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ChE or chemist (a) Estimate the lower heating value of liquid n-butane in kJ/gmo

ID: 518682 • Letter: C

Question

ChE or chemist

(a)        Estimate the lower heating value of liquid n-butane in kJ/gmol and in Btu/lb.

            (b)        Estimate the adiabatic flame temperature if liquid ethanol is burned with 100 % excess air.

                        Assume that the ethanol and air are initially at 25 oC.

For simplicity use the following average specific heats.

                        CO2     51.3 J/gmol-K            H2O    39.8    J/gmol-K

                        O2       33.8 J/gmol-K            N2       31.9    J/gmol-K

Explanation / Answer

a)

Butane combustion is

C4H10 + 6.5O2 = 4CO2 + 5H2O

Gross heating value of C4H10 is 21640 btu/lb

Hvap = 970.4 btu/lb

Lower heating value of C4H10 = gross heating value - xHvap

where x is mole fraction of water in product i.e 4/9

Lower heating value of C4H10 = gross heating value - xHvap

Lower heating value of C4H10 = 21640 - 4/9*970.4

Lower heating value of C4H10 = 21208.71 Btu/lb

1 Btu/lb = 2.326 kJ/kg

Lower heating value of C4H10 = 21208.71 Btu/lb = 49331.46 kJ/kg

49331.46 kJ/kg = 2269.25 kJ/mol

b)

Liquid ethanol is burnt with 100% excess air at 250C i.e. 298K

The air contains 21% O2 and 79% N2 by moles

C2H5OH + 6O2 + 22.57N2 ------> 2CO2 + 3H2O + 3O2 + 22.57N2

Basis: for 1 mole of ethanol at 298K

Standard enthalpy of formation of ethanol is -278 kJ/mol

Standard enthalpy of formation of CO2 is -393.5 kJ/mol

Standard enthalpy of formation of H2O is -242 kJ/mol

At constant pressure, we have,

Hreactants = Hproducts

Hreactants = Enthalpy of formation of reactants

Hreactants = H(C2H5OH) + H(O2) + H(N2)

Hreactants = -278000 + 0 + 0

Hreactants = -278000 J/mol

HProducts = ni[(Hf ) + Cp (Tad - 298)]

HProducts =  ni[(Hf ) + Cp (Tad - 298)]CO2 +  ni[(Hf ) + Cp (Tad - 298)]H2O +  ni[(Hf ) + Cp (Tad - 298)]O2 +  ni[(Hf ) + Cp (Tad - 298)]N2

HProducts = 2[-393500+51.3(Tad - 298)] + 3[-242000+39.8(Tad - 298)] + 3[0+33.8(Tad - 298)] + 22.57[0+31.9(Tad - 298)]

HProducts = -1513000+1043.383(Tad - 298)

Hreactants = Hproducts

-278000 = -1513000+1043.383(Tad - 298)

1235000 = 1043.383(Tad - 298)

(Tad - 298) = 1183.65

Tad = 1481.65 K (Adiabatic temperature)