- Email: [email protected]
CO Raceway Exit Gas Proof
A P P E N D I X
D
CO Raceway Exit Gas Proof Raceway flame temperature calculations of Chapter 14, Raceway Flame Temperature, assume that the carbonaceous portion of raceway exit gas is predominantly CO(g). The objective of this appendix is to show that this is true. Equilibrium and molar balance calculations are used.
D.1 RACEWAY INPUTS AND OUTPUTS
1200 C dry air is blown into the raceway where it reacts with falling 1500 C C(s)-in-coke particles. Reaction of these two hot inputs produces even hotter raceway exit gas B2000 C. First contact of the hot blast’s O2 with descending carbon particles produces CO2(g). This CO2(g) then reacts further with descending C-in-coke of Fig. D.1 to give CO(g) by the following reaction: CO2 g 1 CðsÞ-2CO g
Fig. D.1 describes inputs and outputs of our raceway, with no tuyere injectants.
We postulate that this reaction rapidly reaches equilibrium at the high temperatures in and around the raceway. The equilibrium constant for Reaction (1) is
Solid C-in-coke particles, 1500°C
Blast air: O2(g) + N2(g) 1200°C, 4 bar
Raceway
(D.1)
CO2 ðgÞ1CðsÞ-2COðgÞ
KE
5
2 aECO aECO2 aEC
(D.2)
where KE is the equilibrium constant, unitless and aE is the equilibrium thermodynamic activities of the reactants and product, unitless. The equilibrium activity of C(s) is 1 because it is a pure solid. The equilibrium activities of CO(g) and CO2(g) are
Raceway flame: CO(g) + N2(g), Tflame, 4 bar
E aECO 5 XCO Pt =1
FIGURE D.1 Tuyere raceway with dry blast air (only).
The inputs are 1200 C air and 1500 C falling C(s)-in-coke particles. The objective of this appendix is to show that the product carbonaceous gas is predominantly carbon monoxide.
and
681
E aECO2 5 XCO Pt =1 2
682
APPENDIX D: CO RACEWAY EXIT GAS PROOF
E E where XCO and XCO are the equilibrium mol fractions of CO(g) and CO2(g) in the raceway exit gas, Pt is the absolute pressure in the raceway B4 bar, and 1 is the pressure (bar) at which the thermodynamic activity of a pure ideal gas is 1. With these substitutions, the mol fractions of CO(g) and CO2(g) in the raceway exit gas are related to the equilibrium constant of Eq. (D.2) by the following equation: 2
CO2 ðgÞ1CðsÞ-2COðgÞ
KE
5
E 2 XCO P1t E XCO P1t 1 2
(D.3)
The value of this equilibrium constant at 2000 C is 1.2 3 105 (Appendix E).
Furthermore, because nitrogen does not react in the raceway: raceway exit gas
nN2
D.3 OXYGEN MOLAR BALANCE We now use the raceway’s steady-state oxygen molar balance to calculate kg mol of CO and CO2 in the raceway exit gas. These are then used to determine the raceway’s equilibrium exit gas CO and CO2 mol fractions (Eq. (D.3)). The oxygen molar balance is:
2
input
nO2
5 1 kg mol
(D.4)
All n values in this appendix are based on this 1 kg mol of O2 in blast air. The molar composition of dry air is B79 mol% N2 plus B21 mol% O2 (Appendix B), that is, each kg mol of air contains 0.79 kg mol of N2 and 0.21 kg mol of O2. This leads to: input
nN2
5
79 mol% N2 input nO2 21 mol% O2
5
79 1 kg mol input O2 5 3:76 kg mol input N2 21 (D.5)
5 3:76 kg mol N2 in raceway exit gas (D.6)
D.2 CO(g), CO2(g), AND N2(g) QUANTITIES AND MOL FRACTIONS IN RACEWAY EXIT GAS This section and the next show how we calculate CO and CO2 mol fractions of Eq. (D.1), that is, XCO and XCO . The calculations are begun by specifying that 1 kg mol of O2(g) enters the raceway in blast air, that is;
