Combustion Calculations

1. Combustion Calculations

2. P-1
The flue gas from an industrial furnace has the following
composition by volume
CO2: 11.73%, CO: 0.2 %, SO2: 0.09%, O2: 6.81% and
N2: 81.17 %
Calculate the percentage excess air employed in the
combustion if the loss of carbon in the clinker ash is 1%
of the fuel used and the fuel has the following
composition by weight:
C: 74%, H2: 5 %, O2: 5%, N2: 1%, S: 1%, H2O: 9%, ash
: 5%
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3. P-1
Basis: 100 kg of the fuel
C + O2 => CO2
H2 +1/2 O2 => H2O
S + O2 => SO2
Oxygen Balance:
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• Oxygen required for complete combustion
=74/12 + 5/2 X 1/2 + 1/32=7.447 kg mole
Oxygen present in fuel=5/32=0.157 kg mole
Net oxygen from air= 7.447-0.157=7.29 kg mole

4. P-1
• Carbon balance
Carbon lost in clinker and ash = 1 kg
Carbon burnt= 74-1= 73 kg = 73/12=6.08 kg
Assume z kg moles of flue gas are formed
By carbon balance (0.1173+0.002)*Z=6.08
z=50.96 kg moles
atom
So
N2
N2
N2 in flue gas= 50.96 *0.8117=41.36
from fuel= 1 kg=0.036 kg moles
from air = 41.36-0.036=41.324
Oxygen from air =41.324 * 21/79=10.98 kg
Excess oxygen = 10.98-7.29= 3.69
mole

5. P-1
Excess oxygen = 10.98-7.29= 3.69 kg mole
Percentage excess air used = percentage excess oxygen
used
= excess/theoretical* 100= 3.69/7.29* 100= 50.62

6. P-2
is burnt with 10 % excess
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Octane air. Calculate
Air to fuel
Air to fuel
Weight of
ratio by weight
ratio by volume
dry exhaust gas formed per unit weight of fuel
Moles of oxygen in the exhaust gas per unit weight of fuel
Moles of water vapour in exhaust gas per unit weight of
fuel
Volume of exhaust gas at 1 atmosphere and 260 C per
unit weight of fuel
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The specific gravity of octane may be taken as 0.7

7. P-3
A producer gas with the composition by volume CO: 27.3 %, CO2:5.4%,
O2: 0.6 %, N2 : 66.7% is burnt with 20 % excess air. If the combustion
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is 98% complete, calculate the composition by volume of
Solution:
Basis: 100 kg mole of producer gas burnt
the flue gases
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Oxygen balance:
O2 required for CO combustion= 27.3*0.5=13.65 kg mole
O2 present in fuel = 0.6 kg mole
Net O2 required = 13.65 – 0.6 = 13.05 kg mole
O2
O2
O2
supplied by 20 % excess air = 13.05*1.20= 15.66 kg mole
used for 98% combustion of CO=13.05*0.98=12.8 kg mole
excess = 15.66 -12.8 =2.86 kg mole

8. P-3
Nitrogen balance:
N2 from air = 15.66 * 79/21= 58.91 kg mole
N2 from producer gas = 66.7 kg mole
Total N2 in flue gas = 66.70 + 58.91=125.61 kg mole
CO2
CO2
CO2
balance:
from producer gas 5.4 kg mole
from combustion of CO = 27.3 *0.98= 26.75 kg mole
Total CO2 in flue gas = 5.4 + 26.75 = 32.15 kg mole
CO
CO
CO
balance:
burnt = 26.75 kg mole
left = 27.3 – 26.75 = 0.55 kg mole

9. P-3
Flue gas analysis
Constituent Kg mole Mole %
CO2 32.15 20
CO 0.55 0.34
N2 125.61 78
O2 2.86 1.77
Total 161.17 100

10. P-4
A furnace is fired with a natural gas that consists
entirely of hydrocarbons( no inert or sulphur). The
analysis of the flue gas is 9.5 % CO2, 2.0% O2
and 1.8% CO
What is the molar ratio of net hydrogen to carbon
in the fuel?
What percentage of excess air is being used?
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11. P-4
• Solution:
Basis: 100 kg of dry flue gas
N2 =100 –(9.5+1.8+2.0) =86.7
Oxygen balance:
O2
O2
O2
supplied by air = 86.7 *21/79 =23.05 kg mole
in dry flue gas= 9.5+1.8/2 +2 =12.4 kg mole
unaccounted ( reacted with H2)= 23.05-12.4
=10.65 kg mole
Moles of H2 reacted = 10.65 * 2 =21.3
Amount of carbon = 9.5 + 1.8 = 11.3 kg atoms
Moles of H2/Moles of C= 21.3/11.3 =1.18
Moles of O2 required for complete combustion=moles
=10.65 + 11.3=21.95
Amount of excess O2 = 23.05 -21.95=1.1 kg mole
% excess air = 1.1/21.95*100 = 5
for H2 +moles for C

12. P-5
The exhaust gas from a hydrocarbon fuel oil fired
furnace show 10.2% CO2, 7.9% O2 and 81.9%
N2. Calculate
%excess air used
Kg of dry air supplied per kg of oil burnt
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13. P-6
Determine the flue gas analysis and air/fuel ratio
by weight when a fuel oil with 84.9% carbon,
11.4% hydrogen, 3.2% sulphur, 0.4% oxygen and
0.1% ash by weight is burnt with 20 % excess air.
Assume complete combustion.
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14. P-7
A boiler is fired using 200 kg/hr of a pure saturated
hydrocarbon gas CnHm at atmospheric pressure and 20
C. The dry analysis of the flue gas which leaves the boiler
at atmospheric pressure and 300 C is CO2: 12%, O2: 3%
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and N2 :85%. Estimate the formula of the fuel and total
volumetric flow rate of the gas
Solution:
Basis: 100 kg mole of dry flue gas
Oxygen Balance
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N2 in flue gas = 85 kg mole
O2 supplied by air = 85 *21/79=22.59 kg mole

15. P-7
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O2 reported in
flue gas = O2 in CO2 + O2 as O2
=12 + 3.0 =15.0 kg mole
O2 unaccounted (reacted with H2) =22.59 – 15.0
= 7.59 kg mole
H2 reacted = 7.59 * 2=15.18 kg mole
= 30.36 kg atoms
Amount of carbon = 12 kg atoms
Amount of hydrogen=30.36 kg atoms
Ratio=Atoms of H/Atoms of C=30.36/12=2.53
Paraffin formula=CnH2n+2
(2n+2)/n=2.53
So n =3.77 ~ 4
Hence the fuel is C4H10
Volumetric flow rate =?
Amount of fuel = 200/58 3.45 kg mole
Assuming ideal gas law, volume at 20 C and 1 atmosphere is :
V=nRT/P=3.45*0.08206*293/1 = 82.95 m3/hr

16. P-8
A furnace is fired with a fuel having the volumetric
composition H2: 52%, CH4: 30 %, CO: 8%, C3H6: 3.6%,
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CO2: 2%, O2: 0.4% and rest N2.
• Using a certain quantity of air in excess over stoichiometric.
Complete combustion of the gas is achieved giving a dry
waste gas of 5 m3 per m3 of fuel burned. Estimate
Composition by volume of dry waste gas formed
Per cent excess air used
Weight of water formed per m3 of gas burned
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17. P-8
Solution:
Basis: 100 kg mole of fuel gas
CH4 + 2 O2 => CO2 + 2H2O
CO + 1/2O2 => CO2
C3H6 + 4.5 O2 => 3 CO2 + 3H2O
H2 + ½ O2 => H2O
Oxygen balance
O2
O2
O2
O2
required
required
required
required
for
for
for
for
CO=8*1/2=4 kg mole
CH4=30*2=60 kg mole
C3H6=3.6*4.5=16.2 kg mole
H2 = 52*1/2=26 kg mole
Total O2 required =106.2 kg mole
O2 in fuel =0.4 kg mole

18. P-8
Net O2 required from air =106.2 -0.4 =105.8 kg mole
DRY Flue Gas formed with theoretical air:
CO2 formed=8.0(from CO) + 30.0 (from CH4) + 10.8
C3H6) +2.0(present in fuel)
=50.8 kg mole
N2 from air = 105.8 * 79/21= 398 kg mole
Total N2 = 398 + 4 (present in fuel) = 402 kg mole
(from
Total amount of dry flue gas (with theoretical air) =402+50.8
=452.8 kg mole
Flue gas actually produced=5 m3/m3 of fuel
Total dry flue gas produced=100*5 =500 kg mole
Excess air=500 – 452.8 =47.2 kg mole
Theoretical air =105.8 *100/21=503.81 kg mole

