FURNACE
Mol fr.
O
20.21 N
20.79 20 Kg C
3H
8400 kg Air
CO
2CO H
2O
Kg 44 12
?
Example on incomplete combustion:
20 Kg of Propane (C 3 H 8 ) is burned with 400 kg of air to produce 44 kg of CO 2 and 6 Kg of CO. What was the % excess air ?
C
3H
8+ 5 O
2 3 CO
2+ 4 H
2O ( the complete combustion rxn )
C
3H
8+ 3.5 O
2 3 CO + 4 H
2O ( the incomplete combustion rxn)
O2 theoretical = 0.454 kgmol C3H8
5
kgmol O2 = 2.27 kg mol O2 1 kgmol C3H8O2 entering = 13.36 kgmol air 0.21 kgmol O2 = 2.90 kgmol O2 Kgmol air
% excess air = 100 x O2 entering O2 theoretical O2 theoretical
= 100 x 2.90 kgmol O2 2.27 kgmol O2 = 28%
2.27 kgmol O2
Remember : the basis of the calculation of excess air is complete combustion
Change to kg moles : C 3 H 8 is 0.454 , air is 13.36 CO 2 is 1.00 and CO is 0.215
(Molecular weights in Kg/kgmol: 44, 29, 44, 28 respectively)
Example : 20 Kg of Propane (C 3 H 8 ) is burned incompletely with 200% excess air to produce CO 2 and 14 Kg of CO. What is the analysis of the stack gas produced?
Figure : same as above with different amounts of flows.
Unknown variables: n CO2,P , n H2O,P n N2,P , n O2,P
•Atomic Balances : C, H, O
•Component Balances: N 2 (N 2 is not reacting),
• and total mass balance
Example : 20 Kg of Propane (C 3 H 8 ) is burned
incompletely with 200% excess air to produce CO 2 and 14 Kg of CO. What is the analysis of the stack gas
produced?
Unknown variables: n
CO2,P, n
H2O,Pn
N2,P, n
O2,P•Atomic Balances : C, H, O
•Molecular or component Balances: N
2(N
2is not reacting),
• and total mass balance
FURNACE
Mol fr.
O
20.21 N
20.79 20 Kg C
3H
8200% excess Air
CO
2CO H
2O O
2N
2Kg
? 14 ? ? ?
P
Furnace 20 kg C 3 H 8
200% excess air mol fr O 2 0.21 N 2 0.79
CO 2 ? CO 14 kg H 2 O ? N 2 ? O 2 ?
P
Solution 1 ( using the stoichiometric equations)
Step1: Calculate O
2entering and N
2entering
O
2theoretical = 0.454 kgmol C
3H
85 kgmol O
2= 2.27 kg mol O
21 kgmol C
3H
8O
2entering = 2.27 kgmol O
2+ (200/100) 2.27 kgmol O
2or = 2.27 (1+200/100) kgmol O
2= 2.27 kgmol O
2(1 + 2) = 6.81 kgmol O
2N
2entering =O
2entering x 0.79/0.21 = 25.62 kgmol =
N
2Out
Step2 : calculate from stoichiometry CO
2, H
2O and O
2out CO out = 14 kg/28 = 0.5 kgmol
O
2used up to produce CO = 0.5 kgmolCO 3.5 kgmol O
2= 0.583 kgmol 3 Kgmol CO
C
3H
8reacted to produce CO = 0.5 kgmolCO 1 kgmol C
3H
8= 0.167 kgmol 3 Kgmol CO
C
3H
8reacted to produce CO
2= 0.454 – 0.167 = 0.287 kgmol
C
3H
8+ 5 O
2 3 CO
2+ 4 H
2O ( the complete combustion rxn )
C
3H
8+ 3.5 O
2 3 CO + 4 H
2O ( the partial/incomplete
combustion rxn)
O2 reacted to produce CO2 =
0.287 kgmol C
3H
85 kgmol O
2= 1.435 kgmol 1Kgmol C
3H
8O
2Out = O
2in – O
2that have reacted in both reactions = 6.81 – (1.435 + 0.583) = 4.792 kgmol
CO
2Out = 0.287 x 3 = 0.861 kgmol
1
H
2O Out = 0.287 x 4 + 0.167 x 4 = 1.816 kgmol 1 1
C
3H
8+ 5 O
2 3 CO
2+ 4 H
2O ( the complete combustion rxn )
C
3H
8+3.5 O
2 3 CO + 4 H
2O ( the incomplete combustion rxn )
Component Kg moles % kg mol
CO 2 0.861 2.6
CO 0.500 1.5
H 2 O 1.816 5.4
N 2 25.62 76.3
O 2 4.792 14.3
Total 33.634 100.0
Step3 : Calculate the analysis of the stack gas
Solution 2 Using atomic balances Step1: Same as step 1 above
Step 2 : C, H, O Atomic balances in moles ( note: we do not use the chemical equations)
Given : CO out = 14 kg/28 = 0.5 kgmol
IN = OUT C balance 3 x 0.454 + 0 = 1 x n
CO2+ 1 x n
COH balance 8 x 0.454 + 0 = 2 x n
H2OO balance 0 +2
x6.81 = 2
xn
CO2+ 1
xn
CO+ 1
xn
H2O
+ 2
xn
O2Solve : n
H2O= 1.816 kgmol ; n
CO2= 1.362 – 0.5 = 0.862 kgmol
And nO2 =13.62 – 2 (0.861) – 0.5 – (1.816) = 4.791 kgmol
Step 3: same as step 3 above
Questions:
• When can we use any of the 2 solutions above to solve?
• When can we only use solution 2 to
solve? If the stoichiometric equations are not given
