## Fuel Cell Calculations

Example 10-1 Fuel Flow Rate for 1 Ampere of Current (Conversion Factor Derivation)

What hydrogen flow rate is required to generate 1.0 ampere of current in a fuel cell? (This exercise will generate a very useful conversion factor for subsequent calculations.)

Solution:

For every molecule of hydrogen (H2) that reacts within a fuel cell, two electrons are liberated at the fuel cell anode. This is most easily seen in the PAFC and PEFC because of the simplicity of the anode (fuel) reaction, although the rule of two electrons per diatomic hydrogen molecule (H2) holds true for all fuel cell types. The solution also requires knowledge of the definition of an ampere (A) and an equivalence of electrons.61

61 One equivalence of electrons is 1 g mol of electrons or 6.022 xl023 electrons (Avagadro's number). This quantity of electrons has the charge of 96,487 coulombs (C) (Faraday's constant). Thus, the charge of a single

(PAFC & PEFC anode reaction)

1 coulomb / sec

1 equivalence of e-

96,487 coulombs

1 g mol H2

V 1 hr mH2 = ^0.018655 g^ H2 per a| 12.°158 g 1 f-A^g-1 = 37.605x10"6 or 0.037605

g mol H2 A1000 g

hr kg H2 kA

The result of this calculation, 0.037605 kg H2 per kA (0.08291 lb H2 per kA), is a convenient factor that is often utilized in determining how much fuel must be consumed to supply a desired fuel cell power output as illustrated below.

Example 10-2 Required Fuel Flow Rate for 1 MW Fuel Cell

A 1.0 MWdc fuel cell stack is operated with a cell voltage of 700 mV on pure hydrogen with a fuel utilization, Uf of 80%. (a) How much hydrogen will be consumed in lb/hr? (b) What is the required fuel flow rate? (c) What is the required air flow rate for a 25% oxidant utilization, Uox?

Solution:

(a) We shall simplify the solution of this problem by artificially assuming that the individual fuel cells are arranged in parallel. That is, the fuel cell module voltage is the same as the cell voltage, and the fuel cell module current is equal to the current of an individual fuel cell times the number of fuel cells.

Recalling that power is the product of the voltage and current,

Therefore, the current through the fuel cells can be calculated as

The quantity of hydrogen consumed within the fuel cell is electron is 1.602 x10-19 C. One (1) ampere of current is defined as 1 C/sec.

Note that had we skipped the simplifying assumption that the fuel cells were arranged in parallel, we would have calculated the same hydrogen mass flow answer with a few extra steps. For example, if the fuel cell stacks were composed of 500 cells, then the stack voltage would have been 350 volts [(500 cells)(0.7 v/cell)], and the stack current would have been 2.858 kA [1429 kA / 500 cells]. Because this stack current passes through the 500 cells arranged in series, the hydrogen consumption is calculated as:

mH2,consumed = (2.858 kA)(^^A* H )(5«> cells) = 118.4 ^

Thus, the reader may find it more expedient and less error prone to make the parallel arrangement assumption when determining the mass flow requirement of hydrogen, in spite of the actual arrangement.

(b) Per equation 8-14, the utilization of fuel in a PAFC is defined as

H 2, consumed

Therefore the required fuel flow rate can be calculated as

„ _ 2, consumed _ • ^ ..„„lb H2 H2, in----- 148 0_h""

(c) To determine the air supply requirement, we first observe that the stoichiometric62 ratio of hydrogen to oxygen is 2 to 1 for H2O. Thus, the moles of oxygen required for the fuel cell reaction are determined by n -l 11« 4 lb H 2 V 1 lb mol H2 If 1 lb mol O2 ) lb mol O2

no2, ™d =l1184^Jl 2.0158 lb H2 A 2 lb mol H2 1 = 2938

If a 25% utilization is required, then the air feed must contain four times the oxygen that is consumed,

_ lb mol O2 consumed 11 1 lb mol O2 supplied | _ 118 5 lb molO2 no2, supplied = 12938 h A 0.25 lb mol O2 consumed J = 1185 h~~

Because dry air contains 21% O2 by volume, or by mole percent, the required mass flow rate of dry air is,

