Fuel cell fundamentals-solutions PDF

Preview

Summary

answers...

Description

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 O 2 2

+ 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+ 12 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 12 O2 + 2e− → O2− which is a reduction, or cathode reaction. 2.

1 O + H2 2 2

→ 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 O + H2 O + 2e− → 2(OH )− as the cathode (reducing) reaction. 2 2 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 = 30 L and the weight is 30 kW ∗ 500 = 60 kg. 1 kW W For the fuel tank, you need to hold a quantity of fuel equal to 30 kJ/s ∗

1 hr 3600 s ∗ 300 miles = 540 MJ ∗ 1 hr 60 miles

(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 MJ/L and 8 M J/kg, so the fuel tank must be 1350 MJ ∗ 41MLJ = 337.5 L and 1350 MJ ∗ 81MkgJ = 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 E = ET −

νi Πaprod RT ln i Πaνreact nF

(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 = µiodni + RT ln ai dni + nF ∆φ (4) First, note that:

Q νi n aprod am aN M = RT ln Q νi = RT ln RT ln ai dni = b areact aA a B (5) Rearranging (4) and inserting the above result, you get Q νi P o a µ i dni RT ∆φ = − ln Q νprod (6) − i nF areact nF X

m +ln an −ln a1 −ln ab RT (ln aM M A B)

From the thermodynamic definition of chemical potential, µi ≡ ∂G so that ∂ni Σµiodni = ∆Go where the o in this case denotes reference concentration. We 7

o

relate this term to E by − ∆G = E o , but noting that the o refers only to nF 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: ǫ ≡ ∆G/∆H. For an isothermal reaction, ∆G = ∆H − T ∆S, so ǫ = 1 − T ∆S . If ∆H is going to ∆H 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) + 12 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 ǫ = 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 ) = [∆H io + cpi (T − T o ) − T (S io + cpi ln(T/T o ))] (9) taking out the ∆H o and ∆S o , X o 0 = ∆Hrxn − T ∆S orxn + [cpi (T − T o − T ln(T /T o ))] (10) 8

or

o 0 = ∆Hrxn − T ∆S orxn + (T − T o − T ln(T /T o ))

Substituting numbers,

X

cpi

(11)

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    (P/Po )νP P xνii x RT RT νP −νR Q QP iνi ln ( P/P ) ln E = ET − = ET − o nF nF (P/Po )νR R xiνi 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 )

(15)

Q νi   RT1 x ∆S ∆nG QP iνi = (T1 − To ) − ln (P1 /Po ) E + nF nF x Q R νi i   ∆S RT2 x Eo + (T2 − To ) + ln (P2 /Po )∆nG QP iνi nF nF R xi o

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

−175.06 + 1.5 ∗ 8.314 ∗ ln 1 ∆S + 1.5R ln P1 = 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 hy⇀ H2 O(l) is in equilibrium. To find drogen, and the reaction H2 + 12 O2 ↽ 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 ∆G = ∆Go + RT ln

i Πaνprod

i Πaνreact

(20)

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

a1H2 O a1H2 aO0.5 2

= ∆Go +RT ln

1 = ∆Go −RT ln(1)1.5 (xH2 )(0.21)0.5 (P/Po )1.5 xH2 x0.5 O2 (21)

  ∆Go = ln (0.21)0.5 xH2 RT 10

(22)

xH 2

    1 ∆Go (−237 kJ/mol) = = 5.41·10−42 exp = (2.18) exp (8.314 J/mol · K)(298 K) 0.210.5 RT (23)

so PH2 = 5.41 · 10−42 atm 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 1.23 = 0 −

PH2 /Po (0.21)0.5 RT ln nF 1

(24)

  1.23 ∗ nF = exp − RT

(25)

This gives the same equation xH2 (0.21)

0.5

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

0.75 1 1.23 1.1

ǫ = 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− + 12 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 j ln jo + 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 = 2.3αnF

(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 + 12 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 = 1.30 · 10−5 mol/s · cm2 . For reaction B, the net cm is nF 5·2cm 2 = 2F 2

2

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

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

Moving the negative sign to the right and inverting the argument of the logarithm, RT C ∗ /C o∗ P ] η= (33) ln [ P∗ o∗ CR /C R nF 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 RT aP ln [ ] aR nF

(35)

Πaνi RT ln [ νPi ] nF Πa ...