Fuel Cell FundamentalsSolutions

Timothy P. HolmeRyan O’HayreSuk Won Cha

Whitney ColellaFritz B. Prinz

Solution companion to Fuel Cell Fundamentals.Suggested grading schemes are given after the problem numbers.

1

Chapter 1 solutions

Problem 1.1 (10 points) Possible answers include:FC advantages over other power conversion devices:

1. Potentially higher efficiency.

2. Solid state components have no moving parts, giving higher reliabilityand lower maintenance costs

3. silent operation

4. low emissions

5. fuel cells refuel rather than recharge, which could be faster.

Disadvantages:

1. cost

2. low power density

3. problems of hydrogen storage, production, transport, lower energydensity

4. temperature problems: PEMs can’t start in the cold, high tempera-tures of SOFCs create materials, thermal cycling, and sealing problems

5. water management issues in PEMs.

Applications:

1. portable applications (such as laptops, cellphones, etc.) where theirfast refueling, silent operation, and independent scaling of fuel reser-voir and power make fuel cells an attractive option.

2. Transportation applications where their low emissions and high effi-ciency makes fuel cells an attractive option.

3. Power generation applications where their silent operation, low emis-sions, and high efficiency make fuel cells amenable to siting in cities fordistributed generation (DG) applications, reducing the cost of powerdistribution and possibly making process heat available for combinedpower and heating applications.

2

Problem 1.2 (5 points) Fuel cells have lower power density than en-gines or batteries, but can have large fuel reservoirs, so they are much bettersuited to the high capacity/long runtime applications.

Problem 1.3 (10 points) You can easily tell which reactions are re-duction and which are oxidation by finding which side of the reaction theelectrons appear on.

1. Cu → Cu2+ + 2e− Electrons are liberated, so this is an Oxidationreaction

2. 2H+ + 2e− → H2 Reduction

3. O2− → 12O2 + 2e− Oxidation

4. CH4 + 4O2− → CO2 + 2H2O + 8e− Oxidation

5. O2− + CO → CO2 + 2e− Oxidation

6. 12O2 + H2O + 2e− → 2(OH)− Reduction

7. H2 + 2(OH)− → 2H2O + 2e− Oxidation

Problem 1.4 (15 points) You can write full-cell reactions and thensplit them into the half-cell reactions. You don’t need to be an expertchemist to do this, just use the half-cell reactions given and make sure yourequations balance with species number (ie. that O is conserved) and charge(ie. electrons are conserved).

1. CO+ 12O2 → CO2 is a full-cell reaction common in SOFCs; the half-cell

reactions would be O2− + CO → CO2 + 2e−, which is an Oxidation,or anode, reaction, and 1

2O2 + 2e− → O2− which is a reduction, orcathode reaction.

2. 12O2 +H2 → H2O is a full cell reaction in SOFCs or PEMs, dependingon the circulating ion (O2− or H+, respectively). The half-cell reac-tions would be: 1

2O2 + 2e− + 2H+ → H2O at the cathode of a PEMand H2 → 2H+ + 2e− at the anode of a PEM; or 1

2O2 + 2e− → O2−

at the cathode of an SOFC and O2− + H2 → H2O at the anode of anSOFC.

3. another full cell reaction in an SOFC could be CH4 + 2O2 → CO2 +2H2O with the half cell reactions 8e− + 2O2 → 4O2− as the reducing,or cathode reaction, and CH4 + 4O2− → CO2 + 2H2O + 8e− as theoxidizing, or anode reaction.

3

4. to use the circulating ion (OH)−, you may construct the full cell reac-tion 1

2O2 + H2 + H2O → 2H2O from the half-cell reactions H2 +2(OH)− → 2H2O + 2e− as the oxidizing, or anode reaction, and12O2 + H2O + 2e− → 2(OH)− as the cathode (reducing) reaction.

Problem 1.5 (10 points) From Figure 1.6, H2(l) has a higher volumetricenergy density but lower gravimetric energy density than H2(g) at 7500 PSI.On a big bus, the gravimetric energy density is probably a greater concern,so choose H2(g). Other considerations that could affect the choice includesafety, the amount of hydrogen lost to boil-off, and the cost of liquefactionversus compression.

Problem 1.6 (5 points)

1. Reactant transport–at high current density, there is a depletion effect.Reactants cannot reach active sites quickly enough. The voltage lossresults from a lower concentration of reactants.(ηconc)

2. Electrochemical reaction–voltage loss from the sluggishness of the elec-trochemical reaction (ηact)

3. Ionic conduction–resistance to ion flow in the electrolyte (ηohmic)

4. Product removal–in a PEM, water flooding blocks active reaction sites(ηconc)

Problem 1.7 (20 points) First, multiply the whole reaction by 2 be-cause you can’t formally describe a bond in 1

2O2, but we will later divideby 2 at the end to find the energy of the reaction given. For the reaction2H2 + O2 → 2H2O, H2O has 2 O-H bonds, so E2∗H2O = 2 ∗ 2 ∗ EO−H =4 ∗ 460 kJ/molEO2 = 494 kJ/molE2H2 = 2 ∗ 432 kJ/molThe energy released by the reaction is

12(E2∗H2O − EO2 − E2H2) =

12(4 ∗ 460− 494− 2 ∗ 432) (1)

(in kJ/mol).The energy is −241 kJ/mol . The energy is negative because an energyinput to the system is required to break bonds.

4

Problem 1.8 (20 points) You can see the benefits in a fuel cell of theindependent scaling of the fuel reservoir and the fuel cell stack. Therefore,for this problem, you may compute the volume of the stack and the reservoirindependently.

For the stack, you need to supply 30 kW with a fuel cell that suppliespower at 1 kW/L and 500 W/kg. Therefore, the volume needed is 30 kW ∗1 L

1 kW = 30 L and the weight is 30 kW ∗ 1 kg500 W = 60 kg.

For the fuel tank, you need to hold a quantity of fuel equal to

30 kJ/s ∗ 1 hr

60 miles∗ 3600 s

1 hr∗ 300 miles = 540 MJ (2)

note that 1 W = 1 J/s. Taking into account the 40% efficiency, you needto hold an excess quantity of fuel, 540 MJ/0.40 = 1350 MJ . The hydrogenis compressed to supply 4 MJ/L and 8 MJ/kg, so the fuel tank must be1350 MJ ∗ 1 L

4 MJ = 337.5 L and 1350 MJ ∗ 1 kg8 MJ = 168.75 kg.

The entire system must occupy a volume Vsystem = Vtank + Vcell =337.5 L+30 L = 367.5 L and weighs Wsystem = Wtank+Wcell = 168.75 kg+60 kg = 228.75 kg

Problem 1.9 (5 points) P = V I. See the figure.

5

Figure 1: Sketch of Voltage and Power as a function of current density forthe fuel cell described in problem 1.9.

Chapter 2 Solutions

Problem 2.1 (6 points) When a gas undergoes a volume constriction,possible configurations of the gas are removed. Entropy is a measure ofdisorder, ie. the number of possible configurations a system can assume, sothe entropy of a gas in a smaller volume is lower (given that the tempera-ture of the gas remains constant–entropy is also a function of temperature).Therefore, the entropy change is negative.

Problem 2.2 (6 points) G = H − TS, so for an isothermal reaction(∆T = 0), ∆G = ∆H − T∆S.

(a) if ∆H < 0 and ∆S > 0 then ∆G < 0 and the reaction is spontaneous.(b) in this case, you cannot determine the sign of ∆G unless you are

given the temperature and the size of the changes in entropy and enthalpy.(c) ∆H > 0 and ∆S < 0, so in this case, ∆G > 0 and the reaction is

6

non-spontaneous.(d) again, you cannot make a determination from the information given

Problem 2.3 (6 points) The reaction rate is determined by the acti-vation barrier, and not by the overall energy change of the reaction. Youmay not determine which reaction proceeds faster.

Problem 2.4 (6 points) While the current scales with the amount ofreactants, the voltage does not. The thermodynamic potential comes fromthe energy drop going from products to reactants, which does not scale withreactant amount. Since E = −∆G/nF , you can think of the scaling in ncancelling the scaling in ∆G.

Problem 2.5 (6 points) The Nernst equation

E = ET −RT

nFln

Πaνiprod

Πaνireact

(3)

shows that increasing the activity of the reactants decreases the argumentof the ln, which raises the reversible cell voltage (E) because the ln term isnegative. This, in essence, is Le’Chatlier’s principle.

Problem 2.6 (Not graded) When the reaction is in equilibrium, theelectrochemical potential of the system is zero. Components on the prod-ucts side see the electrical potential φP and on the reactants side see thepotential φR. Note that different species do not experience different electri-cal potentials. Then, if ∆φ is the difference in electrical potential from oneside to the other,

0 =∑

µidni =∑

µoi dni +

∑RT ln aidni + nF∆φ (4)

First, note that:∑RT ln aidni = RT (ln am

M+ln anM−ln a1

A−ln abB) = RT ln

amMan

N

aAabB

= RT ln

∏aνi

prod∏aνi

react

(5)Rearranging (4) and inserting the above result, you get

∆φ = −∑

µoi dni

nF− RT

nFln

∏aνi

prod∏aνi

react

(6)

From the thermodynamic definition of chemical potential, µi ≡ ∂G∂ni

so thatΣµo

i dni = ∆Go where the o in this case denotes reference concentration. We

7

relate this term to E by −∆Go

nF = Eo, but noting that the o refers only toreference concentration so the term may still depend on temperature, werename the quantity ET . Identifying E as the electrical potential across thecell ∆φ, we arrive at the Nernst equation (3).

Problem 2.7 (15 points) Yes, you can have a thermodynamic efficiencygreater than 1. We chose the metric of fuel cell efficiency to be ∆G/∆H,but it is in some sense an arbitrary choice.

As an example in defining efficiencies, consider the efficiency of an electrolyzer–which is a machine that makes hydrogen gas from water using electricity (thisis the exact reverse of a fuel cell, and may be used to generate hydrogen forsome fuel cell applications). An electrolyzer has an efficiency defined to bethe ∆H of reaction (the output is the useful heat energy of hydrogen), di-vided by the energy input ∆G. Therefore, the efficiency of the electrolyzeris the inverse of fuel cell efficiency–so the fuel cell at STP with an efficiencyof 0.83, if ran in reverse as an electrolyzer, would have an efficiency of 1/0.83which is greater than 1.

For a fuel cell, consider the following example: ε ≡ ∆G/∆H. For anisothermal reaction, ∆G = ∆H − T∆S, so ε = 1− T ∆S

∆H . If ∆H is going tobe negative, then you need to make ∆S positive to get an efficiency greaterthan 1. The trick is to use a solid or liquid reactant to make ∆S positivebecause solids and liquids have very low entropy compared to gases. For thefuel cell reaction C(s) + 1

2O2 → CO at 298 K and 1 bar,

∆S = SCO−12SO2−SC = 197.7−0.5∗205.1−5.7 (in J/mol·K) = 89.45 J/mol·K

(7)

∆H = HCO−12HO2−HC = −110.5−0.5∗0−0 (in kJ/mol) = −110.5 kJ/mol

(8)then

ε = 1− T ∆S∆H = 1− (negative number) > 1

Problem 2.8 (15 points) Assuming constant specific heats, ∆H(T ) =∆Ho + cp(T − T o) and ∆S(T ) = ∆So + cp ln(T/T o).

Find the temperature that satisfies:

∆G(T ) = 0 = ∆H(T )−T∆S(T ) =∑

[∆Hoi + cpi(T − T o)− T (So

i + cpi ln(T/T o))](9)

taking out the ∆Ho and ∆So,

0 = ∆Horxn − T∆So

rxn +∑

[cpi(T − T o − T ln(T/T o))] (10)

8

or0 = ∆Ho

rxn − T∆Sorxn + (T − T o − T ln(T/T o))

∑cpi (11)

Substituting numbers,

0 = −41.13 kJ/mol−T (−42.00 J/mol·K)+(T−T o−T ln(T/T o))(3.2 J/mol·K)(12)

A numerical solution (MATLAB or Excel or your graphing calculator workfine) gives T ≈ 1020 K ≈ 747 C . The error of neglecting the dependenceof ∆S and ∆H on temperature led to an answer that was off by about 40degrees in this case. A more sophisticated solution would include the vari-ance of cp with temperature, requiring an iterative solution or an expansionfor cp in terms of T .

Problem 2.9 (a) (10 points) Remember that the effect of temperatureenters into the first term of the Nernst equation. From the Nernst equation,if the reactants and products are ideal,

E = ET −RT

nFln[(P/Po)νP

∏P xνi

i

(P/Po)νR∏

R xνii

]= ET −

RT

nFln[(P/Po)νP−νR

∏P xνi

i∏R xνi

i

](13)

In a reaction, the change in number of moles ∆nG = νP − νR. The temper-ature dependent term is

ET = Eo +∆S

nF(T − To) (14)

For the voltages to be equal

E(T1, P1) = E(T2, P2) (15)

Eo +∆S

nF(T1 − To)−

RT1

nFln[(P1/Po)∆nG

∏P xνi

i∏R xνi

i

]=

Eo +∆S

nF(T2 − To) +

RT2

nFln[(P2/Po)∆nG

∏P xνi

i∏R xνi

i

]Cancelling constant terms that appear on both sides of the equation andsolving for T2,

T1

∆SnF −

RnF ln

[(P1/Po)∆nG

∏P x

νii∏

R xνii

]∆SnF −

RnF ln

[(P2/Po)∆nG

∏P x

νii∏

R xνii

] = T2 (16)

9

Simplifying for the H2/O2 fuel cell with liquid water as product and for purecomponents (xH2 = xO2 = 1) and dropping the Po which is understood tobe 1 atm the expression simplifies to

T1∆S + 1.5R lnP1

∆S + 1.5R lnP2= T2 (17)

(b) (5 points) At STP, using the data from Appendix B,

∆Srxn = SH2O(l)−SH2−0.5SO2 = 69.95−130.86−0.5∗228.3 = −175.06 J/mol·K(18)

If P2 = P1/10 then from the above expression,

T2 = T1∆S + 1.5R lnP1

∆S + 1.5R ln(P1/10)= 298

−175.06 + 1.5 ∗ 8.314 ∗ ln 1−175.06 + 1.5 ∗ 8.314 ∗ ln(1/10)

(19)

Then T2 = 256 K . Note that at this temperature of −17 C, there will beno H2O(g) (so we shouldn’t use the ∆S for water vapor), but there will alsonot be liquid water–the product will be ice! A full solution would use the∆S for liquid water down to 0 C and then the ∆S for ice below that.

To interpret the result, since ∆S is negative, the Nernst voltage decreaseswith an increase in temperature. The decrease in pressure decreases theNernst voltage, so we must compensate by raising the Nernst voltage bydecreasing temperature, therefore T2 < T1.

Problem 2.10 (15 points) Two ways of attacking this problem yieldthe same result. First, you could imagine a box with water, air, and hy-drogen, and the reaction H2 + 1

2O2 H2O(l) is in equilibrium. To findhow much hydrogen is consumed by oxygen, find the equilibrium quantityof hydrogen when oxygen is present. That is to say, find at what PH2 does∆Grxn = 0. From the van’t Hoff isotherm

∆G = ∆Go + RT lnΠaνi

prod

Πaνireact

(20)

assuming air at the cathode (xO2 = 0.21)

∆G = ∆Go+RT lna1

H2O

a1H2

a0.5O2

= ∆Go+RT ln1

(P/Po)1.5xH2x0.5O2

= ∆Go−RT ln(1)1.5(xH2)(0.21)0.5

(21)Solving for xH2 ,

∆Go

RT= ln

[(0.21)0.5xH2

](22)

10

xH2 =1

0.210.5exp

[∆Go

RT

]= (2.18) exp

[(−237 kJ/mol)

(8.314 J/mol ·K)(298 K)

]= 5.41·10−42

(23)so PH2 = 5.41 · 10−42 atm

Note that the partial pressure is very low, because it is very energeticallyfavorable for hydrogen to react with oxygen to form water.

