CACHE Modules on Energy in the Curriculum
Energy Topic: Fuel Cells
Module:
Reactor Design Applied to a Solid Oxide Fuel Cell
Module Author:
Donald J. Chmielewski
Author Affiliation: Center for Electrochemical Science and Engineering
Department of Chemical and Biological Engineering
Illinois Institute of Technology, Chicago, IL 60616
Course:
Reaction Engineering
Text Reference:
Elements of Chemical Reaction Engineering, 2nd Ed.
by H. Scott Fogler (1992).
Primarily section 3.3 (the Stoichiometric Table) with slight
reference to 6.4 (Design of Reactors for Gas-Solid Reactions).
Concept Illustrated: Application of fuel cell reaction kinetics to standard reactor
design equations.
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 the ability to apply the
standard reactor design equations to the reaction kinetics of a fuel cell.
The SOFC reactions are:
Anode:
Cathode:
Overall:
Electron
Flow
(Current)
-
-
e
e
N2
H2
O2
N2
H2
H2
O2
O2-
H2O
H2
H2 H2O
O2O2-
O2
H2 &
H2O
Out
O2
Anode
Cathode
Electrolyte
Figure 1: Reactions within SOFC
Draft 4
Cell Voltage
Air
In
Anode
Gas
Chamber
Cathode
Gas
Chamber
Fuel Cell
N2
H2O
Electric Load
H2
In
O2
O2-
H2O
H2 + O-2  H2O + 2 e1/2O2 + 2 e-  O-2
H2 + 1/2O2  H2O
Air
Out
Figure 2: Flow Diagram for SOFC
-1-
Oct 20, 2009
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 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
In many cases, one can think of a fuel cell as a chemical reactor. The primary difference
is that the rate of reaction term will be a bit more complicated than the kinetic
expressions usually encountered in the chemical engineering literature. In this module we
will use one such rate expression and illustrate its use within the standard reactor design
equations.
As indicated in the background section above, there is a proportional relation between
reaction rate and fuel cell current. If we define ri as the generation rate of species i per
unit area of the cell (typically with units of mole / s cm2) then:
j
 rH 2  rH 2O  2rO 2
2F
(1)
where j is as current density (defined as total current per unit area of the cell). Then,
current density is given by:
jcell = (Ener - Ecell ) / Rint
(2)
where Ecell is the operating voltage of the cell and Rint =  /is the area specific
resistance (is ionic conductivity of the electrolyte (with units of cm-1ohm-1) and  is the
electrolyte thickness). The Nernst voltage is given by:
E ner
1/ 2
RT  PH 2 PO2
 Eo 
ln
2 F  PH 2O




(3)
where Eo = -ΔGºrxn / 2F, ΔGºrxn is the Gibbs free energy of the reaction H2 + ½ O2 
H2O (evaluated at standard pressures and the operating temperature of the cell) and the
Pi’s are the partial pressures at the reaction sites. By substituting Equations 2 and 3 into
Equation 1, we arrive at the following compact expression for the reaction rate:
Draft 4
-2-
Oct 20, 2009
 rH 2

