Question

Liquid propane enters a combustion chamber at 77⁰F and at a rate of 0.75 lbm/min where it is mixed and burned with 150% excess air that enters the combustion chamber at 40⁰F. If the combustion is complete and the exit temperature of the combustion gases is 1800 R, determine (a) the mass flow rate of the air and (b) the rate of heat transfer from the combustion chamber.  

Answer 1

Temperature of propane = 77 oF = 298 K

Mass flow rate of propane = 0.75 lbm / min = 0.34 kg/min

Temperature of air = 40 oF = 277.44 K

Exit temperature = 1800 K

The balance equation for liquid propane and 150% excess air is:

C3H8(l) + 12.5(O2 + 3.76N2)      3CO2 + 4H2O + 7.5O2 + 47N2

(a). The air fuel ratio can be calculated as

AF = Mass of air / Mass of fuel

= (12.5 x 4.76 kmol)(29 kg/kmol) / [ (3 kmol)(12 kg/kmol) + (4 kmol)(2 kg/kmol) ]

= 1725.5 / (36 + 8)

= 39.215 kg air / kg fuel

Mass of air = AF * Mass of fuel

= 39.215 * 0.34 kg/min

= 13.33 kg/min

Mass of air = 29.38 lbm/min