The alternate way to solve the problem is to solve a concentration cellwhere a voltage develops (1.23 V because it is a hydrogen/air system) butEo = 0 because the concentration cell reaction is H2 + O2 → H2 + O2. Inthis formulation, you solve the Nernst equation where the reactants haveactivity 1 because they are pure

1.23 = 0− RT

nFln

PH2/Po(0.21)0.5

1(24)

This gives the same equation

xH2(0.21)0.5 = exp[−1.23 ∗ nF

RT

](25)

Problem 2.11 (5 points) The efficiencies of each part multiply to givethe total efficiency of the cell ε = εthermoεV oltageεfuel. For pure H2/O2 atSTP, εthermo = 0.83. We are given V , so εV oltage = V

E = 0.751.23 . We are given

λ, so εfuel = 1λ = 1

1.1 . Therefore,

ε = 0.83 ∗ 0.751.23

11.1

(26)

ε = 46%

11

Chapter 3 Solutions

Problem 3.1 A. (5 points) Reducing the potential raises the energyof electrons in the electrode. To reduce their energy, electrons leave theelectrode, so the reaction proceeds faster in the forward direction.

B. (5 points) Increasing the potential lowers the energy of electrons inthe electrode, so the reaction is biased in the forward direction.

C. (5 points) We want to increase both reaction rates in the forwarddirection. At the anode (H2 H++2e−) you want to draw electrons to theelectrode, so increase the potential. At the cathode (2H+ + 2e− + 1

2O2 H2O) you want electrons to leave the electrode, so reduce this potential.The overall voltage output falls from both effects.

Figure 2: Schematic of activation voltage losses for problem 3.1

Problem 3.2 (5 points) Yes, it is possible to have a negative GalvaniPotential at one electrode. It means that one half-cell reaction requiresenergy input, and the other results in energy output. So long as the totalpotential adds up to the measured full-cell potential, it is impossible to knowwhat each half-cell potential is!

Problem 3.3 (10 points) Alpha is the charge transfer coefficient, itdescribes whether the “center of the reaction”, or peak of the reaction ac-tivation barrier, falls nearer to one side of the reaction or the other. Inthis figure, note that alpha does not change the final electrochemical energy

12

Figure 3: Schematic of different Galvani potentials problem 3.2

change, only the height of the peak in electrochemical energy.

Problem 3.4 (5 points) The exchange current density is the currentdensity of the forward and reverse reactions at equilibrium (at open circuit).

Problem 3.5 (5 points) (a) The Tafel equation, which holds in theexponential regime, reads:

ηact = a + b log j (27)

and in the exponential regime, the Butler-Volmer equation simplifies to

ηact = − RT

αnFln jo +

RT

αnFln j (28)

Note: to convert between log and ln, use the conversion lnx = 2.3 log x.The terms that go as a logarithm with current density are equal:

b log j =RT

αnFln j ⇒ b/2.3 =

RT

αnF(29)

b = 2.3 RTαnF

(b) Identifying the constant terms, we get

a = − RTαnF ln jo

13

Figure 4: Effect of α on the electrochemical energy pathway for problem3.3.

Problem 3.6 (5 points) The full-cell reaction is CO + 12O2 → CO2.

The half-cell reaction at the anode is O2− + CO → CO2 + 2e− and at thecathode is O2 + 4e− → 2O2−

Problem 3.7 (5 points) The main job of a fuel cell catalyst is to beable to form intermediate strength bonds with reactants and products, ie.to yield a low ∆Gact. Also, it should have a long lifetime, which meansthat it is resistant to poisoning, and does not migrate or agglomerate onthe membrane. The requirements for an effective fuel cell catalyst-electrodestructure are: porosity, a high degree of interconnection between the pores, ahigh effective catalyst area, high TPB density, high electronic conductivity,

14

and high exchange current density. It also must have a long lifetime, meaninghigh mechanical strength and resistance to corrosion. Ideally, it would alsobe cheap and easy to manufacture.

Problem 3.8 (5 points) For reaction A, the net reaction rate in mol/s·cm2 is 5 A

nF ·2cm2 = 2.5A/cm2

2F = 1.30 · 10−5mol/s · cm2. For reaction B, the net

reaction rate is 15 AnF ·5cm2 = 3A/cm2

3F = 1.04 · 10−5mol/s · cm2. Therefore,reaction A has a higher reaction rate.

Problem 3.9 (10 points) At equilibrium, j = 0. The Butler-Volmerequation is

j = joo

(C∗

R

Co∗R

exp [αnFη

RT]−

C∗P

Co∗P

exp [−(1− α)nFη

RT])

= 0 (30)

After canceling the joo term and rearranging, the equation reads

C∗R/Co∗

R

C∗P /Co∗

P

=exp [− (1−α)nFη

RT ]

exp [αnFηRT ]

= exp [(α− 1)nFη

RT− αnFη

RT] = exp [−nFη

RT]

(31)Solving for η,

−η =RT

nFln [

C∗R/Co∗

R

C∗P /Co∗

P

] (32)

Moving the negative sign to the right and inverting the argument of thelogarithm,

η =RT

nFln [

C∗P /Co∗

P

C∗R/Co∗

R

] (33)

The overvoltage, η, is the amount by which the actual voltage is less thanthe standard state voltage

ET − E = η (34)

For an ideal gas, the concentration is the activity, so we have

E = ET −RT

nFln [

aP

aR] (35)

which is almost the Nernst equation

E = ET −RT

nFln [

ΠaνiP

ΠaνiR

] (36)

The difference lies in the ln term, which does not account for a multiplicityof products and reactants. This arises from the fact that the Butler-Volmer

15

equation only accounts for the concentration of the limiting products andreactants. If you inspect the derivation of the BV equation in the text, you’llfind that it is only for one electrode reaction. Since one fuel cell reactionis usually orders of magnitude slower than the other reactions, we typicallyneglect the slugishness of “fast” reactions and only employ the BV equationto find the loss at one electrode. A more rigorous form of the BV equationwould include all of the products and reactants (and you’d have to write oneequation for the anode and one for the cathode) and would reduce exactlyto the Nernst equation.

Problem 3.10 A (2 points) : P = IV = (1 A)(2.5 V ) = 2.5 WB (2 points) : By putting 5 cells together in series, you get an overall

voltage of 5(0.5 V ) = 2.5 V , the necessary voltage.C (3 points) : Note that when you stack cells together in series, the

current does not change. However, the amount of hydrogen required doesincrease. One way to think of it is that when you are required to supply2.5 W of power, you must supply 2.5 J/s of hydrogen, not just the 0.5 J/sper cell. What is going on is that the first fuel cell uses 0.5 J/s of hydrogento raise the voltage of its cathode up to 0.5 V . When this cathode connectedto the anode of the next cell it is raised to a higher voltage, and this celluses 0.5 J/s of hydrogen to raise its output voltage, and so on...

The amount of hydrogen per cell is determined by the current by NH2 =i/nF . The electric charge needed is 1 A ·100 hrs. · 3600 s

1 hr = 360, 000 C. Con-verting to moles of hydrogen, 360, 000 C/2F = 1.866 mol H2. Convertingto grams, 1.866 mol · 2 g

1 mol = 3.73 g H2 per cell. So the fuel cell stack usesfive times that, 18.7 g H2

D (3 points) : From the ideal gas law, V = nRT/P = (1.866 ×5 mol)(0.08205 L · atm/mol ·K)(298K)/(500 atm) = 0.457 L.

V = 457 cm3

For storage in a metal hydride, the 5 wt.% hydrogen gives a storage den-sity of 10 g/cm3 ·0.05 = 0.5 g/cm3. To store 18.7 grams, you need a volumeof (18.7 g)/(0.5 g/cm3) = 37.3 cm3 . This is a significant improvement overstorage at 500 atm!

Problem 3.11 (10 points) The exchange current density is given by

jo = nFC∗

R

Co∗R

e−∆G‡RT (37)

16

Grouping the preexponential terms into a constant that is temperature in-dependent,

jo(T ) = Ce−∆G‡RT (38)

We may multiply and divide the equation by jo(To)

jo(T ) =jo(To)jo(To)

Ce−∆G‡RT = jo(To)e

∆G‡R

( 1To

− 1T

) (39)

This is our equation for jo(T ). To find ∆G‡, first rearrange the generalexpression

R lnjo(T )jo(To)

= ∆G‡(1To− 1

T) (40)

Now, plugging in the numbers given,

∆G‡ =R

( 1To− 1

T )ln

jo(T )jo(To)

= (8.314 J/mol·K)/(1/300 K−1/600 K) ln [10−4/10−8]

(41)

∆G‡ = 45.9 kJ/mol

Problem 3.12 (10 points) The general expression for current densityas a function of concentration is

jo = nFC∗

R

Co∗R

f1e−∆G‡/RT (42)

Multiplying and dividing by jo(Coo∗R ), the current density at a reference

concentration,

jo =jo(Coo∗

R )jo(Coo∗

R )nF

C∗R

Co∗R

f1e−∆G‡/RT (43)

Canceling, we obtain the general expression

jo = jo(Coo∗R )

C∗R

Coo∗R

(44)

We must find the temperature T ′ for which j(T ′, C∗R) = j(T, 10C∗

R)

nF10C∗

R

Co∗R

f1e−∆G‡/RT = nF

C∗R

Co∗R

f1e−∆G‡/RT ′

(45)

Canceling,10e−∆G‡/RT = e−∆G‡/RT ′

(46)

17

ln 10 =∆G‡

R(1/T − 1/T ′) (47)

Solving for T ′,

1/T ′ = 1/T − R

∆G‡ ln 10⇒ T ′ =1

1/T − R∆G‡ ln 10

(48)

Note, the problem should have read: ∆G‡ = 20 kJ/mol. Using this value,

T ′ = 421 K (49)

So the temperature change is ∆T = 121 K .

Problem 3.13 (not graded) The Butler-Volmer equation is

j = Ce−∆G‡

RT

(C∗

R

Co∗R

exp [αnFη

RT]−

C∗P

Co∗P

exp [−(1− α)nFη

RT])

(50)

Doubling the temperature or halving the activation barrier will have thesame effect on the first term in the equation. Doubling the temperature willalso effect the terms in parenthesis. Since the fuel cell is producing current(there is a fixed overvoltage), the first term in parenthesis is larger than thesecond, so changing T will have a larger impact on that term, the forwardcurrent term. So increasing T decreases the contribution from the term inparenthesis, reducing the current. At a fixed overvoltage, there is a largerincrease in current density from halving the activation barrier than fromdoubling temperature.

Problem 3.14 (not graded) As stated in the text, the bond strengthof oxygen is about 8.8 eV , which corresponds to a temperature of T =8.8 eV/kB = 1.0×105 K = 1.0×105 C. Breaking the bond on a Pt catalystrequires 2.3 eV of energy input, which corresponds to a temperature of T =2.3/kB = 2.7× 104 K = 2.6× 104 C. While the Pt catalyst helps greatly,this is still a temperature far too large to allow thermal decomposition ofoxygen. This is why we must supply an overvoltage to the cathode, toprovide an extra driving force for oxygen dissociation.

18

Chapter 4 Solutions

Problem 4.1 (5 points) There must be a voltage gradient (an electricfield) to induce charge to flow. Another way to think about it is the intrinsicresistance of real materials will cause flowing charge to drop in voltage.

Problem 4.2 (5 points) For a given current density, the voltage dropdepends on the ASR, ηohmic = ASR · j. The cell with the lower ASR willbe the cell with the lowest voltage drop. The second cell has an ASR′ =A′ · R′ = 10A · R/9, or an ASR 10

9 higher than the first cell, so the lossesincrease by a factor of 10

9 over the first cell.

Problem 4.3 (5 points) Conductivity is a function of carrier concen-tration and carrier mobility σ = nFcu

Problem 4.4 (5 points) Ion transport in materials is harder becauseions are much more massive than electrons, so µion < µelectron. Also, theconcentration of free electrons in metals is much higher than typical ionconcentrations that can be achieved in electrolytes, celectron > cion. Thetwo effects combine to make ionic conductivity much lower than electronicconductivity.

Problem 4.5 (5 points) Fuel cell electrolytes must have high ionicconductivity, low electronic conductivity, high stability under corroding en-vironments, mechanical strength, low fuel crossover rates. Ideally, they areeasy to manufacture at low cost. The stability requirement is the hardestto fulfill because the membrane must be stable in both the reducing envi-ronment at the anode and oxidizing environment at the cathode.

Problem 4.6 (10 points) There are two differences from the PEMcase: first, the charge carrier has the opposite sign, and second, it flows in theopposite direction, from the cathode to the anode. These two effects combineto make the diagram exactly the same. The ohmic loss still decreases theoutput voltage of the fuel cell, and the oxygen ion is negative it flows up apotential gradient. Note that the graph is exactly the same as the figure inthe text!

Problem 4.7 (5 points) See the figure that shows ohmic and activationlosses.

19

Figure 5: Voltage losses for problem 4.6.

Figure 6: Problem 4.7: the left panel shows only ohmic losses in the elec-trolyte, the right panel shows activation losses at the anode and cathode, aswell as ohmic losses at the electrolyte.

Problem 4.8 (5 points) Find the thickness of a membrane that hasan electric field of 108 V/m with a voltage of 1 V . This can be done bydimensional analysis, or knowing that, with no net charge distribution, theelectric field will be dropped linearly across the membrane. L = V/E =(1 V )/(108 V/m).

L = 10 nm

Problem 4.9 (10 points) We must minimize the voltage loss as afunction of electrolyte thickness. The ohmic loss is given by the current

20

density times the ASR (ASR = R · A = LAρ · A = L

σ ), so ηohmic = Lσ j.

The leakage loss is given to be of the form ηleak = A ln jleak = A ln[B/L].Therefore, the total loss is

ηT = A ln[B/L] + jL

σ(51)

To find the optimal thickness, set the derivative with respect to L equal tozero:

dηT

dL= 0 = − A

L∗ +j

σ(52)

L∗ = Aσ/j

To check that the result makes sense, note that as A increases, the leakagelosses increase, so the optimal thickness must increase to offset the increasedleakage; therefore L∗ is proportional to A. As σ increases, ohmic losses de-crease, so the membrane may be thicker to offset more leakage loss; thereforeL∗ is proportional to σ. Finally, as current density is increased, ohmic lossesincrease, so the optimal membrane thickness should decrease to minimizeohmic loss, so L∗ ∼ 1/j.

Problem 4.10 (5 points) The two resistances simply add, since theyare resistors in series. Therefore, the ohmic loss is

ηohmic = j(ASRelec + ASRelectrolyte) (53)

ηohmic = (0.5 A/cm2)[(0.01 Ω)(5 cm2) + (100 µm)/(0.1 (Ω cm)−1)] (54)

ηohmic = 75 mV

Problem 4.11 Starting with the equation for conductivity,

σ =c(nF )2D

RT(55)

insert the equation for diffusivity,

D = Doe−∆Gact/RT (56)

and you get

σ =c(nF )2Doe

−∆Gact/RT

RT(57)

21

Multiply each side by T and group the temperature independent preexpo-nential factor into a constant σo, and the equation looks very close to thefinal result:

σT =c(nF )2Do

Re−∆Gact/RT = σoe

−∆Gact/RT (58)

The last step requires a little concentration to keep your units straight.If Ea is expressed in units of eV/mole, and ∆Gact is in units of J/mole, andas stated in the text, Ea = ∆Gact/F , then we can convert by noticing somerelations between the fundamental constants: R = kNA and F = qNA.