 P P1 / 2
 E  RT ln  H 2 O2
 o 2 F  PH O
2


 /  2 F


  E cell 




(4)
In section 2.2 of Fogler (1992), the reactor design equations for per volume reaction rates
are given as:
CSTR: V 
FA, o X
 rA ( X )
X
PFR:
dX
 rA ( X )
0
V  FA,o 
Since the reaction rate for a fuel cell is given on a per area basis, the corresponding
reactor design equations are determined to be:
CSTR: Acell 
FH 2,o X
 rH 2 ( X )
X
PFR:
dX
 rH 2 ( X )
0
Acell  FH 2,o 
(See section 6.4 of Fogler (1992) for a similar development. For the PFR case, a cocurrent configuration was assumed with regard to the anode and cathode gas chambers.)
Example Problem Statement:
An SOFC is operated at atmospheric pressure and 973K with the following inlet molar
flow rates (all in mole/s): FH2,o = 5.0 and FH2O,o = 0.5, FO2,o = 30 and FN2,o = 113. At
973K, ΔGºrxn = -194 kJ /mole of H2, ΔHºrxn = -248 kJ / mole of H2 and  = 0.05 cm-1
ohm-1. The electrolyte thickness is 20m. and the cell is operated at 0.75 volts. Assume
the fuel cell is configured as two CSTR gas chambers separated by the electrode /
electrolyte assembly (similar to figure 2).
1) If a 90% conversion of H2 is desired, determine the required electrolyte area, Acell.
2) Using the cell area determined in part 1, determine the resulting H2 conversion if
the cell voltage is changed to 0.8 volts.
Draft 4
-3-
Oct 20, 2009
Example Problem Solution:
Part 1): Defining X as the conversion of H2 (moles reacted per mole fed), we can
construct the following table using the overall reaction: H2 + 1/2O2  H2O.
Table 1: Stoichiometric Table for Example Problem
Species
Feed rate
to reactor
(mole/s)
Change in
reactor
(mole/s)
Effluent rate
from reactor
(mole/s)
Concentration
(mole/m3)
H2
FH2,o
-FH2,oX
FH2,o(1-X)
FH2,o(1-X)/(a)
H2O
H2O FH2,o
+FH2,o X
FH2,o(H2O +X)
FH2,o(H2O +X)/(a)
Total
Anode
Gas
FT,o(a) =
FH2,o+ FH2O,o
0
FT(a) = FT,o(a)
-
O2
FO2,o =
O2 FH2,o
- 1 FH2,o X
FH2,o(O2 - 1 X)
FH2,o(O2 - 1 X)/(c)
N2
FN2,o =
N2 FH2,o
0
FH2,oN2
FH2,o N2 /(c)
Total
Cathode
Gas
FT,o(c) =
FO2,o+ FN2,o
- 1 FH2,o X
FT(c) =
FT,o(c) - 1 FH2,o X
-
FH2O,o =
2
2
2
2
2
where H2O = FH2O,o / FH2,o, O2 and N2 are similarly defined. To determine (a) and (c)
(the volumetric flow rates exiting the anode and cathode chambers, respectively), we
employ Equation 3-41 of Fogler (1992), once for each chamber:
Draft 4

(a)

(c)

(a)
o
 FT( a )

 F (a)
 T ,o
 Po ( a )

 P ( a )

 T ( a )

 T ( a )
 o

 (a)
   o( a )  T

 T (a)

 o
 FT( c )  Po ( c )  T ( c ) 
   ( c )  ( c )  ( c ) 
F  P
 To 
 T ,o 
 (c) 1

 FT ,o  FH 2,o X  T ( c )
(c)
2
 ( c )
 o 
FT(,co)

 To








(c)
o

 ( c )

F
   o( c ) 1  1 H 2,o X  T

 2 F ( c )  T ( c )
T ,o


 o
-4-




Oct 20, 2009
Now recall that the partial pressure of species i in chamber j is given by Pi = CiRT(j),
where Ci is the concentration expression developed in the last column of table 1.
Similarly, Pi,o = (Fi,o /o(j)) RTo(j) is the inlet partial pressure. The pressures of interest are:
PH 2  C H 2 RT ( a ) 
FH 2,o (1  X )

(a)
o
PH 2O  C H 2O RT ( a ) 
T
(a)
To
(a)

RT ( a ) 
FH 2,o ( H 2O  X )

 o( a ) T ( a ) To ( a )
 RT
FH 2,o (1  X )

(a)
1
X)
2
 T ( c )
X  ( c )
 T
 o
FH 2,o ( O 2 
PO 2  C O 2 RT ( c ) 

 o( c ) 1 

1 FH 2,o
2 FT(,co)




(a)
o
RTo
(a)
 PH 2,o (1  X )
 PH 2,o ( H 2O  X )
RT ( c )
1
PO 2,o  O 2 ( O 2  1 X ) PO 2,o ( O 2  1 X )
X)
(
c
)
2
2
2

