2.60J S2020 Homework 3 Solutions

Homework 3 – Solutions Problem 1 a) How much heat is needed to pyrolyze the ammonia? [5 points] First, we start by applying an enthalpy balance (first law), assuming we start from ambient conditions (Tin = T0 = 25C): � = # �% ℎ% − # �% ℎ% '()* (,-. � = /0.5�'̂ ,67 (�):; − �< ) + 1.5�'̂ ,@7 (�):; − �< )A − /ℎB),6@C + �'̂ ,6@C (�%D − �< )A � = /0.5�'̂ ,67 (�):; − �< ) + 1.5�'̂ ,@7 (�):; − �< )A − EℎB),6@C F � = E0.5 × 29.15 × (55) + 1.5 × 28.96 × (55)F − (−45,700 J) = 48.89 kJ b) Calculate the composition of the two streams at the exit of the cell. [6 points | 8 points] Begin by finding the inlet composition on the cathode side. Since the stoichiometry is 2: �̇ S7 ,.,%D = 2 × �̇ @7,T,UV 2 = 1.5 mols/s �̇ 67 ,.,%D = 3.76 × �̇ S7 ,.,%D = 5.64 mols/s The exit conditions on the cathode side are given by: �̇ 67 ,.,):; = �̇ 67 ,.,%D = 5.64 mols/s �̇ S7 ,.,):; = �̇ S7 ,.,%D − �̇ @7 ,-,.)D]:^,* 0.65 × 1.5 = 1.5 − = 1.0125 mols/sec 2 2 �̇ @7 S,.,):; = �̇ @7 ,-,.)D]:^,* = 0.65 × 1.5 = 0.975 mols/sec 1 Likewise, the exit conditions on the anode side are given by: �̇ 67 ,-,):; = �̇ 67 ,.,%D = 0.5 mols/s �̇ @7 ,-,):; = (1 − 0.65) × �̇ @7 ,-,%D = 0.525 mols/s The corresponding exit mole fractions are summarized below: H2 O2 N2 H 2O Cathode 0.1327 0.7394 0.1278 Anode 0.5122 0.4878 - c) Calculate the open circuit voltage of this cell based on the concentrations at the exit. [6 points | 8 points] To calculate the open circuit voltage, we use the Nernst Equation: The equilibrium constant �' for the reaction H2 + 0.5O2 → H2 O at a standard pressure of 1 atm and at 80℃ is calculated based on Table 3.6 in the notes. Using linear interpolation (note this is a simplification), �' = 10fg.fhgi = 2.118 × 10fg (no units). �' (�) = exp l −∆� ) (�) p ℜ� Therefore: Δ�°(� ∗ ) = (8.314 × 353.15) ℜ� ∗ ln �' (� ∗ ) = ln(2.118 × 10fg ) = 1.238 V (2 × 96485.33) � , ℑ- Similarly: 2 Δ�°' (�∗ , � ∗ ) = − �ℜ� ∗ �∗ 0.5 × (8.314 × 353.15) ln l p = ln(2) = 0.00527 V (2 × 96485.33) � , ℑ�< Given the reaction in the cell: 1 �h + �h → �h � 2 The last term is calculated as: €€ „/h  �@7 ,-D)*, �S7 ,.-;ƒ)*, ∏'()* �% U (8.314 × 353.15) ℜ� ∗ ∗) ( = ‚ Δ�°. �% , � = − ln } ln €  (2 × 96485.33) � , ℑ�@7 S,.-;ƒ)*, ∏ � U (,-.; % Δ�°. (�% , � ∗ ) = 0.00576 V Summing up all components: Δ�^-† (�∗ , � ∗ ) = 1.25 V d) Calculate the power delivered by the stack of 200 cells in the proposed application. [2.5 points] � = � × � × � × � = 600 mA × 2000 cm2 × 0.65V × 200 = 156 kW cm2 e) What is the mass flow rate of H2 needed to generate the power? [4 points | 5 points] To supply a current of I = 1200 Amps, the number of moles of electrons needed per second is Ne = I/F (where F is the Faraday number). Since each mole of H2 supplies ne = 2 moles of electrons, we get: �̇ @7 = � �, � = 1,200 = 0.00622 mols/s 2 × 96485.33 For a stack of 200 cells, we will need: �̇ @7 ,.)D]:^,* = 200 × 0.00622 = 1.244 mols/s However, only 65% of H2 is consumed (react and contribute electrons to the current) (see Figure 1). Thus, the total mole flowrate of H2 needed is: 3 �̇ @7 ,]:''%,* = 1.244 