Question

combustion Gases. A fuel containing 86n01% methane (CHal snom ethanescsokPangaven 6m01% propane (C,Met k Ha) ows to a furnace at a rate of 1450 m Jh at 15 C and 15 where it is burned with 8% excess Calculate the molar flow rate of fuel and air into the air. furnace 1450 m 486 Is ess ou nRT 2T

Answer 1

or methane CH4 + 2O2---> C02 +2H20 => 2 moles of oxygen is required for every mole of methane.

for ethane C2H6 +(7/2)O2 ---> 2CO2 + 3H2O => 3.5 moles of O2 is required for every one mole of ethane.

for prpane C3H8 + 5O2----> 3CO2 + 4H2O. 5 moles of O2 is required for every one mole of propane..

for one mole of the mixture 0.86*2 +0.08*3.5+0.06*5 = 2.3 moles of O2 is required..

for every one mole of air,it contains 0.21 moles of oxygen. forr 2.3 moles of oxygen air will be 2.3/0.21 = 10.95 moles of air..

for 8% excess air it will be = 1.08 *10.95 = 11.828 moles of air..

so for every one mole the mixture 11.828 moles of air is taken.

in terms oof volume flow rate its = 1450 *(11.828) = 17151.428 m3/h...

it should be = 17151*(298/288)*(251/101) = 44102 m3. = 4.4*10^4 m3/h.