∆Gact = EaF = Ea(qNA) = Ea(qR

k) (59)

The argument of the exponential in equation 58 becomes:

∆Gact

RT=

Ea(q Rk )

RT=

Eaq

kT(60)

The definition of an electron volt is the energy given to an electron as it israised through a potential of one volt, so 1 eV = q Joules. The factor of qin equation 60 is the conversion from electron volts to Joules. So with theunderstanding that Ea is to be expressed in eV/mol, equation 58 becomes

σT = σoe−Ea/kT (61)

Problem 4.12 We’ll need the saturation water pressure at 70 C and90 C. The empirical fit for psat is

log10psat = −2.1794 + 0.02953T − 9.1837 · 10−5T 2 + 1.4454 · 10−7T 3 (62)

Plugging in T = 70 C,

psat(70 C) = 10−2.1794+0.02953(70)−9.1837·10−5(70)2+1.4454·10−7(70)3 = 10−0.5127 = 0.307 bar(63)

and for T = 90 C,

psat(90 C) = 10−2.1794+0.02953(90)−9.1837·10−5(90)2+1.4454·10−7(90)3 = 10−0.1602 = 0.692 bar(64)

To find the humidity of the exit stream, we need to track the water inflowsand outflows, given some information about the proton flow and inlet hu-midity. First, it is simple to find the hydrogen fluxes. The inlet hydrogen is

22

Figure 7: Problem 4.12: Atom fluxes into the gas channel, into the GDL,and out of the gas channel must sum so that there is no net accumulationof any species in the gas channel.

φH2,in = I/nF = (8 A)/(2∗96485 C/mol) = 4.15 ·10−5 mol/s. The fuel cellcurrent is I = jA = (8 cm2)(0.8 A/cm2) = 6.4 A. The number of hydrogenmolecules (that split into protons to cross the membrane) leaving the gaschannel towards the membrane is

φH2,memb = I/(nF ) = (6.4 A)/(2 ∗ 96485 C/mol) = 3.32 · 10−5 mol/s (65)

The amount of hydrogen that flows out of the gas channel is 8 A− 6.4 A =1.6 A or in mol/s, φH2,out = I/nF = 8.29 · 10−6 mol/s.

To find the water fluxes is slightly more complicated. From the dragcoefficient, the amount of water crossing the membrane is φH2O,memb =α ∗ φH2,memb = 0.8 ∗ 3.32 · 10−5 mol/s = 2.65 · 10−5 mol/s. The inlethumidity tells us that the influx of water is pw

psat= 0.80. Assuming that the

inlet pressure is 1 bar, the mole fraction of water at the inlet is

yw = 0.8psat

Po= 0.8

0.692 bar

1 bar= 0.554 (66)

by knowing the inlet flow of hydrogen, we can determine the inlet flow of

23

waterφH2O,in = φH2,in

0.5541− 0.554

= 5.14 · 10−5 mol/s (67)

Knowing the inlet water and the amount that crosses through the membrane,we can find the outlet water flow

φH2O,out = φH2O,in−φH2O,memb = (5.14·10−5 mol/s)−(2.65·10−5 mol/s) = 2.49·10−5 mol/s(68)

Finally, we can find the activity of water in the exhaust

aw =pw

psat=

yw · Pout

psat(Tout)=

φH2O,out

φH2O,out + φH2,out· Pout

psat(70 C)(69)

Assuming that the pressure drop along the gas channel is negligible,

aw =2.49 · 10−5 mol/s

2.49 · 10−5 mol/s + 8.29 · 10−6 mol/s· 1 bar

0.307 bar= 2.44 (70)

Since the activity is greater than 1, we have liquid water in the exhaust.

Problem 4.13 This problem is quite similar to example 4.4. Followingthe same method, we first find the water content of the membrane by solvingan ODE subject to boundary conditions, then find the conductivity of themembrane as a function of temperature and water content, and finally findthe overpotential by Ohm’s law.

(a) First, the boundary conditions on λ are, from equation 4.34:

λ(aw = 1) = 14;λ(aw = 0.5) = 3.45 (71)

Assuming the current of water through the membrane to be of the formJH2O = α j

2F where α is an unknown constant, we can rearrange equation4.44 to find an equation for the variation of water content through themembrane

dλ

dz=

j

2F

Mm

ρdryDλ(λ)

[2ndrag

λ

22− α

](72)

To find Dλ(λ), we again assume that, since Dλ is a slowly varying functionof λ, it is a constant over the range we consider. Now, we may analyticallysolve the differential equation

dλ

dz= k(

2ndrag

22λ− α) (73)

24

Where k ≡ jMm

2FρdryDλ. The solution is

λ(z) = C exp(kndrag

11z) +

11α

kndrag= Ceκz + α/κ (74)

Where κ ≡ kndrag/11. We determine C from the boundary conditions. Ifthe anode is at z = 0, then for part (a),

λ(0) = C +α

κ= 14⇒ C = 14− α

κ(75)

λ(tm) = Ceκtm +α

κ= 3.45 (76)

[14− α

κ]eκtm +

α

κ= 3.45⇒ α

1κ

[1− eκtm ] = 3.45− 14eκtm (77)

α = κ3.45− 14eκtm

1− eκtm(78)

For lambda, we find

λ(z) = (14− 3.45− 14eκtm

1− eκtm)eκz +

3.45− 14eκtm

1− eκtm(79)

Then the conductivity of the membrane is found from

σ(T, λ) = (0.005193λ− 0.00326) exp [1268 (1/303− 1/T )] (80)

and the resistivity of the membrane is

R =∫ tm

0

dz

σ= e−[1268(1/303−1/T )]

∫ tm

0

dz

(0.005193(Ceκz + α/κ)− 0.00326)(81)

The integral works out to∫dz

aeκz + b=

z

b− ln(b + aeκz)/bκ (82)

so the membrane resistance is

R(T ) =e−[1268(1/303−1/T )]

0.005193ακ − 0.00326

[tm −

1κ

ln0.005193α

κ − 0.00326 + 0.005193Ceκtm

0.005193ακ − 0.00326 + 0.005193C

](83)

To estimate our parameters, we calculate Dλ with λ = 10 to be

Dλ = e2416(1/303−1/T )(2.563−0.33λ+0.0264λ2−0.000671λ3)×10−6 = 3.81×10−6 cm2/s(84)

25

To find κ we also need ndrag = 2.5 λ22 = 1.14, ρdry = 1970 kg/m3 = 1.97 ×

10−3 kg/cm3, Mm = 1 kg/mol

κ =ndrag

11jMm

2FρdryDλ=

1.1411

1 ∗ 12 ∗ 96485 ∗ 1.97× 10−3 ∗ 3.81× 10−6

= 71.3 cm−1

(85)Then we can find alpha to be

α = κ3.45− 14eκtm

1− eκtm= 1.52× 103cm−1 (86)

so that C isC = 14− α/κ = −7.33 (87)

Finally, R(T ) isR(T ) = e−[1268(1/303−1/T )] ∗ .319 (88)

R(80 C) = 0.176 Ωcm2 so the ohmic loss is

R = 0.176 V

(b) In this case, the boundary conditions on λ change

3.45 = C +α

κ(89)

14 = Ceκtm +α

κ(90)

These equations may be solved for alpha

α = κ14− 3.45eκtm

1− eκtm= −277 cm−1 (91)

using our previous value for κ. This gives C = 7.33, so that R(T) is

R(T ) = e−[1268(1/303−1/T )] ∗ .398 (92)

R(80 C) = 0.220 Ωcm2 so the ohmic loss is

R = 0.220 V

This ohmic loss is higher than in part (a), so we can see that humidifyingthe anode is more effective under these conditions.

Problem 4.14 (15 points) (a) The equation for diffusivity is

D =12vo(∆x)2e−

∆GactRT (93)

26

I didn’t subtract points for assuming that the hop distance was the lat-tice constant, 5 A. A more sophisticated analysis takes the crystal struc-ture of cubic ZrO2 into account. Zirconia, in the cubic form, has the flu-orite structure. The oxygen atoms occupy tetrahedral sites in an FCClattice. Those sites are 1/4 of the way along the cube body diagonal,or have positions 1

4(1, 1, 1). The vector between nearest neighbor sites is14(3, 1, 1) − 1

4(1, 1, 1) = 14(2, 0, 0). Therefore, the distance between oxygen

nearest neighbor sites is a/2 = 2.5 A.

D =12(1013 s−1)(2.5 · 10−8 cm)2e−

100 kJ/mol(8.314 J/mol·K)(1273 K) (94)

D = 2.46 · 10−7 cm2/s

(b) The vacancy fraction is

xV ' e−∆Hv/2kT = 0.0105 (95)

the density of oxygen sites is

cO =8

(5 A)3= 6.4 · 1022 atoms/cm3 = 0.106 mol/cm3 (96)

The vacancy concentration is therefore

cV = 1.11−3 mol/cm3

(c) Note: n = 2 since the charge of a vacancy has the opposite charge ofthe ion (O2−). The intrinsic conductivity is

σ =(nF )2cD

RT=

(2 ∗ 96485 C/mol)2(1.11 · 10−3 mol/cm3)(2.46 · 10−7 cm2/s)(8.314 J/mol ·K)(1273 K)

(97)

σ = 9.65 · 10−4(Ω · cm)−1

Problem 4.15 (20 points) First, calculate the carrier concentration,which requires a bit of crystal “stoichiometry”. The crystal is (ZrO2).92(Y2O3).08.If we let the symbol χ stand for a generic cation atom, we have χ1.08O2.08. Ifthis crystal were pure zirconia, with a ratio of one Zr to two O, it would beZr1.08O2.16. Therefore, the 8% doping introduced an absence of 0.08 oxygensfor every 2.16 oxygen sites, or the fractional vacancy level of 0.08

2.16 = 3.70%.

27

This extrinsic doping changes the energetics of intrinsic vacancies, so wemust neglect intrinsic vacancies. At reasonable temperatures, the extrinsicdoping level will dominate. The concentration of oxygen sites is

cO =8

(5 A)3= 6.4 · 1022 atoms/cm3 = 0.106 mol/cm3 (98)

So the concentration of vacancies is

cV = (0.0370)(0.106 mol/cm3) = 3.93 · 10−3 mol/cm3 (99)

Now we may set up two equations with the two unknowns of Do and ∆Gact:

σ(T ) =L

AR(T )=

(nF )2cDoe−∆Gact/RT

RT(100)

Solving for Do, and inserting values for T = 700 K

Do =L

A(47.7 Ω)R(700 K)(e+∆Gact/(8.314 J/mol·K)(700 K))

(nF )2c(101)

Inserting this equation for Do when T = 1000 K, many things cancel andwe are left with:

(700 K)(e+∆Gact/(8.314 J/mol·K)(700 K))(47.7 Ω)

=(1000 K)(e+∆Gact/(8.314 J/mol·K)(1000 K))

(0.680 Ω)(102)

Solving for ∆Gact,

∆Gact =R

11000 −

1700

ln700 K · 0.680 Ω1000 K · 47.7 Ω

(103)

∆Gact = 89.4 kJ/mol

Now, we may solve for Do from equation 101:

Do =100 µm

(1 cm2)(47.7 Ω)(8.314 K/mol ·K)(700 K)(e+(89.4 kJ/mol)/(8.314 J/mol·K)(700 K))

(2 ∗ 96485 C/mol)2(3.93 · 10−3 mol/cm3)(104)

Do = 3.90 · 10−2 cm2/s

To check this result, you may check the diffusivity at a normal operatingtemperature and see that it compares to standard values: at T = 1000 K,we get D = Doe

−∆G/RT ≈ 8 · 10−7 cm2/s, which is in the right ballpark forYSZ.

28

Chapter 5 Solutions

Problem 5.1 (20 points) The use of a different gas in the mixturewill change the effective diffusion constant of oxygen, which changes thelimiting current density jL. Intuitively, since helium is much lighter thannitrogen (MHe MN2), then DO2He DO2N2 . Therefore, jL,He > jL,N2

so ηconc,He < ηconc,N2 . More rigorously:

jL = nFDeff coR

δ(105)

To have a higher jL, a higher Deff is desired. The effective binary diffusionconstant is related to gas properties by:

Dij ∼(pcipcj)1/3(TciTcj)5/12

(TciTcj)b/2

√1

Mi+

1Mj

= (pcipcj)1/3(TciTcj)5/12−b/2

√1

Mi+

1Mj

(106)The relevant properties are Tc,N2 = 126.2 K, Tc,He = 5.2 K, pc,N2 = 33.5 atm, pc,He =2.24 atm,MN2 = 28,MHe = 4,MO2 = 32, b ∼ 1.823. Therefore, to comparediffusion coefficients,

DO2N2

DO2He=

(pc,N2)1/3(Tc,N2)

5/12−b/2√

1MN2

+ 1MO2

(pc,He)1/3(Tc,He)5/12−b/2√

1MHe

+ 1MO2

(107)

DO2N2

DO2He=

(33.5)1/3(126.2)5/12−1.823/2√

128 + 1

32

(2.24)1/3(5.2)5/12−1.823/2√

14 + 1

32

≈ 0.0760.307

(108)

Therefore, the effective binary diffusion constant of oxygen in helium islarger, so the limiting current will be larger in synthetic air. Therefore, theconcentration losses are larger in real air than in synthetic air.

Problem 5.2 (10 points) Most importantly, SOFCs are not plaguedwith the problems of water removal that PEMs are. SOFCs operate at highenough temperature that liquid water is not formed and does not have tobe removed. Second, diffusion is faster at higher temperatures (D ∼ T b) sotemperature aids mass transport and not as much engineering must go intothe flow structures.

29

Problem 5.3 (15 points) The limiting current density is set by thediffusion constant, the diffusion layer thickness, and the bulk concentration.

jL = nFDeff coR

δ(109)

You may ensure high coR by designing flow structures that evenly distribute

reactants. You may decrease the diffusion layer thickness by optimizingelectrode structure (though reaction kinetics places other constraints on theMEA). Finally, you may increase Deff by optimizing fuel cell operatingconditions such as temperature and inlet gas mole fractions (again, therewill be other constraints at work).

Problem 5.4 (25 points) To find the limiting current density, we needto calculate the diffusion constant and the bulk concentration of oxygen.First, the diffusion constant is

DeffO2N2

= ε1.5 a

p

(T√

Tc,O2Tc,N2

)b

(pc,O2pc,N2)1/3(Tc,O2Tc,N2)

5/12

(1

MN2

+1

MO2

)0.5

(110)The relevant properties are a = 2.745×10−4, b = 1.823, Tc,O2 = 154.4 K, Tc,N2 =126.2 K, pc,O2 = 49.7 atm, pc,N2 = 33.5 atm,MO2 = 32,MN2 = 28. Pluggingin,

DeffO2N2

=0.41.5 2.745× 10−4

1 atm

(298 K√

(154.4 K)(126.2 K)

)1.823

∗ ((49.7 atm)(33.5 atm))1/3((154.4 K)(126.2 K))5/12

(128

+132

)0.5

So Deff = 0.0520 cm2/s. Finding the bulk concentration of oxygen is easier.Neglecting the consumption of oxygen as air travels down the flow path, wecan say that the concentration of oxygen in air is 21%. From the ideal gaslaw, we find the concentration in bulk is co

R = 0.21 ∗ NV = 0.21 ∗ P

RT =0.21∗ 1 atm

(0.08205 L·atm/mol·K)(298 K) = 8.59 ·10−3 mol/L = 8.59 ·10−6 mol/cm3.Note that n = 4 because the species we are considering is oxygen, which

transfers 4 electrons per mole. Finally, the limiting current density is

jL = (4 ∗ 96485 C/mol)(0.0520 cm2/s)8.59 · 10−6 mol/cm3

500 · 10−4 cm(111)

jL = 3.44 A/cm2

30

Problem 5.5 (25 points) Using equation 5.23,

ηconc = c lnjL

jL − j(112)

the following plot was generated.

Figure 8: Concentration losses as a function of current density for variousvalues of c for problem 5.5

As you can tell from the plot, for higher fuel cell performance, a lowervalue of c is desired.