19. P-8
Theoretical air =105.8 *100/21=503.81 kg mole
a)
b)
Air
N2
% excess air = 47.2/503.8*100=9.39
Composition of flue gas:
used = 503.81 + 47.2 = 551.01 kg mole
from air = 551.01 *79/100 = 435.29 kg mole
Total N2 = 435.29 + 4.0 =439.29 kg mole
Excess O2 = 47.2 *21/100 =9.91 kg mole
Flue gas analysis: constituent Amount kg
mole
Vol%
CO2
O2
N2
50.8
9.91
439.29
10.16
1.98
87.86
total 500 100

20. P-8
(C) Amount of water formed = 52.0 (from H2) + 60 (from
CH4) + 10.8 ( from C3H6) = 122.8 kg mole = 122.8 *18
=2210.4 kg
Amount of gas burned=100 kg mole=22.4*100 m3 at NTP
Weight of water formed/m3 gas burned=2210.4 kg /2240 m3
=0.987 kg of water/m3 of gas
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21. P-9
The dry flue gas from an oil fired furnace has the
composition of 11.2 % CO2, 5.8% O2, 83% N2. Calculate
% excess air
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Weight of combustion air
Assume fuel has 82% C,
impurities.
Solution:
Basis: 100 kg of oil fired
Oxygen balance:
per kg of oil fired
12 % H2, 3 % S and balance is
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O2
O2
O2
required
required
required
for
for
for
82 kg of carbon=82/12=6.833 kg mole
12 kg of H2 =12/2*0.5=3 kg mole
3 kg of S =0.094 kg mole
total O2 required = 6.833 + 3.0 + 0.094 =9.927 kg mole

22. P-9
Let z kg mole of dry flue gas is formed
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Carbon balance: 0.112* z (out) =6.833 (in)
So z =61.0 kg mole
Amount of N2 in flue gas = 61 * 0.83 = 50.83 kg mole
O2 from air = 50.83 *21/79 =13.46 kg mole
O2 excess = 13.46 – 9.927 =3.533 kg mole
% excess air =% excess O2 = 3.533/9.927*100=35.6%
Amount of combustion air = 50.63*100/79=64.09 kg mole
Molecular weight of air =28.84
Amount of air in kg =64.09*28.84=1848.36 kg
Mass of air in kg/mass of fuel in kg=1848.36/100=18.48

23. P-10
A fuel gas containing 97% methane and 3% N2 by volume
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is burned in boiler furnace with 200% excess air, 85% of
methane goes to CO2, 10% to CO and
unburnt.
Calculate the composition of stack gas
5% remains
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Solution:
Basis: 100 kg mole of fuel gas
CH4 + 2 O2 => CO2 + 2H2O ….. (1)
2 CH4 + 3 O2 => 2 CO + 4 H2O …..(2)
O2 required for complete combustion= 97*2=194 kg mole
O2 supplied by 200 % excess air = 3*194 = 582 kg mole
Reaction (1) is 85% complete and (2) is 10% complete

24. P-10
Methane converted by reaction (1) = 97 *0.85=82.45 kg mole
Methane converted by reaction (2) = 97*0.1=9.7 kg mole
Oxygen used in reaction (1) & 2= 82.5*2+9.7*3/2=179.45 kg mole
stack gas:
CO2 formed = 82.45 kg mole
CO formed = 9.7 kg mole
CH4 unconverted = 5 kg mole
N2 from air = 582 * 79/21=2189.43 kg mole
N2 from fuel = 3.0 kg mole
Total N2 in flue gas =2189.43 + 3.0=2192.43 kg mole
H2O formed = 82.25*2(by reaction 1) + 9.7*4/2(by reaction 2)
=184.3 kg mole
Excess oxygen = 582 – 179.45 = 402.55 kg mole

25. P-10
stack gas analysis
constituent kg mole Mole %
CO2 82.45 2.87
CO 9.70 0.34
CH4 5 0.17
O2 402.55 13.99
N2 2192.43 76.22
H2O 184.30 6.41
Total 2676.28 100