mols/s = 1.914 mols/s 0.65 Assuming a molar mass of 2g/mole for H2, this translates to a H2 mass flowrate of 3.828 grams/sec. f) What is the cooling rate required to keep the fuel cell at the desired temperature? Note that the inlet and exit streams are all at the same temperature as the cell. [4 points | 5 points] Assuming steady state, and ignoring kinetic and potential energy changes, we get the following equation by conserving energy around the fuel cell (the control volume is specified in the figure below). Here �̇.))%D is positive according to the arrow shown in the figure. �̇.))%D + �̇,,. = # �̇ % ℎ% − # �̇ % ℎ% %D,; ):;,; Figure 1 - Control volume to calculate cooling rate �̇,,. = 156 kW from part d. The right-hand side is essentially the heat released from the combustion of 1.244 moles of H2 per second (the number of moles of H2 participating in electrochemical reactions per second, as calculated in part e). This evaluates to 1.244×LHVH2 and �̇������� = 144.8 kW. 4 g) What is the efficiency of the cell? [2.5 points | 3.5 points] We use the HHV value for H2 (since water will be leaving in liquid phase under the specified conditions) and rate of fuel consumed to be more reasonable in the analysis: �š› = Power Out ��-.; 156000 = = = 43.9% Rate of Chemical Energy In �̇ B Δℎ«° ¬,. (1.244 × 285800) h) Is it better efficiency-wise to operate the cell at 10 atm? [8 points] If the fuel cell were operated at 10 atm, the open circuit voltage would be increased by: Δ�°',h (�∗ , � ∗ ) − Δ�°',„ (�∗ , � ∗ ) = − �ℜ� ∗ �h∗ 0.5 × (8.314 × 353.15) 10 ln ® ∗ ¯ = ln l p (2 × 96485.33) � , ℑ�„ 2 = 0.0122 V Therefore, for every 1.5 moles of hydrogen supplied (see Figure 1), we get an additional electric work of ��̇ = �. ��� �(���) = �, ���. � W. Corresponding to 1.5 moles of hydrogen supplied, 7.14 moles of air are supplied (see Figure 1). Thus, the compression work (for an ideal, isothermal compressor operating at � = 80℃ = 353.15 K and a pressure ratio of 5) is: �̇���� = �̇ �� �� ® �∗� ¯ = �. �� × �. ��� × ���. �� × ��(�) = ��. �� kW �∗� Stated differently, we are getting an addition 3.5 kW electric power by supplying 33.74 kW mechanical power to the compressor, which is not desirable. 5 Problem 2 a) Determine the molar flow rate of methane and water fed to the reformer. Note that the partial pressure-based equilibrium constant of the water gas shift reaction at 800 C is Kp = 0.8879. [7 points | 10 points] 6 b) Calculate the flow rate of hydrogen and the current produced by the fuel cell. [3 points | 5 points] c) Determine the heat transfer rate across the reformer/water gas shift reactor. [5 points | 7.5 points] d) Calculate the flow rate of air into the fuel cell. [5 points | 7.5 points] 7 e) Assuming that the product stream (2) of the fuel cell consists of CO, CO2, and H2O, determine the maximum work transfer of the fuel cell. [6 points | 10 points] f) Determine the actual work transfer from the fuel cell, assuming a second law efficiency of 70% for the fuel cell. [2 points | 2.5 points] g) Calculate the operating voltage of the fuel cell. [2 