Problem 5.6 (optional) Using the power law to find the viscosity ofgases,

µN2(T = 800 C) = 16.63 · 10−6 kg/m · s(1073273

)0.67 = 4.16 · 10−5 kg/m · s

µO2(T = 800 C) = 19.19 · 10−6 kg/m · s(1073273

)0.69 = 4.93 · 10−5 kg/m · s

µH2O(T = 800 C) = 11.2 · 10−6 kg/m · s(1073350

)1.15 = 4.06 · 10−5 kg/m · s

Using the molecular weights, viscosity, and mole fractions of each gas, wefind from equation 5.34,

31

Species i Species j Mi/Mj µi/µj Φij xjΦij ΣxjΦij

N2 N2 1 1 1 0.711O2 0.876 0.844 0.919 0.174 0.985H2O 1.555 1.02 1.001 0.100

O2 N2 1.142 1.19 1.094 0.778O2 1 1 1 0.189 1.08H2O 1.776 1.21 1.093 0.109

H2O N2 0.643 0.976 0.980 0.697O2 0.563 0.824 0.902 0.171 0.967H2O 1 1 1 0.1

Equation 5.33 gives the mixture viscosity µmix = 4.29 · 10−5 kg/m · s.The molecular weight of the mixture is Mmix = 27.77 g/mol. The densityis obtained from the ideal gas law:

ρ =p

RT/Mmix=

101325 Pa

(8.314 J/mol ·K)(1073 K)/(0.02777 kg/mol)= 0.315 kg/m3

In a circular pipe, laminar flow holds for Re ∼ 2000, so

Vmax =Reµmix

ρL=

2000 ∗ 4.29 · 10−5 kg/m · s(0.315 kg/m3)(0.001 m)

= 272 m/s (113)

This very fast flow will not be achieved in SOFCs, so flow will always be lam-inar. Laminar flow allows a higher velocity in the case of higher temperaturebecause the density of the gas is diminished at higher temperature.

Problem 5.7 From the current required, 1 A/cm2, you can find theamount of reactant required: JO2/A = j/nF , where JO2 is the flux of oxygenand A is the area.

The limit of laminar flow found in example 5.1 is at the velocity vmax =38.03 m/s. The mole fraction of oxygen carried in the air is xO2 = 0.168.The amount of reactant supplied is therefore JO2 = xO2 vmax where vmax isin units of mol/s.

vmaxxO2

A=

j

nF(114)

To convert the given velocity from m/s to mol/s we use the factor ραwhere ρ is the density in mol/m3 and α is the cross sectional area of the gas

32

channel. Then v = ραv. To find the area that can be supplied with thatamount of reactant, we get

A = xO2vmaxαρnF

j(115)

We are told that the stoichiometric number is 2, and the geometryfrom Example 5.1 has circular cross sections of diameter 1 mm, so thatα = 0.00785 cm2. In addition, ρ = 0.921 kg/m3/(0.02669 kg/mol) =34.51 mol/m3. Therefore,

A = (0.168)(3803 cm/s)(0.00785 cm2)(34.51·10−6 mol/cm3)2 ∗ 96485 C/mol

1 A/cm2

(116)

A = 34 cm2

For a channel of 1 mm to cover an area of 34 cm2, it would have to be3.4 m long! Such a long, thin channel will lead to large pumping losses, sochannels are not usually designed like this. Therefore, channels remain inthe laminar flow regime to very good approximation.

Problem 5.8 The equation developed to describe the oxygen contentas a function of channel length was:

ρO2(x = X) = ρO2(x = 0)−MO2

j

4F

(HC

ShF DO2

+HE

DeffO2

+X

uinHC

)(117)

First, the diffusion constant of O2 in H2O (neglecting multicomponent dif-fusion with N2) is:

DO2H2O =a

p

(T√

Tc,O2Tc,H2O

)b

(pc,O2pc,H2O)1/3(Tc,O2Tc,H2O)5/12

(1

MH2O+

1MO2

)0.5

(118)

DO2H2O =3.64 · 10−4

1

(353√

(154.4)(647.3)

)2.334

((49.7)(217.5))1/3((154.4)(647.3))5/12

(1

18.02+

132

)0.5

(119)Or DO2,H2O = 0.371 cm2/s. To modify for the effective diffusion constant,take into account the porosity: Deff

O2H2O = ε1.5DO2H2O = 0.0940 cm2/s. Theremainder of the values are specified. We take the inlet density of oxygen tobe half of that in the example from section 5.3.3, because that example uses

33

Figure 9: Oxygen density along a fuel cell flow channel for problem 5.8

p = 2 atm (ρO2(x = 0) = 1.2 kg/m3). Therefore, a plot of oxygen densityalong the flow channel can be constructed.

The only tricky part is to make sure you use consistent units. Make surethat distances are in cm or m, and if ρ is in kg/m2, then MO2 should be inkg/mol, not g/mol or kg/kmol.

Problem 5.9 The current in the fuel cell creates water at a flux rate of

JH2O(x, y = C) = MH2Oj(x)2F

(120)

The diffusion from the GDL to the gas channel is (note: water is diffusingin the opposite direction that oxygen diffuses, the positions of the “E” and“C” are switched in this equation compared to the equation in the book)

JdiffH2O(x, y = E) = −Deff

H2O

ρH2O(x, y = E)− ρH2O(x, y = C)HE

(121)

The convective transport from the gas channel to the GDL is given by (youcan check that the direction is correct because a positive flux results whenthe density in the channel is larger than the density at the electrode)

JconvH2O (x, y = E) = −hm[ρH2O(x, y = E)− ρH2O(X, y = channel)] (122)

34

From the steady state conditions JdiffH2O(x, y = E) = JH2O(x, y = C) =

JconvH2O (x, y = E), we get the first relation

−DeffH2O

ρH2O(x, y = E)− ρH2O(x, y = C)HE

= MH2Oj(x)2F

(123)

that we may solve for the desired quantity, the density of water at thecatalyst layer,

ρH2O(x, y = C) = ρH2O(x, y = E) +HE

DeffH2O

MH2Oj(x)2F

(124)

From the second equality in the steady state condition, we obtain the secondrelation

−hm[ρH2O(x, y = E)− ρH2O(X, y = channel)] = MH2Oj(x)2F

(125)

We may make the substitution for hm with the Sherwood number and isolatethe density of water diffusing into the GDL:

ρH2O(x, y = E) = ρH2O(X, y = channel)− HC

ShF DH2OMH2O

j(x)2F

(126)

Plugging this term into equation 124,

ρH2O(x, y = C) = ρH2O(X, y = channel)− HC

ShF DH2OMH2O

j(x)2F

+HE

DeffH2O

MH2Oj(x)2F

(127)Factoring out a term,

ρH2O(x, y = C) = ρH2O(X, y = channel) + MH2Oj(x)2F

[HE

DeffH2O

− HC

ShF DH2O]

(128)The last step is to take care of the x dependence. We’d like the equation

to be in terms of something we can set or measure, like ρH2O(0, y = channel),not ρH2O(X, y = channel). Since the flow streams will likely be humidified,there will be water flowing into the gas channel at a speed uin. Integratingthe flux of water from the GDL to the flow channel over the length of thegas channel is the difference between inlet and outlet water concentrationin the gas channel (note the sign reversal):∫ X

0[JH2O(x, y = E)]dx = −uinHC ρH2O(0, y = channel)+uoutHC ρH2O(X, y = channel)

(129)

35

Alternatively, we can calculate the change in water concentration by inte-grating the water produced by current in the fuel cell.∫ X

0[JH2O(x, y = E)]dx =

MH2O

2F

∫ X

0j(x)dx (130)

These two equations must be equal. Assuming that the velocity of the gasdoes not appreciably change in the channel (uin = uout)

MH2O

2F

∫ X

0j(x)dx = uinHC [ρH2O(X, y = channel)− ρH2O(0, y = channel)]

(131)Again, we make the simplifying assumption that j(x) is a constant, theabove equation becomes

XMH2Oj

2F= uinHC [ρH2O(X, y = channel)− ρH2O(0, y = channel)] (132)

Solving for ρH2O(X, y = channel) and plugging into equation 128, we getthe final result:

ρH2O(x, y = C) = ρH2O(0, y = channel)+X

uinHC

MH2Oj

2F+MH2O

j

2F[

HE

DeffH2O

− HC

ShF DH2O]

(133)or

ρH2O(x, y = C) = ρH2O(0, y = channel)+MH2Oj

2F[

HE

DeffH2O

− HC

ShF DH2O+

X

uinHC]

(134)To check the signs, note that water concentration increases with X, as itshould. Increasing the inflow of water by increasing uin increases the waterthat reaches the catalyst. Decreasing the diffusion layer thickness (HE)increases the amount of water swept away by the gas, so the density ofwater at the catalyst goes down, as it should. As the last check, if theamount of gas is fixed and HC , the channel size, is decreased, the density ofwater increases.

Problem 5.10 Mass concentration effects become important at highcurrent density, so we will use the Tafel approximation to the Butler Volmerequation

j = j00

c∗Rc0∗R

eαnFη/RT (135)

36

If we assume that voltage, and therefore overpotential is constant along theflow channel, the only part that varies in x is the concentration of oxy-gen. Via a control volume analysis, you can find that the ODE governingconcentration along the flow channel is

dc(x)dx

= −j00eαnFη/RT

c0∗R

c(x) (136)

That is, the amount of oxygen flowing into the catalyst layer from the flowchannel is proportional to the concentration in the flow channel. The solu-tion is a decaying exponential

c(x) = Ae−ax (137)

where

a =j00eαnFη/RT

c0∗R

(138)

Applying the boundary condition at the inlet gives the constant A

c(0) = cin = A (139)

Plugging this into equation 5.62, we get

ρO2 |x=X,y=C = ¯ρO2 |x=0,y=channel −MO2

4F

(j(X)hm

+HEj(X)

DeffO2

+a

uinHC

∫ X

0c∗O2

(x)dx

)(140)

= ¯ρO2 |x=0,y=channel −MO2

4F

(j(X)hm

+HEj(X)

DeffO2

+a

uinHC

∫ X

0cine−axdx

)(141)

= ¯ρO2 |x=0,y=channel −MO2

4F

(j(X)hm

+HEj(X)

DeffO2

+acin

uinHC(−e−ax

a|X0 )

)(142)

= ¯ρO2 |x=0,y=channel −MO2

4F

(j(X)hm

+HEj(X)

DeffO2

+cin

uinHC(1− e−aX)

)(143)

= ¯ρO2 |x=0,y=channel[1−MO2

4FuinHC(1− e−aX)]− MO2

4F

(j(X)hm

+HEj(X)

DeffO2

)(144)

37

This says that the oxygen falloff in the channel is exponential rather thanlinear. The falloff is faster for larger overpotential or higher exchange current(faster kinetics meaning faster use of reactant).

38

Chapter 6 Solutions

Problem 6.1 (4 points each) 1. High resistance will lead to a steeplinear section of the curve, this is exhibited in (e)

2. A leakage current will shift the curve left, as in (b)3. Poor kinetics appear as large activation losses, (c)4. Low ohmic resistance will show up as a near-horizontal portion of the

linear section of the curve, (a)5. Reactant starvation leads to large concentration losses (d)

Problem 6.2 (10 points) We intended for students to only considervoltage efficiency. In that case, while both curves show the same VOC andthe same short circuit current, the power (and efficiency) of the SOFC ishigher because the curve lies at higher voltage for every current level.

Another key difference between the SOFC and PEM is that the SOFCwill be operated at much higher temperature. If you consider overall effi-ciency, then let’s find the effect of operating at higher temperature: ε =εthermoεvoltageεfuel and assume that fuel efficiency is roughly the same foreach case, let’s see how thermodynamic efficiency changes:

ε ∼ ∆G

∆H∗VE

=−nF

−nF∗∆G

∆H∗VE

=E

∆H−nF

∗VE

=V

∆H−nF

=V

(∆Ho + ∆T∑

P−R cp,i)/− nF

(145)Remember that ∆Ho < 0. We’ll assume constant specific heats, so thatterm at 298 K is∑

P−R

cp,i = cp,H2O−cp,H2−0.5xp,O2 = 33.6−28.8−.5∗28.9 = −9.7 ≡ cp (146)

The effect of higher temperature (larger ∆T ) is to make the denominatorlarger, making efficiency lower. To get a quantitative feel for the size of theeffect, let’s assume the SOFC is operating at 1000 K and the PEM at 298K.

εSOFC

εPEM=

VSOFC

VPEM∗

∆Ho

−nF∆Ho+∆T cp

−nF

≈ VSOFC

VPEM∗ −285 kJ/mol

−241 kJ/mol − 9.7 J/mol ·K ∗ (1000 K − 298 K)

=VSOFC

VPEM∗ 1.15

Remember that we use values for liquid water for the PEM and water vaporfor the SOFC. That means that ∆HPEM > ∆HSOFC , even though the

39

SOFC is operated at higher temperature. Then both effects combine tomake the SOFC higher efficiency.

Problem 6.3 (15 points) (a) 1. If electrode pores shrink or the elec-trode becomes thicker, the catalytically active area increases because thereare more possible reaction sites.

2. If the ionic resistance increases, active sites further away from theelectrode are less effective because ions will have to be transported throughthicker resistive material. The catalytically active area therefore decreases.

3. Similar to above, if the resistance increases, areas further from thecurrent collector become less active, so active area decreases.

4. The charge transfer resistance is a measure of how effective the cat-alyst is. If the entire catalyst becomes less active, the area does remainsunaffected, though performance decreases.

(b) At high operating temperatures, SOFCs are generally less affectedby mass transport and charge transfer, but they generally have low ionicconductivity. Therefore, the extent to which the electrochemically activearea extends into the electrode is limited by the ionic conductivity of theelectrode.

(c) PEMs are more commonly limited by charge transfer resistance, whileionic conductivity is generally high. Therefore increasing the thickness ofcatalyst layer increases the electrochemically active area.

Problem 6.4 (10 points) The leakage current is larger than the ex-change current density, so we should first assume Tafel kinetics, then checkthe validity of the assumption. Neglecting ohmic and concentration lossesat this low current, the voltage loss from leakage due to activation losses is

η =RT

αnFln

j

j0= 59.1 mV (147)

Using this value for η in the full Butler Volmer equation gives a 1% error inthe current, so this is sufficiently accurate.

Problem 6.5 1. (5 points) Starting from the Nernst equation

E = Eo +∆S

nF(T − To)−

RT

nFln

ΠaνiP

ΠaνiR

(148)

The problem does not state whether liquid water or water vapor is theproduct. This solution is worked for liquid water, but may be easily solved

40

for water vapor product using aH2O(g) 6= 1. The thermodynamic voltage is

E = Eo +1

nF

[∆S(T − To)−RT ln[

1p1

H2p0.5

O2

]

](149)

E = Eo +1

n ∗ F

[∆S(T − To) + RT ln[xH2x

0.5O2

(P

Po)1.5]

](150)

To find ∆S, we need the entropy of liquid water at 330 K, which can befound from the appendix as ∆SH2O(l)

= 77.57 J/mol ·K. Then the entropychange of reaction is

∆Srxn = SH2O(l)−1

2SO2−SH2 = 77.57−0.5∗209.02−133.60 = −160.04 J/mol·K

(151)Inserting values,

E = 1.23 +1

2 ∗ 96485

[−160(330− 298) + 8.314 ∗ 330 ∗ ln[1 ∗ 0.210.5(

11)1.5]

](152)

E = 1.19 V

2. (5 points) In problem set 3 we found the Tafel constants in the Tafelequation

ηact = ac + bc ln j (153)

The constants are

ac = − RT

αnFln jo = − 8.314 ∗ 330

0.5 ∗ 2 ∗ 96485ln 10−3 (154)

ac = 0.196 V

We may use this constant without modification since the conversion betweenln and log was not used. The other constant did use the conversion factorof 2.3, which may be dropped so that

bc =RT

αnF=

8.314 ∗ 3300.5 ∗ 2 ∗ 96485

(155)

bc = 0.0284 V

3. (5 points) The ASR is given by

ASR = L/σ = 0.01/0.1 (156)

41

ASR = 0.1 Ωcm2

4. (5 points) First, the diffusion coefficient is

DO2N2 =a

p

(T√

Tc,O2Tc,N2

)b

(pc,O2pc,N2)1/3(Tc,O2Tc,N2)

5/12

(1

MN2

+1

MO2

)0.5

(157)The relevant properties are a = 2.745×10−4, b = 1.823, Tc,O2 = 154.4 K, Tc,N2 =126.2 K, pc,O2 = 49.7 atm, pc,N2 = 33.5 atm,MO2 = 32,MN2 = 28. Pluggingin,

DO2N2 =2.745× 10−4

1 atm

(330 K√

(154.4 K)(126.2 K)

)1.823

∗ ((49.7 atm)(33.5 atm))1/3((154.4 K)(126.2 K))5/12

(128

+132

)0.5

The diffusion coefficient is DO2N2 = 0.248 cm2/s. The effective diffusioncoefficient is then

DeffO2N2

= 0.201.5 ∗ 0.248 (158)

DeffO2N2

= 0.0222 cm2/s

5. (5 points) The limiting current density is

jL = nFDeff c0R

δ(159)

The bulk concentration of oxygen is

c0R =

N

V=

P

RT=

0.21 atm

0.08205 L · atm/mol ·K ∗ 330 K= 7.75 · 10−6 mol/cm3

(160)Note that n = 4 since we are calculating losses at the cathode (oxygen isthe species of interest) so that we find

jL = 4 ∗ 96485 ∗ 0.0222 ∗ 7.75 · 10−6

0.05(161)

jL = 1.33 A/cm2

42

6. (10 points) The various losses add:

V = E − ηact − ηohmic − ηconc (162)

Neglecting anodic activation losses, we get

V = E − [ac + bc ln(j + jleak)]− j ∗ASR− c lnjL

jL − (j + jleak)(163)

with the values found above, this becomes

V = 1.19− [0.196 + 0.0284 ln(j + 0.005)]− 0.1j − 0.1 ln1.33

1.33− (j + 0.005)(164)

The i− V and i− p plots are shown in the figure.

Figure 10: Voltage and power density as a function of current density forproblem 6.5

Note that since we are using the Tafel form, the activation losses areunderestimated at low current density.

7. (10 points) Finding the maximum power density can be done an-alytically, but as a first guess, we can just determine from the graph thatthe maximum power density is about 0.77 W/cm2 which occurs at a currentdensity of about 1.1 A/cm2.

43

For an analytical solution, we need to find the maximum in the powerdensity curve:

p = V j ⇒ dp

dj= V + j

dV

dj(165)

So the optimum occurs at

dp

dj= 0 =E − [ac + bc ln(j + jleak)]− j ∗ASR− c ln

jL

jL − (j + jleak)+ · · ·

j

(− bc

j + jleak−ASR− c

jL − (j + jleak)

)A numerical solution shows that our guess from the graph was very accurate.The peak power density is pmax = 0.775 W/cm2 at a current density of

j∗ = 1.12 A/cm2 .8. (5 points) If f is the fuel utilization fraction, the overall fuel cell

efficiency is

ε = εthermo ·V

E· f =

∆G

∆HHHV· VE· f (166)

Using values from the appendix to find ∆G, we find

∆Grxn = GH2O(l)−GH2−

12GO2 = −309.15−(−42.91)−0.5∗(−75.07) = −228.59 kJ/mol

(167)We use the voltage at the maximum power point (V = 0.692 V ) to find theoverall efficiency

ε =−228.59−286

· 0.6921.19

· 0.90 (168)

ε = 41.8%

Problem 6.6 Using the fact that the total number of moles is conserved,dxi = −dxj (which also follows from xi + xj = 1), we can find that

Ji + Jj = ρDijdxi

dz+ ρDji

dxj

dz= ρ(Dij −Dji)

dxi

dz(169)

In steady state, there is no net flux, so Ji + Jj = 0. In general, ρ 6= 0 andthere may be concentration gradients, so dxi

dz 6= 0, which means that we musthave Dij = Dji.

44

Problem 6.7 Showing that x1 + x2 + · · · + xN = 1 is equivalent toshowing

∑dxi = 0. Proceeding that way, sum the Stefan-Maxwell equation

over all components to get∑i

dxi

dz=∑

i

RT∑j 6=i

xiJj − xjJi

PDeffij

=RT

P

∑i,j 6=i

xiJj − xjJi

Deffij

(170)

Since Dij = Dji, each term in the sum cancels to give 0, as we require. Toshow that each term cancels, we write out the sum for two species i, j = 1, 2:

∑i,j 6=i

xiJj − xjJi

Deffij

=∑j 6=i

x1Jj − xjJ1

D1j︸ ︷︷ ︸i=1

+x2Jj − xjJ2

D2j︸ ︷︷ ︸i=2

=x1J2 − x2J1

D12︸ ︷︷ ︸i=1,j=2

+x2J1 − x1J2

D21︸ ︷︷ ︸i=2,j=1

(171)

= (x1J2 − x1J2 − x2J1 + x2J1)/D12 = 0 (172)

Problem 6.8 Using the values from the text, the j-V curve is shown be-low. From the figure, the limiting current density is approximately 2.3 A/cm2.

Problem 6.9 The j-V curve at T = 873 K is shown. It is clear from theplot that the ohmic losses increase dramatically at lower temperatures–themaximum current density produced by the fuel cell at this temperature isonly about 0.43 A/cm2!

Problem 6.10 The j-V curves for an electrolyte supported SOFC atT = 1073 K and T = 873 K are shown below. Given the very high ohmiclosses, especially at lower temperatures, it is clear that it is not optimal tosupport an SOFC with a thick electrolyte.

Problem 6.11 (a) The linear approximation to the Butler-Volmer equa-tion is

ηanode =RT

2αF

jp0

j0p=

RT

2αF

jp0

j0pAxH2

(173)

which is similar to equation 6.22 in the book. Using an analogous equationto 6.26 for xH2 , we find

ηanode =RT

2αF

j

j0pA

(xH2 |a − tA jRT

2FpADeffH2,H2O

) (174)

45

Figure 11: IV plot of the SOFC model for problem 6.8

Figure 12: IV plot of the SOFC model at T = 873 K for problem 6.9

46

Figure 13: IV plot of the SOFC model at T = 1073 K for problem 6.10

Figure 14: IV plot of the SOFC model at T = 873 K for problem 6.10

47

where we again implicitly express pC in atm, allowing us to drop the p0.(b) Using anode parameters from table 6.1 in the book, and the rest of

the parameters from table 6.4, the j-V curve may be plotted. Anode lossesin this case are very small.

Figure 15: IV plot of the SOFC model for problem 6.11

Problem 6.12 The i-V curves for an anode supported SOFC accordingto our model are shown in the figure below. In this case, anode lossesremain small, but are not negligible–losses are dominated by ohmic loss.The limiting current density for the cathode is approximately 14.3 A/cm2,and for the anode around 18 A/cm2. In this simple model, the anode lossesdo not blow up as the cathode losses do, because we have taken the linearform for the losses, which breaks down long before the limiting currentdensity.

Problem 6.13 At each value of current density, we must solve for theconstants α∗ and C to find λ(z) in order to find the resistance. Followingthe equations in the text, it is possible to find the complete j-V curve, asis shown below. Important steps along the way are the following equations.

48

Figure 16: IV plot of the anode supported SOFC model for problem 6.12

The conductivity of the membrane is

σ(z) =[0.005193(4.4α∗ + Ce0.000598jz/Dλ)− 0.00326

]e1268( 1

303− 1

T) = s1+s2e

s3z

(175)Where several constants are introduced to simplify notation. Then the re-sistance of the membrane is

Rm =∫ tm

0

dz

σ(z)=∫ tm

0

dz

s1 + s2es3z=[

z

s1− ln(s1 + s2e

s3z)s1s3

]tm

0

(176)

Rm =1

s1s3

(s3tm − ln(s1 + s2e

s3tm) + ln(s1 + s2))

(177)

Note that the ohmic losses are not linear with current, as at highercurrent, the membrane is dried and the resistance increases. As currentincreases, the first and third term in the above equation increase, and Rincreases.

Problem 6.14 (a) The j-V curve for λ = 1.2 is shown below, as are thepower density curves for λ = 1.2 and λ = 2.0. The maximum power densityin the first case is approximately 0.25 W/cm2, so if the pump consumes

49

Figure 17: IV plot of the PEMFC model for problem 6.13

10% of this power, the power delivered is approximately 0.23 W/cm2. Ifλ = 2.0, the max power density is approximately 0.56 W/cm2, so if thepump consumes 20% of this power, the power delivered is approximately0.45 W/cm2, still more than in the first case.

50

Figure 18: IV plot of the SOFC model for λ = 1.2 problem 6.14

Chapter 7 Solutions

Problem 7.1 (5 points) First, to quantitatively determine how “good”a fuel cell is. Second, to determine why it is good or bad.

Problem 7.2 (10 points) There isn’t just one right answer. As longas you give a good argument behind your answer, we accept the followinganswers. Major operational variables include:

1. Temperature: The most important temperature dependence is in theequation for the exchange current density.

j0 = Ae−∆G‡/RT (178)

You can see that the exchange current increases exponentially withtemperature. From the Butler-Volmer equation in the low overpoten-tial regime, ηact ∼ 1/jo, so that the overpotential depends exponen-tially on temperature. In the high overpotential regime, ηact ∼ − ln jo,and the dependence is linear. The Butler-Volmer equation shows a

51

Figure 19: Power density plot of the SOFC model for problem 6.14. λ = 1.2line is blue, λ = 2.0 line is green.

temperature dependence that is less strong:

j = joo

(C∗

R

Co∗R

exp [αnFη

RT]−

C∗P

Co∗P

exp [−(1− α)nFη

RT])

(179)

From this equation, the overpotential is linear in temperature. Also,from the Nernst equation

E = ET −RT

nFln[Πaνi

P

ΠaνiR

](180)

the temperature dependence is linear. For transport, temperaturecomes into play in a ceramic by the equation for conductivity

σ =c(nF )2Doe

−∆GactRT

RT(181)

So the ohmic overpotential depends exponentially on temperature.

2. Gas Pressure: both Nernst and Butler-Volmer are important here too.We showed in class and in an example that the Nernstian pressure

52

dependence is logarithmic, that is to say weak. Also, the transportequation shows logarithmic dependence:

ηconc =RT

αnFln

c0R

c∗R(182)

3. Gas composition: again, Nernst and BV, as well as transport equation182

Other operational variables include the compression force that presses thefuel cell together, the flow rates of gases, etc.

Problem 7.3 (5 points) An EIS measurement gives more detailedinformation, though it is harder to interpret than a current interrupt mea-surement. The current interrupt measurement can be done on larger–higherpower–fuel cells, can be done faster, in parallel with an i-V measurement,and with simpler measurement equipment.

Problem 7.4 (a) (5 points) A current scan from zero current to a highcurrent that is too fast will show a voltage that is too high, because the fuelcell hasn’t reached a steady state. At each current step, the voltage mustdrop down to its steady-state value. The faster scan rate (100 mA/s) willshow a higher voltage, ie. a better performance.

A voltage scan will have the opposite effect. Since you scan from OCV(high voltage) to zero voltage, at a step downwards in voltage, the currentmust ramp up to the new steady-state value. If you step down in voltage toofast, the current will not ramp up to the steady-state value, and you measurea current that is too low. In this case, the faster scan rate (100 mV/s) willshow a current that is too low, ie. a worse performance.

(b) (5 points) The slowest step in the fuel cell reaction will be the mostaffected by a fast scan. As electrochemistry is quite fast in comparison, thegas diffusion is the slowest step in the fuel cell reaction (as shown by thelow frequency of the Warburg element representing mass transport in figure7.12). Thus, the gas diffusion, which dominates at high current density, willbe the most affected by the faster scan rate.

Problem 7.5 Intuition should get students most of the way on theproblem. The circuit diagram is shown in the first figure.

You should inspect the easy, limiting cases of high frequency and lowfrequency first. (1) At high frequency, capacitors act as short circuits, sothe combination of blocking + activated electrode acts as just the resistor

53

Figure 20: One blocking and one activated electrode for problem 7.5.

Rb. The Nyquist plot of this resistor is just a point at Rb. (2) At lowfrequency, capacitors act as open circuits, so we can ignore Ca. The circuitnow acts as a resistor Rb + Ra in series with a capacitor. The Nyquistplot of this circuit is just a vertical line, with the intercept at Rb + Ra. Atintermediate frequencies, we have some of the character of the parallel RCcircuit, which is a semicircle in a Nyquist plot. The amount of the semicircleyou will see depends on the distance between the point Rb and the verticalline at Rb + Ra. We’ll answer that question later.

If you don’t buy that argument, the mathematical treatment is givenhere. The impedance of the circuit is

ZTotal = ZRb+ ZCb

+ZRaZCa

ZRa + ZCa

(183)

Inputting the complex impedances of resistors and capacitors

ZTotal = Rb −j

ωCb+−jRa/Caω

Ra − j/ωCa= Rb −

j

ωCb− jRa

ωCa·

Ra + jωCa

R2a + 1

ω2C2a

(184)

Separating the real and imaginary components of the impedance,

ZRe = Rb +Ra/ω2C2

a

R2a + 1

ω2C2a

= Rb + Ra ·1

1 + (RaωCa)2(185)

−ZIm =1

ωCb+

R2a/ωCa

R2a + 1

ω2C2a

=1

ωCb+

1ωCa

· 11 + (RaωCa)−2

(186)

The low frequency asymptote ω = 0 shows that ZRe = Rb + Ra and ZIm =∞, and the high frequency (ω = ∞) shows that ZRe = Rb and −ZIm =0. This checks with our inspection of the circuit diagram above. Givenan intercept and asymptote, information about the relative magnitudes of

54

the RC components can be used to sketch in the intervening region. Asgiven in the problem, when the RC time constants are well separated, theresolution of each segment should be possible. For the skeptics, a moredetailed treatment follows.

A key question is: is the semicircle visible not? This question is actuallymuch tougher to answer, and probably can’t be found by inspection. If yousee part of a semicircle, that means that −ZIm has a maximum. So, here isthe derivative of −Zim with respect to frequency:

d(−ZIm)dω

= 0 = − 1Cbω2

− 1Caω2

(1

1 + (RaCaω)−2

)+

1ωCa

(− −2(RaCa)−2ω−3

(1 + (RaCaω)−2)2

)(187)

Let χ = (CaRaω)−1, and factor out 1/ω2

0 =1ω2

(− 1

Cb− 1

Ca(1 + χ2)+

2χ2

Ca(1 + χ2)2

)(188)

Since we aren’t interested in the solution ω = ∞, we discard that solutionand clear the denominators:

0 = −(1 + χ2)2

Cb− (1 + χ2)

Ca+

2χ2

Ca(189)

Let C = CbCa

. Now expand and group like terms to get a quadratic in χ2:

0 = (1 + 2χ2 + χ4) + (1 + χ2)C − 2χ2C = χ4 + χ2(2−C) + (1 + C) (190)

The solution is

χ2 =C − 2±

√(C − 2)2 − 4(1 + C)

2(191)

This will have a solution if the determinant is positive, so in order to seemost of our semicircle, we require

(C − 2)2 − 4(1 + C) > 0 =⇒ C2 − 4C + 4− 4− 4C > 0 (192)

Therefore, C2 − 8C > 0 that is, since C > 0 by definition, C > 8. So to seethe peak in the semicircle, Cb > 8Ca, or the blocking R× C must be largerthan the activated R × C. Note that this requirement does not depend onthe relative values of the resistance!

