Appendix Analytical Method Using Equilibrium Constant Greenwood

1. Chemical Equilibrium Constants

Standard free energy of formation at T is defined by Gibbs-Helmholtz as

 T

DG⁰_{f t}= - T∫DH⁰_f/T² dt
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Integrating it, you can obtain

DG⁰_ft = DH⁰_f -DH_t -TDS⁰ (kcal/kgmol)

The value of  DH⁰_f is published in Thermodynamics by Kojima 1969

DH_t and TDS⁰ could also be calculated from specific heat as described in "Power Generation by Biomass Gasification Version 3"

Thus you obtain Standard free energy of formation DG⁰_ft  at T.

When you assume biomass is a pure carbon, you can define following chemical reaction and  chemical equilibrium constants could be calculated for following 6 equations.

Heat of reaction DH⁰  were calculated from standard heat of formation DH⁰_f as shown below.

DH⁰ = (SnDH⁰_f)_prod - (SnDH⁰_f)_feed

Eq. No.1, 2, 4 and 5 having minus H⁰ figures are by definition, exothermic reactions. Eq. No.3 and 4 are endothermic reactions.

Finally you can calculate standard free energy of reaction at T by

DG⁰_t = (SnDG⁰_fti)_prod - (SnDG⁰_fti)_feed

Where n is mol number of chemical equations.

Chemical equilibrium constant K_p could be calculated from standard free energy of reaction DG⁰_t using following thermodynamic equations. Gas constant R=1.987(kcal/kgmol K)

lnK_p = - DG⁰_t/Z R T

K_p = f_prod1 f_prod2/f_feed

As the pressure is almost atmospheric, Z=1、f_i=P_i  Therefore, K_p become as follows;

K_p = P_prod1P_prod2/P_feed

When you plot DG⁰  v.s.T, following diagram will be obtained.

Gibbs Free Energy

Natural logarithms of chemical equilibrium constant, K_p are shown below.

Equilibrium Constants

2. Equilibrium Calculation under Stoichiometric Constaraints (Composition Sheet)

Calculation were conducted for 400, 600, 700, 800 and 900 ^oC. Biomass and oxygen in Air (A) will be converted to a mixture of X_O2, X_CO2, X_CO, X_H2, X_H2O, X_CH4  of Gas. Number of unknown variables are 6. We have 3 equations for C, O, and H element balance as listed below.

If we add following three equilibrium equation K_p1, K_p2, K_p3, we now have 6 algebraic equations. Here, P is total pressure and P_i is partial pressure, y_i is mol fraction and SX_i is a sum of all products gas mol. including nitrogen gas. Eq. No.1 to 2 are independent of total pressure.

P_i = y_iP = X_iP/SX_i

SX_i ＝N + X_O2 + X_CO2+ X_CO + X_H2 + X_CH4 + X_H2O

Here, N is nitrogen gas mol contained in the air feed.(kgmol/h)

N = X_N2 = (79/21)A

In the early stage, from Eq. No. 7 to 12, I have obtained following equation without X_O2.

(X_COX_H2/X_H2O)(P/K_P3) + (X_CO²/X_CO2)(K_P1/K_P2)P + X_CO2 + X_H2O + 3X_CH4-C-O-H-2A-2N = 0

But because there are still 5 unknown variables and there are so many combination of solutions. I could not find correct answer even with Solver of Excel.

Therefore, in the second stage, I have tried to reduce number of unknown variables as far as possible and reached an equation having two variables. Namely X_CO2 and X_CO.

From Eq. No. 7 and 9, you can eliminate X_CH4 and obtain Eq. No. 13.

X_H2O = 0.5H - X_H2 - 2C + 2X_CO2 - 2X_CO                          Eq.No.13

From Eq. No. 8, you can obtain Eq. No. 14.

X_H2O =  2A + O - 2X_O2 -  2X_CO2 - X_CO                             Eq.No.14

From Eq. No. 13 and 14, you can eliminate X_H2O and obtain Eq. No. 15.

0.5H - 2A - O - 2C + 2X_O2 + 4X_CO2 + 3X_CO - X_H2 = 0      Eq. No. 15

From Eq. No. 13 and 14,  you can eliminate P/SX_i  and obtain Eq. No. 16.

K_P2(X_O2/X_CO) = K_P3(X_H2O/X_H2)                                        Eq. No. 16

From Eq. No. 10, you can obtain Eq. No. 17.

X_O2 = X_CO2/K_P1                                                                                        Eq. No. 17

From Eq. No. 16 and 17, you can obtain Eq. No. 18.

 (K_P2/K_P1)(X_CO2/X_CO) = K_P3(X_H2O/X_H2)                         Eq. No. 18

From Eq. No. 14 and 17, you can eliminate X_O2 and obtain Eq. No. 19.

X_H2O =  2A + O - 2X_CO2/K_P1 - 2X_CO2 - X_CO                 Eq. No. 19

From Eq. No. 18, you can obtain Eq. No. 20.

X_H2O =  (K_P2/K_P1)(X_CO2/X_CO)(X_H2/K_P3)                        Eq. No. 20

From Eq. No. 19 and 20, you can eliminate X_H2O and obtain Eq. No. 21.

 (K_P2K_P3/K_P1)(X_CO2X_H2/X_CO) = 2A - O - 2X_CO2(1/K_p1 -1) - X_CO         Eq. No. 21

From Eq. No. 15 and 17, you obtain

      0.5H - 2A - O - 2C + 2X_CO2/K_P1 + 4X_CO2 + 3X_CO - X_H2 = 0    

Then you can derive Eq. No. 22.

X_H2 = 0.5H - 2A - O - 2C + 2X_CO2/(1/K_P1+2) + 3X_CO                     Eq. No. 22

From Eq. No.21, you obtain Eq. No. 23.

X_H2 = (K_P1/(K_P2K_P3))(X_CO/X_CO2)(2A - O - 2X_CO2(1/K_p1 -1) - X_CO)                  Eq. No. 23(

From Eq. No. 22 and 23, you can eliminate X_H2 and obtain  final objective function having two unknown variables.

0.5H - 2A - O - 2C + 2X_CO2(1/K_p1+ 2)) + 3X_CO - (K_p1/(K_p2K_p3))(X_CO/X_CO2)(2A - O - 2X_CO2(1/K_p1-1) - X_CO) = 0

From Eq.No.7 to 12 you can obtain following equation.

0.5H - 2A - O - 2C + 2X_CO2(1/K_p1+ 2)) + 3X_CO - (K_p1/(K_p2K_p3))(X_CO/X_CO2)(2A - O - 2X_CO2(1/K_p1-1) - X_CO)=0

Solver of Excel was used for numerical calculation of the above equation. Once reaction was started air is not needed. Therefore A=0. Constraints of Solver is that All Xi shall be positive figure. Minimum figure of above equation gives a set of  X_CO and X_CO2. Then you can calculate remaining X_i as follows;

From EQ.No.7

X_CH4=C-X_CO2-X_CO

From EQ.No.10

X_O2=X_CO2/K_p1

From EQ.No.8

X_H2O=2A+O-2X_O2-2X_CO2-X_CO

From EQ.No.9

X_H2=0.5H-X_H2O-2X_CH4

January 19, 2010