In order to get maximum calorific 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 the fuel. A burner which has been adjusted for methane as fuel ( with X litre // hour OF CH_(4) and 6 X litre // hour of O_(2)) is to be readjusted for butane C_(4)H_(10). In order to get same calorific output, what should be the rate of supply of butane and oxygen ? Assume that losses due to incomplete combustion etc., are the same for both fuels and that gases behave ideally. Heat of combustion CH_(4)=809kJ mol^(-1), C_(4)H_(10)=2878kJ mol^(-1)

In order to get maximum calorific output,a burner should have an optim

1.	In order to get maximum calorific 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 the fuel. A burner which has been adjusted for methane as fuel ( with X litre // hour OF CH_(4) and 6 X litre // hour of O_(2)) is to be readjusted for butane C_(4)H_(10). In order to get same calorific output, what should be the rate of supply of butane and oxygen ? Assume that losses due to incomplete combustion etc., are the same for both fuels and that gases behave ideally. Heat of combustion CH_(4)=809kJ mol^(-1), C_(4)H_(10)=2878kJ mol^(-1)
Answer» Solution :Given, `{:(,CH_(4),+2O_(2),rarr,CO_(2)+2H_(2)O,,DeltaH=-809kJ mol^(-1)),("Initial volume"//hr(i n litre),X,6X,,,):}` Let the temperature be `T` and assume volume of `1 mol e` of a GAS is `V` litre at this condition. `:. V` litre or `1 mol e CH_(4)` GIVES on combustion `=809kJ` `:' X`litre of `CH_(4)` gives energy on combustion `=(809(X))/(V)kJ` `:' 2878kJ` energy is obtained by `1 mol e `or `V` litre `C_(4)H_(10)` `:. (809(X))/(V)kJ` energy is obtained by `=(809(X)xxV)/(Vxx2878)litre C_(4)H_(10)` `=2.81(X) `litre `C_(4)H_(10)` Thus, butane supplied for same calorific output `=0.281(X)`litre `:. C_(4)H_(10)+(13)/(2)O_(2) rarr4CO_(2)+5H_(2)O,DeltaH=-2878kJ//mol` `Vol. ` of `O_(2)`required `=3xxvol. of O_(2)` for combustion of `C_(4)H_(10)` `=3XX(13)/(2)xxvol. of C_(4)H_(10)` `=3xx(13)/(2)xx0.281(X)=5.48(X) litre O_(2)`