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 $\left( C _6 H _{6}\right)$ , and

(ii) Air and octane vapour $\left( C _8 H _{18}\right)$ .

Given that the enthalpies of combustion at 25°C are :

$C_6 H_6=-3169500$ kJ/mole

$C _8 H_{18}=-5116200$ kJ/mole

Both the above figures are for the case where the water in the products is in the vapour phase.

Question

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 [latex]\left( C _6 H _{6}\right)[/latex], and

(ii) Air and octane vapour [latex]\left( C _8 H _{18}\right)[/latex].

Given that the enthalpies of combustion at 25°C are :

[latex]C_6 H_6=-3169500[/latex] kJ/mole

[latex]C _8 H_{18}=-5116200[/latex] kJ/mole

Both the above figures are for the case where the water in the products is in the vapour phase.

Accepted Answer

(i) Air and benzene vapour :

For benzene, the combustion equation is as follows :

[latex]C _6 H _6(g)+7.5 O _2(g)+7.5\left(\frac{79}{21} ight) N _2(g)=6 CO _2(g)+3 H _2 O (g)+7.5\left(\frac{79}{21} ight) N _2(g)[/latex]

Since the water in the products is in vapour phase, therefore, the given value of enthalpy of combustion corresponds to the lower heating value at constant pressure.

i.e., [latex]( LHV )_p=3169500[/latex] kJ/mole

[latex]( L H V )_{ v } ext { per kg of mixture }=\frac{3169500}{(12 imes 6+6 imes 1)+(7.5 imes 32)+7.5\left(\frac{79}{21} ight)(28)}[/latex]

= [latex]\frac{3169500}{78+240+790}= 2 8 6 1[/latex] kJ/kg.

Now, [latex]( HHV )_p=( LHV )_p+m h_{f g}[/latex]

where [latex]( HHV )_p=[/latex] Higher heating value at constant pressure,

[latex]( LHV )_p=[/latex] Lower heating value at constant pressure,

m = Mass of water formed by combustion.

= 3 × 18 = 54 kg/kg mole of fuel, and

[latex]h_{f g}=[/latex] Latent heat of vapourisation at given temperature per unit mass of water

= 2442 kJ/kg at 25°C.

∴ [latex]( HHV )_p=3169500+54 imes 2442=3301368[/latex] kJ/mole

Thus, [latex]( H H V )_p[/latex] per kg of mixture [latex]=\frac{3301368}{78+240+790}= 2 9 8 0[/latex] kJ/kg.

(ii) Air and octane vapour :

[latex]( LHV )_p=5116200[/latex] kJ/mole of [latex]C _8 H _{18}[/latex]

For octane, the combustion equation is written as follows :

[latex]C _8 H _{18}(g)+12.5 O _2(g) \longrightarrow 8 CO _2(g)+9 H _2 O (g)+12.5\left(\frac{79}{21} ight) N _2(g)[/latex]

[latex]( L H V )_{ p }[/latex] per kg of mixture [latex]=\frac{5116200}{(12 imes 8+18 imes 1)+12.5 imes 32+12.5 imes \frac{79}{21} imes 28}[/latex]

= [latex]\frac{5116200}{114+400+1317}=2794[/latex] kJ/kg.

[latex]( HHV )_p=( LHV )_p+m h_{f g}[/latex]

m = 9 × 18 = 162 kJ/kg mole of fuel

∴ [latex]( HHV )_p=5116200+162 imes 2442=5511804[/latex]

Hence, [latex]( H H V )_p[/latex] per kg of mixture [latex]=\frac{5511804}{114+400+1317}= 3 0 1 0[/latex] kJ/kg.

Question 11.36: Calculate the lower and higher heating values at constant pr...

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 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:

The exhaust from an engine running on benzole was measured with the help of Orsat apparatus. Orsat analysis showed a CO2 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 ...

Verified Answer: