The exhaust from an engine running on benzole was measured with the help of Orsat apparatus. Orsat analysis showed a $CO _2$ content of 12%, but no CO. Assuming that the remainder of the exhaust contains only oxygen and nitrogen, calculate the air-fuel ratio of the engine.

The ultimate analysis of benzole is C = 88% and $H_2=12 \%$ .

Question

The exhaust from an engine running on benzole was measured with the help of Orsat apparatus. Orsat analysis showed a $CO _2$ content of 12%, but no CO. Assuming that the remainder of the exhaust contains only oxygen and nitrogen, calculate the air-fuel ratio of the engine.

The ultimate analysis of benzole is C = 88% and $H_2=12 \%$ .

The ultimate analysis of benzole is C = 88% and [latex]H_2=12 \%[/latex].

Accepted Answer

1 kg of fuel, consisting of 0.88 kg C and 0.12 kg [latex]H_2[/latex], can be written as 0.88/2 moles C and 0.12/2 moles [latex]H_2[/latex]. Therefore, considering 1 mole of dry exhaust gas (D.E.G.) we can write the combustion equation as follows :

[latex]X\left(\frac{0.88}{12} C +\frac{0.12}{2} H _2 ight)+Y O _2+\frac{79}{21} Y N _2 \longrightarrow 0.12 CO _2+a O _2+(0.88-a) N _2+b H _2 O[/latex]

[latex]\left[\begin{array}{l} ext { Let the D. E.G. contain } a ext { moles of } O _2 ext {. The moles of } CO _2 ext { in } 1 ext { mole of D. E.G. are } 0.12 . \ ext { Therefore the D. E.G. contains }(1-a-0.12)=(0.88-a) ext { moles of } N _2 ext {. }\end{array} ight][/latex]

where, X = Mass of fuel per mole D.E.G.,

Y = Moles of [latex]\mathrm{O}_2[/latex] per mole D.E.G.,

a = Moles of excess [latex]\mathrm{O}_2[/latex] per mole D.E.G., and

b = Moles of [latex]H _2 O[/latex] per mole D.E.G.

Now,

Carbon balance : [latex]\frac{0.88}{12} X=0.12[/latex] ∴ X = 1.636

Hydrogen balance : 0.06X = b ∴ b = 0.06 × 1.636 = 0.098

Oxygen balance : 2Y = 2 × 0.12 + 2a + b

or 2Y = 0.24 + 2a + 0.098 ∴ Y = 0.169 + a

Nitrogen balance : [latex]\frac{79}{21} Y=(0.88-a)[/latex] ∴ Y = 0.234 – 0.266a

Equating the expressions for Y gives

0.234 – 0.266a = 0.169 + a ∴ a = 0.0513

i.e., Y = 0.169 + 0.0513 = 0.2203

∴ [latex]\mathrm{O}_2[/latex] supplied = 0.2203 × 32 kg/mole D.E.G.

i.e., Air supplied [latex]=\frac{0.2203 imes 32}{0.233}=30.26[/latex] kg/mole D.E.G.

Since X = 1.636, then, the fuel supplied per mole D.E.G. is 1.636 kg

∴ A/F ratio [latex]=\frac{30.26}{1.636}=18.5 / 1[/latex].

Question 11.29: The exhaust from an engine running on benzole was measured w...

Related Answered Questions

Verified Answer:

The following is the analysis (by weight) of a chemical fuel : Carbon = 60 per cent ; Hydrogen = 20 per cent ; Oxygen = 5 per cent ; Sulphur = 5 per cent and Nitrogen = 10 per cent. Find the stoichiometric amount of air required for complete combustion of this fuel. ...

Verified Answer:

The higher heating value of kerosene at constant volume whose ultimate analysis is 88% and 12% hydrogen, was found to be 45670 kJ/kg. Calculate the other three heating values. ...

Verified Answer:

Calculate the lower and higher heating values at constant pressure per kg of mixture at 25°C, for the stoichiometric mixtures of : (i) Air and benzene vapour (C6H6), and (ii) Air and octane vapour (C8H18). Given that the enthalpies of combustion at 25°C are : C6H6 = – 3169500 kJ/mole ...

Verified Answer:

Calculate the lower heating value of gaseous octane at constant volume if (ΔU)25°C = – 5494977 kJ for the reaction : [C8H18 (gas) + 12.5O2 (gas)] = [8CO2 (gas) + 9H2O (liquid)]. ...

Verified Answer:

ΔH0 for CO at 60°C is given as – 285200 kJ/mole. Calculate ΔH0 at 2500°C given that the enthalpies of gases concerned in kJ/mole are as follows : ...

Verified Answer:

The lower heating value of propane at constant pressure and 25°C is 2044009 kJ per kg mole. Find the higher heating value at constant pressure and at constant volume. ...

Verified Answer:

Calculate ΔU0 in kJ/kg for the combustion of benzene (C6H6) vapour at 25°C given that ΔH0 = – 3169100 kJ/mole and the H2O is in the vapour phase. ...

Verified Answer:

ΔH0 (enthalpy of combustion at reference temperature T0) for benzene vapour (C6H6) at 25°C is – 3301000 kJ/mole with the H2O in the liquid phase. Calculate ΔH0 for the H2O in the vapour phase. ...

Verified Answer:

The analysis of the dry exhaust from an internal combustion engine is as follows : Carbon dioxide (CO2) = 15 per cent Carbon monoxide (CO) = 3 per cent Methane (CH4) = 3 per cent Hydrogen (H2) = 1 per cent Oxygen (O2) = 2 per cent Nitrogen (N2) = 76 per cent Calculate the proportions by mass of ...

Verified Answer: