Fuel Cell FundamentaFuel cell fundamentals-solutionsls-solutions

March 18, 2017 | Author: Bill Chen | Category: N/A
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Fuel Cell Fundamentals Solutions

Timothy P. Holme Ryan O’Hayre Suk Won Cha Whitney Colella Fritz 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 reliability and 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 energy density 4. temperature problems: PEMs can’t start in the cold, high temperatures 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 their fast refueling, silent operation, and independent scaling of fuel reservoir and power make fuel cells an attractive option. 2. Transportation applications where their low emissions and high efficiency makes fuel cells an attractive option. 3. Power generation applications where their silent operation, low emissions, and high efficiency make fuel cells amenable to siting in cities for distributed generation (DG) applications, reducing the cost of power distribution and possibly making process heat available for combined power and heating applications. 2

Problem 1.2 (5 points) Fuel cells have lower power density than engines or batteries, but can have large fuel reservoirs, so they are much better suited to the high capacity/long runtime applications. Problem 1.3 (10 points) You can easily tell which reactions are reduction and which are oxidation by finding which side of the reaction the electrons appear on. 1. Cu → Cu2+ + 2e− Electrons are liberated, so this is an Oxidation reaction 2. 2H + + 2e− → H2 Reduction 3. O2− → 21 O2 + 2e− Oxidation 4. CH4 + 4O2− → CO2 + 2H2 O + 8e− Oxidation 5. O2− + CO → CO2 + 2e− Oxidation 6.

1 2 O2

+ H2 O + 2e− → 2(OH)− Reduction

7. H2 + 2(OH)− → 2H2 O + 2e− Oxidation Problem 1.4 (15 points) You can write full-cell reactions and then split them into the half-cell reactions. You don’t need to be an expert chemist to do this, just use the half-cell reactions given and make sure your equations balance with species number (ie. that O is conserved) and charge (ie. electrons are conserved). 1. CO+ 21 O2 → 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 21 O2 + 2e− → O2− which is a reduction, or cathode reaction. 2.

1 2 O2 + H2

→ H2 O is a full cell reaction in SOFCs or PEMs, depending on the circulating ion (O2− or H + , respectively). The half-cell reactions would be: 12 O2 + 2e− + 2H + → H2 O at the cathode of a PEM and H2 → 2H + + 2e− at the anode of a PEM; or 21 O2 + 2e− → O2− at the cathode of an SOFC and O2− + H2 → H2 O at the anode of an SOFC.

3. another full cell reaction in an SOFC could be CH4 + 2O2 → CO2 + 2H2 O with the half cell reactions 8e− + 2O2 → 4O2− as the reducing, or cathode reaction, and CH4 + 4O2− → CO2 + 2H2 O + 8e− as the oxidizing, or anode reaction. 3

4. to use the circulating ion (OH)− , you may construct the full cell reaction 21 O2 + H2 + H2 O → 2H2 O from the half-cell reactions H2 + 2(OH)− → 2H2 O + 2e− as the oxidizing, or anode reaction, and 1 − − 2 O2 + H2 O + 2e → 2(OH) as the cathode (reducing) reaction. Problem 1.5 (10 points) From Figure 1.6, H2 (l) has a higher volumetric energy 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 include safety, the amount of hydrogen lost to boil-off, and the cost of liquefaction versus 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 loss results from a lower concentration of reactants.(ηconc ) 2. Electrochemical reaction–voltage loss from the sluggishness of the electrochemical 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 because you can’t formally describe a bond in 12 O2 , but we will later divide by 2 at the end to find the energy of the reaction given. For the reaction 2H2 + O2 → 2H2 O, H2 O has 2 O-H bonds, so E2∗H2 O = 2 ∗ 2 ∗ EO−H = 4 ∗ 460 kJ/mol EO2 = 494 kJ/mol E2H2 = 2 ∗ 432 kJ/mol The energy released by the reaction is 1 1 (E2∗H2 O − EO2 − E2H2 ) = (4 ∗ 460 − 494 − 2 ∗ 432) 2 2

(1)

(in kJ/mol). The energy is −241 kJ/mol . The energy is negative because an energy input to the system is required to break bonds. 4

Problem 1.8 (20 points) You can see the benefits in a fuel cell of the independent scaling of the fuel reservoir and the fuel cell stack. Therefore, for this problem, you may compute the volume of the stack and the reservoir independently. For the stack, you need to supply 30 kW with a fuel cell that supplies power at 1 kW/L and 500 W/kg. Therefore, the volume needed is 30 kW ∗ 1 kg 1 L 1 kW = 30 L and the weight is 30 kW ∗ 500 W = 60 kg. For the fuel tank, you need to hold a quantity of fuel equal to 30 kJ/s ∗

1 hr 3600 s ∗ ∗ 300 miles = 540 M J 60 miles 1 hr

(2)

note that 1 W = 1 J/s. Taking into account the 40% efficiency, you need to hold an excess quantity of fuel, 540 M J/0.40 = 1350 M J. The hydrogen is compressed to supply 4 M J/L and 8 M J/kg, so the fuel tank must be 1350 M J ∗ 41MLJ = 337.5 L and 1350 M J ∗ 81 MkgJ = 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 for the 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 of disorder, ie. the number of possible configurations a system can assume, so the entropy of a gas in a smaller volume is lower (given that the temperature 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 − T S, 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 activation barrier, and not by the overall energy change of the reaction. You may not determine which reaction proceeds faster. Problem 2.4 (6 points) While the current scales with the amount of reactants, the voltage does not. The thermodynamic potential comes from the energy drop going from products to reactants, which does not scale with reactant amount. Since E = −∆G/nF , you can think of the scaling in n cancelling the scaling in ∆G. Problem 2.5 (6 points) The Nernst equation i Πaνprod RT E = ET − ln i nF Πaνreact

(3)

shows that increasing the activity of the reactants decreases the argument of the ln, which raises the reversible cell voltage (E) because the ln term is negative. This, in essence, is Le’Chatlier’s principle. Problem 2.6 (Not graded) When the reaction is in equilibrium, the electrochemical potential of the system is zero. Components on the products side see the electrical potential φP and on the reactants side see the potential φR . Note that different species do not experience different electrical potentials. Then, if ∆φ is the difference in electrical potential from one side to the other, X X X 0= µ˜i dni = µoi dni + RT ln ai dni + nF ∆φ (4) First, note that: Q νi n aprod am a RT ln ai dni = = RT ln M bN = RT ln Q νi areact aA aB (5) Rearranging (4) and inserting the above result, you get Q νi P o aprod µi dni RT ∆φ = − − ln Q νi (6) areact nF nF

X

n 1 b RT (ln am M +ln aM −ln aA −ln aB )

∂G From the thermodynamic definition of chemical potential, µi ≡ ∂n so that i o o o Σµi dni = ∆G where the in this case denotes reference concentration. We

7

o

o o refers only to relate this term to E by − ∆G nF = E , but noting that the reference concentration so the term may still depend on temperature, we rename the quantity ET . Identifying E as the electrical potential across the cell ∆φ, we arrive at the Nernst equation (3).

Problem 2.7 (15 points) Yes, you can have a thermodynamic efficiency greater 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 (this is the exact reverse of a fuel cell, and may be used to generate hydrogen for some fuel cell applications). An electrolyzer has an efficiency defined to be the ∆H of reaction (the output is the useful heat energy of hydrogen), divided by the energy input ∆G. Therefore, the efficiency of the electrolyzer is the inverse of fuel cell efficiency–so the fuel cell at STP with an efficiency of 0.83, if ran in reverse as an electrolyzer, would have an efficiency of 1/0.83 which is greater than 1. For a fuel cell, consider the following example:  ≡ ∆G/∆H. For an ∆S isothermal reaction, ∆G = ∆H − T ∆S, so  = 1 − T ∆H . If ∆H is going to be negative, then you need to make ∆S positive to get an efficiency greater than 1. The trick is to use a solid or liquid reactant to make ∆S positive because solids and liquids have very low entropy compared to gases. For the fuel cell reaction C(s) + 21 O2 → CO at 298 K and 1 bar, 1 ∆S = SCO − SO2 −SC = 197.7−0.5∗205.1−5.7 (in J/mol·K) = 89.45 J/mol·K 2 (7) 1 ∆H = HCO − HO2 −HC = −110.5−0.5∗0−0 (in kJ/mol) = −110.5 kJ/mol 2 (8) then ∆S  = 1 − T ∆H = 1 − (negative number) > 1 Problem 2.8 (15 points) Assuming constant specific heats, ∆H(T ) = ∆H o + cp (T − T o ) and ∆S(T ) = ∆S o + cp ln(T /T o ). Find the temperature that satisfies: X ∆G(T ) = 0 = ∆H(T )−T ∆S(T ) = [∆Hio + cpi (T − T o ) − T (Sio + cpi ln(T /T o ))] (9) o o taking out the ∆H and ∆S , X o o 0 = ∆Hrxn − T ∆Srxn + [cpi (T − T o − T ln(T /T o ))] (10) 8

or o o 0 = ∆Hrxn − T ∆Srxn + (T − T o − T ln(T /T o ))

X

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 work fine) gives T ≈ 1020 K ≈ 747 ◦ C . The error of neglecting the dependence of ∆S and ∆H on temperature led to an answer that was off by about 40 degrees in this case. A more sophisticated solution would include the variance of cp with temperature, requiring an iterative solution or an expansion for cp in terms of T . Problem 2.9 (a) (10 points) Remember that the effect of temperature enters into the first term of the Nernst equation. From the Nernst equation, if the reactants and products are ideal, Q Q νi     x (P/Po )νP P xνi i RT RT ν −ν Q νi = ET − E = ET − ln ln (P/Po ) P R QP νi i ν nF (P/Po ) R R xi nF R xi (13) In a reaction, the change in number of moles ∆nG = νP − νR . The temperature dependent term is ET = E o +

∆S (T − To ) nF

(14)

For the voltages to be equal E(T1 , P1 ) = E(T2 , P2 ) ∆S (T1 − To ) − E + nF ∆S Eo + (T2 − To ) + nF o

(15)

Q νi   RT1 ∆nG QP xi ln (P1 /Po ) νi = nF R xi Q νi   RT2 ∆nG QP xi ln (P2 /Po ) νi nF R xi

Cancelling constant terms that appear on both sides of the equation and solving for T2 , h Q νi i ∆S R ∆nG QP xi − ln (P /P ) ν 1 o i nF nF R xi h T1 (16) Q νi i = T 2 ∆S R ∆nG QP xνi − ln (P /P ) 2 o nF nF x i R

9

i

Simplifying for the H2 /O2 fuel cell with liquid water as product and for pure components (xH2 = xO2 = 1) and dropping the Po which is understood to be 1 atm the expression simplifies to T1

∆S + 1.5R ln P1 = T2 ∆S + 1.5R ln P2

(17)

(b) (5 points) At STP, using the data from Appendix B, ∆Srxn = SH2 O(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 ln P1 −175.06 + 1.5 ∗ 8.314 ∗ ln 1 = 298 (19) ∆S + 1.5R ln(P1 /10) −175.06 + 1.5 ∗ 8.314 ∗ ln(1/10)

Then T2 = 256 K . Note that at this temperature of −17 ◦ C, there will be no H2 O(g) (so we shouldn’t use the ∆S for water vapor), but there will also not 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 decreases with an increase in temperature. The decrease in pressure decreases the Nernst voltage, so we must compensate by raising the Nernst voltage by decreasing temperature, therefore T2 < T1 . Problem 2.10 (15 points) Two ways of attacking this problem yield the same result. First, you could imagine a box with water, air, and hydrogen, and the reaction H2 + 21 O2 * ) H2 O(l) is in equilibrium. To find how much hydrogen is consumed by oxygen, find the equilibrium quantity of hydrogen when oxygen is present. That is to say, find at what PH2 does ∆Grxn = 0. From the van’t Hoff isotherm o

∆G = ∆G + RT ln

i Πaνprod i Πaνreact

(20)

assuming air at the cathode (xO2 = 0.21) ∆G = ∆Go +RT ln Solving for xH2 ,

a1H2 O 1 o 1.5 0.5 = ∆Go +RT ln 1 0.5 0.5 = ∆G −RT ln(1) (xH2 )(0.21) 1.5 aH2 aO2 (P/Po ) xH2 xO 2 (21)   ∆Go = ln (0.21)0.5 xH2 RT 10

(22)