RTo 

1




F
1 FH 2,o 
 O 2  xO 2 ,o X
1  1 H 2 , o X 
 o( c ) 1 
X
(c)
(c)
2



2
2
F
F
T ,o
T ,o




FH 2,o ( O 2 
where xO2,0 is the mole fraction of O2 and the cathode inlet and Pi,o is the partial pressure
of species i at the inlet. These can now be substituted into the previous reaction rate
expression (Equation 4):

 P P1 / 2
 E  RT ln  H 2 O2
 o 2 F  PH O
2

 rH 2 ( X )  
 /  2 F
Eo 



  E cell 




1/ 2
1/ 2
RT  (1  X ) PO 2,o ( O 2  X / 2) 
ln 
  E cell
2 F  ( H 2O  X )  O 2  xO 2,o X / 2 1 / 2 
 /  2 F
(5)
1/ 2
1/ 2
 194000  8.314  973  (1  X ) 0.21 (6  X / 2) 
ln 
 0.75


1/ 2 
 2  96485  2  96485  (0.1  X ) 6  0.21  X / 2 

0.00002 / 5  2  96485
Draft 4
-5-
Oct 20, 2009
Then evaluation of equation 5 at X = 0.9 gives
1.005 
 rH 2 ( X  0.9) 
8.314  973  (1  0.9) 0.211 / 2 (6  0.9 / 2)1 / 2 
ln 
  0.75
2  96485  (0.1  0.9) 6  0.21  0.9 / 21 / 2 
0.00002 / 5  2  96485
 0.1617 moles / m 2 s
Finally, we combine with the CSTR design equation to find the required area.
Acell 

