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Chapter 9

Fuel Cells

Problem Solutions

252 Page 2 of 2 Prob. Sol. 9.1 Fund. of Renewable Energy Processes

Most of the data above are given in the problem statement, but we have to look up the molecular masses of the materials under discussion. At least, we have to look up the atomic mass of aluminum. We may remember that the dalton is defined as 1/32 of the molecular mass of O 2 , so that the mass of the latter is exactly 32.00 daltons. Once we know the masses of Al and O 2 , we can easily calculate that of Al 2 O 3. The free energy is ∆G = ∆H − T ∆S

where T ∆S is the heat that must appear to compensate the change in entropy owing to the reaction. The reaction involved in the formation of aluminum oxide from its elements is 4 Al + 3 O 2 = 2 Al 2 O 3.

At RTP, the entropy of aluminum is 1.05 kJ K−^1 kg−^1 , as given in the problem statement. Since the atomic mass of Al is 27.0, the entropy is

1. 05 × 27 = 28.4 kJ K−^1 kmole−^1. In a similar manner, the entropy of O 2 was given as 6.41 kJ K−^1 kg−^1 or 6. 41 × 32 = 205.1 kJ K−^1 kmole−^1. Consequently, the entropy of the reactants is 4× 28 .4+3× 205 .1 = 729. 0 kJ K−^1. On the other side of the equation, the entropy of the products is 2 × 51 .0 = 102 kJ K−^1. Considering only the matter balance, the reaction causes an entropy re- duction of 729. 0 − 102 .0 = 627 kJ K−^1 or 627. 0 /2 = 313.5 kJ K−^1 kmole−^1 (of aluminum oxide). This contradicts the second law of thermodynamics, and a certain amount, Q, of heat must evolve. Assuming that the reaction proceeds isothermally (heat is removed as fast as it is created), then Q = T ∆S = 298 × 313 , 500 = 93. 4 × 106 joules per kilomole of oxide or 93.4 MJ per kilomole of oxide. The enthalpy change owing to the reaction is 1670 MJ per kilomole of oxide. Of this, 1670 − 93 = 1577 is free energy and can (theoretically) be transformed into electricity.

The standard free energy of aluminum oxide is -1.577 GJ per kmole.

Solution of Problem 9.

Fund. of Renewable Energy Processes Prob. Sol. 9.2 Page 1 of 1 253

Prob 9.2 Daniel cells used to be popular last century as a source of electricity, especially in telegraph systems. These cells consisted of a container divided into two compartments by a mem- brane permeable to ions. In one compartment, a zinc electrode was dipped in a zinc sulfate solution and, in the other, a copper electrode in a copper sulfate solution. The zinc oxidizes (i.e., loses electrons): Zn → Zn++^ + 2 e− The zinc is eroded, going into the solution in the form of ions. At the other electrode, the copper is reduced (i.e., the copper ions in the sulfate accept electrons and the copper from the sulfate plates out onto the copper electrode):

Cu++^ + 2 e−^ → Cu The cell will deliver current until it runs out of either zinc or sulfate, whichever is less. Assume the cell has 95 g of zinc and 450 ml of a 0.1 M CuSO 4 solution. M stands for molarity: moles of solute per liter of solu- tion. How long can this cell deliver a current of 2 A to a load? ..................................................................................................................... A 2 ampere current corresponds to the flow of 2 coulombs every second or 2/ 1. 6 × 10 −^19 = 1. 25 × 1019 electrons per second. This is 1. 25 × 1019 / 6. 02 × 1026 = 2. 08 × 10 −^8 kmoles of electrons per second. Since, in the reaction of interest, 1 kmole of zinc or 1 kmole of copper sulfate corresponds to 2 kmoles of electrons, the consumption of each of these “fuels” is 1. 04 × 10 −^8 kmoles per second. The atomic masses involved (in daltons) are Zn 65. Cu 63. S 32. O 16. CuSO 4 63 .5 + 32 + 4 × 16 = 159. 5 . The zinc consumption rate is 1. 04 × 10 −^8 × 65 .4 = 6. 8 × 10 −^7 kg/s. 0.095 kg of zinc will last 0. 095 / 6. 8 × 10 −^7 = 140, 000 s or 38.8 h. The CuSO 4 consumption rate is 1. 04 × 10 −^8 × 159 .5 = 1. 66 × 10 −^6 kg/s. We have 450 ml of 0.1 M solution of this salt. This solution contains

1. 1 × 159 .5 kg of copper sulfate per cubic meter of water. Hence, our 450 ml contain 0.00718 kg of the salt. At the indicated consumption rate, this will last 0. 00718 / 166 × 10 −^6 = 4325 s or 72.1 min. The cell will run out of copper sulfate much before it runs out of zinc. A “fuel cell” could be made by feeding in additional copper sulfate solution.

The limiting substance in the cell is copper sulfate which will run out in 72.1 minutes.

Solution of Problem 9.

Fund. of Renewable Energy Processes Prob. Sol. 9.3 Page 2 of 3 255

The energy, WL, delivered to the load is the generated energy minus the internal losses:

WL = |∆G|N − (4N N 0 × 1. 6 × 10 −^19 /t)^2 Rt = 1 MWh = 3. 6 × 109 J.

Introducing the values of ∆G, N 0 , and R,

3. 94 × 108 N − 1. 49 × 1014 N 2 /t = 3. 6 × 109.

From this, t = 3. 76 × 105 N 2 /(N − 9 .14),

dt dN

= 3. 76 × 105

[

2 N

(N − 9 .14)

N

N − 9. 14

2 )]

N = 18. 3 kmoles,

t = 13. 7 × 106 seconds or 159 days.

The current during the discharge is

4 N N 0 × 1. 6 × 10 −^19 t

4 × 18. 3 × 6. 02 × 1026 × 1. 6 × 10 −^19

13. 7 × 106

= 515 A.

The internal resistance of the cell is R = 0.001 Ω, and the total resis- tance is RL + R. The current delivered is, as we saw, 515 A, therefore,

R + RL =

Voc I

RL = 0. 001 Ω.

The overall reaction is C+O 2 →CO 2 -393.5 MJ.

The ideal open-circuit voltage is 1.024 V.

The entropy change owing to the material balance in the reaction is +3.02 kJ K−^1 (kmole of CO 2 )−^1.

To deliver 1 MWh, 18.3 kilomoles of carbon are needed.

The load resistance must be 1 milliohm.

Solution of Problem 9.

256 Page 3 of 3 Prob. Sol. 9.3 Fund. of Renewable Energy Processes

See a simpler solution below!

To transfer 1 MWh (3. 6 × 109 J) in minimum time, means using max- imum power. This is accomplished when RL = R = 0.001 Ω. The current that does this is

IL =

VL

R + RL

= 512 A.

The load voltage is

VL = ILRL = 512 × 0 .001 = 0. 512 V.

The power in the load is

PL = VLIL = 0. 512 × 512 = 262. 1 W.

The time necessary to deliver 3. 6 × 109 J at the rate of 262.1 J/s is

t =

3. 6 × 109

= 13. 7 × 106 s.

The energy that the fuel cell must deliver is 2 × 3. 6 × 109 = 7. 2 × 109 J because the internal loss is equal to the power in the load. 1 kilomole of carbon delivers 394.5 MJ of electricity. Thus, to deliver

7. 2 × 109 J, the number of kilomoles of carbon must be

N =

7. 2 × 109

394. 5 × 106

= 18. 25 kmoles.

These are the same results as before.

Solution of Problem 9.