    1 ∆Go (−237 kJ/mol) xH2 = exp = (2.18) exp = 5.41·10−42 0.210.5 RT (8.314 J/mol · K)(298 K) (23) −42 atm so PH2 = 5.41 · 10 Note that the partial pressure is very low, because it is very energetically favorable for hydrogen to react with oxygen to form water. The alternate way to solve the problem is to solve a concentration cell where a voltage develops (1.23 V because it is a hydrogen/air system) but E o = 0 because the concentration cell reaction is H2 + O2 → H2 + O2 . In this formulation, you solve the Nernst equation where the reactants have activity 1 because they are pure RT PH2 /Po (0.21)0.5 ln nF 1

(24)

  1.23 ∗ nF xH2 (0.21)0.5 = exp − RT

(25)

1.23 = 0 − This gives the same equation

Problem 2.11 (5 points) The efficiencies of each part multiply to give the total efficiency of the cell  = thermo V oltage f uel . For pure H2 /O2 at STP, thermo = 0.83. We are given V , so V oltage = VE = 0.75 1.23 . We are given 1 1 λ, so f uel = λ = 1.1 . Therefore,  = 0.83 ∗

0.75 1 1.23 1.1

 = 46%

11

(26)

Chapter 3 Solutions Problem 3.1 A. (5 points) Reducing the potential raises the energy of electrons in the electrode. To reduce their energy, electrons leave the electrode, so the reaction proceeds faster in the forward direction. B. (5 points) Increasing the potential lowers the energy of electrons in the electrode, so the reaction is biased in the forward direction. C. (5 points) We want to increase both reaction rates in the forward direction. At the anode (H2 * ) H + +2e− ) you want to draw electrons to the electrode, so increase the potential. At the cathode (2H + + 2e− + 21 O2 * ) H2 O) 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 Galvani Potential at one electrode. It means that one half-cell reaction requires energy input, and the other results in energy output. So long as the total potential adds up to the measured full-cell potential, it is impossible to know what each half-cell potential is! Problem 3.3 (10 points) Alpha is the charge transfer coefficient, it describes whether the “center of the reaction”, or peak of the reaction activation barrier, falls nearer to one side of the reaction or the other. In this 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 current density of the forward and reverse reactions at equilibrium (at open circuit). Problem 3.5 (5 points) (a) The Tafel equation, which holds in the exponential regime, reads: ηact = a + b log j

(27)

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

RT RT ln jo + ln j αnF αnF

(28)

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

RT RT ln j ⇒ b/2.3 = αnF αnF RT b = 2.3 αnF

(b) Identifying the constant terms, we get RT a = − αnF ln jo

13

(29)

Figure 4: Effect of α on the electrochemical energy pathway for problem 3.3. Problem 3.6 (5 points) The full-cell reaction is CO + 21 O2 → CO2 . The half-cell reaction at the anode is O2− + CO → CO2 + 2e− and at the cathode is O2 + 4e− → 2O2− Problem 3.7 (5 points) The main job of a fuel cell catalyst is to be able to form intermediate strength bonds with reactants and products, ie. to yield a low ∆Gact . Also, it should have a long lifetime, which means that it is resistant to poisoning, and does not migrate or agglomerate on the membrane. The requirements for an effective fuel cell catalyst-electrode structure are: porosity, a high degree of interconnection between the pores, a high effective catalyst area, high TPB density, high electronic conductivity, 14

and high exchange current density. It also must have a long lifetime, meaning high mechanical strength and resistance to corrosion. Ideally, it would also be cheap and easy to manufacture. Problem 3.8 (5 points) For reaction A, the net reaction rate in mol/s· 2.5A/cm2 A is nF5·2cm = 1.30 · 10−5 mol/s · cm2 . For reaction B, the net 2 = 2F

cm2

2

3A/cm A reaction rate is nF15·5cm = 1.04 · 10−5 mol/s · cm2 . Therefore, 2 = 3F reaction A has a higher reaction rate.

Problem 3.9 (10 points) At equilibrium, j = 0. The Butler-Volmer equation is  ∗  CR CP∗ αnF η (1 − α)nF η o j = jo ] =0 (30) o∗ exp [ RT ] − C o∗ exp [− CR RT P After canceling the joo term and rearranging, the equation reads η ∗ /C o∗ ] exp [− (1−α)nF CR (α − 1)nF η αnF η nF η R RT = exp [ − ] = exp [− ] = ∗ o∗ αnF η CP /CP RT RT RT exp [ RT ] (31) Solving for η, o∗ C ∗ /CR RT −η = ln [ R ] (32) nF CP∗ /CPo∗

Moving the negative sign to the right and inverting the argument of the logarithm, C ∗ /C o∗ RT η= ln [ P∗ Po∗ ] (33) nF CR /CR The overvoltage, η, is the amount by which the actual voltage is less than the standard state voltage ET − E = η (34) For an ideal gas, the concentration is the activity, so we have aP RT ln [ ] nF aR

(35)

Πaνi RT ln [ Pνi ] nF ΠaR

(36)

E = ET − which is almost the Nernst equation E = ET −

The difference lies in the ln term, which does not account for a multiplicity of products and reactants. This arises from the fact that the Butler-Volmer 15

equation only accounts for the concentration of the limiting products and reactants. If you inspect the derivation of the BV equation in the text, you’ll find that it is only for one electrode reaction. Since one fuel cell reaction is usually orders of magnitude slower than the other reactions, we typically neglect the slugishness of “fast” reactions and only employ the BV equation to find the loss at one electrode. A more rigorous form of the BV equation would include all of the products and reactants (and you’d have to write one equation for the anode and one for the cathode) and would reduce exactly to the Nernst equation. Problem 3.10 A (2 points) : P = IV = (1 A)(2.5 V ) = 2.5 W B (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 does increase. One way to think of it is that when you are required to supply 2.5 W of power, you must supply 2.5 J/s of hydrogen, not just the 0.5 J/s per cell. What is going on is that the first fuel cell uses 0.5 J/s of hydrogen to raise the voltage of its cathode up to 0.5 V . When this cathode connected to the anode of the next cell it is raised to a higher voltage, and this cell uses 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 N˙ H2 = s i/nF . The electric charge needed is 1 A · 100 hrs. · 3600 1 hr = 360, 000 C. Converting to moles of hydrogen, 360, 000 C/2F = 1.866 mol H2 . Converting g to grams, 1.866 mol · 1 2mol = 3.73 g H2 per cell. So the fuel cell stack uses five 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 density of 10 g/cm3 · 0.05 = 0.5 g/cm3 . To store 18.7 grams, you need a volume of (18.7 g)/(0.5 g/cm3 ) = 37.3 cm3 . This is a significant improvement over storage at 500 atm! Problem 3.11 (10 points) The exchange current density is given by jo = nF

∗ ∆G‡ CR e− RT o∗ CR

16

(37)

Grouping the preexponential terms into a constant that is temperature independent, jo (T ) = Ce−

∆G‡ RT

(38)

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

∆G‡ 1 1 jo (To ) − ∆G‡ Ce RT = jo (To )e R ( To − T ) jo (To )

(39)

This is our equation for jo (T ). To find ∆G‡ , first rearrange the general expression 1 1 jo (T ) = ∆G‡ ( − ) (40) R ln jo (To ) To T Now, plugging in the numbers given, ∆G‡ =

( T1o

R −

1 T)

ln

jo (T ) = (8.314 J/mol·K)/(1/300 K−1/600 K) ln [10−4 /10−8 ] jo (To ) (41) ∆G‡ = 45.9 kJ/mol

Problem 3.12 (10 points) The general expression for current density as a function of concentration is jo = nF

∗ CR ‡ f1 e−∆G /RT o∗ CR

(42)

oo∗ ), the current density at a reference Multiplying and dividing by jo (CR concentration, ∗ oo∗ ) CR jo (CR ‡ jo = nF f1 e−∆G /RT (43) oo∗ o∗ jo (CR ) CR

Canceling, we obtain the general expression oo∗ jo = jo (CR )

∗ CR oo∗ CR

(44)

∗ ) = j(T, 10C ∗ ) We must find the temperature T 0 for which j(T 0 , CR R

nF

∗ ∗ 10CR CR ‡ 0 −∆G‡ /RT f e = nF f1 e−∆G /RT 1 o∗ o∗ CR CR

(45)

Canceling, ‡ /RT

10e−∆G

‡ /RT 0

= e−∆G

17

(46)

ln 10 =

∆G‡ (1/T − 1/T 0 ) R

(47)

Solving for T 0 , 1/T 0 = 1/T −

1 R ln 10 ⇒ T 0 = R ‡ ∆G 1/T − ∆G ‡ ln 10

(48)

Note, the problem should have read: ∆G‡ = 20 kJ/mol. Using this value, T 0 = 421 K

(49)

So the temperature change is ∆T = 121 K . Problem 3.13 (not graded) The Butler-Volmer equation is   ∗ −∆G‡ CP∗ CR αnF η (1 − α)nF η RT j = Ce ] o∗ exp [ RT ] − C o∗ exp [− CR RT P

(50)

Doubling the temperature or halving the activation barrier will have the same effect on the first term in the equation. Doubling the temperature will also 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 the second, so changing T will have a larger impact on that term, the forward current term. So increasing T decreases the contribution from the term in parenthesis, reducing the current. At a fixed overvoltage, there is a larger increase in current density from halving the activation barrier than from doubling temperature. Problem 3.14 (not graded) As stated in the text, the bond strength of 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 catalyst requires 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 of oxygen. This is why we must supply an overvoltage to the cathode, to provide an extra driving force for oxygen dissociation.

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Chapter 4 Solutions

Problem 4.1 (5 points) There must be a voltage gradient (an electric field) to induce charge to flow. Another way to think about it is the intrinsic resistance of real materials will cause flowing charge to drop in voltage. Problem 4.2 (5 points) For a given current density, the voltage drop depends on the ASR, ηohmic = ASR · j. The cell with the lower ASR will be the cell with the lowest voltage drop. The second cell has an ASR0 = A0 · R0 = 10A · R/9, or an ASR 10 9 higher than the first cell, so the losses 10 increase by a factor of 9 over the first cell. Problem 4.3 (5 points) Conductivity is a function of carrier concentration and carrier mobility σ = nF cu Problem 4.4 (5 points) Ion transport in materials is harder because ions are much more massive than electrons, so µion < µelectron . Also, the concentration of free electrons in metals is much higher than typical ion concentrations that can be achieved in electrolytes, celectron > cion . The two effects combine to make ionic conductivity much lower than electronic conductivity. Problem 4.5 (5 points) Fuel cell electrolytes must have high ionic conductivity, low electronic conductivity, high stability under corroding environments, mechanical strength, low fuel crossover rates. Ideally, they are easy to manufacture at low cost. The stability requirement is the hardest to fulfill because the membrane must be stable in both the reducing environment at the anode and oxidizing environment at the cathode. Problem 4.6 (10 points) There are two differences from the PEM case: first, the charge carrier has the opposite sign, and second, it flows in the opposite direction, from the cathode to the anode. These two effects combine to make the diagram exactly the same. The ohmic loss still decreases the output voltage of the fuel cell, and the oxygen ion is negative it flows up a potential gradient. Note that the graph is exactly the same as the figure in the text! Problem 4.7 (5 points) See the figure that shows ohmic and activation losses.

19

Figure 5: Voltage losses for problem 4.6.

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

Problem 4.8 (5 points) Find the thickness of a membrane that has an electric field of 108 V /m with a voltage of 1 V . This can be done by dimensional analysis, or knowing that, with no net charge distribution, the electric 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 a function of electrolyte thickness. The ohmic loss is given by the current 20

L density times the ASR (ASR = R · A = A ρ · A = Lσ ), 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] + j

L σ

(51)

To find the optimal thickness, set the derivative with respect to L equal to zero: j dηT A =0=− ∗ + (52) dL L σ L∗ = Aσ/j To check that the result makes sense, note that as A increases, the leakage losses increase, so the optimal thickness must increase to offset the increased leakage; therefore L∗ is proportional to A. As σ increases, ohmic losses decrease, so the membrane may be thicker to offset more leakage loss; therefore L∗ is proportional to σ. Finally, as current density is increased, ohmic losses increase, so the optimal membrane thickness should decrease to minimize ohmic loss, so L∗ ∼ 1/j. Problem 4.10 (5 points) The two resistances simply add, since they are 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 )2 D RT

(55)

insert the equation for diffusivity, D = Do e−∆Gact /RT

(56)

and you get σ=

c(nF )2 Do e−∆Gact /RT RT 21

(57)

Multiply each side by T and group the temperature independent preexponential factor into a constant σo , and the equation looks very close to the final result: σT =

c(nF )2 Do −∆Gact /RT e = σo e−∆Gact /RT R

(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, and as stated in the text, Ea = ∆Gact /F , then we can convert by noticing some relations between the fundamental constants: R = kNA and F = qNA . ∆Gact = Ea F = Ea (qNA ) = Ea (q

R ) k

(59)

The argument of the exponential in equation 58 becomes: Ea (q R Ea q ∆Gact k) = = RT RT kT

(60)

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

(61)

Problem 4.12 We’ll need the saturation water pressure at 70 ◦ C and 90 The empirical fit for psat is ◦ C.

log10 psat = −2.1794 + 0.02953T − 9.1837 · 10−5 T 2 + 1.4454 · 10−7 T 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 inflows and outflows, given some information about the proton flow and inlet humidity. 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 accumulation of any species in the gas channel.