. ,, „lb mol O2 supplied ^ ( 1 lb mol air | ( 29.0 lb dry air ^ ^ , „„lb dry air m supplied =1 118.5-2——-II-1|---1 = 16,400---

air, supplied 1 h Jl0.21 lb mol O2 Jl 1 lb mol of airJ h

### Example 10-3 PAFC Effluent Composition

A PAFC, operating on reformed natural gas (900 lb/hr) and air, has a fuel and oxidant utilization of 86% and 70% respectively. With the fuel and oxidant composition and molecular weights listed below, (a) How much hydrogen will be consumed in lb mol/hr? (b) How much oxygen is consumed in lb mol/hr? (c) What is the required air flow rate in lb mol/hr and lb/hr? (d) How much water is generated? (e) What is the composition of the effluent (spent) fuel and air streams in mol %?

62 The stoichiometric ratio is the ratio of atoms in a given molecule.

 Fuel Data mol % CH4 4.0 CO 0.4 CO2 17.6 H2 75.0 H2O 3.0 Total 100.0 MW 10.55

Air Data

mol %, dry

mol %, wet

H2O

0.00

1.00

N2

79.00

78.21

O2

21.00

20.79

Total

100.00

100.00

MW

28.85

(a) Before determining the lb mol/hr of hydrogen, we will first determine the molar fuel flow.

n fuel, supplied

I „„„ lb fuel I ( 1 lb mol fuel | „ „ lb mol fuel

_____lb mol fuelY 75 lb mol H 2 || 86 lb mol H2 consumed | lb mol H

^H, consumed =1 8529-1-II —-— II ——--— I = 55.01

h A 100 lb mol fuel J K 100 lb mol H 2 supplied

(b) To determine how much oxygen is consumed, it is useful to note the overall fuel cell reaction

H2 (g) + / O2 (g) ^ H2O (g), therefore, lbmolH^li / lb mol O2 ) 2751 lb mol O2

(c) The required air flow will be determined on a wet air basis, thus n -I 2751lb mol O2 |f 100 lb mol O2 supplied ji 100 lb mol wet air j _ 18901 lbmolwetair nair,required ' h A70 lb mol O2 consumedJl 20.79 lb mol O2 " '

1 , ™ „, lb mol wet air ) ( 28.74 lb wet air ) , „ „lb wet air mair, required = I 189.01-T-II ^-]-— I = 5,433-T-

(d) Per the overall fuel cell reaction above, the water generated is equal to the moles of hydrogen consumed, nH2O generated nH2 consumed 55.01

lb mol H2

(e) The composition of the fuel is developed in the table below, by working from the left to right. The composition is determined by converting the composition to moles, accounting for the fuel cell reaction, and converting back to the desired units, mol %. (Note: mol % is essentially equivalent to volume % for low pressure gases.)

 mol % lb mol/hr mol % Gas FC inlet FC inlet FC reaction FC outlet FC outlet CH4 4.0 3.41 3.41 11.27 CO 0.4 0.34 0.34 1.13 CO2 17.6 15.01 15.01 49.58 H2 75.0 63.97 -55.01 8.96 29.58 H2O 3.0 2.56 2.56 8.45 Total 100.0 85.29 -55.01 30.28 100.00

In the PAFC, only the moles of hydrogen change on the anode (fuel) side of the fuel cell. The other fuel gas constituents simply pass through to the anode exit. These inert gases act to dilute the hydrogen, and as such will lower the cell voltage. Thus, it is always desirable to minimize these diluents as much as possible. For example, to reform natural gas, significant quantities of steam are typically added to maximize the reforming reactions. The wet reformer effluent would commonly have a water composition of 30 to 50%. The reformate gas utilized in this example has been "dried" to only 3% moisture via condensation in a contact cooler.

The spent oxidant composition is calculated in a similar manner. We note that in both the PAFC and PEFC the water is generated on the cathode (air) side. This can be seen from the cathode reaction listed below and the following table listing the fuel cell reaction quantities.

