Butane C4H10 and dry air are fed to a combustion reactor at

Butane (C4H10) and dry air are fed to a combustion reactor at a steady rate of 53.9 mols butane and 2500.0 mols dry air. Assume the butane reacts completely to produce CO2 and H20. The reactants enter the reactor at 25°C and the products leave the reactor at 1022.4°C. Calculate the percent excess air. Number Using the information below, find the heat interaction Number kW AH at 298 K (kJ/mol) FCP (kJ/kmol K) lf Species 3.5R N2 3.5R -125.5 12R 4H10 -393.5 3.5R CO2 -241.8 3.5R H20

Solution

Accroding to the balanced equation

53.9 kmol/s C4H10 requires 350.35 mol/s O2

Which means 350.35*4.76= 1667.66 mol/s

Excess Air = 2500- 1667.66 = 837.34 mol/s

Excess Air Percentage = 837.34/1667.66 = 49.91%

Total mass of the products = 2634.75 mol/s

Amt of heat carried away by air = mCdT = 76469.25 kW

Heat released by combustion =(53.9*-125.5 -(4*53.9*-393.5 + 5*53.9*-241..8)) = 143239.25 kW

Net heat interaction = 143239.25- 76469.25 = 66770 kW