258 Page 2 of 5 Prob. Sol. 9.4 Fund. of Renewable Energy Processes

b. How much electric energy will an ideal fuel cell (using metha- nol and air) produce per kg of fuel? ..................................................................................................................... Ideally, the electric energy generated by a fuel cell is equal to the free energy change owing to the reaction:

We =

−∆G

J/kg (methanol)

The water is produced by the fuel cell in in vapor form. Thus, the ∆G of the reaction would be -685.3MJ/kmole if the oxygen were at RTP. But the problem requires air as the oxidant, a gas in which the oxygen pressure is only 0.2 atm. The remaining fluids are still at 1 atmosphere. The free energy of each kilomole of oxygen is decreased (in absolute value) by an amount RT ln 0.2 = 4.0 MJ. Since there is 1.5 kilomoles of oxygen for each kilomole of methanol, the free energy of the reaction is now

∆G = − 685 .3 + 1. 5 × 4 .0 = − 679. 3 MJ/kmole or 21. 2 MJ/kg.

In an ideal fuel cell, 1 kg of methanol yields 21.2 MJ of electricity.

c. How much heat does the cell reject? ..................................................................................................................... The amount of heat an ideal fuel cell absorbs is

Wheat = |∆H| − |∆G| = 638. 5 − 679 .3 = 40. 9 MJ/kmole

or

1. 28 MJ/kg of methanol.

Notice that this ideal fuel cell operates endothermically, i.e., it absorbs heat from the environment, instead of rejecting it.

The cell absorbs 1.28 MJ of heat per kmole of methanol used.

d. A practical OTTO cycle engine has an efficiency of, say, 20%, while a practical methanol fuel cell may have an efficiency of 60% (this is the efficiency of the practical cell compared with that of the ideal cell). If a methanol fueled IC car has a high- way performance of 10 km per liter, what is the performance of the fuel cell car assuming that all the other characteristics of the cars are identical? ..................................................................................................................... The energy delivered by a practical IC car per kilogram of methanol burned is

WIC = 0. 2 × 21 .1 = 4. 2 MJ/kg,

Solution of Problem 9.

Fund. of Renewable Energy Processes Prob. Sol. 9.4 Page 3 of 5 259

while the energy delivered by a practical fuel-cell is

WF C = 0. 6 × 21 .2 = 12. 7 MJ/kg.

It follows that the highway performance of the fuel-cell car will be

12. 7
13. 2

× 10 = 30. 3 km/liter.

The fuel cell car has a kilometrage of 30.3 km/liter.

e. If you drive 2000 km per month and a gallon of methanol costs $2.40, how much do you save in fuel per year when you use the fuel cell version compared with the IC version? Can you think of other savings besides that in fuel? ..................................................................................................................... To drive 12 × 2000 = 24, 000 km per year, the fuel used will be: For the IC car: 24 , 000 /10 = 2400 liters. For the FC car: 24 , 000 / 30 .3 = 792 liters. The fuel related annual savings of the FC car relative to the IC car is

(2400 − 792) ×

gallon

×

gallon 3 .78 liter

= $1020/year.

Yearly fuel savings are a modest $1020/year

There would be some additional savings:

1. The FC car uses no oil.
2. The city mileage would be much closer to the highway mileage because of the simplicity of using regenerative braking in which the energy for decelerating the car is returned to the battery†
3. At a stop sign, the IC car consumes fuel while idling. The FC car consumes no energy when stopped.
4. The FC car requires no smog inspection, You save the cost of inspec- tion.

f. Suppose you get a ten year loan such that the yearly repay- ments of principal plus interest are 18% of the initial amount borrowed. By how much can the initial cost of the fuel-cell car exceed that of the IC car for you to break even? Assume that after 10 years the car is totally depreciated. ..................................................................................................................... Let PIC and PF C be, respectively, the price of the IC and that of the FC car.

† The FC car would probably have a small battery for braking and fast acceleration.

Solution of Problem 9.