(a) (10 points) When R×C is much smaller for the parallel RC than forthe series RC, the entire semicircle can be seen before the ω = 0 asymptote

55

Figure 21: EIS plot for 7.5(a), when RaCa < RbCb

is approached. A MATLAB plot of the EIS curve for some chosen values ofR and C is shown here.

(b) (10 points) When R × C for the parallel RC is much larger thanthe series RC, the semicircle is only beginning before the asymptote is ap-proached. Only the first linear part of the semicircle can be seen. A MAT-LAB plot of the EIS curve is shown here.

(c) (10 points) The circuit diagram is shown here.The impedance of the circuit is

Z =− jR1

ωC1

R1 − jωC1

+ R2 +− j

ωC2

[R3 + σ√

ω(1− j) tanh(δ

√jωD )]

− jωC2

+ R3 + σ√ω(1− j) tanh(δ

√jωD )

(193)

To simplify notation, make the following definitions: χ−11 ≡ ωC1R1, χ−1

2 ≡ωC2R3, and W ≡ σ

R3√

ωtanh(δ

√jωD ). Clearing out the imaginary denomi-

nators,

Z = R1χ1χ1 − j

1 + χ21

+ R2 + R3χ2−j[1 + 2W + χ2W + 2W 2 + jχ2(1 + W )]

(1 + W )2 + (χ2 + W )2(194)

56

Figure 22: EIS plot for 7.5(b), when RaCa > RbCb

Figure 23: Circuit diagram for problem 7.5

The real impedance is

ZRe = R1χ2

1

1 + χ21

+ R2 + R3χ2χ2(1 + W )

(1 + W )2 + (χ2 + W )2(195)

and the imaginary impedance is

−ZIm =R1χ1

1 + χ21

+ R3χ21 + W (2 + χ2 + 2W )(1 + W )2 + (χ2 + W )2

(196)

The high frequency asymptote (χ1 = χ2 = 0) gives

ZRe = R2; −ZIm = 0 (197)

57

which could be confirmed by inspection of the circuit diagram. The lowfrequency (χ1 = χ2 =∞) gives

ZRe = R1 + R2 + R3(1 + W ); −ZIm = R3W (198)

Which makes sense because the definition of W has R3 in the denominator.A program such as MATLAB can plot the results for given values for theconstants. A plot that uses numbers from Table 7.2 is shown.

Figure 24: EIS plot generated for problem 7.5

Problem 7.6 (10 points) Porosity refers to the amount of free spacein a structure, but for a structure to be permeable, the holes need to go theentire way through a material. A sketch of a highly-porous material withlow permeability (in the direction in the plane of the page) is shown.

Problem 7.7 (15 points) In Example 7.1, we found the values at point(b): η

(b)ohmic = 0.10 V and η

(b)act = 0.30 V .

At point (c), the current appears to be 1.75 A. From the left interceptof the Zreal axis, the ohmic resistance is 0.10 Ω. The ohmic overvoltage is

η(c)ohmic = iRohmic = (1.75 A)(0.10 Ω) (199)

58

Figure 25: The highly-porous material with low permeability for problem7.6.

η(c)ohmic = 0.175 V

Remember that Rf is the diameter of the semicircle, not the intercept ofthe semicircle. Then from the right intercept of the Zreal axis minus the leftintercept, Rf = 0.20 Ω, so the activation overvoltage is

η(c)act = iRf = (1.75 A)(0.20 Ω) (200)

η(c)act = 0.350 V

We may fit the data to the equation to extract the remaining desiredvalues. If you plug in numbers first, since i(b) = 1, things simplify and thesolution is shorter. Or, solving the general case:

η(b)act =

RT

αnFln

i(b)

io(201)

and

η(c)act =

RT

αnFln

i(c)

io⇒ α =

RT

nFη(c)act

lni(c)

io(202)

Substituting α into equation 201

η(b)act = η

(c)act

ln[i(b)/io]ln[i(c)/io]

(203)

Isolating the io term

η(b)act

η(c)act

ln i(c) − ln i(b) = ln io

(η

(b)act

η(c)act

− 1

)(204)

59

Therefore, io is

io = exp

η(b)act

η(c)act

ln i(c) − ln i(b)

η(b)act/η

(c)act − 1

= exp

[0.300.35 ln(1.75)− ln(1)

0.30/0.35− 1

](205)

io = 0.0348 A

Substituting this value back into equation 202 gives alpha

α =(8.314)(300)

(2)(96485)(0.35)ln

1.750.0348

(206)

α = 0.145

Problem 7.8 (10 points) The scan rate allows us to convert froman i-V plot to an i-t plot. To find the total charge Qh, remember thatcharge equals current times time, so the area under the i-t curve gives thecharge. Estimating the width of the peak to be 400 mV × 1

10 mV/s = 40 sand the peak of the curve to be about 40 µA, the area under the curve isapproximately

Qh ≈ (40 s)(40 µA) = 1600 µC (207)

Then the active catalyst area coefficient is

Ac =Qh

Qm ∗Ageometric=

1600 µC

210 µC/cm2 ∗ (.1 cm)2(208)

Ac = 760

60

Chapter 8 Solutions

Problem 8.1 (5 points)(a) Ni is used in high temperature fuel cellsas a catalyst. While it is not as good a catalyst as some noble metals, it isfar cheaper. The position of Ni in the periodic table is above Pt and Pd,both good catalysts, showing that its electronic structure is well suited tocatalysis.

(b) YSZ is mixed with Ni in SOFCs primarily to add ionic conductivity.In addition, it adds thermal expansion compatibility, mechanical stability,and maintains the high porosity of the electrode.

(c) Cr is added to Ni in MCFCs to maintain porosity and high surfacearea.

Problem 8.2 (5 points) As long as a convincing argument is provided,we recommend accepting either of the answers:

1) High temperature fuel cells benefit from faster reaction kinetics, andtherefore require lower noble metal loadings as catalysts. As cost is a pri-mary barrier to widespread fuel cell adoption and noble metals add greatlyto the cost of low temperature fuel cells, the lower noble metal loading ofhigh temperature fuel cells is the largest advantage.

2) High temperature fuel cells benefit from fuel flexibility. Whereas lowtemperature fuel cells are poisoned by CO presence in the ppm level, hightemperature fuel cells use CO as a fuel. Since generation of neat hydrogenis currently very costly, in the foreseeable future, the only economical waytowards a hydrogen economy is via reforming of hydrocarbons. In that case,fuel flexibility is an enabling feature, as impurities such as CO are presentin large levels in reformed fuel.

Note: alleviation of water clogging issues is also an important advantageof high temperature fuel cells. The availability of high-quality waste heatis a nice perk, but is not a main advantage of high temperature fuel cells.Most of the present research on high temperature fuel cells concerns howto lower the operating temperature, which shows that people consider thewaste heat to be of lesser importance.

Problem 8.3 (10 points) See the figure below.

Problem 8.4 (5 points)(a) The stack power is 25 kW . That poweris produced by 40 cells, each with a cell area of 6000 cm2. The area-based

61

Figure 26: Fuel cell in problem 8.3

power density is therefore

25 kW/40 cells

6000 cm2/cell(209)

104 mW/cm2

(b) Each cell has a volume

120 cm× 81.4 cm× 0.65 cm = 6349.2 cm3 (210)

. Then the volumetric power density of the stack is

25 kW

40 cells× 6349.2 cm3/cell(211)

98.4 mW/cm3

Problem 8.5 (5 points) Assuming the hydrogen is stored at roomtemperature, the quantity of hydrogen on board the vehicle is

N =PV

RT=

350 atm ∗ 156.6 L

0.08205 L·atmmol·K ∗ 298 K

= 2.24 kmol (212)

The energy stored in this fuel is (using the HHV of hydrogen, 286 kJ/mol)

2.24 kmol ∗ 286 kJ/mol = 641 MJ (213)

62

If this energy is used continually over a span of 4.3 hours (430 km/100 km/hr),the power output is

641 MJ

4.3 hr ∗ 3600s/hr= 41.4 kW (214)

If the fuel cell is 55% efficient, the fuel cell produces 22.8 kW during opera-tion under these conditions.

Problem 8.6 (10 points)(a) Carnot efficiency is given by

εCarnot = 1− Tc

Th(215)

The efficiency is εCarnot = 1− 3731073

εCarnot = 65.2%

(b) If the SOFC is 55% efficient, then 45% of the input energy is rejectedas heat. This heat is available to the heat engine that has efficiency ofεHE = 0.6εCarnot = 39.1%. The efficiency of the combined cycle is εTot =εSOFC + (1− εSOFC)εHE = 0.55 + 0.45× 0.391.

εTot = 72.6%

63

Chapter 9 Solutions

Problem 9.1 (5 points) Fuel flows up the center of the donut and in theporous anode. Air flows in around the sides of the porous cathode . Sealingprevents mixing of air and fuel. One advantage of the structure is that itfacilitates oxygen transport: usually oxygen is the limiting reactant. As itflows through the cathode, it gets depleted, but it has access to the mostreactive area around the edges where it is least depleted. As the oxygen getsdepleted along the cathode, it accesses less area so that the depletion is notas detrimental to performance. The figure shows a sketch of the structure,the inset shows the gas flow.

Figure 27: One possible configuration for a donut shaped fuel cell for prob-lem 9.1

Problem 9.2(5 points) The DOE canceled the on-board fuel reformingproject because current technologies are inadequate and there is no promis-ing path forward. There is no interest from industry, and no reason tosuspect that a future FCV (fuel cell vehicle) with on-board reforming willoutperform future gasoline-electric hybrids. Since the great interest in fuelcell cars lies in their zero emissions capability, we do not perceive a hugeimpetus to implement FCV’s that run on gasoline.

Problem 9.3(10 points) (a) The heat generation comes from ineffi-

64

ciency. The thermodynamic efficiency at STP is 83%, and we must findthe voltage efficiency εV = 0.7/1.23 = 57%. So the fuel cell efficiency is0.83 ∗ 0.57 = 47%. Since the fuel cell generates 1000 W of electricity at47% efficiency, the input fuel must flow at a rate of 1000 W/.47 = 2128 W ,so it must generate 2128 W ∗ .53 = 1128 W of heat. The heat rejection is1128 W .

(b) The parasitic power consumed by the cooling system is 1128 W/25 =45.1 W .

Problem 9.4(10 points) The molar mass of methanol is 32.0 g/mol, sothe molar density of methanol is 0.79 g/cm3

32 g/mol = 0.0247 mol/cm3. The molar

mass of water is 18 g/mol, so the molar density of water is 1.0 g/cm3

18 g/mol =0.0556 mol/cm3. If we assume that the volume of methanol and water donot change upon mixing, we must find x, the number of moles of water andmethanol that will fill one liter.

1 L =x

55.6 mol/L+

x

24.7 mol/L=⇒ x = (

155.6 mol

+1

24.7 mol)−1 = 17.1 mol

(216)Next, we find the HHV of methanol. A methanol combustion reaction is

CH3OH +32O2 → CO2 + 2H2O (217)

So that the HHV of methanol is obtained by

∆HHHV =∆HCO2f + 2∆H

H2O(l)f −∆H

CH3OH(l)f − 3

2∆HO2

f

=− 393.51 + 2(−285.83)− (−238.4− 32∗ 0) = −727 kJ/mol

Where the value ∆HCH3OH(l)f = −238.4 kJ/mol was found from the NIST

Webbook (a good resource if you don’t know it). In kWh, this is 727 kJ/mol3600 s/hr =

0.202 kWh/mol. Finally, we know how much fuel we have, so the fuel tankcapacity is 17.1 mol ∗ 0.202 kWh/mol = 3.45 kWh. For our 2 liter system,we have an energy density of 1.72 kWh/L . The small difference betweenthis value and the quoted value may be explained if the ∆Hf numbers wereobtained from different sources.

Problem 9.5(10 points) Effectiveness is defined in the text as

effectiveness =% conversion of carrier energy to electricity

% conversion of neat H2 to electricity(218)

65

We can rewrite this as

effectiveness =(% conversion of carrier energy to “dirty′′ H2)×% conversion of “dirty′′ H2 to electricity

% conversion of neat H2 to electricity

Using the numbers given, we find

effectiveness = 0.75 ∗ 0.8 = 0.6 (219)

So that the effectiveness of our system is 60% .

Problem 9.6(10 points) (a) Given the constraint that the total poweris specified to be P , we say that the power density of the fuel cell is pFC =PxV . Now, the efficiency explicitly written as a function of x is

ε(x) = A−BP

xV(220)

Now we may write equation 9.9 in the book, for total energy of the systemas a function of x, as

E = (1− x)V eF (A−BP

xV) (221)

Setting the derivative with respect to x equal to zero, we find

dE

dx= BV eF

P

x2V− V eF A = 0 (222)

By inspection of this equation, we can see that the extremum we will obtainwill be a maximum. Solving for x,

P

x2B = V A =⇒ x =

√B

A

P

V(223)

So for a fuel cell with higher efficiency (lower B or larger A), the volume wedevote to the fuel cell decreases. For larger power requirements, we devotemore volume to the fuel cell, and for larger volume allowance, we devote asmaller fraction of the volume to the fuel cell.

(b) Plugging in numbers,

x =

√0.003 L/W ∗ 500 W

.7 ∗ 100 L(224)

So x = 0.146 . We devote about 15% of the system to the fuel cell. For apower density of pFC = 500 W

0.146∗100 L = 34 W/L = 0.034 W/cm3, a reasonablepower density, we’d operate a fuel cell at an efficiency of 0.7− 0.003 W/L ∗34 W/L = 0.6, which is also reasonable.

66

Chapter 10 Solutions

Problem 10.1 The four primary sub-systems of a fuel cell system are 1)the fuel cell sub-systems, 2) the fuel processing sub-system, 3) the thermalmanagement sub-system, and 4) the power electronics sub-system.

For example, a fuel reformer in the fuel processing sub-system dependson the thermal management sub-system to provide heat from other systemcomponents for an endothermic reaction at the reformer or to extract heatfrom an exothermic reaction at the reformer.

As another example, the fuel cell sub-system depends on the power elec-tronics sub-system to regulate the output voltage of the system. As shownby the shape of the fuel cell’s polarization curve, the fuel cells voltage de-clines at higher currents and higher powers. Converters enable the systemto provide a constant voltage.

As another example, the heat produced by an afterburner downstreamof the fuel cell stack depends on the anode utilization of the stack. Athigh anode utilizations, less anode off-gas will be available for downstreamcombustion. Instead of a downstream afterburner, a downstream micro-turbine could be used to recapture energy in the anode off-gas.

One way fuel cell systems can be integrated is to use excess heat fromexothermic chemical reactions to preheat inlet fuel, oxidant, and water.

Problem 10.22) The Pressure Swing Absorption Unit (PSA) is a gaspurification unit that uses a physical mechanism to remove contaminants.From a hydrogen-rich gas stream containing CO, HCs, CO2 and otherspecies, a PSA unit can produce a 99.99% pure hydrogen stream. Oneof its adsorbent beds adsorbs all non-hydrogen species because of the rela-tively higher molecular weight of these species as compared with hydrogen.The “swing” in PSA refers to the “switching” between two adsorbent beds.When one adsorbent bed is being regenerated, the other is adsorbing non-hydrogen species. When one bed is full and the other bed is regenerated,the hydrogen-rich stream is re-diverted (“swings”) to the regenerated bed.

Problem 10.3 An exothermic reaction is defined as a chemical reactionthat releases heat. An endothermic reaction is a chemical reaction thatabsorbs heat.

