Question

Consider ethanol (C2H5OH, or C2H6O) burning with air. The lower heating value (LHV) of ethanol is given as 26,700 kJ/kg. Compute the higher heating value (HHV) of ethanol in kJ/kg from the given LHV value and the enthalpy of vaporization of water (44,010 kJ/kmol).

Answer 1

Answer

29564 kJ/kg

Explanation

Balanced equation for combustion of ethanol is

C2H5OH(l) + 3O2(g) -----> 2CO2(g) + 3H2O(g)

n_H2O,out/ n_{fuel ,in} = 3/1 =3

HHV = LHV + H_v( n_H2O,in/ n_{fuel, in} )

H_v = Heat of vaporization of water, 44..010kJ/mol =

LHV = 1230.3kJ/mol

HHV = 1230.3 kJ/mol + (44.010kJ/mol × 3 )

HHV = 1230.3 kJ /mol + 132.03kJ/mol

HHV = 1362.3 kJ/mol

HHV = 29564 kJ/kg