Question

A fuel oil contains 86 % carbon, 13% hydrogen and sulfur 1% by weight. Its heating value is reported to be 46000 kJ/kg. Determine the higher and lower heating values and enthalpy of formation on a mole basis.

Answer 1

Sol" The fuel composition is, Element wt% Normalize with respect to a C 86 8 12 = Tod 7 & 7.17 = 1 13 & t = 13€ 7.17 = 1.8. 32 = 0.03 7.17 -0.004 The fuel composition is Chics 50.004, and formula weight is, its Mee 12+ (1.8) (1) + (0:004) (32) -me a 13.9 Heating values are most commonly reported as the higher heating value ; thus Theks () --HH V = -44000 kJ/ke The molas entholby of combustion is, shcHS = Me shchs = (1309) (-44000 'T/ai A hGHS = -616, 600 J/molok Combustion of this fuel Proceeds according to : CHI S0.004 t 1.454.0.2 — >C02 +0.004 S0, 10:91:20 Thus, 0 g mol of water is generated of fuel (carbon) burned. for each mole.

Date: __ Page no: (2) The latent heat of vaporization of water at 298k is shiv (298k) - 44000 J/mol A hcl = ahors to-go ohr 2-611,600 + 0.90 x (44,000) Ž -572,000 J (mol c ) 1 The enthalpy of formation of this fuel is determined using lower heating values, fenthalpy of formation; sheCHI.850.004 (T.) - Dhe.coo(I.) +0.004shf,502 (1) 40-90 the H2O (Ti) - 1.4 540 hf 0, (T) - AhoL(T) From Book data, nheco, (298k) = -394, 088 the so2a-297, 269 Dhe,H2O --2429174 the CH1.8 $0.004(1.) --394;088 +0.004(-297269) +0-90(-242,174) -1.454 (0) - (-572,000)! Lo hp, CH1.85 0.004 = -41, 234 (molot