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Fuel Cells. Fuel cells are electrochemical devices used for the continuous production of direct-current by electrochemically reacting a fuel with an oxidant.The free energy of the oxidation of the fuel is converted directly into electrical energy (no combustion).Fuel cells could be considered primary batteries.As direct conversion devices fuel cells are not subject to the efficiency limitation of the Carnot cycle..
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1. Optimal Design of Current Take Off Bus Bars for Tubular Solid Oxide Fuel Cells
11. Objectives
Determine the ohmic power loss in the anode and cathode bus bars for any desired location of the power take-off.
Determine the location of the power take-off for minimum power loss.
Inputs
Axial current density profile for the stack, as a fourth order polinomical.
Dimensional and materials property data for the bus bar.
Dimensional data for the SOFC fuel electrode.
Constraint
Cross sectional area of the bus bar is constant along the axis.
12. Current Density on the Bus
The current density (mA/cm2) for the stack is represented by a fourth order polinomial:
jS = A1x14 + B1x13 + C1x12 + D1x1 + E1 = f1(x) mA cm-2 1)
To obtain the corresponding equation for the bus bar equation 1 is multiplied by a factor F where:
2)
When the length of the bus bar is equal to the active length of the cell:
3)
13. Area Ratio (F) Values for Two SOFC Cell Types
Geometry Area Ratio Factor
4) Figure 1
5)
Figure 2
14. Model
Figure 3 shows the integration of the current density on the bus (j (x) mA/cm2) over a length (x) of the bus to form a current Ix.
6) The power dP loss as the current Ix passes thru an element of the bus of length dx and resistance dR is:
7)
Combining equations 6 and 7:
8)
The total power loss to the current take-off at x = ?1 is:
9)
15. Model Continued
Figure 4 shows integration of the current density from both ends of the bus to the current take-off point.
Because f1(x) requires that x be measured
from the cell closed end, when integrating
from the open end we must form f2(x).
f2(x) is the inverse of f1(x) and is obtained
by replacing x in the fourth order polinomial
by ? -x.
Then
In general
10)
When the power lead is attached at the fuel entry end of the SOFC stack the upper limit ?1 in the first integral is replaced by ? and the second integral becomes zero.
When the power lead is attached at the fuel exit end of the SOFC stack the upper limit ?- ?1 in the second integral is replaced by ? and the first integral becomes zero.
16. Formation of f2(x)
When integrating from the “closed end”
f1(x) = Ax14 + Bx13 + Cx12 + Dx1 + E 11)
When integrating from the “open end”
f2(x) = A(? -x2)4 + B(? -x2)3 + C(? -x2)2 + D(? -x2) + E 12)
After expanding the (? -x2)n terms
f2(x) = A2x24 + B2x23 + C2x22 + D2x2 + E2 13)
A2 = A1 14)
B2 = -4 ?A1 – B1 15)
C2 = 6 ?2A1 + 3 ?B1 + C1 16)
D2 = -4 ?3A1 - 3 ?2B1 - 2 ?C1 – D1 17)
E2 = ?4A1 + ?3B1 + ?2C1 + ?D1 + E1 18)
17. Evaluation of the Integrals
19)
20)
21)
22)
By a similar means
23)
18. In equations 23 and 24 the coefficients with n = 1 and n = 2 respectively are as follows:
24) 29)
25) 30)
26) 31)
27) 32)
28)
19. Summary of Ohmic Power Loss Calculation
Input data
The following coefficients are obtained from a fourth order polinomial fit of current density Js (ma/cm2) for the stack as a function of x (cm) measured from the closed end of the cells.
A mA cm-6
B mA cm-5
C mA cm-4
D mA cm-3
E mA cm-2
For the bus bar material the following dimensional and property data is provided.
? Ohmic resistivity ?cm
w Width of bus bar cm
t Thickness of bus bar cm
p Fuel electrode perimeter cm
? Total length of bus bar cm
?1 Length from “closed end” to power take-off point cm
20. Summary of Ohmic Power Loss Calculation Continued
Determination of Coefficients and Integrals
F is determined from p and w using equation 4 or 5 depending upon cell type.
Is determined
A2, B2, C2, D2, are E2 are determined from A1, B1, C1, D1, and E1 using equations 14 through 18
With n=1, K1,1 through K1,9 are determined and with n=2 K2,1 through K2,5 are determined, using equations 24 through 32
Int1 is determined from K1,1 through K1,9 and ?1 using equation 27
Int2 is determined from K2,1 through K2,9 ? and ?1 using equation 23
21. Summary of Power Loss Calculation Continued
Assembly of Power Loss Equation
With the coefficient and the integrals Int1 and Int2 determined
Total Power Loss 33)
22. Optimal Power Take-Off Position
To obtain the value of ?1 (power take of position measured from the closed end) which is associated with minimum power loss, we set the first derivative of the equation for P with respect to ?1 equal to zero. This yields:
O = 11K1,1?110+10K1,2?19+9K1,3?18+8K1,4?17+7K1,5?16+6K1,6?15+5K1,7?14+4K1,8?13+3K1,9?12
-11K2,1(?- ?1)10-10K2,2(?- ?1)9-9K2,3(?- ?1)8-8K2,4(?- ?1)7-7K2,5(?- ?1)6 34)
-6K2,6(?- ?1)5-5K2,7(?- ?1)4-4K2,8(?- ?1)3-3K2,9(?- ?1)2
If attempts to obtain the solution for ?1 using an equation solver such as “Goal Seek” and “Solver” adopted the following methodology can be adopted.
23. Optimal Power Take-Off Position continued
With equations 22 for Int1 and 23 for Int2 used in equation 34, differentiation with respect to ?1 of the equation for power loss P yields.
11K1,1?110+10K1,2?19+9K1,3?18+8K1,4?17+7K1,5?16+6K1,6?15+5K1,7?14+4K1,8?13+3K1,9?12
-11K2,1(?- ?1)10-10K2,2(?- ?1)9-9K2,3(?- ?1)8-8K2,4(?- ?1)7-7K2,5(?- ?1)6
-6K2,6(?- ?1)5-5K2,7(?- ?1)4-4K2,8(?- ?1)3-3K2,9 (?- ?1)2 35)
The value is used along with the following ?1 values
to create nine values of dP/d?1.
24. Optimal Power Take-Off Position continued
A third order polinomial can be fitted to the resulting values of dP/d?,
i.e., 36)
With a, b, c, and d now determined either “Goal Seek” or Solver can be used to solve for
?IC (the critical value of ?1) in equation 38.
O = a?I,C3+b?I,C2+c?I,C+d 37)
The solution “?I,C” is the location of the power take-off that gives minimum power loss within the bus bar. Intuitively this position is associated with minimal perturbation of the “natural” current density profile within the generator.