φH2 ,in = I/nF = (8 A)/(2 ∗ 96485 C/mol) = 4.15 · 10−5 mol/s. The fuel cell current is I = jA = (8 cm2 )(0.8 A/cm2 ) = 6.4 A. The number of hydrogen molecules (that split into protons to cross the membrane) leaving the gas channel 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 drag coefficient, the amount of water crossing the membrane is φH2 O,memb = α ∗ φH2 ,memb = 0.8 ∗ 3.32 · 10−5 mol/s = 2.65 · 10−5 mol/s. The inlet w humidity tells us that the influx of water is ppsat = 0.80. Assuming that the inlet pressure is 1 bar, the mole fraction of water at the inlet is yw = 0.8

psat 0.692 bar = 0.8 = 0.554 Po 1 bar

(66)

by knowing the inlet flow of hydrogen, we can determine the inlet flow of 23

water

0.554 = 5.14 · 10−5 mol/s (67) 1 − 0.554 Knowing the inlet water and the amount that crosses through the membrane, we can find the outlet water flow φH2 O,in = φH2 ,in

φH2 O,out = φH2 O,in −φH2 O,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 =

φH2 O,out yw · Pout Pout pw · = = psat psat (Tout ) φH2 O,out + φH2 ,out psat (70 ◦ C)

(69)

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

aw =

2.49 · 10−5 mol/s 1 bar · = 2.44 −5 −6 2.49 · 10 mol/s + 8.29 · 10 mol/s 0.307 bar

(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. Following the same method, we first find the water content of the membrane by solving an ODE subject to boundary conditions, then find the conductivity of the membrane as a function of temperature and water content, and finally find the 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 form j JH2 O = α 2F where α is an unknown constant, we can rearrange equation 4.44 to find an equation for the variation of water content through the membrane   λ dλ j Mm = 2ndrag −α (72) dz 2F ρdry Dλ (λ) 22 To find Dλ (λ), we again assume that, since Dλ is a slowly varying function of λ, it is a constant over the range we consider. Now, we may analytically solve the differential equation 2ndrag dλ = k( λ − α) dz 22 24

(73)

Where k ≡

jMm 2F ρdry Dλ .

The solution is

λ(z) = C exp(k

ndrag 11α z) + = Ceκz + α/κ 11 kndrag

(74)

Where κ ≡ kndrag /11. We determine C from the boundary conditions. If the anode is at z = 0, then for part (a), α α = 14 ⇒ C = 14 − κ κ α κtm λ(tm ) = Ce + = 3.45 κ 1 α κtm α + = 3.45 ⇒ α [1 − eκtm ] = 3.45 − 14eκtm [14 − ]e κ κ κ 3.45 − 14eκtm α=κ 1 − eκtm λ(0) = C +

(75) (76) (77) (78)

For lambda, we find λ(z) = (14 −

3.45 − 14eκtm κz 3.45 − 14eκtm )e + 1 − eκtm 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 Z tm Z tm dz dz −[1268(1/303−1/T )] =e R= κz + α/κ) − 0.00326) σ (0.005193(Ce 0 0 (81) The integral works out to Z dz z = − ln(b + aeκz )/bκ (82) κz ae + b b so the membrane resistance is   e−[1268(1/303−1/T )] 1 0.005193 ακ − 0.00326 + 0.005193Ceκtm tm − ln R(T ) = 0.005193 ακ − 0.00326 κ 0.005193 ακ − 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

λ To find κ we also need ndrag = 2.5 22 = 1.14, ρdry = 1970 kg/m3 = 1.97 × −3 3 10 kg/cm , Mm = 1 kg/mol

ndrag jMm 1∗1 1.14 = = 71.3 cm−1 11 2F ρdry Dλ 11 2 ∗ 96485 ∗ 1.97 × 10−3 ∗ 3.81 × 10−6 (85) Then we can find alpha to be

κ=

α=κ

3.45 − 14eκtm = 1.52 × 103 cm−1 1 − eκtm

(86)

so that C is C = 14 − α/κ = −7.33

(87)

R(T ) = e−[1268(1/303−1/T )] ∗ .319

(88)

Finally, R(T ) is 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 α κ α + κ

3.45 = C + 14 = Ceκtm

(89) (90)

These equations may be solved for alpha α=κ

14 − 3.45eκtm = −277 cm−1 1 − eκtm

(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 humidifying the anode is more effective under these conditions. Problem 4.14 (15 points) (a) The equation for diffusivity is ∆Gact 1 D = vo (∆x)2 e− RT 2

26

(93)

I didn’t subtract points for assuming that the hop distance was the lattice constant, 5 ˚ A. A more sophisticated analysis takes the crystal structure of cubic ZrO2 into account. Zirconia, in the cubic form, has the fluorite structure. The oxygen atoms occupy tetrahedral sites in an FCC lattice. Those sites are 1/4 of the way along the cube body diagonal, or have positions 14 (1, 1, 1). The vector between nearest neighbor sites is 1 1 1 4 (3, 1, 1) − 4 (1, 1, 1) = 4 (2, 0, 0). Therefore, the distance between oxygen nearest neighbor sites is a/2 = 2.5 ˚ A. 1 − D = (1013 s−1 )(2.5 · 10−8 cm)2 e (8.314 2

100 kJ/mol 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 = 6.4 · 1022 atoms/cm3 = 0.106 mol/cm3 (5 ˚ A)3

(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 of the ion (O2− ). The intrinsic conductivity is σ=

(nF )2 cD (2 ∗ 96485 C/mol)2 (1.11 · 10−3 mol/cm3 )(2.46 · 10−7 cm2 /s) = RT (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 (Y2 O3 ).08 . If we let the symbol χ stand for a generic cation atom, we have χ1.08 O2.08 . If this crystal were pure zirconia, with a ratio of one Zr to two O, it would be Zr1.08 O2.16 . Therefore, the 8% doping introduced an absence of 0.08 oxygens 0.08 for every 2.16 oxygen sites, or the fractional vacancy level of 2.16 = 3.70%. 27

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

8 = 6.4 · 1022 atoms/cm3 = 0.106 mol/cm3 (5 ˚ A)3

(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 ) =

(nF )2 cDo e−∆Gact /RT L = AR(T ) RT

(100)

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

L R(700 K)(e+∆Gact /(8.314 A(47.7 Ω) (nF )2 c

J/mol·K)(700 K) )

(101)

Inserting this equation for Do when T = 1000 K, many things cancel and we are left with: (700 K)(e+∆Gact /(8.314 J/mol·K)(700 (47.7 Ω)

K) )

=

(1000 K)(e+∆Gact /(8.314 J/mol·K)(1000 (0.680 Ω) (102)

K) )

Solving for ∆Gact , ∆Gact =

1 1000

R −

1 700

ln

700 K · 0.680 Ω 1000 K · 47.7 Ω

(103)

∆Gact = 89.4 kJ/mol Now, we may solve for Do from equation 101: Do =

100 µm (8.314 K/mol · K)(700 K)(e+(89.4 kJ/mol)/(8.314 J/mol·K)(700 (1 cm2 )(47.7 Ω) (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 operating temperature and see that it compares to standard values: at T = 1000 K, we get D = Do e−∆G/RT ≈ 8 · 10−7 cm2 /s, which is in the right ballpark for YSZ. 28

K) )

Chapter 5 Solutions Problem 5.1 (20 points) The use of a different gas in the mixture will change the effective diffusion constant of oxygen, which changes the limiting current density jL . Intuitively, since helium is much lighter than nitrogen (MHe  MN2 ), then DO2 He  DO2 N2 . Therefore, jL,He > jL,N2 so ηconc,He < ηconc,N2 . More rigorously: jL = nF Def f

coR δ

(105)

To have a higher jL , a higher Def f is desired. The effective binary diffusion constant is related to gas properties by: s s (pci pcj )1/3 (Tci Tcj )5/12 1 1 1 1 + = (pci pcj )1/3 (Tci Tcj )5/12−b/2 + Dij ∼ b/2 Mi Mj Mi Mj (Tci Tcj ) (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 compare diffusion coefficients, q 1 1 1/3 (T 5/12−b/2 (p ) ) c,N c,N 2 2 MN2 + MO2 DO2 N2 q = (107) 1 DO2 He (pc,He )1/3 (Tc,He )5/12−b/2 + 1 MHe

DO2 N2 DO2 He

M O2

q 1 1 + 32 (33.5)1/3 (126.2)5/12−1.823/2 28 0.076 q ≈ = 0.307 1 1 (2.24)1/3 (5.2)5/12−1.823/2 4 + 32

(108)

Therefore, the effective binary diffusion constant of oxygen in helium is larger, so the limiting current will be larger in synthetic air. Therefore, the concentration losses are larger in real air than in synthetic air. Problem 5.2 (10 points) Most importantly, SOFCs are not plagued with the problems of water removal that PEMs are. SOFCs operate at high enough temperature that liquid water is not formed and does not have to be removed. Second, diffusion is faster at higher temperatures (D ∼ T b ) so temperature aids mass transport and not as much engineering must go into the flow structures.

29

Problem 5.3 (15 points) The limiting current density is set by the diffusion constant, the diffusion layer thickness, and the bulk concentration. jL = nF Def f

coR δ

(109)

You may ensure high coR by designing flow structures that evenly distribute reactants. You may decrease the diffusion layer thickness by optimizing electrode structure (though reaction kinetics places other constraints on the MEA). Finally, you may increase Def f by optimizing fuel cell operating conditions such as temperature and inlet gas mole fractions (again, there will be other constraints at work). Problem 5.4 (25 points) To find the limiting current density, we need to calculate the diffusion constant and the bulk concentration of oxygen. First, the diffusion constant is !b 0.5  T 1 a 1 ef f 1.5 1/3 5/12 p + DO2 N2 =  (pc,O2 pc,N2 ) (Tc,O2 Tc,N2 ) p MN 2 MO 2 Tc,O2 Tc,N2 (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. Plugging in, !1.823 −4 298 K ef f 1.5 2.745 × 10 p DO2 N2 =0.4 1 atm (154.4 K)(126.2 K)   1 0.5 1 1/3 5/12 + ∗ ((49.7 atm)(33.5 atm)) ((154.4 K)(126.2 K)) 28 32 So Def f = 0.0520 cm2 /s. Finding the bulk concentration of oxygen is easier. Neglecting the consumption of oxygen as air travels down the flow path, we can say that the concentration of oxygen in air is 21%. From the ideal gas P law, we find the concentration in bulk is coR = 0.21 ∗ N V = 0.21 ∗ RT = 1 atm −3 mol/L = 8.59 · 10−6 mol/cm3 . 0.21 ∗ (0.08205 L·atm/mol·K)(298 K) = 8.59 · 10 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

jL = 3.44 A/cm2 30

(111)

Problem 5.5 (25 points) Using equation 5.23, ηconc = c ln

jL jL − j

(112)

the following plot was generated.