FH 2,o X
(6)
 rH 2 ( X )
5  0.9
 27.8 m 2
 rH 2 (0.9)
At this point we can also calculate the amount of power being produced. We start by
determining current density:
j  rH 2  2 F  0.1617  2  96485  3.12  10 4 A / m 2
Then the total power is found to be
Pcell = Ecell Icell = Ecell j Acell = 0.75 x 3.12x104 x 27.8 = 651 kW
Part 2): If cell voltage is changed to 0.8 the resulting H2 conversion can be determined
either graphically or via a numeric search.
Graphical Method: A plot of Acell(X) for part 1) is given in figure 3 (the curve denoted as
Ecell = 0.75). The solution of Acell = 27.8 m2 is found as the point corresponding to a
conversion of 0.9. The second curve (Ecell = 0.8) was generated by changing Ecell in
equation 5 and substitution into equation 6. Then based on the second curve, the
conversion corresponding to an area of 27.8 m2 is found to be about 0.79. The following
Matlab code was used to generate the plots of figure 3.
clear
FH2o=5.0; FH2Oo=0.5; FO2o=30; FN2o=113;
ThetaH2O=FH2Oo/FH2o; ThetaO2=FO2o/FH2o; xO2o=FO2o/(FO2o+FN2o);
Rbar=0.00002/5; Eo=194000/(2*96485); R=8.314; T=973;
Ecell=0.75;
for ii=1:999
X=ii/1000;
XX_75(ii)=X;
Draft 4
-6-
Oct 20, 2009
rH2=-((Eo-Ecell+(R*T/(2*96485))*log( ...
((1-X)/(ThetaH2O+X))*(0.21*(ThetaO2-X/2)/ ...
(ThetaO2-xO2o*X/2))^0.5))/(Rbar*2*96485))
Acell_75(ii)=FH2o*X/-rH2;
end
Ecell=0.8;
for ii=1:999
X=ii/1000;
XX_8(ii)=X;
rH2=-((Eo-Ecell+(R*T/(2*96485))*log( ...
((1-X)/(ThetaH2O+X))*(0.21*(ThetaO2-X/2)/ ...
(ThetaO2-xO2o*X/2))^0.5))/(Rbar*2*96485));
Acell_8(ii)=FH2o*X/-rH2;
end
plot(XX_75,Acell_75,XX_8,Acell_8)
axis([0 0.95 0 30])
30
25
2
Cell Area (m )
20
15
Ecell = 0.8
10
5
Ecell =0.75
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Conversion
Figure 3: SOFC Cell Area as a function of conversion for the CSTR case.
Numeric Method: The following Matlab code calculates the area required to achieve 90%
H2 conversion if Ecell = 0.75 (the solution to part 1).
clear
FH2o=5.0; FH2Oo=0.5; FO2o=30; FN2o=113;
ThetaH2O=FH2Oo/FH2o; ThetaO2=FO2o/FH2o; xO2o=FO2o/(FO2o+FN2o);
Rbar=0.00002/5; Eo=194000/(2*96485); R=8.314; T=973;
Draft 4
-7-
Oct 20, 2009
Ecell=0.75; X=0.9;
rH2=-((Eo-Ecell+(R*T/(2*96485))*log( ((1-X)/(ThetaH2O+X)) ...
*(0.21*(ThetaO2-X/2)/(ThetaO2-xO2o*X/2))^0.5))/(Rbar*2*96485))
Acell=FH2o*X/-rH2
j=-rH2*(2*96485)
P=Ecell*j*Acell
To determine the conversion assuming Acell = 27.8 m2 and Ecell = 0.8 we employ the
following bisection search algorithm. Using this code the solution is found to be: X =
0.793.
clear
FH2o=5.0; FH2Oo=0.5; FO2o=30; FN2o=113;
ThetaH2O=FH2Oo/FH2o; ThetaO2=FO2o/FH2o; xO2o=FO2o/(FO2o+FN2o);
Rbar=0.00002/5; Eo=194000/(2*96485); R=8.314; T=973;
Acell=27.8; Ecell=0.8;
Xmin=0; Xmax=1; Xguess=(Xmin+Xmax)/2;
for i=1:10
X=Xguess;
f_of_Xguess=FH2o*X/((Eo-Ecell+(R*T/(2*96485))*log( ...
((1-X)/(ThetaH2O+X))*(0.21*(ThetaO2-X/2)/ ...
(ThetaO2-xO2o*X/2))^0.5))/(Rbar*2*96485)) - Acell;
if f_of_Xguess > 0
Xmax=Xguess;
else
Xmin=Xguess;
end
Xguess=(Xmin+Xmax)/2
end
Bisection Search: The bisection search algorithm is a fairly standard technique to obtain
roots of a nonlinear equation. A quick search of the internet will yield numerous
descriptions of the scheme. Here is the basic idea: Assume you are given a continuous
function f(x) along with an interval of interest xmin < x < xmax. If f(xmin) < 0 and f(xmax) > 0,
then there must be some value of x between xmin and xmax where the function is equal to
zero. This zero crossing is the point we seek. Now bisect the interval xguess = (xmin +
xmax)/2. If f(xguess) > 0 then the zero crossing must be to the left of xguess. In this case, xmax
should be replaced with xguess. If f(xguess) < 0 then xmax should be replaced with xguess. In
either case the interval reduced by half. Repeating this process will again reduce the
interval by half. Repeating the process 10 times will reduce the interval by 210=1024. In
such a case, the difference between the actual zero crossing and the final xguess will be at
most 0.00098 (clearly sufficient accuracy of this application).
Draft 4
-8-
Oct 20, 2009
Home Problem Statement:
1) In many cases the flow of air through the cathode chamber serves the dual
purpose of cooling the fuel cell as well as providing oxygen to the reaction. In
these cases the conversion of oxygen will be quite low, due to the very high flow
rate. Using the specifications of the example problem, compare the part 1 results
with the case of assuming no change in oxygen partial pressure.
2) Repeat the calculations of the example problem assuming a co-current PFR
configuration. (Hint: Use numeric integration to make plots of Acell as a function
of conversion. Then identify the required values visually. Alternatively, one could
develop a non-graphical solution, again based on the bisection search.)
Draft 4
-9-
Oct 20, 2009