Question

Determine what is the temperature of the hot combustion gas that is produced in the engine of a butanol (C4H9OH)-burning bus that uses ambient air for the combustion. Tip: write the balanced stoichiometric equation for the combustion of butanol assuming air is composed of 80% N2 and 20% O2 by volume, and derive the energy balance of the combustion. Atomic weights: C-12, H-1, O-16, N-14. For butanol: cp = 2 kJ/kg/°C; cv = 1.75 kJ/kg/°C; heat released by combustion, or ΔH = 60 MJ/kg.

Answer 1

C₄H₉OH+6 O₂-->4CO₂+5H₂O

Basis : 1 mole of C4H9OH

1 mole of C4H9OH correspond to 74 gms of C4H9OH

Moles of oxygen required =6 moles

Moles of air required= 6/0.2 =30 mole

Moles of Nitrogen = moles of air supplied- moles of oxygen =30-6= 24 moles

Moles of CO2 =4 mass of CO2= 4*44= 176 gms moles of H2O=5 , mass of water = 5*18=90

Mass of N2= 24*28=672 gms

Assuming products are supplied in stoichiometric ratios,

Heat of combustion = 60,000,000 J/kg= 60,000 j/g

For 74 gms of C4H9OH it is 60,000*74=4440000 Joules

Heat of combustion= Enthalpy change of N2 + Enthalpy change of CO2 + enthalpy change of water

4440000 = Mass of N2*Cp of N2* (T2-T1) + Mass of CO2* Cp of CO2 (T2-T1)+ Mass of water vapor* Cp of water *(T2-T1)

4440000 = 672*1.040*(T2-298.15) + 176*0.763*(T2-298.15) + 90*1.864*(T2-298.15)

T1 is considered at 25 deg.c which makes the enthalpy of reactants= 0

All the heat is assumed to be transferred to the gases only

4440000=(T2-298.15) {672*1.040+ 176*0.763+90*1.864) = 1000.93

T2-298.15= 4440000/1000.93 =4435.874

T2= 4435.875+298.15 =4734.025K