!/2O2 + 2H+ + 2e" ^ H2O (PAFC & PEFC cathode reaction)

 mol % lb mol/hr mol % Gas FC inlet FC inlet FC reaction FC outlet FC outlet H2O 1.00 1.89 55.01 56.90 26.28 N2 78.21 147.82 147.82 58.27 O2 20.79 39.30 -27.51 11.79 5.44 Total 100.00 189.01 27.51 216.51 100.00

Example 10-4 MCFC Effluent Composition - Ignoring the Water Gas Shift Reaction

An MCFC operating on 1000 lb/hr of fuel gas and a 70% air/30% CO2 oxidant has a fuel and oxidant utilization of 75% and 50% respectively. With the fuel and oxidant composition and molecular weights listed below, (a) How much hydrogen will be consumed in lb mol/hr? (b) How much oxygen is consumed in lb mol/hr? (c) What are the required air and oxidant flow rates in lb mol/hr? (d) How much CO2 is transferred from the cathode to the anode? (e) What is the composition of the effluent (spent) fuel and oxidant streams in mol % (ignoring the water gas shift equilibrium)?

 Fuel Data mol % CH4 0.0 CO 0.0 CO2 20.0 H2 80.0 HO 0.0 Total 100.0 MW 10.42

Oxidant Data

Air

Air + CO2

mol %, wet

mol %, wet

CO2

0.00

30.00

H2O

1.00

0.70

N2

78.21

54.75

O2

20.79

14.55

Total

100.00

100.00

MW

28.74

(a) Before determining the lb mol/hr of hydrogen, we will first determine the molar fuel flow.

it™«lb fuel i (1 lb mol fuel ) „, „ „ lb mol fuel

i ^ lb mol fuel | ( 80 lb mol H2 ) ( 75 lb mol H 2 consumed ) lb mol H 2

H2 consumed i h jy 100 lb mol fuel) 1100 lb mol H 2 supplied) h

(b) To determine how much oxygen is consumed, it is useful to note the overall fuel cell reaction,

(c) The required air flow will be determined on a wet air basis; thus n -I 2881 lb mol ®2 |f 100 lb mol O2 supplied ji 100 lb mol wet air | _277. lbmolwetair nair, required " ^ ' h J150 lb mol O2 consumed JI 20.79 lb mol O2 " '

The oxidant flow rate will be calculated knowing that air is 70% of the total oxidant flow.

, „„„,, lb mol wet airV100 lb mol oxidant ] „ lb mol oxidant noxidant, required = I 277U-T-II ^-]-- : I = 395.86

(d) Per the overall fuel cell reaction presented below, the quantity of CO2 transferred from the cathode to the anode side of the fuel cell equals the moles of hydrogen consumed,

H2, anode + 2 O2, cathode + CO2, cathode ^ H2O, anode + CO2, anode ; therefore lb mol H2

(e) The composition of the fuel effluent is developed in the table below, by working from the left to right. The composition is determined by converting the composition to moles, accounting for the fuel cell reaction, and converting back to the desired units, mol %.

 mol % lb mol/hr mol % Gas FC inlet FC inlet FC reaction FC outlet FC outlet CH4 0.0 0.00 0.00 0.00 CO 0.0 0.00 0.00 0.00 CO2 20.0 19.20 57.61 76.82 50.00 H2 80.0 76.82 -57.61 19.20 12.50 H2O 0.0 0.00 57.61 57.61 37.50 Total 100.0 96.02 -57.61 153.63 100.00

The spent oxidant composition is calculated in a similar manner. We note that in the MCFC, both oxygen and carbon dioxide are consumed on the cathode (air) side. This can be seen from the cathode reaction listed below and the following table listing the fuel cell reaction quantities.

 mol % lb mol/hr mol % Gas FC inlet FC inlet FC reaction FC outlet FC outlet CO2 30.00 83.13 -57.61 25.52 13.38 H2O 0.70 1.94 1.94 1.02 N2 54.70 151.71 151.71 79.56 O2 14.6 40.33 -28.81 11.52 6.04 Total 100.00 277.11 -86.42 190.69 100.00

Example 10-5 MCFC Effluent Composition - Accounting for the Water Gas Shift Reaction

For the above example, determine the composition of the effluent (spent) fuel stream in mol % including the effect of the water gas shift equilibrium. Assume an effluent temperature of 1200°F.

Solution:

The solution to this problem picks up where we left off in Example 10-4 above. For convenience, the water gas shift reaction is presented below:

The double headed arrow is used in the field of chemistry to indicate that a reaction is an equilibrium reaction. That is, the reaction does not proceed completely to the left or to the right.

