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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. |
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Answer» (i) Higher heating value at constant pressure, (HHV)P : The combustion reaction for propane is written as C3H8 + 5O2 = 3CO2 + 4H2O Now (HHV)p = (LHV)p + mhfg where, HHV = Higher heating value at constant pressure, LHV = Lower heating value, m = Mass of water formed by combustion = 4 × 18 = 72 kg per kg mole, and hfg = Latent heat of vaporisation at given temperature per unit mass of water = 2443 kJ/kg at 25°C. ∴ (HHV)p = 2044009 + 72(2442) = 2219833 kJ/kg. (ii) Higher heating value at constant volume, (HHV)v : Now (∆U) = ∆H – ∆nR0T or – (HHV)v = – (HHV)p – ∆nR0 T or (HHV)v = (HHV)p + ∆nR0 T where R0 = universal gas constant = 8.3143 kJ/kg mol K ∆n = nP – nR [np = number of moles of gaseous products] [nR = number of moles of gaseous reactants] Now, the reaction for higher heating value is C3H8 + 5O2 = 3CO2 + 4H2O (liquid) ∆n = 3 – (1 + 5) = – 3 (HHV)v = 2219905 – 3(8.3143)(25 + 273) = 2212472 kJ/kg. |
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