Thermodynamics of CeO 2 thermochemical fuel production - Supporting Information

Transcription

1 Thermodynamics of CeO 2 thermochemical fuel production - Supporting Information B. Bulfin, F. Call, M. Lange, O. Lübben, C. Sattler, R. Pitz-Paal, I. V. Shvets Abstract In this supporting document the equilibrium oxygen partial pressure as a function of temperature in 1 bar of both and CO 2 is calculated. In addition the minimum quantities of both sweep gas and oxidiser for a given cycle are calculated using equilibrium considerations. Some select results for CO 2 splitting are also given. Nomenclature δ Oxygen stoichiometry H Change in enthalpy S Change in entropy G Change in Gibbs free energy K a T rd T ox T δ rd δ ox Equilibrium constant Reduction temperature Oxidation temperature T rd T ox Reduced stoichiometry Oxidized stoichiometry Corresponding author. Telephone: bulfinb@tcd.ie Preprint submitted to ACS Energy and Fuels http: // pubs. acs. org/ doi/ abs/ / ef ? journalcod

2 δ eq Equilibrium yield P O2 Oxygen partial pressure P fo2 Final P O2 during reduction C p x i n i Specific heat capacity Mole fraction of species i Moles of species i α Fraction completed for oxidation reaction a i Activity of species i R H 2 Equilibrium ratio between H 2 and in contact with CeO 2 δ 1. Oxygen partial pressure in steam Specific heat capacities are given in table 2 and standard Gibbs energy and enthalpy of formation are given in table 1. H O 2 H f = [kjmol 1 ] G f = [kjmol 1 ] CO 2 CO+ 1 2 O 2 H f = [kjmol 1 ] G f = [kjmol 1 ] Table 1: Standard change in enthalpy and Gibbs free energy at 298 K and 1 bar (SATP) Oxygen partial pressure in oxidiser H O 2 (1) The equilibrium constant for the water splitting reaction is K ws (T ) = x H 2 x 0.5 O 2, (2) x H2 O 2

3 where x represents the mole fractions of each species, and K ws is the equilibrium constant. The value of the equilibrium constant at SATP can be calculated using the equation ( ) G Kws f = exp, (3) RT and the standard change in Gibbs energy given in table 1. The temperature dependence of the equilibrium constant is given by ( ) K ws (T ) T H(T ) ln = dt. (4) Kws 298 K RT 2 The temperature dependence of the change in enthalpy for water splitting can be determined from the energy required to raise the temperature of the steam compared to that of the products, which is H(T ) = H f T 298 K C ph2 O (T ) C p H2 (T ) 1 2 C p O2 (T ) dt. (5) Polynomial fits of the specific heat capacities are given in table 2. C pn2 (T ) T T T T 4 C po2 (T ) T T T 3 C ph2 (T ) T T T 3 C pceo2 (T ) T T T T 4 C (T ) ph2 O T T T T 4 C pco2 (T ) T T T T 4 C pco (T ) T T T T 4 Table 2: Polynomial fits of specific heat capacities C p [J mol 1 K 1 ]. They were fit from K and the polynomial value was always within 1% of the literature value [1, 2] Using equations 3, 4 and 5, K ws (T ) can now be calculated for a given temperature. From the balanced reaction given in equation 1 the mole fractions of hydrogen and steam can be defined as x H2 = 2x O2 and x H2 O = 1 3x O2. These can be substituted into equation 2 to get K ws (T ) = 2x1.5 O 2 1 3x O2. (6) 3

4 Given that x O2 at the temperatures considered (T < 2000 K) should be quite low, solving this condition iteratively once is enough to accurately determine x O2. ( K ws (T ) = 2x 1.5 O 2 x O2 (T ) = K ws (T ) 1 ) 1.5 3x O 2 (7) Log P O CO T C Figure 1: -Log(P O2 ) vs. temperature in 1 bar of steam and CO 2. The same procedure can be followed to compute the oxygen mole fraction in CO 2 as a function of temperature. The oxygen partial pressure is then simply the mole fraction multiplied by the total pressure Minimum quantity of sweep gas As discussed in the manuscript the minimum quantity of sweep gas n N2 can be calculated by assuming that all of the oxygen released during reduction is at equilibrium conditions. By rearranging the equation for the equilibrium vacancy concentration (equation 5 from the manuscript), the equilibrium P O2 can be expressed as P O2 (δ, T ) = ( ( ) 8700(0.35 δ) exp kjmol 1 and the mole fraction of released oxygen is then δ RT x O2 (δ, T ) = P O 2 (δ, T ) 10 6 P P 4 ) [bar], (8), (9)

