
502 Fuel Cells
From the over energy balance equation, we get
TT
R
TT
R
Q
a
ce ch,a ce ch,c
c
gen
−
+
−
=
TT
ce ce
−
+
−
=×××
−
300
400
35 10 10 10
.
Solving for the cathode–electrolyte interface temperatures:
q
T
R
T
R
ce
gen
ch,a
a
ch,c
c
=
++
+
∑∑
T
ce
=
×+ +
+
35 10
300
0 2937
400
0 2178
1
0 2937
1
0 2178
3
.
..
T
ce
= 795.1687°C.
Heat transfer to the anode and cathode side is computed as follows:
q
TT
hA
L
kA
L
kA
a
ce ch,a
a
e
=
−
++
=
−
1
795 1687 300
02
.
.
9937
= 1685.9676 W
and
TT
hA
L
kA
c
ce ch,c
c
=
−
+
=
−
=
1
795 1687 400
0 21784
18
.
.
114 0318
.
Temperature at the anode–electrolyte is computed as
q
TT
hA
L
kA
a
ae ch,a
a
=
−
+
1
9676
300
.
=
−
T
ae
T
ae
= ...