Figure 8: Concentration losses as a function of current density for various values of c for problem 5.5 As you can tell from the plot, for higher fuel cell performance, a lower value of c is desired. Problem 5.6 (optional) Using the power law to find the viscosity of gases, µN2 (T = 800 ◦ C) = 16.63 · 10−6 kg/m · s(

1073 0.67 ) = 4.16 · 10−5 kg/m · s 273

1073 0.69 ) = 4.93 · 10−5 kg/m · s 273 1073 1.15 µH2 O (T = 800 ◦ C) = 11.2 · 10−6 kg/m · s( ) = 4.06 · 10−5 kg/m · s 350 Using the molecular weights, viscosity, and mole fractions of each gas, we find from equation 5.34, µO2 (T = 800 ◦ C) = 19.19 · 10−6 kg/m · s(

31

Species i

Species j

Mi /Mj

µi /µj

Φij

xj Φij

Σxj Φij

N2

N2 O2 H2 O

1 0.876 1.555

1 0.844 1.02

1 0.919 1.001

0.711 0.174 0.100

0.985

N2 O2 H2 O

1.142 1 1.776

1.19 1 1.21

1.094 1 1.093

0.778 0.189 0.109

1.08

O2

H2 O

N2 0.643 0.976 0.980 0.697 0.563 0.824 0.902 0.171 0.967 O2 H2 O 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 density is obtained from the ideal gas law: ρ=

p 101325 P a = = 0.315 kg/m3 RT /Mmix (8.314 J/mol · K)(1073 K)/(0.02777 kg/mol)

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

Reµmix 2000 ∗ 4.29 · 10−5 kg/m · s = = 272 m/s ρL (0.315 kg/m3 )(0.001 m)

(113)

This very fast flow will not be achieved in SOFCs, so flow will always be laminar. Laminar flow allows a higher velocity in the case of higher temperature because the density of the gas is diminished at higher temperature. Problem 5.7 From the current required, 1 A/cm2 , you can find the amount of reactant required: JO2 /A = j/nF , where JO2 is the flux of oxygen and 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 v˜max where v˜max is in units of mol/s. j v˜max xO2 = (114) A nF To convert the given velocity from m/s to mol/s we use the factor ρα where 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 that amount of reactant, we get A = xO2 vmax αρ

nF j

(115)

We are told that the stoichiometric number is 2, and the geometry from Example 5.1 has circular cross sections of diameter 1 mm, so that α = 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 be 3.4 m long! Such a long, thin channel will lead to large pumping losses, so channels are not usually designed like this. Therefore, channels remain in the laminar flow regime to very good approximation. Problem 5.8 The equation developed to describe the oxygen content as a function of channel length was: ! j HC HE X ρO2 (x = X) = ρO2 (x = 0) − MO2 + ef f + (117) 4F ShF DO2 uin HC D O2

First, the diffusion constant of O2 in H2 O (neglecting multicomponent diffusion with N2 ) is: DO2 H2 O

a = p

T p Tc,O2 Tc,H2 O

DO2 H2 O

3.64 · 10−4 = 1

!b 1/3

(pc,O2 pc,H2 O )

5/12

(Tc,O2 Tc,H2 O )



1 + MH2 O M O2 (118) 1

0.5

!2.334

353

1/3

p

(154.4)(647.3)

((49.7)(217.5))

((154.4)(647.3))

(119) Or DO2 ,H2 O = 0.371 cm2 /s. To modify for the effective diffusion constant, ef f take into account the porosity: DO = 1.5 DO2 H2 O = 0.0940 cm2 /s. The 2 H2 O remainder of the values are specified. We take the inlet density of oxygen to be half of that in the example from section 5.3.3, because that example uses 33

5/12



1 1 + 18.02 32

0.5

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 density along the flow channel can be constructed. The only tricky part is to make sure you use consistent units. Make sure that distances are in cm or m, and if ρ is in kg/m2 , then MO2 should be in kg/mol, not g/mol or kg/kmol. Problem 5.9 The current in the fuel cell creates water at a flux rate of j(x) JˆH2 O (x, y = C) = MH2 O 2F

(120)

The diffusion from the GDL to the gas channel is (note: water is diffusing in 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) dif f ef f ρH2 O (x, y = E) − ρH2 O (x, y = C) JˆH (x, y = E) = −DH 2O 2O HE

(121)

The convective transport from the gas channel to the GDL is given by (you can check that the direction is correct because a positive flux results when the density in the channel is larger than the density at the electrode) conv JˆH (x, y = E) = −hm [ρH2 O (x, y = E) − ρ¯H2 O (X, y = channel)] 2O

34

(122)

dif f From the steady state conditions JˆH (x, y = E) = JˆH2 O (x, y = C) = 2O conv (x, y = E), we get the first relation JˆH 2O ef f −DH 2O

j(x) ρH2 O (x, y = E) − ρH2 O (x, y = C) = MH2 O HE 2F

(123)

that we may solve for the desired quantity, the density of water at the catalyst layer, ρH2 O (x, y = C) = ρH2 O (x, y = E) +

HE ef f DH 2O

MH2 O

j(x) 2F

(124)

From the second equality in the steady state condition, we obtain the second relation −hm [ρH2 O (x, y = E) − ρ¯H2 O (X, y = channel)] = MH2 O

j(x) 2F

(125)

We may make the substitution for hm with the Sherwood number and isolate the density of water diffusing into the GDL: ρH2 O (x, y = E) = ρ¯H2 O (X, y = channel) −

HC j(x) MH2 O ShF DH2 O 2F

(126)

Plugging this term into equation 124, ρH2 O (x, y = C) = ρ¯H2 O (X, y = channel)−

j(x) j(x) HC HE MH2 O + ef f MH2 O ShF DH2 O 2F D 2F H2 O (127)

Factoring out a term, HC j(x) HE [ ef f − ] 2F D ShF DH2 O H2 O (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 ρ¯H2 O (0, y = channel), not ρ¯H2 O (X, y = channel). Since the flow streams will likely be humidified, there will be water flowing into the gas channel at a speed uin . Integrating the flux of water from the GDL to the flow channel over the length of the gas channel is the difference between inlet and outlet water concentration in the gas channel (note the sign reversal): Z X [JˆH2 O (x, y = E)]dx = −uin HC ρ¯H2 O (0, y = channel)+uout HC ρ¯H2 O (X, y = channel) ρH2 O (x, y = C) = ρ¯H2 O (X, y = channel) + MH2 O

0

(129) 35

Alternatively, we can calculate the change in water concentration by integrating the water produced by current in the fuel cell. Z 0

X

MH2 O [JˆH2 O (x, y = E)]dx = 2F

Z

X

j(x)dx

(130)

0

These two equations must be equal. Assuming that the velocity of the gas does not appreciably change in the channel (uin = uout ) MH2 O 2F

Z

X

j(x)dx = uin HC [¯ ρH2 O (X, y = channel) − ρ¯H2 O (0, y = channel)] 0

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

MH2 O j = uin HC [¯ ρH2 O (X, y = channel) − ρ¯H2 O (0, y = channel)] (132) 2F

Solving for ρ¯H2 O (X, y = channel) and plugging into equation 128, we get the final result: ρH2 O (x, y = C) = ρ¯H2 O (0, y = channel)+

HC X MH2 O j j HE +MH2 O [ ef f − ] uin HC 2F 2F D ShF DH2 O H2 O (133)

or j HE HC X [ ef f − + ] 2F D ShF DH2 O uin HC H2 O (134) To check the signs, note that water concentration increases with X, as it should. Increasing the inflow of water by increasing uin increases the water that reaches the catalyst. Decreasing the diffusion layer thickness (HE ) increases the amount of water swept away by the gas, so the density of water at the catalyst goes down, as it should. As the last check, if the amount of gas is fixed and HC , the channel size, is decreased, the density of water increases. ρH2 O (x, y = C) = ρ¯H2 O (0, y = channel)+MH2 O

Problem 5.10 Mass concentration effects become important at high current density, so we will use the Tafel approximation to the Butler Volmer equation c∗R αnF η/RT j = j00 0∗ e (135) cR 36

If we assume that voltage, and therefore overpotential is constant along the flow channel, the only part that varies in x is the concentration of oxygen. Via a control volume analysis, you can find that the ODE governing concentration along the flow channel is dc(x) j 0 eαnF η/RT = − 0 0∗ c(x) dx cR

(136)

That is, the amount of oxygen flowing into the catalyst layer from the flow channel is proportional to the concentration in the flow channel. The solution is a decaying exponential c(x) = Ae−ax where a=

(137)

j00 eαnF η/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

MO2 = ρ¯O2 |x=0,y=channel − 4F

= ρ¯O2 |x=0,y=channel −

MO2 4F

a j(X) HE j(X) + + ef f hm uin HC DO2 j(X) HE j(X) a + + ef f hm uin HC DO2

Z 0

X

! c∗O2 (x)dx

(140) ! Z X cin e−ax dx 0

(141) MO2 = ρ¯O2 |x=0,y=channel − 4F

= ρ¯O2 |x=0,y=channel −

MO2 4F

! j(X) HE j(X) acin e−ax X + + (− | ) ef f hm uin HC a 0 DO 2 (142) ! j(X) HE j(X) cin + + (1 − e−aX ) hm uin HC Def f O2

(143) MO2 M O2 = ρ¯O2 |x=0,y=channel [1 − (1 − e−aX )] − 4F uin HC 4F

37

j(X) HE j(X) + ef f hm DO 2 (144)

!

This says that the oxygen falloff in the channel is exponential rather than linear. 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 steep linear 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 consider voltage efficiency. In that case, while both curves show the same VOC and the same short circuit current, the power (and efficiency) of the SOFC is higher because the curve lies at higher voltage for every current level. Another key difference between the SOFC and PEM is that the SOFC will be operated at much higher temperature. If you consider overall efficiency, then let’s find the effect of operating at higher temperature:  = thermo voltage f uel and assume that fuel efficiency is roughly the same for each case, let’s see how thermodynamic efficiency changes: ∆G V −nF ∆G V E V V V P ∗ = ∗ ∗ = ∆H ∗ = ∆H = o + ∆T ∆H E −nF ∆H E E (∆H P −R cp,i )/ − nF −nF −nF (145) Remember that ∆H o < 0. We’ll assume constant specific heats, so that term at 298 K is X cp,i = cp,H2 O −cp,H2 −0.5xp,O2 = 33.6−28.8−.5∗28.9 = −9.7 ≡ c¯p (146) ∼

P −R

The effect of higher temperature (larger ∆T ) is to make the denominator larger, making efficiency lower. To get a quantitative feel for the size of the effect, let’s assume the SOFC is operating at 1000 K and the PEM at 298 K. SOF C VSOF C = ∗ P EM VP EM =

∆H o −nF ∆H o +∆T c¯p −nF



VSOF C −285 kJ/mol ∗ VP EM −241 kJ/mol − 9.7 J/mol · K ∗ (1000 K − 298 K)

VSOF C ∗ 1.15 VP EM

Remember that we use values for liquid water for the PEM and water vapor for the SOFC. That means that ∆HP EM > ∆HSOF C , even though the 39

SOFC is operated at higher temperature. Then both effects combine to make the SOFC higher efficiency. Problem 6.3 (15 points) (a) 1. If electrode pores shrink or the electrode becomes thicker, the catalytically active area increases because there are more possible reaction sites. 2. If the ionic resistance increases, active sites further away from the electrode are less effective because ions will have to be transported through thicker resistive material. The catalytically active area therefore decreases. 3. Similar to above, if the resistance increases, areas further from the current collector become less active, so active area decreases. 4. The charge transfer resistance is a measure of how effective the catalyst is. If the entire catalyst becomes less active, the area does remains unaffected, though performance decreases. (b) At high operating temperatures, SOFCs are generally less affected by mass transport and charge transfer, but they generally have low ionic conductivity. Therefore, the extent to which the electrochemically active area extends into the electrode is limited by the ionic conductivity of the electrode. (c) PEMs are more commonly limited by charge transfer resistance, while ionic conductivity is generally high. Therefore increasing the thickness of catalyst layer increases the electrochemically active area. Problem 6.4 (10 points) The leakage current is larger than the exchange current density, so we should first assume Tafel kinetics, then check the validity of the assumption. Neglecting ohmic and concentration losses at this low current, the voltage loss from leakage due to activation losses is η=

RT j ln = 59.1 mV αnF j0

(147)

Using this value for η in the full Butler Volmer equation gives a 1% error in the current, so this is sufficiently accurate. Problem 6.5 1. (5 points) Starting from the Nernst equation ΠaνPi ∆S RT E=E + (T − To ) − ln nF nF ΠaνRi o

(148)

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

40

for water vapor product using aH2 O(g) 6= 1. The thermodynamic voltage is # " 1 1 o (149) E=E + ∆S(T − To ) − RT ln[ 1 0.5 ] nF pH2 pO2   1 0.5 P 1.5 E=E + ∆S(T − To ) + RT ln[xH2 xO2 ( ) ] n∗F Po o

(150)