input
5 nN2
2 3 kg mol O kg mol O 6 7 5 4 in raceway 5 in input O2 exit gas CO 2 3 2 3 kg mol O kg mol O 6 7 6 7 1 4 in raceway 5 1 4 in raceway 5 exit gas O2 exit gas CO2
(D.7)
The equilibrium amount of O2 in raceway exit gas is very small (Appendix F) so that Eq. (D.7) simplifies to:
3 2 3 2 kg mol O kg mol O kg mol O 4 4 5 5 in raceway 1 in raceway 5 (D.8) in input O2 exit gas CO exit gas CO2
One kilogram mole of O2 contains 2 kg mol of O. One kilogram mole of CO2 also contains 2 kg mol of O. One kilogram mole of CO, however, contains only 1 kg mol of O so that: 2
3 kg mol O 4 in 1 kg mol 5 5 2 kg mol input O2 of input O2 2 3 2 3 kg mol CO kg mol O in 4 4 5 1 kg mol of 5 1 in raceway 5 exit gas raceway exit gas CO
683
APPENDIX D: CO RACEWAY EXIT GAS PROOF
and
raceway exit gas
ð2Þ nN2
2
3 2 3 kg mol CO2 kg mol O 4 5 4 in 1 kg mol of 5 2 in raceway 5 raceway exit gas CO2 exit gas
the total amount of raceway gas is; raceway exit gas
nT
from which Eq. (D.8) becomes; 2 3 2 3 kg mol CO kg mol CO2 kg mol 4 4 5 2 5 1 in raceway 1 2 in raceway 5 input O2 exit gas exit gas
kg mol and, because input is specified as 1 kg mol; O2 2
raceway exit gas
5 nCO2
2
3
raceway exit gas
nT
raceway exit gas
5 2 2 nCO2
1 3:76
raceway exit gas
(D.14)
so that; raceway exit gas
(D.10)
Þ 1 3:76
5 5:76 2 nCO2
3
kg mol CO kg mol CO2 2 1 5 1 4 in raceway 5 1 2 4 in raceway 5 exit gas exit gas
raceway exit gas
1 ð2 2 2 nCO2
or combining right side terms;
(D.9)
5 3:76
raceway exit gas
XCO2
5
nCO2
raceway exit gas nT
raceway exit gas
5
nCO2
raceway exit gas
5:76 nCO2
(D.15)
which we simplify to; raceway exit gas
2 5 1 nCO
or subtracting 22
Likewise; raceway exit gas
1 2 nCO2 o
n raceway exit gas 2 nCO2
raceway exit gas nCO2
51
(D.11)
from both sides;
raceway exit gas
raceway exit gas
raceway exit gas
XCO
5
nCO
raceway exit gas nT
5
2 2 nCO2 5:76
raceway exit gas nCO2
(D.16)
raceway exit gas nCO
or switching sides; raceway exit gas
nCO
raceway exit gas
5 2 2 nCO2
D.4 CALCULATING CO(g) AND CO2(g) MOL FRACTIONS FOR EQUILIBRIUM CONSTANT EQ. (D.3) The mol fraction of CO in the raceway exit gas is; raceway exit gas
raceway exit gas
XCO
5
D.5 EQUILIBRIUM MOLE FRACTIONS
(D.12)
nCO
raceway exit gas
nT
exit gas where nraceway is the total mol of exit gas, T that is;
Returning to Eq. (D.3); CO2 ðgÞ1CðsÞ-2COðgÞ
KE
5
and applying Eqs. (D.15) and (D.16) gives; 2 E XCO P1t g ð Þ CO2 ðgÞ1CðsÞ-2COðgÞ KE 5 1:2 3 105 5 Pt E XCO 1 1 2 ðgÞ 2 2 raceway exit gas raceway exit gas 22 nCO 22 nCO Pt 2 2 raceway exit gas 1 raceway exit gas 5:76nCO 5:76nCO P 5 raceway2 exit gas t (D.17) 5 raceway2 exit gas nCO nCO 1 P t 2 2 raceway exit gas 1 raceway exit gas 5:76nCO
2
raceway exit gas nT
raceway exit gas 5 nCO2
raceway exit gas 1 nCO
raceway exit gas 1 nN2
(D.13)
or because, by Equation D.12 and D.6: ð1Þ
raceway exit gas nCO
52 2
raceway exit gas nCO2
E 2 XCO P1t 5 1:2 3 105 Pt E 1 XCO 1 2
5:76nCO
2
from which, with 4 bar absolute pressure (Pt 5 4) in the raceway (Section D.1): raceway exit gas
nCO2
5 2:2 1025 kg mol CO2
684
APPENDIX D: CO RACEWAY EXIT GAS PROOF
and because; raceway exit gas
nCO
Another result of the calculations is that raceway exit gas
5 2 2 nCO2 5 B2 kg mol CO
So the carbonaceous raceway exit gas is virtually all CO. Additional calculations show that this true for all temperatures above 1500 C.
raceway exit gas
E XCO 5
nCO
nT
5
2 5 0:35 2 1 3:76
This value is used in Appendix G.
