InterviewSolution
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InterviewSolution
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In order to get maximum calorific output a burner should have an optimum fuel to oxygen ratio which corresponds to three times as much oxygen as is required theorectically for complete combusion of the fuel A burner which has been adjused for methane as fuel (with `x L h^(-1)` of `CH_(4)` and `6x Lh^(-1) of CO_(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 to butane and oxygen? Assume that losses due to incomplete combustion etc are the same for both fuels and that the gases behave ideally Heats of combusion `CH_(4) =809 kJ mol^(-1),C_(4)H_(10) =2878 kJmol^(-1)` . |
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Answer» In `1h, xL` of `CH_(4)` required `6x L` of `O_(2)` `DeltaH` (combustion) of `CH_(4) = 809 kJ mol^(-1)` `rArr (809)/(24.48) kJ mol^(-1) at 1.0 atm` and `25^(@)C` `DeltaH` (combustion) of `C_(4)H_(10) = 2878 kJ mol^(-1)` `rArr (2878)/(24.48) kJ L^(-1) at 1.0 atm` and `25^(@)C` `xL of CH_(4)` produces `809. x//24.48 kJ` Now this much enegry has to be provided by burning of `C_(4)H_(10)`. `rArr (809)/(24.48) (x) kJ` will be provided by `((809)/(24.48)(x)) xx ((24.48)/(2878)) - (0.28x)L of C_(4)H_(10)` `C_(4)H_(10) +13//2O_(2) rarr 4CO_(2) +5H_(2)O` `rArr 1mol C_(4)H_(10) = 13//2 mol O_(2)` `rArr (3xx13//2)` times of `O_(2)` is required per mol. `rArr` Rate of `O_(2)` per hour `=(0.28x) xx (39//2)` `=(5.48x) L O_(2)` |
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