5 where the 10 6 P is to account for the 1 ppm O 2 already in the sweep gas. The total number of moles of gas n gas (Q N2 + Q O2 ) per mole of vacancies created is then given by n gas (δ, T ) δ = 1 2x O2 (δ, T ), (10) where 2x O2 accounts for there being half a mole of O 2 per mole of vacancies. In addition, n gas will be at a minimum if the sweep gas is only used to remove the oxygen when at the upper temperature T rd, as the equilibrium P O2 increases with temperature. The minimum n gas, is then given by the normalised integral δrd n gas(δ,t ) δ n gas = ox(α) dδ δ, (11) α δ eq and thus the minimum number of moles of sweep gas is n N2 = n gas n O2, (12) where n O2 = α δ eq. The maximum average mole fraction of oxygen in the 2 product stream is then x O2 = α δ eq 2n gas. (13) It is necessary to introduce an additional condition on the value of P O2. If P O2 P, it must be set equal to the total pressure. In other words, in the case where no sweep gas is needed, the oxygen is just removed at the operating pressure of the reactor as it heats up, avoiding errors in the value of x O2. { PO2 (δ, T ) : P P O2 (δ, T ) = O2 (δ, T ) < P (14) P : P O2 (δ, T ) P The maximum average oxygen mole fraction is obtained in the same way except that in equation 9 the expression for P O2 is given by equation 14. The value x O2 is very important as it determines the amount of sweep gas needed per mole of fuel produced, and hence can greatly effect the efficiency. In figure 2, the results for the cycles with T < 500 stop abruptly as the cycle becomes impossible when δ eq < 0. For reduction operating at a total pressure of 1 bar, x O2 decreases with decreasing P fo2. The energy cost of producing nitrogen is quite high and thus cycles with low x O2 (< 10 2 ) will 5

6 4 Log x O P T = Log P fo 2 Figure 2: A plot of log( x O2 ) vs. log(p fo2 ) with T rd = 1500 C. The solid lines are for a range of values of T (0, 100, 200 and 500 C) with P = 1 bar. The dashed line is a plot of log( x O2 ) vs. log(p ), with P fo2 = 10 5 bar and T rd = 1500 C. suffer from very poor efficiency. In order to obtain low P fo2 it may be more sensible to pump the system to lower pressures, which reduces the amount of sweep gas needed. This is illustrated by the dashed line in figure 2. It can be seen that with P = 1 bar isothermal cycles have minimum x O2 of 10 3, which will result in very poor efficiency. This appears to agree with the experimental demonstration of the isothermal cycle performed by Venstrom et al. [3], where ceria was reduced at 1500 C using a sweep gas flow rate of 581 ml min 1 g 1 for 9.5 min resulting in a yield of δ = 0.01 (see figure 5 in their manuscript), which corresponds to an x O2 of approximately Here the maximum x O2 was calculated to be , for a cycle with T rd = 1500 C, α = 0.66, T = 0, with a yield α δ eq = 0.01 (these values were determined from the data presented in the Venstrom paper) and using CO 2 as the oxidiser. This maximum value is, as expected, greater than that obtained in practice by Venstrom et al. by a factor of 1.7. The maximum value is larger, but also comparable to the value obtained in practice offers good support for the calculation Minimum quantity of oxidiser Here the amount of required to oxidise the ceria is calculated. First we must calculated the equilibrium mole fraction of H 2 for steam in equilibrium with CeO 2 δ. 6

7 The reactions equations are given by the following. H O 2 (15) 1 CeO 2 δox 1 CeO 2 δrd + 1 δ rd δ ox δ rd δ ox 2 O 2 (16) 1 CeO 2 δrd + 1 CeO 2 δox + H 2 δ rd δ ox δ rd δ ox (17) The Gibbs free energies must add up to that of water splitting at any given temperature giving G ws (T ) = G rd (T ) + G ox (T ). (18) Now using the equation G = RT ln(k) and the equilibrium constants K rd = (a O 2 ) 1 2 (a CeO2 δrd ) 1 δ (a CeO2 δox ) 1 δ K ox = (a H 2 )(a CeO2 δox ) 1 δ, (19) (a H2 )(a CeO2 δrd ) 1 δ we can write equation 18 as ( (ao2 ) 1 2 (a H2 )(a CeO2 δrd G ws (T ) = RT ln ) 1 δ (a ) ) 1 δ CeO2 δox. (20) (a H2 O)(a CeO2 δox ) 1 δ (a CeO2 δrd ) 1 δ Simplifying and subbing in the partial pressures for the activities gives ( ) P H2 ln = G ws(t ox ) 1 RT 2 ln(p O 2 (δ, T ox )), (21) P H2 O which can be re-arranged to give the equilibrium ratio between H 2 and as ( ) R H 2 (δ, T ox ) = P H 2 G ws (T ox ) = exp 1 P H2 O RT ox 2 ln(p O 2 (δ, T ox )). (22) The value of G ws (T ) can be calculated using K ws (T ) from the previous section, and the formula G = RT ln(k). Then using this value and the equilibrium oxygen partial pressure given in equation 8 R H 2 can be calculated. 7

