Question

Propane is mixed with 20% excess air (propane air mixture is initially at 25 °C and 2.0 atm) and combusted at atmospheric pressure. It is assumed that the heat transfer between the chamber and the surroundings is negligible. In addition, you may assume that heat capacities of the gases are constant and that the reaction proceeds to 100% conversion. (30 points) a) Determine the flame temperature of the burner. b) Determine the flame temperature if 25% of the heat generated by combustion was lost through heat transfer to the surroundings? While doing the calculations you may assume that the heat capacities of the gases are constant. The heat capacities of the gases of interest are shown below. Gas Heat capacity Nitrogen 29.12 J/(mol*k) Oxygen 29.39 J/(mol*k) Propane 74.92 J/(mol*k) Carbon dioxide 37.14 J/(mol*k) Water 33.57 J/(mol*k)

Answer 1

propane is mixed with 20% excess in air, hence propane air mixture is initailly at 25^oC and 2.0 atm

exposive limits of propane gas = 2.15 % as lower limit

9.6% as upper limit

ignition temperature of propane gas = 493^oC - 604^oC

a) percent excess air = 20%

percent available heat = total heat delivered x 100

total heat into furnace

hence , propane gas flame temperature of burner = 1980^oC

b) while doing the calculations assume that the heat capacities of the gase are constant.

the heat capacities of the gases of interest

gas heat capacity nitrogen= 29.12 J/mol K

gas heat capacity of oxygen = 29.39  J/mol K

gas heat capacity of Propane = 74.92  J/mol K

gas heat capacity of Carbon di oxide = 37.14  J/mol K

gas heat capacity of Wate = 33.57  J/mol K

given system is adiabatic in condition, hence the flame temperatures are constant, unless the heat generated by combustion was lost through heat transfer to the surroundingss