Instead, the reaction proceeds to an equilibrium point, where both "products" and "reactants" remain. The equilibrium composition is dependent upon both the initial composition and final temperature. Fortunately, the equilibrium concentrations can be determined by a temperature dependent equilibrium constant, K, and the following equation.

At 1200°F, the equilibrium constant is 1.96763. A check of the compositions from the preceding example shows that those concentration levels are not in equilibrium.

[CO2 ][H2 ] _ [°-5°][°-125] , , 1967 [CO][H2O] [0.0 [0.375] '

Because the numerator contains the products of the reaction and the denominator contains the reactants, it is clear that the reaction needs to proceed more towards the reactants. We shall equilibrate this equation, by introducing a variable x, to represent the extent of the reaction to proceed to the right and rewriting the equilibrium equation as:

K= [CO2 ][H 2 ] _ [°-50 + x] [C-125 + x] 1967 [CO][H2O] [0.0- x][0.375-x] '

This can be solved by trial and error or algebraically. First, we'll demonstrate the trial and error solution, by guessing that x is -0.1. That is, the reaction should "move" to the left as written (more CO and H2O). This yields the following:

K= [CO2 ][H 2 ] _ [°50 + -M [°.125 + -M [0-4Q [C-025] 021()

Equilibrium constants can be calculated from fundamental chemical data such as Gibbs free energy, or can be determined from temperature dependent tables or charts for common reactions. One such table has been published by Girdler Catalysts (1). The following algorithm fits this temperature dependent data to within 5% for 800 to 1800°F, or within 1% for 1000 to 1450°F: Kp= e(4,276/T -3 961). Kp(1200°F or 922K) equals 1.967.

The value of x of -0.1 was in the right direction, but apparently too large. We shall now guess x is -0.05.

= [CÛ2 ][H2 ] = [0.50 + -0.05][0.125 +-0.05] _ [0.45][0.075] _ 158§ [CO] [H2O] [0.0--0.05 [0.375--0.05] [0.05][0.425] '

We could continue this simple trial and error procedure until we guessed that x is -0.0445, which yields:

_ [CO2][H2]_ [0.50 +-0.0445][0.125 +-0.0445] _ [0.4555][0.0805] _ [CO][H2O] " [0.0--0.0445[0-375- -0.0445] " [0.0445][0.4195] " '

These concentrations are now in equilibrium. The following table summarizes the effect of accounting for the water gas shift equilibrium.

 mol % lb mol/hr, assuming 100 lb mol/hr basis mol % Gas FC outlet w/o shift. FC outlet w/o shift effect of shift rxn FC outlet in shift equil. FC outlet in shift equil. CO 0.00 0.00 4.45 4.45 4.45 CO2 50.00 50.00 -4.45 45.55 45.55 H2 12.50 12.50 -4.45 8.05 8.05 H2O 37.50 37.50 4.45 41.95 41.95 Total 100.0 100.00 0.00 100.00 100.00

Alternately, one could have solved this problem algebraically as follows:

k[CO - x][h2O - x] = [CO2 + x] [h2 + x] ,which can be expanded as

K |x2 -([CO] + [H2O])x + [CO] [H2O]|= x2 +([CO2] + [H2])x + [CO2] [H2], which can be combined to

(1_JK) x2 + {[CO2 ]+[H ] + K([CO] + [H2O])jx + {[CO2 ][H2]-[C0][^20]IC}

This is in the standard quadratic form of:

which can be solved by the quadratic formula:

Substituting the appropriate values for K and the concentrations yields two roots of -0.0445 and 1.454. We throw out the larger root because it is a nonsensical root. This larger root "wants to" react more CO and H2O than are initially present. When using the quadratic formula, the user will throw out all roots greater than 1 or less than -1. The remaining root of -0.0445 is precisely what was developed by our previous trial and error exercise.

Example 10-6 SOFC Effluent Composition - Accounting for Shift and Reforming Reactions

An SOFC is operating on 100 % methane (CH4) and a fuel utilization of 85%. (a) What is the composition of the effluent (spent) fuel in mol %? Assume that the methane is completely reformed within the fuel cell, and the moisture required for reforming is supplied by internal recirculation.