To find ∆S, we need the entropy of liquid water at 330 K, which can be found from the appendix as ∆SH2 O(l) = 77.57 J/mol · K. Then the entropy change of reaction is 1 ∆Srxn = SH2 O(l) − SO2 −SH2 = 77.57−0.5∗209.02−133.60 = −160.04 J/mol·K 2 (151) Inserting values,   1 0.5 1 1.5 −160(330 − 298) + 8.314 ∗ 330 ∗ ln[1 ∗ 0.21 ( ) ] E = 1.23 + 2 ∗ 96485 1 (152) E = 1.19 V 2. (5 points) In problem set 3 we found the Tafel constants in the Tafel equation ηact = ac + bc ln j (153) The constants are ac = −

RT 8.314 ∗ 330 ln jo = − ln 10−3 αnF 0.5 ∗ 2 ∗ 96485

(154)

ac = 0.196 V We may use this constant without modification since the conversion between ln and log was not used. The other constant did use the conversion factor of 2.3, which may be dropped so that bc =

RT 8.314 ∗ 330 = αnF 0.5 ∗ 2 ∗ 96485

(155)

bc = 0.0284 V 3. (5 points) The ASR is given by ASR = L/σ = 0.01/0.1 41

(156)

ASR = 0.1 Ωcm2 4. (5 points) First, the diffusion coefficient is !b

0.5 1 1 p DO2 N2 (pc,O2 pc,N2 ) (Tc,O2 Tc,N2 ) + MN 2 MO 2 Tc,O2 Tc,N2 (157) −4 The relevant properties are a = 2.745×10 , 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. Plugging in, a = p

DO2 N2

T

2.745 × 10−4 = 1 atm

1/3

5/12

!1.823

330 K p



(154.4 K)(126.2 K) 1/3

∗ ((49.7 atm)(33.5 atm))

5/12

((154.4 K)(126.2 K))



1 1 + 28 32

0.5

The diffusion coefficient is DO2 N2 = 0.248 cm2 /s. The effective diffusion coefficient is then ef f DO = 0.201.5 ∗ 0.248 (158) 2 N2 ef f DO = 0.0222 cm2 /s 2 N2

5. (5 points) The limiting current density is jL = nF Def f

c0R δ

(159)

The bulk concentration of oxygen is N P 0.21 atm = = = 7.75 · 10−6 mol/cm3 V RT 0.08205 L · atm/mol · K ∗ 330 K (160) Note that n = 4 since we are calculating losses at the cathode (oxygen is the species of interest) so that we find c0R =

jL = 4 ∗ 96485 ∗ 0.0222 ∗

7.75 · 10−6 0.05

jL = 1.33 A/cm2

42

(161)

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 ln

jL 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 ln

1.33 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 for problem 6.5 Note that since we are using the Tafel form, the activation losses are underestimated at low current density. 7. (10 points) Finding the maximum power density can be done analytically, but as a first guess, we can just determine from the graph that the maximum power density is about 0.77 W/cm2 which occurs at a current density of about 1.1 A/cm2 . 43

For an analytical solution, we need to find the maximum in the power density curve: dp dV p=Vj ⇒ =V +j (165) dj dj So the optimum occurs at jL dp = 0 =E − [ac + bc ln(j + jleak )] − j ∗ ASR − c ln + ··· dj jL − (j + jleak )   bc c j − − ASR − j + jleak 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 V ∆G V  = thermo · · f = · ·f (166) E ∆HHHV E Using values from the appendix to find ∆G, we find 1 ∆Grxn = GH2 O(l) −GH2 − GO2 = −309.15−(−42.91)−0.5∗(−75.07) = −228.59 kJ/mol 2 (167) We use the voltage at the maximum power point (V = 0.692 V ) to find the overall efficiency −228.59 0.692 = · · 0.90 (168) −286 1.19  = 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 = ρDij

dxj dxi dxi + ρDji = ρ(Dij − Dji ) dz dz dz

(169)

In steady state, there is no net flux, so Ji + Jj = 0. In general, ρ 6= 0 and i there may be concentration gradients, so dx dz 6= 0, which means that we must have Dij = Dji .

44

Problem 6.7 Showing that x1 + x2 + · · · + xN = 1 is equivalent to P showing dxi = 0. Proceeding that way, sum the Stefan-Maxwell equation over all components to get X dxi i

dz

=

X

RT

i

X xi Jj − xj Ji ef f P Dij

j6=i

=

RT X xi Jj − xj Ji ef f P Dij i,j6=i

(170)

Since Dij = Dji , each term in the sum cancels to give 0, as we require. To show that each term cancels, we write out the sum for two species i, j = 1, 2:   X xi Jj − xj Ji i,j6=i

ef f Dij

=

 X  x1 Jj − xj J1 x2 Jj − xj J2  x1 J2 − x2 J1 x2 J1 − x1 J2 + +  = D1j D2j D12 D21   j6=i | {z } | {z } | {z } | {z } i=1

i=1,j=2

i=2

i=2,j=1

(171) = (x1 J2 − x1 J2 − x2 J1 + x2 J1 )/D12 = 0

(172)

Problem 6.8 Using the values from the text, the j-V curve is shown below. 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 the plot that the ohmic losses increase dramatically at lower temperatures–the maximum current density produced by the fuel cell at this temperature is only about 0.43 A/cm2 ! Problem 6.10 The j-V curves for an electrolyte supported SOFC at T = 1073 K and T = 873 K are shown below. Given the very high ohmic losses, especially at lower temperatures, it is clear that it is not optimal to support an SOFC with a thick electrolyte. Problem 6.11 (a) The linear approximation to the Butler-Volmer equation is RT jp0 RT jp0 ηanode = = (173) 2αF j0 p 2αF j0 pA xH2 which is similar to equation 6.22 in the book. Using an analogous equation to 6.26 for xH2 , we find ηanode =

RT 2αF

j  j0

pA

xH2 |a −

tA

jRT ef f 2F pA DH ,H 2

45

 2O

(174)

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 losses in 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 according to our model are shown in the figure below. In this case, anode losses remain 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 losses do not blow up as the cathode losses do, because we have taken the linear form for the losses, which breaks down long before the limiting current density. Problem 6.13 At each value of current density, we must solve for the constants α∗ and C to find λ(z) in order to find the resistance. Following the equations in the text, it is possible to find the complete j-V curve, as is 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 h i 1 1 σ(z) = 0.005193(4.4α∗ + Ce0.000598jz/Dλ ) − 0.00326 e1268( 303 − T ) = s1 +s2 es3 z (175) Where several constants are introduced to simplify notation. Then the resistance of the membrane is   Z tm Z tm dz dz z ln(s1 + s2 es3 z ) tm Rm = = = − (176) σ(z) s1 + s2 es3 z s1 s1 s3 0 0 0 Rm =

 1 s3 tm − ln(s1 + s2 es3 tm ) + ln(s1 + s2 ) s1 s3

(177)

Note that the ohmic losses are not linear with current, as at higher current, the membrane is dried and the resistance increases. As current increases, the first and third term in the above equation increase, and R increases. Problem 6.14 (a) The j-V curve for λ = 1.2 is shown below, as are the power density curves for λ = 1.2 and λ = 2.0. The maximum power density in 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 . If λ = 2.0, the max power density is approximately 0.56 W/cm2 , so if the pump consumes 20% of this power, the power delivered is approximately 0.45 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 long as you give a good argument behind your answer, we accept the following answers. Major operational variables include: 1. Temperature: The most important temperature dependence is in the equation for the exchange current density. ‡ /RT

j0 = Ae−∆G

(178)

You can see that the exchange current increases exponentially with temperature. From the Butler-Volmer equation in the low overpotential regime, ηact ∼ 1/jo , so that the overpotential depends exponentially 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 = 1.2 line is blue, λ = 2.0 line is green. temperature dependence that is less strong:  ∗  CR CP∗ αnF η (1 − α)nF η o j = jo ] o∗ exp [ RT ] − C o∗ exp [− CR RT P

(179)

From this equation, the overpotential is linear in temperature. Also, from the Nernst equation  νi  ΠaP RT ln (180) E = ET − nF ΠaνRi the temperature dependence is linear. For transport, temperature comes into play in a ceramic by the equation for conductivity c(nF )2 Do e− σ= RT

∆Gact 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 transport equation shows logarithmic dependence: ηconc =

c0 RT ln R αnF c∗R

(182)

3. Gas composition: again, Nernst and BV, as well as transport equation 182 Other operational variables include the compression force that presses the fuel cell together, the flow rates of gases, etc. Problem 7.3 (5 points) An EIS measurement gives more detailed information, though it is harder to interpret than a current interrupt measurement. The current interrupt measurement can be done on larger–higher power–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 high current that is too fast will show a voltage that is too high, because the fuel cell hasn’t reached a steady state. At each current step, the voltage must drop down to its steady-state value. The faster scan rate (100 mA/s) will show 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 current must ramp up to the new steady-state value. If you step down in voltage too fast, the current will not ramp up to the steady-state value, and you measure a current that is too low. In this case, the faster scan rate (100 mV /s) will show a current that is too low, ie. a worse performance. (b) (5 points) The slowest step in the fuel cell reaction will be the most affected by a fast scan. As electrochemistry is quite fast in comparison, the gas diffusion is the slowest step in the fuel cell reaction (as shown by the low frequency of the Warburg element representing mass transport in figure 7.12). Thus, the gas diffusion, which dominates at high current density, will be the most affected by the faster scan rate. Problem 7.5 Intuition should get students most of the way on the problem. The circuit diagram is shown in the first figure. You should inspect the easy, limiting cases of high frequency and low frequency first. (1) At high frequency, capacitors act as short circuits, so the 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 low frequency, capacitors act as open circuits, so we can ignore Ca . The circuit now acts as a resistor Rb + Ra in series with a capacitor. The Nyquist plot of this circuit is just a vertical line, with the intercept at Rb + Ra . At intermediate frequencies, we have some of the character of the parallel RC circuit, which is a semicircle in a Nyquist plot. The amount of the semicircle you will see depends on the distance between the point Rb and the vertical line at Rb + Ra . We’ll answer that question later. If you don’t buy that argument, the mathematical treatment is given here. The impedance of the circuit is ZT otal = ZRb + ZCb +

ZRa ZCa ZRa + ZCa

(183)

Inputting the complex impedances of resistors and capacitors ZT otal = Rb −

j j −jRa /Ca ω j jRa Ra + ωCa (184) + = Rb − − · 2 ωCb Ra − j/ωCa ωCb ωCa Ra + ω21C 2 a

Separating the real and imaginary components of the impedance, ZRe = Rb +

Ra /ω 2 Ca2 1 = Rb + Ra · 1 2 2 1 + (R Ra + ω 2 C 2 a ωCa )

(185)

a

−ZIm =

1 R2 /ωCa 1 1 1 + 2a = + · 1 ωCb Ra + ω2 C 2 ωCb ωCa 1 + (Ra ωCa )−2

(186)

a

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. Given an intercept and asymptote, information about the relative magnitudes of 54

the RC components can be used to sketch in the intervening region. As given in the problem, when the RC time constants are well separated, the resolution of each segment should be possible. For the skeptics, a more detailed treatment follows. A key question is: is the semicircle visible not? This question is actually much tougher to answer, and probably can’t be found by inspection. If you see part of a semicircle, that means that −ZIm has a maximum. So, here is the derivative of −Zim with respect to frequency:     d(−ZIm ) 1 1 −2(Ra Ca )−2 ω −3 1 1 + − =0=− − dω Cb ω 2 Ca ω 2 1 + (Ra Ca ω)−2 ωCa (1 + (Ra Ca ω)−2 )2 (187) Let χ = (Ca Ra ω)−1 , and factor out 1/ω 2   1 1 1 2χ2 0= 2 − − + (188) ω Cb Ca (1 + χ2 ) Ca (1 + χ2 )2 Since we aren’t interested in the solution ω = ∞, we discard that solution and clear the denominators: 0=− Let C =

Cb Ca .

