Page 521 - Advanced thermodynamics for engineers

514 CHAPTER 21 FUEL CELLS

Solution
This question requires the application of Eqn (21.42), giving

8 9
> !>
>
<T < p rH 2 O =
>
E ¼ ln K p r ln
p
z i F > 1=2 >
> >
: p rH 2 rO 2 ;
From tables at 500 K, K pr ¼ 7.92127 10 22 bar 1/2 . Hence the open circuit potential is
( !)
8:3143 500 22 1=2 40 1 1=2
E ¼ ln 7:92127 10 1 ln
2 96487 40 40 1=2
¼ 1:176 V
The maximum work output from the fuel cell is based on Eqn (21.14) is
0
W _ ¼ _ mDG ¼ _ mf 239081:7 86346 ð0:5 ½ 95544Šþ½ 58472ŠÞg
T
¼ _ mð 219:18Þ MJ=kmol H 2
Also, at 500 K the lower enthalpy of reaction of the process is 244.02 MJ/kmol, which gives a
higher enthalpy of reaction of
3 3

Q p ¼ 244:02 10 þ 18 1831 ¼ 277:0 10 ¼ 277:0MJ=kmol H 2
This is equal to the change of enthalpy of the working fluid as it passes through the cell, and hence
applying the steady flow energy equation to the fuel cell gives
_ _
Q W ¼ _ mðDhÞ
_
_
giving Q ¼ _ mðDhÞþ W ¼ 277:0 þ 219:18 ¼ 57:82 MJ=kmol H 2
Thus the efficiency of the fuel cell, based on the higher enthalpy of reaction, is
219:18
h ¼ ¼ 0:7913
277:0
If it was evaluated from the lower enthalpy of reaction, which is usually used to calculate the
efficiency of power plant, then the value would be
219:18
h ¼ ¼ 0:8983:
244
Example 3
The operatingpressure of the fuel cell in example 2 is changed to 80 bar. Calculate the change in emf.
Solution
If the pressure is raised to 80 bar, this only affects the pressure term because K pr ¼ f(T), and hence

(" 1=2 !# " 1=2 !#)
<T 80 1 40 1
E 80 E 40 ¼ ln K p r ln ln K p r ln
z i F 80 80 1=2 40 40 1=2
<T 1=2
¼ ln 2 ¼ 0:0075 V
z i F
giving E 80 ¼ 1:176 þ 0:0075 ¼ 1:183 V: