   n Hˆ
i
i
outlet
i
i
inlet
 0.1 (241kJ / mole H 2 )
 n H 2 ,out 25.93  n H 2O ,out 33.32   n N 2 ,out 25.53  n O2 ,out 27.12 
 n H 2 ,in 22.85  n N 2 ,in 22.23  n O2 ,in 23.61
 24.1 kJ / s
 0.0525.93  0.1033.32   1.6025.53  0.3827.12 
 0.1522.85  1.6022.23  0.4323.61
 17.466 kJ / sec
Thus, 17,466 watts of power are delivered to the load. Dividing this by the 19,297 amps
of current, we find the cell voltage to be 0.905 volts.
Draft 2
-5-
July 31, 2007