Endothermic: steam reformingExothermic: oxidation of hydrogen fuel in a fuel cell, partial oxidation,

water gas shift reaction, selective methanation, selective oxidation, combus-

67

tion of fuel cell exhaust gasesNeither: 1) a chemical reaction takes place but heat is neither released

nor absorbed: autothermal reforming, 2) no chemical reaction takes placebut heat may be released: hydrogen separation via palladium membranes,pressure swing adsoption, condensing water vapor to a liquid, expansion ofhydrogen gas (hydrogen heats up as a result of the Joule-Thompson Effect),compression of natural gas (in practice, releases heat due to frictional lossesat the compressor)

Problem 10.4 A process diagram for a fuel cell system for a scooter isincluded in the accompanying figure.

Figure 28: Process diagram for a fuel cell system for a fuel cell scooter.(Problem 10.4)

Problem 10.5 A process diagram for a fuel cell system for a scooterwith metal hydride storage is shown. Metal hydride storage tanks are com-posed of a special granular metal that absorbs and releases hydrogen like asponge absorbs and releases water. Depending on the metal hydride mate-rial, it may need to be pressurized to charge it with hydrogen and heated todischarge the hydrogen. The process of charging and discharging the metalhydride requires a more complicated thermal management sub-system. One

68

possible design is sketched in the figure: Heat from the fuel cell stack isused to heat the hydride to release hydrogen. Coolant air is used to cool thehydride so that it can absorb hydrogen. Because hydrogen absorption is anexothermic reaction, heat must be continuously removed from the alloy bed.The rate at which a metal hydride can absorb or release hydrogen is primar-ily dependent upon the rate at which the alloy can release or absorb heat,respectively. Although not shown in the figure, the thermal managementsub-system becomes more complicated than the one shown if the fuel celloperates at a lower temperature (for example, 80C for a PEM) than theminimum temperature required for hydrogen release in the metal hydride(for example, 150C for a NaAlH4 metal hydride).

Figure 29: Process diagram for a fuel cell system for a fuel cell scooter withmetal hydride hydrogen storage. (Problem 10.5)

The absorption reaction with the metal hydride depends on the pressureand temperature of the hydrogen gas. If the gas pressure is above a certainequilibrium pressure at a certain temperature, the metal absorbs hydrogen.If the hydrogen gas pressure is below the equilibrium pressure, the metalhydride releases its hydrogen.

Problem 10.6 For partial oxidation, all of the products are CO and

69

H2. A balanced reaction is

C8H18 + a(O2 + 3.76N2)→ bCO + cH2 + 3.76aN2 (225)

C8H18 + 4(O2 + 3.76N2)→ 8CO + 9H2 + 15.04N2 (226)

The hydrogen yield is then

yH2 = 9H2/(8CO + 9H2 + 15.04N2) = 28%. (227)

However, if all of the CO in the product stream was shifted to hydrogen viathe water gas shift reaction,

CO + H2O → CO2 + H2 (228)

then the hydrogen yield could be greater. The overall reaction would be

C8H18 + 4(O2 + 3.76N2) + 8H2O → 8CO2 + 17H2 + 15.04N2 (229)

The hydrogen yield is then

yH2 = 17H2/(8CO2 + 17H2 + 15.04N2) = 42% (230)

The WGS reaction can increase the hydrogen yield by 14%. (This calculationdoes not directly consider the addition energy input required to vaporizeliquid water for the WGS reaction.)

Problem 10.7 We do a back-of-the envelope calculation based on theexample problems to produce a reasonable estimate at STP. For steam re-forming, for a 100% efficient transfer of heat between a methane burner anda steam generator, the mass/moles/volume of methane needed for combus-tion is at a minimum about 31.5% of the moles/mass/volume of methaneconsumed by the steam reformer (Example Problem 1.6) at STP. Assum-ing only 72% efficient heat transfer, then the moles of methane needed forcombustion is at a minimum about (31.5%/0.72 =) 43.8% of the moles ofmethane consumed by the steam reformer at STP. Under these conditions,the fuel reformer efficiency in terms of HHV is

εR =∆HHHV,H2

∆HHHV,Fuel=

4 mol H2(284 kJ/mol H2)1.438 mol CH4(880 kJ/mol CH4)

= 89.8% (231)

A more exact calculation can be conducted for reactants entering andproducts leaving at 1000 K. ∆hCH4(1000 K) = −36.62 kJ/mol, ∆hCO2(1000 K) =−360.11 kJ/mol, ∆hH2O(g)

(1000 K) = −215.83 kJ/mol, ∆hO2(1000 K) =

70

27.71 kJ/mol, so the enthalpy of methane combustion is ∆hrxn,CH4 =−810.57 kJ/mol. The enthalpy of hydrogen combustion is (∆hH2(1000 K) =20.68 kJ/mol) ∆hrxn,H2 = −250.37 kJ/mol. Then the fuel reformer effi-ciency is

εFR =4 mol H2(250.37 kJ/mol H2)

1.438 mol CH4(810.57 kJ/mol CH4)= 85.9% (232)

Problem 10.8 We conduct a back-of-the envelope calculation based onthe example problems to produce a reasonable estimate. We ignore theenthalpy of formation of nitrogen because it does not chemically react. Theautothermal reforming reaction is

CxHy +zH2O(l) +(x−z/2)O2 ↔ xCO2 +(z+y/2)H2 → CO,CO2,H2,H2O(233)

The value for the steam-to-carbon (S/C) ratio, here shown as z/x, shouldbe chosen such that the reaction is energy neutral, neither exothermic norendothermic. For autothermal reforming of propane, the only variable is zin the reaction:

C3H8 + zH2O(l) + (3− z/2)O2 ↔ 3CO2 + (z + 4)H2 (234)

We must find for what z the ∆hrxn = 0. You can find the equation for z tobe:

z(∆hH2 −∆hH2O +12∆hO2) = ∆hC3H8 + 3∆hO2 − 3∆hCO2 (235)

z =−12 kJ/mol + 3 ∗ 22 kJ/mol − 3 ∗ (−360 kJ/mol)

20 kJ/mol − (−216 kJ/mol) + 1222 kJ/mol

= 4.6 (236)

The steam to carbon ratio is then z/x = 4.6/3 = 1.5. The reformer wouldproduce 4.6 + 4 = 9.6 moles of H2 per mole of fuel.

Problem 10.9Steam reformingThe heat of reaction for this steam reforming reaction at 1000 K is

∆hrxn = 190.89 kJ/mol. The heat of reaction for the complete combustionof methane at 1000 K is ∆hrxn = −800.57 kJ/mol. Therefore, assumingall of the reactants are preheated to 1000K, only 190/800 = 0.24 moles ofmethane must be combusted to provide energy for every 1 mole of methane

71

that goes through the endothermic steam reforming process. The reactionstoichiometry is

1.24CH4 + 2H2O + 0.48O2 → 1.24CO2 + 4H2 + 0.48H2O (237)

These reactions are not usually combined in the same reactor. Becausecombustion is usually done in a separate reactor from the steam reforming,the products of combustion CO2 and H2O and the gas N2 do not dilutethe hydrogen yield in the product gas. The hydrogen yield is then yH2 =4 H2/(1CO2 + 4H2) = 80%, as shown in the example problem. The ratio ofhydrogen produced per unit of methane consumed is 4/1.24 = 3.23.

Partial OxidationThe hydrogen yield for the partial oxidation reaction is the same as

shown in the text. For partial oxidation, all of the products are CO and H2.A balanced reaction is

CH4 + 0.5(O2 + 3.76N2)→ CO + 2H2 + 1.88N2 (238)

The hydrogen yield is then yH2 = 2H2/(1CO + 2H2 + 1.88N2) = 41% Theratio of hydrogen produced per unit of methane consumed is 2. However, ifall of the CO in the product stream was shifted to hydrogen via the watergas shift reaction then the hydrogen yield could be greater. The overallreaction would be

CH4 + 0.5(O2 + 3.76N2) + H2O → CO2 + 3H2 + 1.88N2 (239)

The hydrogen yield is then yH2 = 3H2/(1CO2 + 3H2 + 1.88N2) = 51%.The WGS reaction can increase the hydrogen yield by 10%. The ratio ofhydrogen produced per unit of methane consumed is 3. (This calculationdoes not directly consider the addition energy input required to vaporizeliquid water for the WGS reaction.) The autothermal reforming reaction formethane is

Autothermal ReformingWe have the reaction

CH4 + zH2O(g) + (1− z/2)O2 → CO2 + (z + 2)H2 (240)

We need to find the value of z for which the enthalpy of this reaction iszero at 1000 K (and assuming water enters as a gas at 1000 K), which givesan equation for z:

0 = ∆hCO2 + (z + 2)∆hH2 −∆hCH4 − z∆hH2O − (1− z/2)∆hO2 (241)

72

Using the values at 1000 K (in kJ/mol) ∆hCH4 = −36.62, ∆hH2O = −215.83,∆hO2 = 22.71, ∆hCO2 = −360.11, ∆hH2 = 20.68, we find z = 1.22.

The number of moles of N2 involved is 3.76 ∗ (1− 1.22/2) = 1.46 moles.The hydrogen yield is then yH2 = 3.23H2/(1CO2+3.23H2+1.46N2) = 57%.The ratio of hydrogen produced per unit of methane consumed is 3.23. Thisis the same ratio as with steam reforming, but with steam reforming, thehydrogen yield is higher due to the separation of the combustion reactionand the steam reforming reaction in separate reactors.

Problem 10.10 From example 10.8, the heat available to heat the build-ing is 11.6 kW . We need to find the volume of space that can be heated,assuming heat is lost through conduction through the walls. Then the heattransfer is

Q = kA∆T

l(242)

Where l is the wall thickness, which we’ll assume to be 10 cm. The thermalconductivity of the walls, k, we can take to be 0.1 W/mK, which is beapproximately the value for wood or some types of concrete. Then thesurface area of the heated room is

A =Ql

k∆T=

(11.6 kW )(0.1 m)(0.1 W/mK)(23 K)

= 504 m2 (243)

If the building is cubic, the volume of the building is V = (A/6)3/2 = 770 m3.Alternatively, if the building is 8 ft high, standard for a one story buildingin the US, the building may have a square footprint of 18.1 m per side(assuming no heat is lost through the ground, it is only lost through theroof and sides).

Problem 10.11 The T-H diagram looks similar to that for the condenserin figure 10.11.

As in table 10.9, we assume that half of the waste heat from the fuel cellexits via the cathode exhaust gas. Then the heat flow in the exhaust thatneeds to be removed by the condenser is QFC/2 + Qc, where QFC is theheat produced by the fuel cell and Qc is the heat needed to condense water,given by the latent heat of condensation of water. Convection removes anamount of heat

Q = hA(Th − Tc) (244)

where A is the surface area of the condenser in contact with the air and h isa function of the velocity of the air flowing past the condenser. To removingenough heat from the exhaust to condense the water, one must design the

73

Figure 30: T-H diagram of a condenser. (Problem 10.11)

condenser to have an area large enough that Q = QFC/2 + Qc even for lowvelocity.

If we assume that the fuel cell operates at a voltage of 0.5 V and has arange of 2 hours at maximum power, then the rate of hydrogen use is

NH2 =P/V

2F= 0.010 mol/s (245)

The rate of water production is NH2O = NH2 , and the amount of waterproduced is

NH2O = (0.010 mol/s)(2 ∗ 3600 s) = 74.6 mol = 1.34 kg (246)

The volume of this water, at STP, is 1.34 L.

Problem 10.12 An alternative heat exchanger network configurationthat increases the pinch is to use two parallel cold streams. The first absorbsheat in series from the fuel cell, the aftercooler, the selective oxidation unit,and the post-shift converter. The second absorbs heat from the condenser.The following figures show the new T-H diagrams and the new pinch pointtemperatures. The pinch no longer occurs in the condenser, and the pinchin the overall system increases. The pinch is greater than ten degrees over arange of mass flow rates of water in each cold stream. The last figure showsthis range of the ratio of mass flow rates in one cold stream (that flowsthrough the fuel cell, aftercooler, selective oxidation unit, and post-shiftconverter) vs. the other cold stream (flowing through the condenser).

74

This configuration manages the cooling loop by placing four thermalsources in series with each other and altogether in parallel with respect tothe condenser, a fifth thermal source. The primary advantages of this designare 1) its ability to increase the pinch point temperature and 2) under certaindesign conditions, this configuration can capture much of the waste heat.The disadvantages of this design are 1) greater system complexity than withrespect to the first configuration examined and 2) more complex controlcompared with initial configuration.

Figure 31: Temperature change of the cold and hot streams and four thermalsources, as a function of enthalpy change of the two streams. (Problem10.12)

Analysis of Pinch Points for Configuration 3 at Full Power (6kWe)A model for the new parallel configuration can be made to visualize

the change in enthalpy versus flow stream temperature. Figures 31 and32 show the results. The change in enthalpy refers to the aggregate levelof enthalpy being exchanged between the hot stream and the cold stream,and the temperatures shown at each enthalpy point are the inlet and outlettemperatures for the thermal source. Figure 31 shows the temperature vs.enthalpy plot for the four thermal sources in series on one side of the parallelloop: the fuel cell, aftercooler, selox, and post shift sources. Figure 32 showsthe temperature vs. enthalpy plot for the condenser on the other side of the

75

Figure 32: Temperature change of the cold and hot streams in the condenser,as a function of enthalpy change of the two streams. (Problem 10.12)

parallel loop. The combined total flow rate for both streams is the massflow rate needed to capture all the heat available so as to increase the initialinlet temperature for the domestic cooling stream from 25 C to the desiredoutlet temperature of 80 C. The flow rates chosen for each of the streams inparallel are based on the mass flow ratio of 0.58 (first parallel stream) to 0.42(second parallel stream), which allows each parallel stream to independentlyachieve an exit temperature of 80 C.

Figures 31 and 32 show each parallel stream’s pinch point temperature,the minimum temperature difference between hot and cold streams for effec-tive heat transfer. From figure 31, one can see that the first parallel stream’spinch point temperature is approximately 3500 W cumulative heat load inthe aftercooler at a temperature 8 C. Figure 32 shows the second parallelstream’s pinch point temperature to be approximately 1800 W cumulativeheat load in the condenser at a temperature of 18 C.

Keeping the combined total mass flow rate of water to the domestic cool-ing loop constant at the same rate, the ratios of the flows in the two parallelbranches should be varied to find the mass flow rate ratio at which thepinch point in the system is maximized. Figure 33 shows the results of thisanalysis, by plotting the minimum pinch point temperature for each ther-mal source with respect to the ratio of mass flow of the first parallel stream(containing the fuel cell, aftercooler, selox, and post shift) with respect tothe total flow rate. The intersection of the aftercooler pinch point curve andthe condenser pinch point curve shows that the pinch point for the system

76

can be maximized to a temperature difference of 12.5 C by operating at amass flow ratio of 0.64. If the mass flow ratio range is maintained between0.60 and 0.67, the pinch point temperature will fall at or above 10 C.

Figure 33: Pinch point analysis considering all five thermal sources as afunction of the two stream’s flow rates. Results are for a power of 6 kWe.The overall pinch point follows the black line. (Problem 10.12)

To understand this result consider that at one extreme, the flow ratio islimited to above a certain minimum by the pinch point in the aftercooler(shown by the square symbols in figure 33) to maximize the pinch pointtemperature in the aftercooler’s hot and cold streams. At the other extreme,the flow ratio is limited to below a certain maximum by the pinch point in thecondenser (shown by the starred crosses), because the condenser’s coolingwater must remain below 100 C to avoid vaporization. The compositesketch of the system’s overall pinch point range is shown as the Overall PinchPoint curve (shown by a single curve), which is the composite minimum ofthe aftercooler pinch point and condenser pinch point curves.

