InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
In order to get maximum calorific output, a burner should have an optimum fuel value to oxygen ratio which corresponds to 3 times as much oxygen as it 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 6x litre `//` hour of `O_(2))` is to be readjusted for butane, `C_(4)H_(10)`. In order to get the same calorific output, what should be the rate of supply of butane and oxygen?Assume that the losses due to incomplete combustion etc. are the same for both the fuels and that the gases behave ideally. Heats of combustion `:` `CH_(4) = 809 kJ // mol,C_(4) H_(10) =2878 kJ // mol` |
|
Answer» Calculation of `C_(4)H_(10)` `CH_(4)+ 2O_(2) rarr CO_(2) + 2H_(2)O, DeltaH = - 809 kJ mol^(-1)` Initial volume `//` hr (L) x 6x Suppose the volume of 1 mole of the gas under the given conditions is V litre. Thus, V litre (1 mole)of `CH_(4)` give energy on combustion `= 809 kJ` `:. ` x litre of `CH_(4)` will give energy on combustion `= (809 x)/( V ) kJ` Now, 2878 kJ of energy is obtained from 1 moleor V litre of `C_(4) H_(10)` `:. ( 809x)/( V) kJ` of energy will be obtained from `C_(4)H_(10) = ( 809 x xx V )/( V xx 2878) = 0.281 x ` litre Thus, butane supplied for same calorific output `= 0.281 x` litre `//` hr. Calculation of`O_(2)` `C_(4)H_(10) + ( 13)/( 2) O_(2) rarr 4CO_(2) + 5H_(2)O , DeltaH = - 2878 kJ mol^(-1)` Volume of `O_(2)` required `= 2 xx` volume of `O_(2)` for combustion of `C_(4)H_(10)` (Given ) `= 3 xx ( 13)/(2) xx ` Vol. of `C_(4) H_(10)= 3 xx ( 13)/(2) xx 0.28 x = 5.48 x `litre `//` hr |
|