points | 2.5 points] 8 Problem 3 a) Determine the mole fractions at the anode and cathode. What is the theoretical open circuit voltage of this cell? [12 points] Anode: Start by using the information we have for water vapor: �%D = �Æ �Æ �< = �]-; �]-; → ���� = �%D l �]-; p = 0.2 × 0.4708 = 0.09416 �< Given the molar ratio of H2 to CO2: ��� = E1 − �@7 S F × 4 = 0.7247 5 ���� = E1 − �@7 S F × 1 = 0.1812 5 Cathode: Start by using the information we have for water vapor: �%D = �Æ �Æ �< = �]-; �]-; → ���� = �%D l �]-; p = 0.4 × 0.4708 = 0.18832 �< Given the molar ratio of N2 to O2: ��� = E1 − �@7 S F × 1 = 0.1705 4.76 ��� = E1 − �@7 S F × 3.76 = 0.6412 4.76 9 Theoretical Open Circuit Voltage: We do this calculation based on inlet conditions – assuming each stream is pure: Taking T0 = 298 K, the properties @ (353K and 1) bar are calculated as follows: ℎ% = �'̂ (�% − �< ) + ℎB),% �% �% = �%) + �'̂ ln l p �< �% = ℎ% − �% �% H2 O2 H2O Enthalpy (J/mol) 1592.8 1625.25 -239978.55 Entropy ( J/mol-K) 135.5851 210.157 194.5243 Gibbs free energy (J/mol) -46268.8 -72560.2 -308645.6 In this case, the reaction occurring within the fuel cell is: 1 �h + �h → �h � 2 „/h Δ�SÎ �@7 ,-D)*, �S7 ,.-;ƒ)*, −Δ�¬< (353K) �� = + ln ‚ = 1.178 V 2� 2� �@7 S,.-;ƒ)*, b) Determine What is the open circuit (thermodynamic) efficiency based on the lower heating value of H2? (LHV of H2 is 120.1 MJ/kg) [2 points] Δ�Ð@Ñ = (120.1 MJ/kg) × (2.016 kg/kmol) = 242,121.6 J/mol ηÔÕ = Δ�¬ 2F(Δ�SÎ ) = = 93.9% Δ�Ð@Ñ Δ�Ð@Ñ c) How many individual cells are needed to supply the required power for the vehicle? [2 points] ����� = �., ��� �., = ����� 80,000 W = = 205 cells A ��� 2 1 2 × 0.6 � × 650 cm cm 10 d) What is the total molar flow rate of oxygen? [2 points] � = �, ��S7 ,.)D]:^,* = �̇ S7 ,.)D]:^,* = ����� � ����� 80,000 W = = 0.3455 mol/s � (�, � ) 4 × 0.6 � × 96485 C/mol Given the air stoichiometry is 2: �̇ S7 ,%D,; = 2 × 0.3455 = 0.69 mol/s The corresponding molar flow rates of N2 and H2O at the inlet are: �̇ 67 ,%D = 3.76 × �̇ S7 ,%D,; = 2.59 mol/s �̇ @7 S,%D �@7 S,%D = �̇ S7 ,%D �S7 ,%D → �̇ @7 S,%D = l 0.18832 p × 0.69 = 0.762 mol/s 0.1705 e) What is the composition of the air-side stream at the exit of the stack, expressed in terms of the mole fractions of N2, O2, and H2O? [5 points] �̇ @7 S,'()* = 2�̇ S7 ,.)D]:^,* = 0.69 mol/s �̇ @7 S,,†%; = �̇ @7 S,%D,; + �̇ @7 S,'()* = 1.452 mol/s �̇ S7 ,,†%; = �̇ S7 ,%D,; − �̇ S7 ,.)D]:^,* = 0.3445 mol/s �̇ 67 ,,†%; = �̇ 67 ,%D,; = 2.59 mol/s Therefore: �@7 S = �S7 = �̇ @7 S,,†%; = 0.331 �̇ @7 S,,†%; + �̇ S7 ,,†%; + �̇ 67 ,,†%; �̇ S7 ,,†%; = 0.0785 �̇ @7 S,,†%; + �̇ S7 ,,†%; + �̇ 67 ,,†%; �67 = �̇ 67 ,,†%; = 0.591 �̇ @7 S,,†%; + �̇ S7 ,,†%; + �̇ 67 ,,†%; 11 f) When the design exit air relative humidity is 90%, does the current operating conditions satisfy the design target? [2 points] �= �@7 S �@7 S,]-; = 0.331 = 70% 0.4708 Therefore, the relative humidity is lower than the design target. g) What is the composition at the anode side exit stream? [6 points] The molar fraction is independent of the total number of moles (provided the composition