Fund. of Renewable Energy Processes Prob. Sol. 9.5 Page 1 of 2 261

Prob 9.5 An inventor wants to build a hydrogen manometer based on the dependence of the output voltage of a fuel cell on the pressure of the reactants. Take a H 2 /O 2 fuel cell operating at 298 K. Assume that it produces water vapor and that the open circuit voltage is that of an ideal cell. The oxygen pressure is maintained at a constant 0.1 MPa while the hydrogen pressure, pH 2 , is the quantity to be measured.

a. What is the output voltage when pH 2 is 0.1 MPa? ..................................................................................................................... For the reaction 2H 2 + O 2 → 2H 2 O (g),

the free energy change is − 228 .6 MJ/kmole of H 2 O at RTP. Since for each kilomole of H 2 O, 2 kilomoles of electrons are involved, the reversible voltage of the fuel cell is

V =

|∆G|

neqN 0

228. 6 × 106

1. 6 × 10 −^19 × 2 × 6. 02 × 10 −^26

= 1. 185 V.

The output voltage at 0.1 MPa is 1.185 V.

b. What is the output voltage when pH 2 is 1 MPa? ..................................................................................................................... The pressure of the oxygen is unchanged but the hydrogen pressure is increased and the energy necessary to increase this pressure isothermally is free energy and will appear in the electric output of the cell. The energy is

Wpress = RT ln pH 2 pH (^20)

= 8314× 298 ×ln

= 5. 70 × 106 J/kmole of water.

The free energy change of the reaction is increased (in absolute value) by the above amount.

|∆G| = |∆G 0 | + 5. 70 × 106 = 234. 3 × 106 J/kmole of water. V =

234. 3 × 106

1. 6 × 10 −^19 × 2 × 6. 02 × 10 −^26

= 1. 214 V.

When the hydrogen pressure is raised to 1 MPa, the reversible voltage rises to 1.214 V. Notice that the voltage is not very sensitive to pressure.

c. Develop an expression showing the rate of change of voltage with pH 2. What is this rate of change when pH 2 is 0.1 MPa? .....................................................................................................................

V =

|∆G|

neqN 0

Solution of Problem 9.

262 Page 2 of 2 Prob. Sol. 9.5 Fund. of Renewable Energy Processes

dV dpH 2

neqN 0

d dp

|∆G| =

neqN 0

d dp

[

|∆G 0 | + RT 0 ln

pH 2 pH (^20)

)]

RT 0

neqN 0

d dp

[

ln

pH 2 pH (^20)

)]

RT 0

neqN 0 pH 2

=

8314 × 298

2 × 1. 6 × 10 −^19 × 6. 02 × 10 −^26 pH 2

pH 2

V/Pa.

At pH 2 = 0.1 MPa, dV dpH 2

= 1. 28 × 10 −^7 V/Pa.

At 0.1 MPa, the rate of change of voltage with pressure is a minute 0.128 microvolts per pascal.

..................................................................................................................... d. The output voltage of the cell is sensitive to temperature. Assume that a ±10% uncertainty in pressure measurement can be tolerated (when the pressure is around 1 MPa). In other words, assume that when a voltage corresponding to 1 MPa and 298 K is read, the actual pressure is 0.9 MPa because the temperature of the gases is no longer 298 K. What is the maximum tolerable temperature variation? ..................................................................................................................... From the Gibbs-Helmholtz equation: dV dT

V − (^) n∆eqNH 0 T

1. 214 − 241.^8 ×^10

6 2 × 1. 6 × 10 −^9 × 6. 02 × 1026 298 =

= − 138 × 10 −^6 V/K.

The problem requires that the change in voltage, ∆V owing to the pres- sure error, ∆p, be equal to the change in voltage owing to the temperature error. (^) dV

dp

∆p =

dV dT

∆T

or

∆T =

dV /dp dV /dT

∆T =

1. 28 × 10 −^8

138 × 10 −^6

∆p = 92. 6 × 10 −^6 ∆p.