Problem 10.13 Reading off figure 10.11 the location of the pinch is theaftercooler. To make this quantitative, we need to find first the temperatureat which the water in the first stream condenses–this is where the pinch point

77

occurs. As in example 10.10, we equate the heat transfer of the entire processwith the heat transfer to a stream of liquid water and a stream with watervapor to solve for the condensation temperature Tc:

Q = (mcp)1(Th − Tc) + (mcp)2(Tc − Tl) (247)

Tc =Q− (mcp)1Th + (mcp)2Tl

(mcp)1 − (mcp)2(248)

Using the data for mcp and the high and low temperatures from table 10.9,we solve to find Tc = 84.78 C. This is the temperature of the hot stream atthe pinch point. Next, we need to find the temperature of the cold streamat this point, so we need to find what H point on the curve this temperaturecorresponds to:

dH = (mcp)1(Tc − Tl) = (276 W/C)(84.78C − 60 C) = 6838 W (249)

We can now find the temperature of the cold stream

Tb − Tbi =dH

mcp=

6838 W

143 W/K⇒ Tb = 72.82C (250)

Therefore, the pinch point is 84.78− 72.82 ≈ 12 C.

78

Chapter 11 Solutions

Problem 11.1 The primary steps of Process Chain Analysis (PCA) areas follows:

1. Research and develop an understanding of the supply chain from rawmaterial production to end use.

2. Sketch a supply chain showing important processes and primary massand energy flows.

3. Identify the “bottleneck” processes, which consume the largest amountsof energy or which produce the largest quantities of harmful emissions(or both).

4. Analyze the energy and mass flows in the supply chain using a controlvolume analysis and the principles of conservation of mass and energy.

5. Having analyzed the individual processes within the supply chain, eval-uate the entire supply chain as a single control volume. Aggregate netenergy and emission flows for the chain.

6. Quantify the environmental impacts of these net flows, for example,in terms of human health impacts, external costs, and potential forglobal warming.

7. Compare the net change in energy flows, emissions, and environmentalimpacts of one supply chain with another.

8. Rate the environmental performance of each supply chain against theothers.

9. Repeat analysis for an expanded, more detailed number of processesin the supply chain.

Problem 11.2 Several gases and particles are understood to have awarming effect on the Earth, a good summary is given in figure 11.5. Section11.3.3 describes the following effects from gases and particles:

Anthropogenic greenhouse gases include CO2, CH4, H2O, and nitrousoxide (N2O). Greenhouse gases selectively absorb infrared radiation, andthen re-emit this radiation partly back towards the Earth’s surface.

79

In addition to these gases, certain particles also have a warming effecton the Earth, but through a different mechanism. Dark-colored particles,such as soot, absorb sunlight, re-emit this energy as infrared radiation, andtherefore also warm the Earth. Dark-colored particles that contribute toglobal warming include black carbon (BC). The warming effect of blackcarbon is enhanced by Organic Matter (OM), which focuses additional lightonto black carbon. The center portion of Figure 11.5 shows the warmingmechanism of dark-colored particles. Figure 11.5 shows that these gasesand particles re-emit infrared radiation towards the Earth’s surface to causewarming; they also re-emit infrared away from the Earth.

In contrast, light-colored particles reflect sunlight and have a coolingeffect. Light-colored particles that cool the Earth include sulfates (SULF)and nitrates (NIT). SULF also attract water, which reflects light as well.Emitted gases that have a cooling effect include sulfur oxides (SOx), nitro-gen oxides (NOx), and non-methane organic compounds or volatile organiccompounds (VOC). These gases react in the atmosphere and convert to par-ticles, which are mostly light in color. SOx converts to SULF, NOx convertsto NIT, and VOC convert to organics that are mostly light in color. Theright portion of Figure 11.5 shows the cooling mechanism of light-coloredparticles.

Problem 11.3 Section 11.4 describes the six primary emissions thatcreate air pollution: ozone, carbon monoxide, nitrogen oxides, particulatematter, sulfur oxides, and volatile organic compounds (VOC). VOC are non-methane organic compounds, such as the higher hydrocarbons. Some ofthese compounds are air pollutants themselves. Others react with chemicalsto produce air pollution. Effects of air pollution on human health can includerespiratory illness, pulmonary illness, damage to the central nervous system,cancer, and increased mortality.

Four of the more important air pollutants known to affect human healthinclude CO, NO2, O3, and PM10. CO is known to cause increased inci-dents of headaches, hospitalization, and mortality. NO2 is known to causeincreased irritation to the sinuses, throat and eyes. O3 increases the inci-dents of asthma attacks, eye irritation, lower respiratory illness, upper res-piratory illness, and other conditions. PM10 is known to increase incidentsof asthma attacks, respiratory restriction, chronic illness, and morality. (SeeTable 11.3)

Problem 11.4 A National Emission Inventory (NEI) is a database of acountry’s emissions from various sources, which records the type of emission,

80

its quantity, the location of the emission, and its time of release. Nationalemission inventories vary in the depth of material they provide. The USEPA’s inventory records emissions from vehicles, power plants, factories,chemical processing plants, industrial facilities and other sources. Emissionsmay be recorded on an annual, monthly, or even daily basis. Since thelocation of stationary facilities is known, the location of these emissions isknown. The location of emissions from mobile sources (such as vehicles)may be estimated based on the location of the sale of the fuel.

Problem 11.5 Leaked hydrogen can combust with oxygen in air if acertain quantity of hydrogen accrues in high enough concentrations at highenough temperatures. The self-ignition temperature of hydrogen is 858 Kand its ignition limits in air are between 4% and 75% at STP. Withoutproper ventilation, a closed indoor area could accrue high concentrations ofhydrogen that could lead to ignition.

Problem 11.6 As described in section 11.3.5, the mechanisms throughwhich hydrogen might contribute to global warming and air pollution arevery complex and still the subject of intense research.

One mechanism through which released H2 might increase global warm-ing is by indirectly increasing the concentration of the greenhouse gas CH4.In the troposphere (lower atmosphere), H2 reacts with the hydroxyl radical(OH), according to the reaction

H2 + OH → H2O + H (251)

If H2 did not consume OH in this reaction, OH might otherwise reduce thepresence of CH4 via the reaction

CH4 + OH → CH3 + H2O (252)

The net effect of hydrogen then is to reduce the concentration of OH, whichincreases the concentration of CH4, a greenhouse gas. However, numerousother chemical reactions must also be considered.

One mechanism through which released H2 might increase one type ofair pollutant is through a series of chemical reactions that enhance the con-centration of O3. In the troposphere, H2 might increase O3 by increasingthe concentration of atomic hydrogen (H). After several years in the atmo-sphere, molecular hydrogen decays to atomic hydrogen in the presence ofthe hydroxyl radical (OH), via the reaction

H2 + OH → H2O + H (253)

81

Atomic hydrogen (H) could then react with oxygen in air in the presenceof photon energy (hν) from light to increase O3 in the presence of M, anymolecule in the air that is neither created nor destroyed during the reactionbut that absorbs energy from the reaction.

H + O2 + M → HO2 + M (254)

NO + HO2 → NO2 + OH (255)

NO2 + hν → NO + O (256)

O + O2 + M → O3 + M (257)

However, other sets of reactions must also be considered, with a focuson their net effect on air pollution. The net effect of these reactions may beable to be determined with computer simulations of chemical reactions in theatmosphere (atmospheric models). As we learned in PCA, these simulationsshould model not the mere addition or subtraction of an individual chemicalcomponent, but rather the net change in emissions among different scenariosto be accurate.

Problem 11.7 One possible answer is to compare and contrast differentenergy conversion scenarios using Process Chain Analysis (PCA) and datafrom an NEI. For example, one may examine the biological generation ofhydrogen via different mechanisms. Some researchers have concluded thatthe energy input requirements are too great to merit development of someof these. This assertion could be examined. A research proposal abstractshould be no more than a page in length and address a technically literatebut non-expert audience.

Problem 11.8 We can perform this calculation from the following equa-tion (11.14)

CO2,equivalent =mCO2 + 23mCH4 + 296mN2O + α(mOM,2.5 + mBC,2.5)(258)

− β[mSULF,2.5 + mNIT,2.5 + 0.40mSOX+ 0.10mNOX

+ 0.05mV OC ]

where m is the mass of each species emitted, with, for example, mOM,2.5

indicating the mass of organic matter 2.5 microns in diameter and less. Thecoefficient α can range between 95 and 191. The coefficient β can rangebetween 19 and 39. We will perform the calculation for a low case and a

82

high case, where for the high case, α = 191 and β = 19. For the low case,α = 95 and β = 39.

We use the data in table 11.2; please note that for mV OC you mayuse the entry for total non-methane organics. Further, the problem asksonly to consider organic gas and particulate matter. Then we find for thehigh case mCO2,equivalent = 7.35× 108 metric tons/yr, and for the low case,mCO2,equivalent = 3.98× 108 metric tons/yr.

Problem 11.9 We first need to find how much fuel is consumed by afleet of fuel cell vehicles. The fuel consumption rate can be found from theconsumption of gasoline in the current fleet after adjusting for the relativeefficiencies of the vehicles.

mH2C =VMT

Fh(259)

whereFh =

barMgvfVc∆HHHV,hεh

ρg∆HHHV,gεg(260)

We are given that εh/εg = 2, so using the numbers from example 11.2 wefind

Fh =(17.11 mi/gal)(264 gal/m3)(142 MJ/kg)

(750 kg/m3)(47.3 MJ/kg)2 = 36.16 mi/kg (261)

The vehicle miles per year is VMT = 2.68×1012 mi/yr so we find that annualfuel consumption is mH2C = 7.41 × 1010 kg/yr. If the hydrogen leakage is2%, the annual hydrogen production must be 7.56× 1010 kg/yr.

According to the Example 10.3, in the steam reforming reaction approxi-mately 4 moles of H2 are produced for every one mole of CH4. If the heat forthis endothermic steam reforming reaction is provided by the waste heat ofanother process, no additional CH4 need be consumed. If CH4 is combustedto provide heat for the reaction, some additional CH4 is consumed. In ex-ample 10.6, a back-of-the envelope calculation shows that approximately31.5% excess CH4 must be combusted to provide sufficient heat for the re-action. Then, for every 4 moles of H2 produced, approximately 1.315 molesof CH4 may be consumed. (The actual ratio depends on the design of thefuel reformer, the supply of external heat to the reformer from other sources,and the efficiency of the reformer.) Then the mass of CH4 needed for thishydrogen production is

mH2

1.3154

16 g/mol CH4

2 g/mol H2= 1.99× 1011 kg/yr (262)

83

Considering the 1% methane leakage, actual production must be 2.01 ×1011 kg/yr.

To compare this to current production, we must assume that natural gasis simply CH4, and find that the US current annual production is 23.14 ×1015 BTU/yr. To find this in kg, convert

(23.14× 1015BTU/yr)1055 J

1 BTU∗ 1 kg CH4

5.55× 107J= 4.40× 1011kg/yr (263)

So the amount we are considering for a hydrogen fleet is 46% of current USproduction.

The CO2 equivalent of the leaked methane is calculated from equation258. The CO2 equivalent of the leaked methane is 4.63 × 1012 of CO2

equivalent/yr.The external cost of the leaked methane can be estimated from the ex-

ternal cost of global warming, estimated as $0.026 and $0.067 per kg of CO2

equivalent (section 11.3.7). Based on these estimates, the external cost ofthe leaked methane (1% of the total) is between $1.2 and $3.1 billion/yr.

Problem 11.10 The EPA gives a Natural Gas Hydrogen Fuel Cell Ve-hicle (HFCV) Scenario as an alternative to the base case. In both scenarios,they predict CO2 equivalent emissions.

The base case gives CO2 equivalent estimates (low and high) of 5.33×109

and 5.86 × 109 tons/yr. The HFCV scenario estimates are 4.58 × 109 and5.08 × 109 tons/yr. Therefore, reasonable estimates for the change in CO2

equivalent can range from a reduction of 12% to 15% of all anthropogenicsources in the natural hydrogen fuel cell vehicle case (compared with the1999 fleet). The actual reduction depends on choices pertaining to the sce-nario, such as the type of reformer and the energy requirements required forstoring hydrogen on the vehicle (gas or liquid). See the following table.

84

Species HFCV and Natural Gas GasesCarbon Monoxide (CO) -55.29%Nitrogen Oxides (NOx) as NO2 -33.18%OrganicsParaffins (PAR) -27.10%Olefins (OLE) -32.43%Ethylene (C2H4) -24.86%Formaldehyde (HCHO) -19.68%Higher aldehydes (ALD2) -51.98%Toluene (TOL) -16.71%Xylene (XYL) -27.69%Isoprene (ISOP) -49.54%Total Non-Methane Organics -26.24%Methane (CH4) 20.88%Sulfur Oxides (SOx) as SO2 2.05%Ammonia (NH3) -5.25%Particulate MatterOrganic Matter (OM2.5) -1.78%Black Carbon (BC2.5) -15.08%Sulfate (SULF2.5) -0.32%Nitrate (NIT2.5) -0.82%Other (OTH2.5) -0.15%Total PM2.5 -1.26%

Organic Matter (OM10) -1.18%Black Carbon (BC10) -10.87%Sulfate (SULF10) -0.35%Nitrate (NIT10) -0.38%Other (OTH10) -0.03%Total PM10 -0.41%SpeciesCarbon Dioxide (CO2) -15.03%Water (H2O) 1.97%CO2,equivalent (low) -14.00%CO2,equivalent (high) -13.25%

Problem 11.11 This is an open-ended problem for which answers willvary. Reasonable electrical efficiencies for fuel cell systems are between 40%and 60%, and the efficiency of hydrogen generation can approach 1 (seeexample 10.6). The efficiency of current US electricity generation is ap-

85

proximately 32%. Therefore, the efficiency could potentially almost double.Emission reductions are estimated to be significant, based on emission esti-mates shown in the text in Table 11.1.

Problem 11.12 This is an open-ended problem for which answers willvary. One fuel cell manufacturer produces fuel cell systems that have alreadyachieved electrical efficiencies of 50% in combination with heat recoveryefficiencies close to 40%. If 90% of the heating value of the fuel could beusefully used, the efficiency of current US electricity and heat generationcombined could approximately triple. (The average boiler/furnace in theUS achieves an efficiency of about 80%. State-of-the-art condensing boilersachieve efficiencies in the low to mid 90% range.)

Under this combined-heat-and-power scenario, emission reductions areestimated to be even more significant than those shown in the text in Table11.1.

Problem 11.13 This is an open-ended problem for which answers willvary. Of all of the links in the upstream supply chains, one of the mostimportant is the energy required to put hydrogen into a form in which itcan be stored onboard a vehicle. For liquid H2 storage, approximately 30%of the heating value of the fuel is consumed in the energy required to coolH2 down to a liquid. For gaseous H2 storage, approximately 10% of the heatvalue of the fuel is required for hydrogen compression. In this case, one canassume the energy required to run the hydrogen compressors is provided byelectricity from the grid.

These upstream electricity generating plants can be assumed to have thesame mix of plant as shown in the text in Figure 11.2. Based on this scenario,a reasonable set of results for the change in emissions is shown in the tableaccompanying problem 11.10. The estimates in the table also assume 1%CH4 leakage. The change in CO2 equivalent is also shown in the table.All estimates with respect to total anthropogenic emissions. From theseemission estimates, one can calculate the change in health effects, externalcosts due to air pollution, and external costs due to global warming.

First, as an example, we will calculate the health effects from this change.Using the data in table 11.2 in the text for emissions from on-road vehicles,and table 11.4 showing the monetized health effects, combined with the datain the table of problem 11.10, we can find the change in health effects fromthis switch. The low estimate is a reduction in costs of $36.8 million, andthe high estimate is a savings of $570 million.

86

Problem 11.14 This is an open ended problem for which answers willvary.

Problem 11.15 In example 11.2, we find the quantity of hydrogen con-sumed by a hydrogen fuel cell vehicle fleet to be approximately 57.1 MT ofH2/year. If 1% of H2 that was produced leaked, approximately 0.6 MT ofH2/year would leak. According to table 1.2, conventional on-road vehiclesproduce 0.16 MT of H2/yr. Therefore, this hydrogen fleet might almostquadruple H2 emissions from vehicles.

87