(1 + χ2 )2 (1 + χ2 ) 2χ2 − + Cb Ca Ca

(189)

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

0 = (1 + 2χ2 + χ4 ) + (1 + χ2 )C − 2χ2 C = χ4 + χ2 (2 − C) + (1 + C) (190) The solution is χ2 =

C −2±

p (C − 2)2 − 4(1 + C) 2

(191)

This will have a solution if the determinant is positive, so in order to see most of our semicircle, we require (C − 2)2 − 4(1 + C) > 0 =⇒ C 2 − 4C + 4 − 4 − 4C > 0

(192)

Therefore, C 2 − 8C > 0 that is, since C > 0 by definition, C > 8. So to see the peak in the semicircle, Cb > 8Ca , or the blocking R × C must be larger than the activated R × C. Note that this requirement does not depend on the relative values of the resistance! (a) (10 points) When R×C is much smaller for the parallel RC than for the series RC, the entire semicircle can be seen before the ω = 0 asymptote 55

Figure 21: EIS plot for 7.5(a), when Ra Ca < Rb Cb

is approached. A MATLAB plot of the EIS curve for some chosen values of R and C is shown here. (b) (10 points) When R × C for the parallel RC is much larger than the series RC, the semicircle is only beginning before the asymptote is approached. Only the first linear part of the semicircle can be seen. A MATLAB plot of the EIS curve is shown here. (c) (10 points) The circuit diagram is shown here. The impedance of the circuit is  q  j σ √ − ωC2 R3 + ω (1 − j) tanh(δ jω jR1 D) − ωC 1 q Z= + R + (193) 2 j j jω σ R1 − ωC √ − + R + (1 − j) tanh(δ ) 1 3 ωC2 D ω −1 To simplify notation, make theq following definitions: χ−1 1 ≡ ωC1 R1 , χ2 ≡ ωC2 R3 , and W ≡ R σ√ω tanh(δ jω D ). Clearing out the imaginary denomi3 nators,

Z = R1 χ1

χ1 − j −j[1 + 2W + χ2 W + 2W 2 + jχ2 (1 + W )] + R + R χ 2 3 2 (1 + W )2 + (χ2 + W )2 1 + χ21 (194)

56

Figure 22: EIS plot for 7.5(b), when Ra Ca > Rb Cb

Figure 23: Circuit diagram for problem 7.5 The real impedance is ZRe = R1

χ21 χ2 (1 + W ) + R2 + R3 χ2 (1 + W )2 + (χ2 + W )2 1 + χ21

(195)

and the imaginary impedance is −ZIm =

R1 χ1 1 + W (2 + χ2 + 2W ) + R3 χ2 (1 + W )2 + (χ2 + W )2 1 + χ21

(196)

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

ZRe = R2 ;

57

(197)

which could be confirmed by inspection of the circuit diagram. The low frequency (χ1 = χ2 = ∞) gives ZRe = R1 + R2 + R3 (1 + W );

−ZIm = R3 W

(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 the constants. 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 space in a structure, but for a structure to be permeable, the holes need to go the entire way through a material. A sketch of a highly-porous material with low 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) (b) (b): ηohmic = 0.10 V and ηact = 0.30 V . At point (c), the current appears to be 1.75 A. From the left intercept of the Zreal axis, the ohmic resistance is 0.10 Ω. The ohmic overvoltage is (c)

ηohmic = iRohmic = (1.75 A)(0.10 Ω) 58

(199)

Figure 25: The highly-porous material with low permeability for problem 7.6.

(c)

ηohmic = 0.175 V Remember that Rf is the diameter of the semicircle, not the intercept of the semicircle. Then from the right intercept of the Zreal axis minus the left intercept, Rf = 0.20 Ω, so the activation overvoltage is (c)

ηact = iRf = (1.75 A)(0.20 Ω)

(200)

(c)

ηact = 0.350 V We may fit the data to the equation to extract the remaining desired values. If you plug in numbers first, since i(b) = 1, things simplify and the solution is shorter. Or, solving the general case: RT i(b) ln αnF io

(201)

RT i(c) RT i(c) ln ⇒α= ln (c) αnF io io nF ηact

(202)

(b)

ηact = and (c)

ηact =

Substituting α into equation 201 (b)

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

ηact = ηact

(203)

Isolating the io term (b)

ηact (c) ηact

(b)

ln i(c) − ln i(b) = ln io

59

ηact (c)

ηact

! −1

(204)

Therefore, io is   io = exp 

(b)

ηact (c) ηact

ln i(c) − ln i(b)

(b) (c) ηact /ηact

−1



"

  = exp

0.30 0.35

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=

1.75 (8.314)(300) ln (2)(96485)(0.35) 0.0348

(206)

α = 0.145 Problem 7.8 (10 points) The scan rate allows us to convert from an i-V plot to an i-t plot. To find the total charge Qh , remember that charge equals current times time, so the area under the i-t curve gives the 1 charge. Estimating the width of the peak to be 400 mV × 10 mV /s = 40 s and the peak of the curve to be about 40 µA, the area under the curve is approximately Qh ≈ (40 s)(40 µA) = 1600 µC (207) Then the active catalyst area coefficient is Ac =

Qh 1600 µC = Qm ∗ Ageometric 210 µC/cm2 ∗ (.1 cm)2 Ac = 760

60

(208)

Chapter 8 Solutions Problem 8.1 (5 points)(a) Ni is used in high temperature fuel cells as a catalyst. While it is not as good a catalyst as some noble metals, it is far 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 to catalysis. (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 surface area. 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, and therefore require lower noble metal loadings as catalysts. As cost is a primary barrier to widespread fuel cell adoption and noble metals add greatly to the cost of low temperature fuel cells, the lower noble metal loading of high temperature fuel cells is the largest advantage. 2) High temperature fuel cells benefit from fuel flexibility. Whereas low temperature fuel cells are poisoned by CO presence in the ppm level, high temperature fuel cells use CO as a fuel. Since generation of neat hydrogen is currently very costly, in the foreseeable future, the only economical way towards a hydrogen economy is via reforming of hydrocarbons. In that case, fuel flexibility is an enabling feature, as impurities such as CO are present in large levels in reformed fuel. Note: alleviation of water clogging issues is also an important advantage of high temperature fuel cells. The availability of high-quality waste heat is 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 how to lower the operating temperature, which shows that people consider the waste 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 power is 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 room temperature, the quantity of hydrogen on board the vehicle is N=

PV 350 atm ∗ 156.6 L = = 2.24 kmol L·atm RT 0.08205 mol·K ∗ 298 K

(212)

The energy stored in this fuel is (using the HHV of hydrogen, 286 kJ/mol) 2.24 kmol ∗ 286 kJ/mol = 641 M J 62

(213)

If this energy is used continually over a span of 4.3 hours (430 km/100 km/hr), the power output is 641 M J = 41.4 kW 4.3 hr ∗ 3600s/hr

(214)

If the fuel cell is 55% efficient, the fuel cell produces 22.8 kW during operation under these conditions. Problem 8.6 (10 points)(a) Carnot efficiency is given by Carnot = 1 − The efficiency is Carnot = 1 −

Tc Th

(215)

373 1073

Carnot = 65.2% (b) If the SOFC is 55% efficient, then 45% of the input energy is rejected as heat. This heat is available to the heat engine that has efficiency of HE = 0.6Carnot = 39.1%. The efficiency of the combined cycle is T ot = SOF C + (1 − SOF C )HE = 0.55 + 0.45 × 0.391. T ot = 72.6%

63

Chapter 9 Solutions Problem 9.1 (5 points) Fuel flows up the center of the donut and in the porous anode. Air flows in around the sides of the porous cathode . Sealing prevents mixing of air and fuel. One advantage of the structure is that it facilitates oxygen transport: usually oxygen is the limiting reactant. As it flows through the cathode, it gets depleted, but it has access to the most reactive area around the edges where it is least depleted. As the oxygen gets depleted along the cathode, it accesses less area so that the depletion is not as 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 problem 9.1 Problem 9.2(5 points) The DOE canceled the on-board fuel reforming project because current technologies are inadequate and there is no promising path forward. There is no interest from industry, and no reason to suspect that a future FCV (fuel cell vehicle) with on-board reforming will outperform future gasoline-electric hybrids. Since the great interest in fuel cell cars lies in their zero emissions capability, we do not perceive a huge impetus to implement FCV’s that run on gasoline. Problem 9.3(10 points) (a) The heat generation comes from ineffi64

ciency. The thermodynamic efficiency at STP is 83%, and we must find the voltage efficiency V = 0.7/1.23 = 57%. So the fuel cell efficiency is 0.83 ∗ 0.57 = 47%. Since the fuel cell generates 1000 W of electricity at 47% 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 is 1128 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, so g/cm3 3 the molar density of methanol is 0.79 32 g/mol = 0.0247 mol/cm . The molar 3

g/cm mass of water is 18 g/mol, so the molar density of water is 1.0 18 g/mol = 0.0556 mol/cm3 . If we assume that the volume of methanol and water do not change upon mixing, we must find x, the number of moles of water and methanol that will fill one liter. x x 1 1 1L= + =⇒ x = ( + )−1 = 17.1 mol 55.6 mol/L 24.7 mol/L 55.6 mol 24.7 mol (216) Next, we find the HHV of methanol. A methanol combustion reaction is

3 CH3 OH + O2 → CO2 + 2H2 O 2

(217)

So that the HHV of methanol is obtained by 3 − ∆HfO2 2 3 = − 393.51 + 2(−285.83) − (−238.4 − ∗ 0) = −727 kJ/mol 2 H2 O(l)

∆HHHV =∆HfCO2 + 2∆Hf

CH3 OH(l)

Where the value ∆Hf

CH3 OH(l)

− ∆Hf

= −238.4 kJ/mol was found from the NIST

kJ/mol Webbook (a good resource if you don’t know it). In kW h, this is 727 3600 s/hr = 0.202 kW h/mol. Finally, we know how much fuel we have, so the fuel tank capacity is 17.1 mol ∗ 0.202 kW h/mol = 3.45 kW h. For our 2 liter system, we have an energy density of 1.72 kWh/L . The small difference between this value and the quoted value may be explained if the ∆Hf numbers were obtained from different sources.

Problem 9.5(10 points) Effectiveness is defined in the text as ef f ectiveness =

% conversion of carrier energy to electricity % conversion of neat H2 to electricity 65

(218)

We can rewrite this as ef f ectiveness =(% conversion of carrier energy to “dirty 00 H2 )× % conversion of “dirty 00 H2 to electricity % conversion of neat H2 to electricity Using the numbers given, we find ef f ectiveness = 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 power is specified to be P , we say that the power density of the fuel cell is pF C = P xV . Now, the efficiency explicitly written as a function of x is P (220) xV Now we may write equation 9.9 in the book, for total energy of the system as a function of x, as (x) = A − B

P ) xV Setting the derivative with respect to x equal to zero, we find E = (1 − x)V eF (A − B

(221)

P dE = BV eF 2 − V eF A = 0 (222) dx x V By inspection of this equation, we can see that the extremum we will obtain will be a maximum. Solving for x, r P BP B = V A =⇒ x = (223) 2 x AV So for a fuel cell with higher efficiency (lower B or larger A), the volume we devote to the fuel cell decreases. For larger power requirements, we devote more volume to the fuel cell, and for larger volume allowance, we devote a smaller fraction of the volume to the fuel cell. (b) Plugging in numbers, r 0.003 L/W ∗ 500 W x= (224) .7 ∗ 100 L So x = 0.146 . We devote about 15% of the system to the fuel cell. For a 500 W 3 power density of pF C = 0.146∗100 L = 34 W/L = 0.034 W/cm , a reasonable power 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 thermal management sub-system, and 4) the power electronics sub-system. For example, a fuel reformer in the fuel processing sub-system depends on the thermal management sub-system to provide heat from other system components for an endothermic reaction at the reformer or to extract heat from an exothermic reaction at the reformer. As another example, the fuel cell sub-system depends on the power electronics sub-system to regulate the output voltage of the system. As shown by the shape of the fuel cell’s polarization curve, the fuel cells voltage declines at higher currents and higher powers. Converters enable the system to provide a constant voltage. As another example, the heat produced by an afterburner downstream of the fuel cell stack depends on the anode utilization of the stack. At high anode utilizations, less anode off-gas will be available for downstream combustion. Instead of a downstream afterburner, a downstream microturbine could be used to recapture energy in the anode off-gas. One way fuel cell systems can be integrated is to use excess heat from exothermic chemical reactions to preheat inlet fuel, oxidant, and water. Problem 10.22) The Pressure Swing Absorption Unit (PSA) is a gas purification unit that uses a physical mechanism to remove contaminants. From a hydrogen-rich gas stream containing CO, HCs, CO2 and other species, a PSA unit can produce a 99.99% pure hydrogen stream. One of its adsorbent beds adsorbs all non-hydrogen species because of the relatively 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 nonhydrogen 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 reaction that releases heat. An endothermic reaction is a chemical reaction that absorbs heat. Endothermic: steam reforming Exothermic: oxidation of hydrogen fuel in a fuel cell, partial oxidation, water gas shift reaction, selective methanation, selective oxidation, combus67

tion of fuel cell exhaust gases Neither: 1) a chemical reaction takes place but heat is neither released nor absorbed: autothermal reforming, 2) no chemical reaction takes place but heat may be released: hydrogen separation via palladium membranes, pressure swing adsoption, condensing water vapor to a liquid, expansion of hydrogen gas (hydrogen heats up as a result of the Joule-Thompson Effect), compression of natural gas (in practice, releases heat due to frictional losses at the compressor) Problem 10.4 A process diagram for a fuel cell system for a scooter is included 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 scooter with metal hydride storage is shown. Metal hydride storage tanks are composed of a special granular metal that absorbs and releases hydrogen like a sponge absorbs and releases water. Depending on the metal hydride material, it may need to be pressurized to charge it with hydrogen and heated to discharge the hydrogen. The process of charging and discharging the metal hydride requires a more complicated thermal management sub-system. One 68

possible design is sketched in the figure: Heat from the fuel cell stack is used to heat the hydride to release hydrogen. Coolant air is used to cool the hydride so that it can absorb hydrogen. Because hydrogen absorption is an exothermic reaction, heat must be continuously removed from the alloy bed. The rate at which a metal hydride can absorb or release hydrogen is primarily dependent upon the rate at which the alloy can release or absorb heat, respectively. Although not shown in the figure, the thermal management sub-system becomes more complicated than the one shown if the fuel cell operates at a lower temperature (for example, 80◦ C for a PEM) than the minimum temperature required for hydrogen release in the metal hydride (for example, 150◦ C for a N aAlH4 metal hydride).