We note that the value of dV /dp at 1 MPa is 1. 28 × 10 −^8 V/Pa. To keep the pressure reading error within 10% (10 × 105 at 1 MPa), the temperature must be kept within ± 9. 3 K.

The temperature of the device must be kept within 9 C of 298 K.

The proposed manometer is not a good idea—large changes in pressure cause only very small changes in the observed voltage.

Solution of Problem 9.

264 Page 1 of 3 Prob. Sol. 9.7 Fund. of Renewable Energy Processes

Prob 9.7 A fuel cell has the reactions: ANODE: 2 A 2 = 4 A++^ + 8 e−. (1) CATHODE: 4 A++^ + B 2 + 8 e−^ = 2 A 2 B (2) All data are at RTP. The overall reaction, 2 A 2 + B 2 = 2 A 2 B, releases 300 MJ per kmole of A 2 B product in a calorimeter. The entropies of the different substances are: A 2 : 200 kJ K−^1 kmole−^1 , B 2 : 400 kJ K−^1 kmole−^1 , A 2 B : 150 kJ K−^1 kmole−^1. A 2 and B 2 are gases, whereas A 2 B is liquid. a. What is the voltage of an ideal fuel cell that uses the above reaction at RTP? ..................................................................................................................... The overall reaction of the fuel-cell is

2A 2 + B 2 → 2A 2 B.

The reactant entropy is 2 × 200 + 400 = 800 kJ/K; that of the product is 2 × 150 = 300 kJ/K. The entropies associated with the materials in the reaction decreased by 500 kJ/K for each 2 kmoles of the product, A 2 B or

∆S = 250 kJ K−^1 per kmole of A 2 B.

To satisfy the second law of thermodynamics (under reversible condi- tions), an amount, T ∆S, of heat must be generated:

T ∆S = 298 × 250 = 74. 5 MJ per kmole of A 2 B.

The change in free energy owing to the reaction is

∆G = ∆H − T ∆S = −300 + 74.5 = − 225. 5 MJ per kmole of A 2 B.

8 electrons circulate in the load for every 2 molecules of A 2 B formed, or 4 electrons/molecule:

V =

−∆G

neqN 0

4 × 1. 6 × 10 −^19 × 6. 022 × 1026

= 0. 584 V.

The reversible voltage of the cell is 0.584 V.

b. Estimate the voltage at standard pressure and 50 C. .....................................................................................................................

Solution of Problem 9.

Fund. of Renewable Energy Processes Prob. Sol. 9.7 Page 2 of 3 265

The rate of change of the reversible voltage with temperature (at con- stant pressure) is

( ∂V ∂T

p

V + (^) n∆e qNH 0 T

− 300 × 106

4 × 6. 022 × 1026 × 1. 6 × 10 −^19

= − 0. 649 × 10 −^3 V/K.

V50 C = 0. 584 − 0. 649 × 103 (50 − 25) = 0. 568 V.

At 50 C, the reversible voltage of the cell is 0.568 V.

c. How much heat does the ideal fuel cell produce per kilomole of A 2 B, at RTP? .....................................................................................................................

The ideal fuel-cell will generate 74.5 MJ of heat per kilomole of A 2 B produced.

d. What is the voltage of the cell if the gases are delivered to it at 100 MPa? The operating temperature is 25 C. ..................................................................................................................... The energy of isothermal compression is free energy and will, therefore, affect the voltage. Compress 2 kilomoles of A 2 from 0.1 MPa to 100 MPa: WcomprA 2 = 2RT ln

p p 0

= 2 × 8314 × 298 ln 10^3 = 34. 2 MJ.

Compress 1 kilomole of B 2 from 0.1 MPa to 100 MPa: WcomprB 2 = 1 × 8314 × 298 ln 10^3 = 17. 1 MJ.