D
CO Raceway Exit Gas Proof Raceway flame temperature calculations of Chapter 14, Raceway Flame Temperature, assume that the carbonaceous portion of raceway exit gas is predominantly CO(g). The objective of this appendix is to show that this is true. Equilibrium and molar balance calculations are used.
D.1 RACEWAY INPUTS AND OUTPUTS
1200 C dry air is blown into the raceway where it reacts with falling 1500 C C(s)-in-coke particles. Reaction of these two hot inputs produces even hotter raceway exit gas B2000 C. First contact of the hot blast’s O2 with descending carbon particles produces CO2(g). This CO2(g) then reacts further with descending C-in-coke of Fig. D.1 to give CO(g) by the following reaction: CO2 g 1 CðsÞ-2CO g
Fig. D.1 describes inputs and outputs of our raceway, with no tuyere injectants.
We postulate that this reaction rapidly reaches equilibrium at the high temperatures in and around the raceway. The equilibrium constant for Reaction (1) is
Solid C-in-coke particles, 1500°C
Blast air: O2(g) + N2(g) 1200°C, 4 bar
Raceway
(D.1)
CO2 ðgÞ1CðsÞ-2COðgÞ
KE
5
2 aECO aECO2 aEC
(D.2)
where KE is the equilibrium constant, unitless and aE is the equilibrium thermodynamic activities of the reactants and product, unitless. The equilibrium activity of C(s) is 1 because it is a pure solid. The equilibrium activities of CO(g) and CO2(g) are
Raceway flame: CO(g) + N2(g), Tflame, 4 bar
E aECO 5 XCO Pt =1
FIGURE D.1 Tuyere raceway with dry blast air (only).
The inputs are 1200 C air and 1500 C falling C(s)-in-coke particles. The objective of this appendix is to show that the product carbonaceous gas is predominantly carbon monoxide.
and
681
E aECO2 5 XCO Pt =1 2
682
APPENDIX D: CO RACEWAY EXIT GAS PROOF
E E where XCO and XCO are the equilibrium mol fractions of CO(g) and CO2(g) in the raceway exit gas, Pt is the absolute pressure in the raceway B4 bar, and 1 is the pressure (bar) at which the thermodynamic activity of a pure ideal gas is 1. With these substitutions, the mol fractions of CO(g) and CO2(g) in the raceway exit gas are related to the equilibrium constant of Eq. (D.2) by the following equation: 2
CO2 ðgÞ1CðsÞ-2COðgÞ
KE
5
E 2 XCO P1t E XCO P1t 1 2
(D.3)
The value of this equilibrium constant at 2000 C is 1.2 3 105 (Appendix E).
Furthermore, because nitrogen does not react in the raceway: raceway exit gas
nN2
D.3 OXYGEN MOLAR BALANCE We now use the raceway’s steady-state oxygen molar balance to calculate kg mol of CO and CO2 in the raceway exit gas. These are then used to determine the raceway’s equilibrium exit gas CO and CO2 mol fractions (Eq. (D.3)). The oxygen molar balance is:
2
input
nO2
5 1 kg mol
(D.4)
All n values in this appendix are based on this 1 kg mol of O2 in blast air. The molar composition of dry air is B79 mol% N2 plus B21 mol% O2 (Appendix B), that is, each kg mol of air contains 0.79 kg mol of N2 and 0.21 kg mol of O2. This leads to: input
nN2
5
79 mol% N2 input nO2 21 mol% O2
5
79 1 kg mol input O2 5 3:76 kg mol input N2 21 (D.5)
5 3:76 kg mol N2 in raceway exit gas (D.6)
D.2 CO(g), CO2(g), AND N2(g) QUANTITIES AND MOL FRACTIONS IN RACEWAY EXIT GAS This section and the next show how we calculate CO and CO2 mol fractions of Eq. (D.1), that is, XCO and XCO . The calculations are begun by specifying that 1 kg mol of O2(g) enters the raceway in blast air, that is;
input
5 nN2
