Module 8. Fuels
Lesson 20
NUMERICAL PROBLEMS
20.1 Numerical Problems
20.1.1 A petrol sample was found to have 86% carbon and 14% hydrogen by mass. When used in an engine the air supply is 90% of that theoretically required for complete combustion. Assuming that all the hydrogen is burnt and that the carbon burns to carbon monoxide and carbon dioxide so that there is no free carbon left, calculate the percentage analysis of dry exhaust gases by volume.
Solution
For complete combustion of 1 kg of Fuel
Amount of O_{2} required and CO_{2} produced can be calculated as shown in table 20.1
Table 20.1
Fuel Element 
Mass (kg) 
O_{2} (Kg) 
Combustion product (Kg) 
C 
0.86 Kg 


H_{2} 
0.14 kg 
0.14 ×
8 = 1.12 
0.14 ×
9 H_{2}O 
Total O_{2}= 3.41 Kg
O_{2} supplied in actual is 90% = 3.41 × 0.9
= 3.069 kg
As O_{2} is less, so CO will also form
Let 𝓍 part of total carbon content changes to CO
And (1𝓍) part of total carbon contents converts to CO_{2}
Hence for incomplete combustion the tabular form of mass of elements is given as in table 20.2
Table 20.2
Element Mass (kg) 
O_{2} needed (kg) 
Combustion product (kg) 
C
= 0.86
𝓍 


C = 0.86 (1 𝓍) 


H_{2} = 0.14 

9 × 0.14 of H_{2}O 
Total O_{2} required for incomplete combustion
Or
Putting the value of 𝓍
Table 20.3 Conversion of mass analysis to volume analysis
Element 
Mass ( kg) 
Molecular mass (M) 
Relative volume 
Percentage Volume

CO 
0.6 
28 


CO_{2} 
2.207 
44 


N_{2} 
10.274 
28 


20.1.2 The percentage composition by mass of a sample of coal is C = 90, H_{2} = 3.5, O_{2} = 3.0, N_{2} = 1.0, S = 0.5, the remaining being ash. Estimate the minimum mass of air required for the complete combustion of 1 kg of this fuel and the composition of dry products of composition, by volume, if 50% excess air is supplied.
Solution
For complete combustion of one kg of fuel, the amount of oxygen required and amount of products of combustion formed are calculated as shown in table 20.4
Table 20.4
Element Mass. (kg) 
O_{2} Required in (kg) 
Dry Products (kg) 
C = 0.9 


H_{2} = 0.035 
0.035 × 8 = 0.28 
0.0359 × 9 =.315 H_{2}O 
O_{2 }= 0.03 
= 0.03 
 
N_{2} = 0.01 
= zero 
0.01 N_{2} 
S = 0.005 
0.005 × 1 = 0.005 
0.005 × 2 =.01 SO_{2} 
Total O_{2} = 2.655kg
(a)
So
minimum mass of air required for complete combustion
(b)Actual air supplied = 1.5 × 11.54 = 17.3 kg
Due to excess air, quantity of N_{2} and O_{2} will increase in the products of combustion.
Total mass of N_{2} in dry product of combustion
= N_{2} present in air supplied +N_{2} present in
fuel
Total mass of O_{2} in dry product of combustion
= wt of O_{2} in excess air supplied
Total wt of CO_{2 }is as already calculated in task_{ }1 = 3.3 kg
Total wt of SO_{2 }is as already calculated in task_{ }1 = 0.01 kg
(ii) The percentage composition by volume can be found out as follows:
Table 20.5 Conversion of mass analysis to volume analysis