8 The equilibrium mole fraction of H 2 is x H2 1 = x H2 x H2 O + x H2 + x O2 = x H2 x H2 O x H2 O x H2 O + x H 2 x H2 O + x O 2 x H2 O R H R H 2, (23) since x H2 O + x H2 + x O2 = 1 and x O2 << x H2 O + x H2. To get the mole fraction of hydrogen produced by the reaction with ceria, the equilibrium hydrogen in a pure steam atmosphere must also be subtracted giving x H2 (δ, T ox ) = R H 2 (δ, T ox ) 1 + R H 2 (δ, T ox ) x H 2 (T ox ) eq, (24) where x H2 (T ox ) eq is the equilibrium mole fraction of hydrogen in an isolated atmosphere at temperature T ox. The minimum number of moles of oxidiser required per mole of fuel produced is then n H2 O(δ, T ox ) δ = 1 x H2 (δ, T ox ), (25) with the total number of moles of oxidiser required to cycle one mole of ceria given by δrd n H2 O(δ,T ox) δ n H2 O = ox(α) dδ δ. (26) α δ eq Figure 3 shows how x H2 depends on T. For isothermal oxidation hundreds of moles of oxidiser are required to produce a mole of fuel, but this decreases rapidly for increasing T. Decreasing the value of P fo2 also decreases the amount of oxidiser required, and where the lines abruptly stop is where operation becomes impossible as δ eq < 0. Stopping the oxidation at 95 % completion is also shown to greatly reduce the amount of oxidiser required. The calculation was also performed for an oxidiser of CO 2, the results of which are given in figure 3. Again we can compare the results to the experimental demonstration of the isothermal cycle performed by Venstrom et al. [3], where ceria was oxidised in CO 2 at 1500 C using an oxidiser flow rate of 297 ml min 1 g 1 for 0.9 min, resulting in a yield of α δ eq = 0.01, which corresponds to an x CO of approximately Here the maximum x CO was calculated to be , for a cycle with T rd = 1500 C, α = 0.66, T = 0, with a yield α δ eq = As before the value is larger 8

9 Log x H P fo Α = T C Figure 3: A plot of -log( x H2 ) vs. the change in temperature T. The solid lines are different values of P fo2 (10 1, 10 3 and 10 5 bar) with T rd = 1500 C. The dashed line shows the effect of stopping the oxidation reaction before equilibrium at 95% completion, with T rd = 1500 C and P fo2 = 10 5 bar. than that obtained in practice as expected, and the numbers are comparable which again offers support for the calculation. The low concentration of CO is very impractical as it must be separated from the product stream, which is not a trivial task. Together with sweep gas energy costs, these technical demands will make ambient pressure isothermal operation both impractical and very inefficient CO 2 splitting The same calculation shown above can be made to determine R CO. There CO 2 are a number of other differences to consider when calculating the efficiency of CO 2 splitting. The value of δ ox needs to be changed throughout the calculations, as CeO 2 has a lower equilibrium vacancy concentration in CO 2 compared to water. In addition the heat required to sustain the oxidation reaction becomes ( ) Tox Q ox = n CO2 (1 ɛ gas ) C pco2 (T )dt, (27) T amb where n CO2 = 1 x CO. 9

10 Log x CO P fo Α = T C Figure 4: A plot of -log( x CO ) vs. the change in temperature T. The solid lines are different values of P fo2 (10 1, 10 3 and 10 5 bar) with T rd = 1500 C. The dashed line shows the effect of stopping the oxidation reaction before equilibrium at 95 % completion, with T rd = 1500 C and P fo2 = 10 5 bar. The calculation of x CO, was performed the same as that for x H2, except and δ ox in CO 2. This resulted in less oxidiser which can be seen using R CO CO 2 by comparing figure 4 to figure 3. In general this gave slightly greater fuel production efficiency, which can be seen in figure 5. This is neglecting the fact that the product stream is a mixture of CO 2 and CO which needs to be separated. References [1] Bonnie J. McBride, Sanford Gordon, and Martin A. Reno. Thermodynamic data for fifty reference elements. NASA Technical Paper, 3287, [2] L. B. Pankratz and R. V. Mrazek. Thermodynamic properties of elements and oxides. US Bureau of Mines Bulletin, 672, [3] Luke J Venstrom, Robert M De Smith, Yong Hao, Sossina M Haile, and Jane H Davidson. Efficient splitting of co2 in an isothermal redox cycle based on ceria. Energy & Fuels, 28(4): ,

11 Ηfuel T rd Log P fo 2 Figure 5: A plot of the η fuel vs. log(p fo2 ) for a range of values of T rd ( and 1600 C), with P = 1 bar, T = 500 C, α = 0.95 and using CO 2 as the oxidiser. 11