Question

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.

Answer 1

Combustion of 1 kg of fuel produces the following products :

CO₂ = \(\cfrac{44}{12}\) × 0.88 = 3.23 kg

H₂O = \(\cfrac{18}{2}\) × 0.12 = 1.08 kg

At 25°C : 

(u_g – u_f) 

u_fg = 2304 kJ/kg 

h_fg = 2442 kJ/kg 

(i) (LHV)_v : 

(LHV)_v = (HHV)_v – m(u_g – u_f) 

= 45670 – 1.08 × 2304 = 43182 kJ/kg 

Hence (LHV)_v = 43182 kJ/kg

(ii) (HHV)_p, (LHV)_p : 

The combustion equation is written as follows :

= \(\left(\cfrac{3.23}{44}-\cfrac{3.31}{32}\right)\)

[ n_p = number of moles of gaseous product]

[ n_g = number of moles of gaseous reactants]

Since in case of higher heating value, H₂O will appear in liquid phase

(HHV)_p = 45670 – \(\left(\cfrac{3.23}{44}-\cfrac{3.31}{32}\right)\) × 8.3143 × (25 + 273)

= 45744 kJ/kg.

(LHV)_p = (HHV)_p – 1.08 × 2442 = 45774 – 1.08 × 2442

= 43107 kJ/kg.