IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

ashgirl

1.

in order to get maximum calorific value output, a burner should have an optimum fuel to oxygen ratio which corresponds to 3 times as much oxygen as is required theoretically for complete combustion of fuel. A burner which has been adjusted for methane as fuel ( with x litre/hour of CH4 and 6x litre/hour of O2) is to be readjusted for butane, C4H10. in order to get the same calorific output wat shud be the rate of supply of butane and oxygen?

assume that the losses due to incomplete combustion etc. are the same for both fuels and that the gases behave ideally.

heats of combustion:

CH4= 809 kJ/mol C4H10=2978 kJ/mol

ashgirl

somebody plz help!!!!!!!!!!!

Rates assured!!

ashgirl

oh.....plz...............do help me!!!

newarnav

got it ! IIT JEE 1993(the toughest one )

here we go..

CH₄+2O₂= CO2+2H2O

from here we see that for each volume(V) of CH4 we should supply 2V O2 "theoriticaly" but 3*2V=6V practically for the complete comb.

now

we know that Heat generated per sec.

(

H comb)X (gas supplied per sec )

=k (

H comb)X (gas supplied per sec )

for the two gas CH4 n C4H10 "Heat generated per sec" is same hence we get

(

H comb of CH4)X (CH45 supplied per sec )=(

H comb of C4H10)X (C4H10 supplied per sec )

from above we get

Butane supplied per sec = x*809/2978---------------------------(1)

now for combstion of butane we have

C4H10+13/2O2= 4CO2+5H2O

ie. we should supply 13/2V oxygen for each V of Butane

and thus practicaly 3*13/2V of oxygen

but from (1) we have V= x*809/2978

thus rate of supply of oxygen =3*x*809/2978

calculating we get

Butane supplied per sec =0.271x L /sec

and Oxygen=13/2*0.271 =5.29x L/sec

I must say I've done this before and hope this is right again

m i rit ?

(typed too much ? hena ?)

ashgirl

thanx a lot!!

it was urgent!!

u r correct!!

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