Solution:

(a) There are many different ways to approach this problem, some of which may seem rather complex because of the simultaneous reactions (fuel cell, reforming and water gas shift reactions) and the recycle stream supplying moisture required for the reforming reaction. We shall simplify the solution to this problem by focusing on the fuel cell exit condition. Because we have drawn the box of interest at a point after the recycle, this will allow us to ignore the recycle stream and to deal with the reactions in steps and not simultaneously.

First, we shall write the relevant reactions:

Fuel Feed

Recycle

Recycle

Fuel Feed Point of Interest

(Steam Reforming Reaction) (Fuel Cell Reaction) (Water Gas Shift Reaction)

Next we shall combine the reforming reaction and the fuel cell reaction into an overall reaction for that portion of the fuel that is utilized within the fuel cell (i.e., 85% ). The combined reaction is developed by adding the steam reforming reaction to 4 times the fuel cell reaction. The factor of four allows the hydrogen molecules to drop out of the resulting equation because it is fully utilized.

CH4, anode + 2O2, cathode ^ 2H2O, anode + CO2, anode

(Steam Reforming Reaction) (Fuel Cell Reaction) (Combined Reforming and FC Reactions)

For the 15% of the fuel that is not utilized in the cell reaction we shall simply employ the reforming reaction. To the resulting gas composition, we will then impose the water gas shift equilibrium.

For ease of calculation, we shall assume a 100 lb/hr basis for the methane.

-fuel, supplied

= lioo*CHi

1 lb mol CH4 16.043 lb CH4

Thus, 85%, or 5.30 lb mol CH4 /h, will be reformed and consumed by the fuel cell. The remainder will be reformed but not consumed by the fuel cell reaction. We will summarize these changes in the following table.

 mol % lb mol/hr mol % Gas FC inlet FC inlet Ref / FC rxn Reforming FC outlet FC outlet CH4 100.0 6.23 -5.30 -0.93 0.00 0.00 CO 0.0 0.00 0.00 0.00 0.00 0.00 CO2 0.0 0.00 5.30 0.93 6.23 33.33 H2 0.0 0.00 0.00 3.74 3.74 20.00 H2O 0.0 0.00 10.60 -1.87 8.73 46.67 Total 100.0 6.23 10.60 1.87 18.70 100.00

Now, we have created an artificial solution that reflects only two out of three reactions. We shall now apply the water gas shift reaction to determine the true exit composition. We shall apply the quadratic equation listed in Example 10-5 to determine how far the reaction will proceed, where x is the extent of the reaction in the forward direction as written.

§CO2 ] + [h2 ] + k([c0] + [h20])}= 0.3333 + 0.2000 + 0.574*(0.00 + 0.4667) = .8012

c = {[CO2][H2]-[CO] [H20]k| = (0.3333)(0.20) - (0.00)(0.4667)(0.574) = 0.0666

- b +Vb2 - 4ac = - 0.8012 (0.8012)2 - 4(0.426)(0.0666)

Again we throw out the value greater than 1, or less than -1, leaving -0.0873. The following table summarizes the effect of accounting for the water gas shift equilibrium.

 mol % lb mol/hr, assuming 100 lb mol/hr basis mol % Gas FC outlet w/o shift. FC outlet w/o shift Effect of shift rxn FC outlet in shift equil. FC outlet in shift equil. CO 0.00 0.00 -(-8.73) 8.73 8.73 CO2 33.33 33.33 -(-8.73) 24.61 24.61 H2 20.00 20.00 -8.73 11.27 11.27 H2O 46.67 46.67 -8.73 55.39 55.39 Total 100.00 100.00 0.00 100.00 100.00

Example 10-7 Generic Fuel Cell - Determine the Required Cell Area, and Number of Stacks

Given a desired output of 2.0 MWdc, and the desired operating point of 600 mV and 2 2 400 mA/cm , (a) How much fuel cell area is needed? (b) Assuming a cell area of 1.00 m per cell and 280 cells per stack, how many stacks are needed for this 2.0 MW unit?

Solution:

(a) Recalling again that power is the product of the voltage and current, we first determine the total current for fuel cell as

v 1 MWA 1 W A1000 Ay

Because each individual fuel cell will operate at 400 mA/cm , we determine the total area required as,

b) The number of required stacks and cells are calculated simply as

(lm2per cell) 110,000 cm

(833 cells)

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