Figure 29: Process diagram for a fuel cell system for a fuel cell scooter with metal hydride hydrogen storage. (Problem 10.5) The absorption reaction with the metal hydride depends on the pressure and temperature of the hydrogen gas. If the gas pressure is above a certain equilibrium pressure at a certain temperature, the metal absorbs hydrogen. If the hydrogen gas pressure is below the equilibrium pressure, the metal hydride releases its hydrogen. Problem 10.6 For partial oxidation, all of the products are CO and

69

H2 . A balanced reaction is C8 H18 + a(O2 + 3.76N2 ) → bCO + cH2 + 3.76aN2

(225)

C8 H18 + 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 via the water gas shift reaction, CO + H2 O → CO2 + H2

(228)

then the hydrogen yield could be greater. The overall reaction would be C8 H18 + 4(O2 + 3.76N2 ) + 8H2 O → 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 calculation does not directly consider the addition energy input required to vaporize liquid water for the WGS reaction.) Problem 10.7 We do a back-of-the envelope calculation based on the example problems to produce a reasonable estimate at STP. For steam reforming, for a 100% efficient transfer of heat between a methane burner and a steam generator, the mass/moles/volume of methane needed for combustion is at a minimum about 31.5% of the moles/mass/volume of methane consumed by the steam reformer (Example Problem 1.6) at STP. Assuming only 72% efficient heat transfer, then the moles of methane needed for combustion is at a minimum about (31.5%/0.72 =) 43.8% of the moles of methane consumed by the steam reformer at STP. Under these conditions, the fuel reformer efficiency in terms of HHV is R =

∆HHHV,H2 4 mol H2 (284 kJ/mol H2 ) = = 89.8% ∆HHHV,F uel 1.438 mol CH4 (880 kJ/mol CH4 )

(231)

A more exact calculation can be conducted for reactants entering and ˆ CH (1000 K) = −36.62 kJ/mol, ∆h ˆ CO (1000 K) = products leaving at 1000 K. ∆h 4 2 ˆ ˆ −360.11 kJ/mol, ∆hH2 O(g) (1000 K) = −215.83 kJ/mol, ∆hO2 (1000 K) = 70

ˆ rxn,CH = 27.71 kJ/mol, so the enthalpy of methane combustion is ∆h 4 ˆ H (1000 K) = −810.57 kJ/mol. The enthalpy of hydrogen combustion is (∆h 2 ˆ rxn,H = −250.37 kJ/mol. Then the fuel reformer effi20.68 kJ/mol) ∆h 2 ciency is F R =

4 mol H2 (250.37 kJ/mol H2 ) = 85.9% 1.438 mol CH4 (810.57 kJ/mol CH4 )

(232)

Problem 10.8 We conduct a back-of-the envelope calculation based on the example problems to produce a reasonable estimate. We ignore the enthalpy of formation of nitrogen because it does not chemically react. The autothermal reforming reaction is Cx Hy + zH2 O(l) + (x − z/2)O2 ↔ xCO2 + (z + y/2)H2 → CO, CO2 , H2 , H2 O (233) The value for the steam-to-carbon (S/C) ratio, here shown as z/x, should be chosen such that the reaction is energy neutral, neither exothermic nor endothermic. For autothermal reforming of propane, the only variable is z in the reaction: C3 H8 + zH2 O(l) + (3 − z/2)O2 ↔ 3CO2 + (z + 4)H2

(234)

ˆ rxn = 0. You can find the equation for z to We must find for what z the ∆h be: ˆ H − ∆h ˆ H O + 1 ∆h ˆ O ) = ∆h ˆ C H + 3∆h ˆ O − 3∆h ˆ CO z(∆h 2 2 2 3 8 2 2 2 z=

−12 kJ/mol + 3 ∗ 22 kJ/mol − 3 ∗ (−360 kJ/mol) = 4.6 20 kJ/mol − (−216 kJ/mol) + 12 22 kJ/mol

(235) (236)

The steam to carbon ratio is then z/x = 4.6/3 = 1.5. The reformer would produce 4.6 + 4 = 9.6 moles of H2 per mole of fuel. Problem 10.9 Steam reforming The heat of reaction for this steam reforming reaction at 1000 K is ˆ ∆hrxn = 190.89 kJ/mol. The heat of reaction for the complete combustion ˆ rxn = −800.57 kJ/mol. Therefore, assuming of methane at 1000 K is ∆h all of the reactants are preheated to 1000K, only 190/800 = 0.24 moles of methane must be combusted to provide energy for every 1 mole of methane

71

that goes through the endothermic steam reforming process. The reaction stoichiometry is 1.24CH4 + 2H2 O + 0.48O2 → 1.24CO2 + 4H2 + 0.48H2 O

(237)

These reactions are not usually combined in the same reactor. Because combustion is usually done in a separate reactor from the steam reforming, the products of combustion CO2 and H2 O and the gas N2 do not dilute the 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 of hydrogen produced per unit of methane consumed is 4/1.24 = 3.23. Partial Oxidation The 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% The ratio of hydrogen produced per unit of methane consumed is 2. However, if all of the CO in the product stream was shifted to hydrogen via the water gas shift reaction then the hydrogen yield could be greater. The overall reaction would be CH4 + 0.5(O2 + 3.76N2 ) + H2 O → 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 of hydrogen produced per unit of methane consumed is 3. (This calculation does not directly consider the addition energy input required to vaporize liquid water for the WGS reaction.) The autothermal reforming reaction for methane is Autothermal Reforming We have the reaction CH4 + zH2 O(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 is zero at 1000 K (and assuming water enters as a gas at 1000 K), which gives an equation for z: ˆ CO + (z + 2)∆h ˆ H − ∆h ˆ CH − z∆h ˆ H O − (1 − z/2)∆h ˆO 0 = ∆h 2 2 4 2 2 72

(241)

ˆ H O = −215.83, ˆ CH = −36.62, ∆h Using the values at 1000 K (in kJ/mol) ∆h 2 4 ˆ ˆ ˆ ∆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. This is the same ratio as with steam reforming, but with steam reforming, the hydrogen yield is higher due to the separation of the combustion reaction and the steam reforming reaction in separate reactors. Problem 10.10 From example 10.8, the heat available to heat the building 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 heat transfer is ∆T Q = kA (242) l Where l is the wall thickness, which we’ll assume to be 10 cm. The thermal conductivity of the walls, k, we can take to be 0.1 W/mK, which is be approximately the value for wood or some types of concrete. Then the surface area of the heated room is A=

Ql (11.6 kW )(0.1 m) = = 504 m2 k∆T (0.1 W/mK)(23 K)

(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 f t high, standard for a one story building in 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 the roof and sides). Problem 10.11 The T-H diagram looks similar to that for the condenser in figure 10.11. As in table 10.9, we assume that half of the waste heat from the fuel cell exits via the cathode exhaust gas. Then the heat flow in the exhaust that needs to be removed by the condenser is QF C /2 + Qc , where QF C is the heat 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 an amount of heat Q˙ = hA(Th − Tc ) (244) where A is the surface area of the condenser in contact with the air and h is a function of the velocity of the air flowing past the condenser. To removing enough 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˙ = QF C /2 + Qc even for low velocity. If we assume that the fuel cell operates at a voltage of 0.5 V and has a range of 2 hours at maximum power, then the rate of hydrogen use is P/V N˙ H2 = = 0.010 mol/s 2F

(245)

The rate of water production is N˙ H2 O = N˙ H2 , and the amount of water produced is NH2 O = (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 configuration that increases the pinch is to use two parallel cold streams. The first absorbs heat 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 point temperatures. The pinch no longer occurs in the condenser, and the pinch in the overall system increases. The pinch is greater than ten degrees over a range of mass flow rates of water in each cold stream. The last figure shows this range of the ratio of mass flow rates in one cold stream (that flows through the fuel cell, aftercooler, selective oxidation unit, and post-shift converter) vs. the other cold stream (flowing through the condenser). 74

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

Figure 31: Temperature change of the cold and hot streams and four thermal sources, as a function of enthalpy change of the two streams. (Problem 10.12) 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 and 32 show the results. The change in enthalpy refers to the aggregate level of enthalpy being exchanged between the hot stream and the cold stream, and the temperatures shown at each enthalpy point are the inlet and outlet temperatures for the thermal source. Figure 31 shows the temperature vs. enthalpy plot for the four thermal sources in series on one side of the parallel loop: the fuel cell, aftercooler, selox, and post shift sources. Figure 32 shows the 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 mass flow rate needed to capture all the heat available so as to increase the initial inlet temperature for the domestic cooling stream from 25 ◦ C to the desired outlet temperature of 80 ◦ C. The flow rates chosen for each of the streams in parallel 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 independently achieve 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 effective heat transfer. From figure 31, one can see that the first parallel stream’s pinch point temperature is approximately 3500 W cumulative heat load in the aftercooler at a temperature 8 ◦ C. Figure 32 shows the second parallel stream’s pinch point temperature to be approximately 1800 W cumulative heat load in the condenser at a temperature of 18 ◦ C. Keeping the combined total mass flow rate of water to the domestic cooling loop constant at the same rate, the ratios of the flows in the two parallel branches should be varied to find the mass flow rate ratio at which the pinch point in the system is maximized. Figure 33 shows the results of this analysis, by plotting the minimum pinch point temperature for each thermal 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 to the total flow rate. The intersection of the aftercooler pinch point curve and the 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 a mass flow ratio of 0.64. If the mass flow ratio range is maintained between 0.60 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 a function of the two stream’s flow rates. Results are for a power of 6 kW e. The overall pinch point follows the black line. (Problem 10.12) To understand this result consider that at one extreme, the flow ratio is limited to above a certain minimum by the pinch point in the aftercooler (shown by the square symbols in figure 33) to maximize the pinch point temperature 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 the condenser (shown by the starred crosses), because the condenser’s cooling water must remain below 100 ◦ C to avoid vaporization. The composite sketch of the system’s overall pinch point range is shown as the Overall Pinch Point curve (shown by a single curve), which is the composite minimum of the aftercooler pinch point and condenser pinch point curves. Problem 10.13 Reading off figure 10.11 the location of the pinch is the aftercooler. To make this quantitative, we need to find first the temperature at 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 process with the heat transfer to a stream of liquid water and a stream with water vapor to solve for the condensation temperature Tc : Q = (mc ˙ p )1 (Th − Tc ) + (mc ˙ p )2 (Tc − Tl ) Tc =

Q − (mc ˙ p )1 Th + (mc ˙ p )2 Tl (mc ˙ p )1 − (mc ˙ p )2

(247) (248)