Mass

Mol wt

Relative volume

Percentage volume

CO_{2} 
3.3 
44 


N_{2} 
13.33 
28 
0.4761 
80.33 
O_{2} 
1.327 
32 
0.0414 
6.98 
SO_{2} 
0.01 
64 
0.000156 
0.0002 
(Ans)
20.1.3 A boiler in a power station was fired with coal having an ultimate composition by mass of C = 84%, H_{2} = 4.5%, O_{2} = 4.0% and the rest noncombustibles. The calorific value of coal is 33900 kJ/kg. The coal was burnt with 85% excess air and mean temperature of flue gases leaving the boiler was found to be at 315.5^{o}C. When the ambient temperature in boiler house was 32.2^{o}C. Assuming specific heat of flue gases as 1.005 kJ/kgK determine:
(i) kg of air actually supplied per kg coal burnt
(ii) kg of flue gases produced per kg fuel burnt
(iii) The percentage of energy of coal that is carried away by flue gases per kg of coal burnt.
Sol:
(a) Theoretical air required for complete combustion of 1kg of coal is calculated as shown in table 20.6
Table 20.6
Element mass (kg) 
O_{2} Required (kg) 
Products Formed (kg) 
C = 0.84 


H_{2} = 0.045 
0.045 × 8 = 0.36 
0.0450.84 × 9 = .405 H_{2}O 
O_{2} = 0.04 
= 0.04 


Total O_{2} = 2.56kg 

Actual
air supplied (85% excess) = 11.13 × 1.85
= 120.6 kg (Ans)
(b) (1) Mass of N_{2} in products of combustion = Mass of N_{2} in the air supplied
= 15.86 kg
(2) Mass of O_{2} in combustion product = Mass of O_{2} in excess air
= 2.18 kg
Total mass of flue gases = CO_{2} +N_{2} + O_{2 }+ H_{2}O
= 3.08 + 15.86 + 2.18 + 0.405 H_{2}O (Steam)
= 21.12 kg + 0.405 H_{2}O
= 21.525 kg
However mass of dry flue gases=21.12 kg
(3) Heat taken away by flue gases per kg of coal
= Mass of dry products × Sp. Heat × Rise in temperature
= 21.12 × 1.005 × (315.5 32.2)
= 6010 kJ
Total heat produced by 1 kg of coal = 33900 KJ
= 17.73 % (Ans)
20.1.4 A fuel with composition C_{10}H_{22} is burnt using an air fuel ration of 13:1 by mass. Determine the complete volumetric analysis of the products of combustion, assuming that the whole amount of hydrogen burns to form water vapour and there is neither any free oxygen nor any free carbon left. The carbon is burning to CO_{2} and CO. Air contains 77% of Nitrogen and 23% oxygen by mass.
Solution
To calculate the chemically correct amount of air required, the chemical reaction is given as
C_{10}H_{22 }+ O_{2 }= CO_{2 }+ H_{2}O
After balancing the equation it becomes
2 C_{10}H_{22 }+ 31 O_{2 }= 20 CO_{2 }+ 22 H_{2}O
2 × 142 + 31 × 32 = 20 × 44 + 22 × 18
From this
equation we know that for complete combustion of one kg of C_{10}H_{22},
requirement of O_{2} will be
Stoichiometric
quantity of air will be
But Actual quantity of air supplied=13 kg/ kg of fuel.
Thus air supplied is less than that required and this less quantity is equal to (15.1913) kg per kg of fuel.
It is given that hydrogen burn completely and only some of the carbon remains deficient of oxygen and so converts to CO.
1 Kg of C requires Kg of Oxygen to produce CO_{2}.
And 1 Kg of C requires Kg of Oxygen to produce CO.
Thus for incomplete combustion to CO, one kg carbon requires only half of oxygen or kg less oxygen.
So amount of carbon converting to CO due to 0.5307 kg of less oxygen.
From the formula C_{10}H_{22}, amount of carbon and amount of hydrogen in one Kg of fuel will be
So Amount of Carbon converting to CO_{2} will be
= Total Carbon − carbon converting to CO
= 0.845 − 0.378
= 0.467 kg
Thus mass of CO_{2} formed = 0.467 = 1.712 kg
Mass of CO formed = 0.378 = 0.882 kg
Mass of water formed = 0.155 × 9 = 1.395 kg
Mass of Nitrogen formed = 13 = 10.01 kg
Table 20.7 Volumetric analysis of products of combustion
Element 
Element wt kg 
Molecular wt 
Relative volume 
Percentage Volume 
CO_{2} 
1.712 
44 


CO 
0.882 
28 


H_{2}O 
1.395 
18 


N_{2} 
10.01 
28 





Total=0.5055 
= (Ans) 