2 3 kg mol O kg mol O 6 7 5 4 in raceway 5 in input O2 exit gas CO 2 3 2 3 kg mol O kg mol O 6 7 6 7 1 4 in raceway 5 1 4 in raceway 5 exit gas O2 exit gas CO2
(D.7)
The equilibrium amount of O2 in raceway exit gas is very small (Appendix F) so that Eq. (D.7) simplifies to:
3 2 3 2 kg mol O kg mol O kg mol O 4 4 5 5 in raceway 1 in raceway 5 (D.8) in input O2 exit gas CO exit gas CO2
One kilogram mole of O2 contains 2 kg mol of O. One kilogram mole of CO2 also contains 2 kg mol of O. One kilogram mole of CO, however, contains only 1 kg mol of O so that: 2
3 kg mol O 4 in 1 kg mol 5 5 2 kg mol input O2 of input O2 2 3 2 3 kg mol CO kg mol O in 4 4 5 1 kg mol of 5 1 in raceway 5 exit gas raceway exit gas CO
683
APPENDIX D: CO RACEWAY EXIT GAS PROOF
and
raceway exit gas
ð2Þ nN2
2
3 2 3 kg mol CO2 kg mol O 4 5 4 in 1 kg mol of 5 2 in raceway 5 raceway exit gas CO2 exit gas
the total amount of raceway gas is; raceway exit gas
nT
from which Eq. (D.8) becomes; 2 3 2 3 kg mol CO kg mol CO2 kg mol 4 4 5 2 5 1 in raceway 1 2 in raceway 5 input O2 exit gas exit gas
kg mol and, because input is specified as 1 kg mol; O2 2
raceway exit gas
5 nCO2
2
3
raceway exit gas
nT
raceway exit gas
5 2 2 nCO2
1 3:76
raceway exit gas
(D.14)
so that; raceway exit gas
(D.10)
Þ 1 3:76
5 5:76 2 nCO2
3
kg mol CO kg mol CO2 2 1 5 1 4 in raceway 5 1 2 4 in raceway 5 exit gas exit gas
raceway exit gas
1 ð2 2 2 nCO2
or combining right side terms;
(D.9)
5 3:76
raceway exit gas
XCO2
5
nCO2
raceway exit gas nT
raceway exit gas
5
nCO2
raceway exit gas
5:76 nCO2
(D.15)
which we simplify to; raceway exit gas
2 5 1 nCO
or subtracting 22
Likewise; raceway exit gas
1 2 nCO2 o
n raceway exit gas 2 nCO2
raceway exit gas nCO2
51
(D.11)
from both sides;
raceway exit gas
raceway exit gas
raceway exit gas
XCO
5
nCO
raceway exit gas nT
5
2 2 nCO2 5:76
raceway exit gas nCO2
(D.16)
raceway exit gas nCO
or switching sides; raceway exit gas
nCO
raceway exit gas
5 2 2 nCO2
D.4 CALCULATING CO(g) AND CO2(g) MOL FRACTIONS FOR EQUILIBRIUM CONSTANT EQ. (D.3) The mol fraction of CO in the raceway exit gas is; raceway exit gas
raceway exit gas
XCO
5
D.5 EQUILIBRIUM MOLE FRACTIONS
(D.12)
nCO
raceway exit gas
nT
exit gas where nraceway is the total mol of exit gas, T that is;
Returning to Eq. (D.3); CO2 ðgÞ1CðsÞ-2COðgÞ
KE
5
and applying Eqs. (D.15) and (D.16) gives; 2 E XCO P1t g ð Þ CO2 ðgÞ1CðsÞ-2COðgÞ KE 5 1:2 3 105 5 Pt E XCO 1 1 2 ðgÞ 2 2 raceway exit gas raceway exit gas 22 nCO 22 nCO Pt 2 2 raceway exit gas 1 raceway exit gas 5:76nCO 5:76nCO P 5 raceway2 exit gas t (D.17) 5 raceway2 exit gas nCO nCO 1 P t 2 2 raceway exit gas 1 raceway exit gas 5:76nCO
2
raceway exit gas nT
raceway exit gas 5 nCO2
raceway exit gas 1 nCO
raceway exit gas 1 nN2
(D.13)
or because, by Equation D.12 and D.6: ð1Þ
raceway exit gas nCO
52 2
raceway exit gas nCO2
E 2 XCO P1t 5 1:2 3 105 Pt E 1 XCO 1 2
5:76nCO
2
from which, with 4 bar absolute pressure (Pt 5 4) in the raceway (Section D.1): raceway exit gas
nCO2
5 2:2 1025 kg mol CO2
684
APPENDIX D: CO RACEWAY EXIT GAS PROOF
and because; raceway exit gas
nCO
Another result of the calculations is that raceway exit gas
5 2 2 nCO2 5 B2 kg mol CO
So the carbonaceous raceway exit gas is virtually all CO. Additional calculations show that this true for all temperatures above 1500 C.
raceway exit gas
E XCO 5
nCO
nT
5
2 5 0:35 2 1 3:76
This value is used in Appendix G.