The total energy used up in compressing the input gases is Wcompr = 34.2 + 17.1 = 51. 3 MJ/per 2 kmoles. Per kilomole, Wcompr = 25.6 MJ. Since the product is liquid, it is not affected by the pressure change. The energy used in compressing the gases isothermally is added to the free energy of the reaction or, correspondingly, causes a voltage increase of

Vcompr =

25. 6 × 106

neqN 0

= 0. 066 V,

hence, V25 C, 100 MPa = 0.584 + 0.066 = 0. 651 V.

At 25 C, 100 MPa, the reversible voltage of the cell is 0.651 V.

Solution of Problem 9.

Fund. of Renewable Energy Processes Prob. Sol. 9.8 Page 1 of 4 267

Prob 9.8 Owing to its ceramic electrolyte, a fuel cell can operate at 827 C. Pure oxygen is used as oxidizer. The gases are fed to the cell at a pressure of 1 atmosphere. Use the data below:

∆h

◦ f ∆g ◦ f γ^ s ◦ MJ MJ kJ K−^1 /kmol /kmol /kmol CO(g) -110.54 -137.28 1.363 197. CO2(g) -393.51 -394.38 1.207 213. O2(g) 0 0 1.341 205.

The values of γ are those appropriate for the the 25 C to 827 C interval. We want to examine the influence of temperature on the per- formance of an ideal fuel cell. a. Calculate the reversible voltage and the efficiency of the above (ideal) cell at both 25 C and 827 C. ..................................................................................................................... The overall reaction is

CO + 12 O 2 → CO 2.

At 25 C (RTP conditions) the ∆g of the reaction is

∆gREACT ION = ∆gCO 2 − ∆gCO − 12 ∆gO 2 = − 394. 38 − (− 137 .28) − 12 × 0 = − 257. 10 MJ/kmol.

Since oxygen is divalent, 4 electrons will circulate in the load for each O 2 molecule. Hence, there will be 2 electrons per CO 2 molecule.

Vrev =

|∆gREACT ION | neqN 0

Vrev =

257. 10 × 106

2 × 1. 6 × 10 −^19 × 6. 02 × 10 −^26

= 1. 334 V.

The reversible voltage of the cell, at RTP, is 1.334 V.

At 25 C (RTP conditions) the ∆h of the reaction is

∆hREACT ION = ∆hCO 2 − ∆hCO − 12 ∆hO 2 = − 393. 51 − (− 110 .54) − 12 × 0 = 282. 97 MJ/kmol.

Solution of Problem 9.

268 Page 2 of 4 Prob. Sol. 9.8 Fund. of Renewable Energy Processes

The efficiency is

η = ∆g ∆h

The efficiency at RTP is 90.8%

As a first step in estimating the conditions at 827 C, we have to calcu- late the values of ∆hf and ∆gf for 827 C (1100 K) and 1 atmos. For this, we need to know the values of cp.

cp = R

γ − 1

This yields

J K−^1 kmol−^1 CO 31, CO 2 48, O 2 32,

The enthalpy of formation at the new temperature is

∆hf = ∆h

◦ f +^ cp∆T.

resulting in

∆hfCO = − 110. 54 × 106 +31, 213 ×(1100−298) = − 85. 50 × 106 MJ/kmol,

∆hfCO 2 =− 393. 51 × 106 +48, 496 ×(1100−298) =− 354. 62 × 106 MJ/kmol,

and

∆hfO 2 = 0 + 32, 690 × (1100 − 298) = 26. 22 × 106 MJ/kmol.

The free energy at the new temperature is

∆gf = ∆g◦ f + (cp − s◦)∆T − T cp ln

T

T 0.

resulting in

∆gfCO = − 137. 28 × 106 + (31. 213 − 197 .5) × 103 × (1100 − 298)

− 1100 × 31. 213 × 103 × ln

= − 315. 52 × 106 MJ/kmol.

Solution of Problem 9.