remains constant). Take the composition to be that of the anode inlet (Part a) such that ∑�% = 1 mols of mixture (at anode inlet), and account for changes. Given the fuel utilization, we know that: j= n!H2,inlet - n!H2,exit n!H 2,inlet = 0.9 n!H2,exit = 0.1n!H2,inlet Similarly by mass conservation in the anode: n!H 2O ,exit = n!H 2O ,inlet = n!CO2 ,exit = n!CO2 ,inlet = X H 2O X H2 X CO2 X H2 n!H 2 ,inlet n!H2 ,inlet By the definition of molar fraction: X H2 = X CO2 = X H 2O = n!H2 ,exit n!H2 ,exit + n!CO2 ,exit + n!H 2O,exit n!CO2 ,exit n!H2 ,exit + n!CO2 ,exit + n!H 2O ,exit n!H2O ,exit n!H2 ,exit + n!CO2 ,exit + n!H 2O ,exit = 0.07247 = 0.208 0.07247 + 0.1812 + 0.09416 = 0.1812 = 0.521 0.07247 + 0.1812 + 0.09416 = 0.09416 = 0.271 0.07247 + 0.1812 + 0.09416 12 h) What is the molar flow rate of the fuel stream at the inlet? [3 points] � = �, ��@7 ,.)D]:^,* = �̇ @7 ,.)D]:^,* = ����� � ����� 80,000 W = = 0.691 mol/s � (�, � ) 2 × 0.6 � × 96485 C/mol Accounting for fuel utilization: �̇ @7 ,%D,; = �̇ @7 ,.)D]:^,* = 0.7678 mol/s 0.9 i) What is the first law efficiency of the cell? [2 points] hI = generated power 80kW 80 = = = 43.37% chemical energy in LHVH 2 ´ n!H 2,inlet 240.2´ 0.7678 j) What is the cooling rate required to keep the fuel cell at 80C? [4 points] 0 = å n!i hi - å n!i hi + Q! -W! in out Q! = W! + å n!i hi - å n!i hi out ( ) ( in ) ( Q! = W! + n!H2O, produced hH2O - n!H2 ,consumed hH2 - n!O2 ,consumed hO2 ) Q! = 80 + 0.691´(-239.98) - ( 0.691´1.59 ) - (0.3455´1.63 ) = -87.49 kW 13 k) Derive an expression for the exit relative humidity: [15 points] By definition: fexit = n!H 2O ,exit p0 n!H 2O ,exit + n! N2 ,exit + n!O2 ,exit pH 2O , sat Expanding the terms: n!O2 ,consumed = Power 4VF n! N2 ,exit = n! N2 ,inlet = 3.76l n!O2 ,consumed n!O2 ,exit = n!O2 ,inlet - n!O2 ,consumed = ( l -1) n!O2 ,consumed To find the last term, we start by defining a ratio of vapor pressures: �=® �@7S,%D �%D �@7 S,]-; ¯=® ¯ �67 ßS7 ,%D �< − �%D �@7 S,]-; Likewise, using the definition of partial pressure: �=® �@7 S,%D �@ S,%D ¯=® 7 ¯ �67 ßS7 ,%D �67 ßS7 ,%D For one mole of oxygen entering, this becomes: �=/ �@7S,%D A → 4.76 �@7 S,%D = 4.76� (per mole of Oh ) Accounting for the total amount of oxygen entering: �̇ @7 S,%D = 4.76���̇ S7 ,.)D]:^,* Also we know that: n! H 2O, produced = 2n!O2 ,consumed Therefore: n!H 2O,exit = n!H 2O ,inlet + n!H 2O, produced = 4.76ly n!O2 ,consumed + 2n!O2 ,consumed 14 Substituting back into our starting equation: fexit = fexit = = n!H 2O ,exit p0 n!H 2O ,exit + n! N2 ,exit + n!O2 ,exit pH 2O , sat 4.76ly + 2 p0 4.76ly + 2 + 3.76l + ( l -1) pH 2O , sat 4.76ly + 2 p0 ly + 0.42 p0 = 4.76ly + 4.76l +1 pH 2O , sat l (y + 1) + 0.21 pH 2O , sat l) What should be the relative humidity of the inlet air if the relative humidity of the exit air is 90%? [5 points] We start by solving for the ratio of vapor pressures: fexit = 0.9 = y= 2y + 0.42 1 2y + 2.21 0.4708 finlet pH O,sat = 0.448 p - finlet pH O,sat 2 0 2 Using the definition of the �: 1 y = p0 - finlet pH2O,sat finlet pH O,sat 2 = 1 1 -1 = finlet 0.4708 0.448 finlet = 0.657 15