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.

Question

Accepted Answer

Combustion of 1 kg of fuel produces the following products :

[latex]CO _2=\frac{44}{12} imes 0.88=3.23[/latex] kg

[latex]H _2 O =\frac{18}{2} imes 0.12=1.08[/latex] kg

At 25°C : [latex]\left(u_g-u_f ight)[/latex] i.e., [latex]u_{f g}=2304[/latex] kJ/kg

[latex]h_{f g}=2442[/latex] kJ/kg

(i) [latex]( L H V )_{ v }[/latex] :

[latex]( LHV )_v=( HHV )_v-m\left(u_g-u_f ight)[/latex]

= 45670 – 1.08 × 2304 = 43182 kJ/kg

Hence [latex]( L H V )_v=48182[/latex] kJ/kg.

(ii) [latex]( H H V )_p, \quad( L H V )_p[/latex] :

The combustion equation is written as follows :

[latex]1 ext { mole fuel }+\frac{x}{32} O _2 \longrightarrow \frac{3.23}{44} CO _2+\frac{1.08}{18} H _2 O[/latex]

i.e., [latex]\frac{x}{32}=\frac{3.23}{44}+\frac{1.08}{18 imes 2}[/latex]

or x = 3.31 kg

i.e., 1 kg fuel + 3.31 kg [latex]O _2=3.23 CO _2+1.08 H _2 O[/latex]

Also, [latex]\Delta H=\Delta U+\Delta n R_0 T[/latex]

i.e., [latex]-( HHV )_p=-( HHV )_v+\Delta n R_0 T[/latex]

or [latex]( HHV )_p=( HHV )_v-\Delta n R_0 T[/latex]

where [latex]\Delta n=n_p-n_R[/latex]

= [latex]\left(\frac{3.23}{44}-\frac{3.31}{32} ight) \quad\left[\begin{array}{l}n_p= ext { number of moles of gaseous products } \n_R= ext { number of moles of gaseous reactants }\end{array} ight][/latex]

Since in case of higher heating value, [latex]H _2 O[/latex] will appear in liquid phase

[latex]( H H V )_{ p }=45670-\left(\frac{3.23}{44}-\frac{3.31}{32} ight) imes 8.3143 imes(25+273)[/latex]

= 45744 kJ/kg.

[latex]( L H V )_{ p }=( HHV )_p-1.08 imes 2442=45774-1.08 imes 2442[/latex]

= 43107 kJ/kg.

Question 11.37: The higher heating value of kerosene at constant volume whos...

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