CACHE Modules on Energy in the Curriculum
Fuel Cells
Module 2 (Final Draft): Material 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 4.7
Concept Illustrated: Material balances on a reactive process with complex geometry;
Extension of balance and stoichiometry concepts to electrons.
Problem Motivation: 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 the design of an
SOFC is an understanding of the fuel and oxidant utilization as well as the amount of
current generated.
The SOFC reactions are:
H2 + O-2  H2O + 2 e1/2O2 + 2 e-  O-2
H2 + 1/2O2  H2O
Electron
Flow
(Current)
e-
eH2
N2
O2
O
H2O
Anode:
Cathode:
Overall:
O2
H2
H2
O2
O2-
H2O
H2O
H2 H2O
O2O
2-
Air
In
Anode
Gas
Chamber
Cathode
Gas
Chamber
Fuel Cell
N2
H2
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 (moles of electrons/s) to electrical current
(coulombs/s or amps), one would use Faraday’s constant: F  96,485 coulombs / mole of
electrons. The primary objective of a fuel cell is to deliver energy to the electric load. To
calculate the energy delivery rate (also know as power) one would multiply the current
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times the cell voltage: Power = Current · Voltage. (Recall the unit conversions:
coulomb  volt  Joule and Joule / s  Watt ).
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Problem Information
Example Problem Statement: A SOFC is operated with an inlet flow of 20 g/s of pure
hydrogen and an inlet flow of 1450 g/s of air. If the fuel utilization is 50%, then
determine the following:
1) The mass flow rate out of the anode gas chamber.
2) The mass flow rate out of the cathode gas chamber and the oxygen utilization.
3) The current through the electric load.
4) The power delivered to the load if the cell voltage was 0.8 volts.
The term utilization is synonymous with the percent conversion, as defined in Section 4.6
of Felder and Rousseau (2000).
20 g/s H2
Anode Gas Chamber
1450 g/s Air
Cathode Gas Chamber
Figure 3: Flow Diagram for Example Problem
Example Problem Solution:
1) To determine the mass flow out of the anode we will need to know the flow rate of H2
in, the rate of H2 consumption and the rate of H2O production. Then application of a
material balance around the anode will yield the flow out of the anode. Although a mass
balance is possible, we will go the route of a mole balance. As such we begin by
converting the anode inlet mass flow rate into a molar flow:
20 g H 2 fed 1 mole H 2 10 mole H 2 fed


s
2 g H2
s
The amount of H2 consumed is determined as
0.5 mole H 2 reacted 10 mole H 2 fed 5 mole H 2 reacted


mole of H 2 fed
s
s
Thus, 5 moles/s of H2 will be converted to H2O. The remaining 5 moles/s of H2 will then
exit with the generated steam. Converting back to mass flows, we have:
5 mole H 2 exiting
2 g H2
10 g H 2 exiting


s
1 mole H 2
s
5 mole H 2 O exiting 18 g H 2 O
90 g H 2 O exiting


s
1 mole H 2 O
s
for a total of 100 g/s exiting the anode.
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2) To determine the mass flow out of the cathode, we will again apply a material balance,
but this time around the cathode chamber. Once again we will need the flow of O2 and
the rate of O2 consumption, which through a balance will give the flow rate out.
Assuming the mass fraction of air is 76.7% N2 and 23.3% O2, we find the cathode inlet
molar flows to be:
1450 g of air fed 0.767 g N 2 mole N 2 39.7 mole N 2 fed



s
g of air
28 g N 2
s
1450 g of air fed 0.233 g O2 mole O2 10.6 mole O2 fed



s
g of air
32 g O2
s
From the stoichiometry of the overall reaction, we find that the 5 moles/s of H2 converted
in the anode must be matched by 2.5 moles/s of O2 consumed in the cathode. Since 10.6
moles/s of O2 are fed and 2.5 mole/s are consumed, 8.1 moles/s of O2 must exit. Thus, the
oxygen utilization (i.e., conversion of oxygen) is calculated as:
Utilizatio n  100 
2.5 moles of O2 reacted
 23.6%
10.6 moles of O2 fed
Converting back to mass flow, we have:
8.1 mole O2 exiting 32 g O2
259 g O2 exiting


s
1 mole O2
s
Adding this to the inert flow of N2 (1450 g air/s x 0.767 = 1112 g N2/s), gives a total of
1371 g/s exiting the cathode chamber.
3) Looking at the anode stoichiometry, we find that 2 moles of electrons are sent to the
load for every mole of hydrogen consumed. Thus, the electron flow is 10 moles/s. If we
now employ Faraday’s constant, F  96,485 coulombs / mole of electrons, we find
10 mole e  96,485 coulombs 964,850 coulombs


s
s
mole e 
which is equal to a current of 964,850 amps (1 amp = 1 coulomb/sec).
4) Application of the relation: Power = Current · Voltage, yields
964,850 coulombs
770,000 coulomb volts
 0.8 volts 
s
s
which is equal to a power of 770,000 J/s or 0.77 mega-watts (MW).
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Home Problem Statement:
A SOFC is operated with an anode exit flow of 2 g/s (hydrogen and water vapor) and an
inlet of pure hydrogen. If the cell is operated at 0.75 volts and delivers 10kW of power,
determine the following:
1) The mass flow rate into the anode gas chamber.
2) The mass flow rate into the cathode chamber if 20% oxygen utilization is desired.
m a ,in
Anode Gas Chamber
m c ,in
Cathode Gas Chamber
m a ,out  2 g/s
m c ,out
Figure 4: Flow Diagram for Home Problem
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