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 O₂ required and CO₂ produced can be calculated as  shown in table 20.1    

Table 20.1

Fuel Element	Mass (kg)	O2 (Kg)	Combustion product (Kg)
C	0.86 Kg
H2	0.14 kg	0.14 × 8 = 1.12	0.14 × 9 H2O

Total O₂= 3.41 Kg

O₂ supplied in actual is 90% = 3.41 × 0.9

= 3.069 kg

As O₂ is less, so CO will also form

Let 𝓍 part of total carbon content changes to CO

And (1-𝓍) part of total carbon contents converts to CO₂

Hence for incomplete combustion the tabular form of mass of elements is given as in table 20.2

Table 20.2

Element Mass (kg)	O2 needed (kg)	Combustion product (kg)
C = 0.86 𝓍
C = 0.86 (1- 𝓍)
H2 = 0.14		9 × 0.14 of H2O

Total O₂ 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
CO2	2.207	44
N2	10.274	28

20.1.2 The percentage composition by mass of a sample of coal is C = 90, H₂ = 3.5, O₂ = 3.0, N₂ = 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)	O2 Required in (kg)	Dry Products (kg)
C = 0.9
H2 = 0.035	0.035 × 8 = 0.28	0.0359 × 9 =.315 H2O
O2 = 0.03	= -0.03	-
N2 = 0.01	= zero	0.01 N2
S = 0.005	0.005 × 1 = 0.005	0.005 × 2 =.01      SO2

Total O₂ = 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 N₂ and O₂ will increase in the products of combustion.

Total mass of N₂ in dry product of combustion

                     = N₂ present in air supplied +N₂ present in fuel         

Total mass of O₂ in dry product of combustion

                     = wt of O₂ in excess air supplied                       

Total wt of CO₂ is as already calculated in task 1 = 3.3 kg

Total wt of SO₂ 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
CO2	3.3	44
N2	13.33	28	0.4761	80.33
O2	1.327	32	0.0414	6.98
SO2	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%, H₂ = 4.5%, O₂ = 4.0% and the rest non-combustibles. 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^oC. When the ambient temperature in boiler house was 32.2^oC. 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)	O2 Required (kg)	Products Formed (kg)
C = 0.84
H2 = 0.045	0.045 × 8 = 0.36	0.0450.84 × 9 = .405   H2O
O2 = 0.04	= -0.04
	Total O2 = 2.56kg

Actual air supplied (85% excess) = 11.13 × 1.85

= 120.6 kg                  (Ans)

(b) (1) Mass of N₂ in products of combustion = Mass of N₂ in the air supplied

= 15.86 kg

(2)  Mass of O₂ in combustion product = Mass of O₂ in excess air

= 2.18 kg

 Total mass of flue gases = CO₂ +N₂ + O₂ + H₂O

                                        = 3.08 + 15.86 + 2.18 + 0.405 H₂O (Steam)

                                        = 21.12 kg + 0.405 H₂O

                                        = 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 C₁₀H₂₂ 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 CO₂ 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

C₁₀H₂₂ + O₂ = CO₂ + H₂O

After balancing the equation it becomes

 2 C₁₀H₂₂ + 31 O₂ = 20 CO₂ + 22 H₂O

 2 × 142 + 31 × 32 = 20 × 44 + 22 × 18

From this equation we know that for complete combustion of one kg of C₁₀H₂₂, requirement of O₂ 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 (15.19-13) 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 CO₂.

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 C₁₀H₂₂, amount of carbon and amount of hydrogen in one Kg of fuel will be

So Amount of Carbon converting to CO₂ will be

= Total Carbon − carbon converting to CO

= 0.845 − 0.378

= 0.467 kg

Thus mass of CO₂ 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
CO2	1.712	44
CO	0.882	28
H2O	1.395	18
N2	10.01	28
			Total=0.5055	= (Ans)