Using the data for mc ˙ p 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 at the pinch point. Next, we need to find the temperature of the cold stream at this point, so we need to find what H point on the curve this temperature corresponds to: dH = (mc ˙ p )1 (Tc − Tl ) = (276 W/◦ C)(84.78◦ C − 60 ◦ C) = 6838 W (249) We can now find the temperature of the cold stream Tb − Tbi =

dH 6838 W = ⇒ Tb = 72.82◦ C mc ˙ p 143 W/K

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

78

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Chapter 11 Solutions

Problem 11.1 The primary steps of Process Chain Analysis (PCA) are as follows: 1. Research and develop an understanding of the supply chain from raw material production to end use. 2. Sketch a supply chain showing important processes and primary mass and energy flows. 3. Identify the “bottleneck” processes, which consume the largest amounts of 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 control volume analysis and the principles of conservation of mass and energy. 5. Having analyzed the individual processes within the supply chain, evaluate the entire supply chain as a single control volume. Aggregate net energy 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 for global warming. 7. Compare the net change in energy flows, emissions, and environmental impacts of one supply chain with another. 8. Rate the environmental performance of each supply chain against the others. 9. Repeat analysis for an expanded, more detailed number of processes in the supply chain. Problem 11.2 Several gases and particles are understood to have a warming effect on the Earth, a good summary is given in figure 11.5. Section 11.3.3 describes the following effects from gases and particles: Anthropogenic greenhouse gases include CO2 , CH4 , H2 O, and nitrous oxide (N2 O). Greenhouse gases selectively absorb infrared radiation, and then re-emit this radiation partly back towards the Earth’s surface. 79

In addition to these gases, certain particles also have a warming effect on the Earth, but through a different mechanism. Dark-colored particles, such as soot, absorb sunlight, re-emit this energy as infrared radiation, and therefore also warm the Earth. Dark-colored particles that contribute to global warming include black carbon (BC). The warming effect of black carbon is enhanced by Organic Matter (OM), which focuses additional light onto black carbon. The center portion of Figure 11.5 shows the warming mechanism of dark-colored particles. Figure 11.5 shows that these gases and particles re-emit infrared radiation towards the Earth’s surface to cause warming; they also re-emit infrared away from the Earth. In contrast, light-colored particles reflect sunlight and have a cooling effect. 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 ), nitrogen oxides (N Ox ), and non-methane organic compounds or volatile organic compounds (VOC). These gases react in the atmosphere and convert to particles, which are mostly light in color. SOx converts to SULF, N Ox converts to NIT, and VOC convert to organics that are mostly light in color. The right portion of Figure 11.5 shows the cooling mechanism of light-colored particles. Problem 11.3 Section 11.4 describes the six primary emissions that create air pollution: ozone, carbon monoxide, nitrogen oxides, particulate matter, sulfur oxides, and volatile organic compounds (VOC). VOC are nonmethane organic compounds, such as the higher hydrocarbons. Some of these compounds are air pollutants themselves. Others react with chemicals to produce air pollution. Effects of air pollution on human health can include respiratory illness, pulmonary illness, damage to the central nervous system, cancer, and increased mortality. Four of the more important air pollutants known to affect human health include CO, N O2 , O3 , and P M10 . CO is known to cause increased incidents of headaches, hospitalization, and mortality. N O2 is known to cause increased irritation to the sinuses, throat and eyes. O3 increases the incidents of asthma attacks, eye irritation, lower respiratory illness, upper respiratory illness, and other conditions. P M10 is known to increase incidents of asthma attacks, respiratory restriction, chronic illness, and morality. (See Table 11.3) Problem 11.4 A National Emission Inventory (NEI) is a database of a country’s emissions from various sources, which records the type of emission,

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its quantity, the location of the emission, and its time of release. National emission inventories vary in the depth of material they provide. The US EPA’s inventory records emissions from vehicles, power plants, factories, chemical processing plants, industrial facilities and other sources. Emissions may be recorded on an annual, monthly, or even daily basis. Since the location of stationary facilities is known, the location of these emissions is known. 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 a certain quantity of hydrogen accrues in high enough concentrations at high enough temperatures. The self-ignition temperature of hydrogen is 858 K and its ignition limits in air are between 4% and 75% at STP. Without proper ventilation, a closed indoor area could accrue high concentrations of hydrogen that could lead to ignition. Problem 11.6 As described in section 11.3.5, the mechanisms through which hydrogen might contribute to global warming and air pollution are very complex and still the subject of intense research. One mechanism through which released H2 might increase global warming 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 → H2 O + H

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If H2 did not consume OH in this reaction, OH might otherwise reduce the presence of CH4 via the reaction CH4 + OH → CH3 + H2 O

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The net effect of hydrogen then is to reduce the concentration of OH, which increases the concentration of CH4 , a greenhouse gas. However, numerous other chemical reactions must also be considered. One mechanism through which released H2 might increase one type of air pollutant is through a series of chemical reactions that enhance the concentration of O3 . In the troposphere, H2 might increase O3 by increasing the concentration of atomic hydrogen (H). After several years in the atmosphere, molecular hydrogen decays to atomic hydrogen in the presence of the hydroxyl radical (OH), via the reaction H2 + OH → H2 O + H 81

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Atomic hydrogen (H) could then react with oxygen in air in the presence of photon energy (hν) from light to increase O3 in the presence of M, any molecule in the air that is neither created nor destroyed during the reaction but that absorbs energy from the reaction. H + O2 + M → HO2 + M

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N O + HO2 → N O2 + OH

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N O2 + hν → N O + O

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O + O2 + M → O3 + M

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However, other sets of reactions must also be considered, with a focus on their net effect on air pollution. The net effect of these reactions may be able to be determined with computer simulations of chemical reactions in the atmosphere (atmospheric models). As we learned in PCA, these simulations should model not the mere addition or subtraction of an individual chemical component, but rather the net change in emissions among different scenarios to be accurate. Problem 11.7 One possible answer is to compare and contrast different energy conversion scenarios using Process Chain Analysis (PCA) and data from an NEI. For example, one may examine the biological generation of hydrogen via different mechanisms. Some researchers have concluded that the energy input requirements are too great to merit development of some of these. This assertion could be examined. A research proposal abstract should be no more than a page in length and address a technically literate but non-expert audience. Problem 11.8 We can perform this calculation from the following equation (11.14)

CO2,equivalent =mCO2 + 23mCH4 + 296mN2 O + α(mOM,2.5 + mBC,2.5 ) (258) − β[mSU LF,2.5 + mN IT,2.5 + 0.40mSOX + 0.10mN OX + 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. The coefficient α can range between 95 and 191. The coefficient β can range between 19 and 39. We will perform the calculation for a low case and a

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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 may use the entry for total non-methane organics. Further, the problem asks only to consider organic gas and particulate matter. Then we find for the high 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 a fleet of fuel cell vehicles. The fuel consumption rate can be found from the consumption of gasoline in the current fleet after adjusting for the relative efficiencies of the vehicles. VM T m ˙ H2 C = (259) Fh where Fh =

barMgvf Vc ∆HHHV,h h ρg ∆HHHV,g g

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We are given that h /g = 2, so using the numbers from example 11.2 we find Fh =

(17.11 mi/gal)(264 gal/m3 )(142 M J/kg) 2 = 36.16 mi/kg (750 kg/m3 )(47.3 M J/kg)

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The vehicle miles per year is VM T = 2.68×1012 mi/yr so we find that annual fuel consumption is m ˙ H2 C = 7.41 × 1010 kg/yr. If the hydrogen leakage is 2%, the annual hydrogen production must be 7.56 × 1010 kg/yr. According to the Example 10.3, in the steam reforming reaction approximately 4 moles of H2 are produced for every one mole of CH4 . If the heat for this endothermic steam reforming reaction is provided by the waste heat of another process, no additional CH4 need be consumed. If CH4 is combusted to provide heat for the reaction, some additional CH4 is consumed. In example 10.6, a back-of-the envelope calculation shows that approximately 31.5% excess CH4 must be combusted to provide sufficient heat for the reaction. Then, for every 4 moles of H2 produced, approximately 1.315 moles of CH4 may be consumed. (The actual ratio depends on the design of the fuel 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 this hydrogen production is m ˙ H2

1.315 16 g/mol CH4 = 1.99 × 1011 kg/yr 4 2 g/mol H2 83

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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 gas is simply CH4 , and find that the US current annual production is 23.14 × 1015 BT U/yr. To find this in kg, convert (23.14 × 1015 BT U/yr)

1055 J 1 kg CH4 ∗ = 4.40 × 1011 kg/yr 1 BT U 5.55 × 107 J

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So the amount we are considering for a hydrogen fleet is 46% of current US production. The CO2 equivalent of the leaked methane is calculated from equation 258. 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 external 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 of the 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 Vehicle (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 and 5.08 × 109 tons/yr. Therefore, reasonable estimates for the change in CO2 equivalent can range from a reduction of 12% to 15% of all anthropogenic sources in the natural hydrogen fuel cell vehicle case (compared with the 1999 fleet). The actual reduction depends on choices pertaining to the scenario, such as the type of reformer and the energy requirements required for storing hydrogen on the vehicle (gas or liquid). See the following table.

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Species Carbon Monoxide (CO) Nitrogen Oxides (NOx) as N O2 Organics Paraffins (PAR) Olefins (OLE) Ethylene (C2 H4 ) Formaldehyde (HCHO) Higher aldehydes (ALD2) Toluene (TOL) Xylene (XYL) Isoprene (ISOP) Total Non-Methane Organics Methane (CH4 ) Sulfur Oxides (SOx) as SO2 Ammonia (N H3 ) Particulate Matter Organic Matter (OM2.5 ) Black Carbon (BC2.5 ) Sulfate (SU LF2.5 ) Nitrate (N IT2.5 ) Other (OT H2.5 ) Total P M2.5 Organic Matter (OM10 ) Black Carbon (BC10 ) Sulfate (SU LF10 ) Nitrate (N IT10 ) Other (OT H10 ) Total P M10 Species Carbon Dioxide (CO2 ) Water (H2 O) CO2,equivalent (low) CO2,equivalent (high)

HFCV and Natural Gas Gases -55.29% -33.18% -27.10% -32.43% -24.86% -19.68% -51.98% -16.71% -27.69% -49.54% -26.24% 20.88% 2.05% -5.25% -1.78% -15.08% -0.32% -0.82% -0.15% -1.26% -1.18% -10.87% -0.35% -0.38% -0.03% -0.41% -15.03% 1.97% -14.00% -13.25%

Problem 11.11 This is an open-ended problem for which answers will vary. Reasonable electrical efficiencies for fuel cell systems are between 40% and 60%, and the efficiency of hydrogen generation can approach 1 (see example 10.6). The efficiency of current US electricity generation is ap85

proximately 32%. Therefore, the efficiency could potentially almost double. Emission reductions are estimated to be significant, based on emission estimates shown in the text in Table 11.1. Problem 11.12 This is an open-ended problem for which answers will vary. One fuel cell manufacturer produces fuel cell systems that have already achieved electrical efficiencies of 50% in combination with heat recovery efficiencies close to 40%. If 90% of the heating value of the fuel could be usefully used, the efficiency of current US electricity and heat generation combined could approximately triple. (The average boiler/furnace in the US achieves an efficiency of about 80%. State-of-the-art condensing boilers achieve efficiencies in the low to mid 90% range.) Under this combined-heat-and-power scenario, emission reductions are estimated to be even more significant than those shown in the text in Table 11.1. Problem 11.13 This is an open-ended problem for which answers will vary. Of all of the links in the upstream supply chains, one of the most important is the energy required to put hydrogen into a form in which it can 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 cool H2 down to a liquid. For gaseous H2 storage, approximately 10% of the heat value of the fuel is required for hydrogen compression. In this case, one can assume the energy required to run the hydrogen compressors is provided by electricity from the grid. These upstream electricity generating plants can be assumed to have the same 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 table accompanying 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 these emission estimates, one can calculate the change in health effects, external costs 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 data in the table of problem 11.10, we can find the change in health effects from this switch. The low estimate is a reduction in costs of $36.8 million, and the high estimate is a savings of $570 million.

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Problem 11.14 This is an open ended problem for which answers will vary. Problem 11.15 In example 11.2, we find the quantity of hydrogen consumed by a hydrogen fuel cell vehicle fleet to be approximately 57.1 MT of H2 /year. If 1% of H2 that was produced leaked, approximately 0.6 MT of H2 /year would leak. According to table 1.2, conventional on-road vehicles produce 0.16 MT of H2 /yr. Therefore, this hydrogen fleet might almost quadruple H2 emissions from vehicles.

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