Thermodynamics and Thermochemistry 2 Question 22
22. 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 $\mathrm{CH}{4}$ and $6 x$ litre/hour of $\mathrm{O}{2}$ ) is to be readjusted for butane, $\mathrm{C}{4} \mathrm{H}{10}$.
In order to get the 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 the gases behave ideally. Heats of combustions:
$\mathrm{CH}{4}=-809 \mathrm{~kJ} / \mathrm{mol}, \mathrm{C}{4} \mathrm{H}_{10}=-2878 \mathrm{~kJ} / \mathrm{mol}$
(1993, 3M)
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Solution:
- At same temperature and pressure, equal volumes contain equal moles of gases.
Let $1.0 \mathrm{~L}$ of $\mathrm{CH}_{4}$ contain ’ $n$ ’ mol
$\Rightarrow \quad x \mathrm{~L}$ of $\mathrm{CH}_{4}$ contain $n x$ mol
$\Rightarrow$ Heat evolved in combustion by $x \mathrm{~L} \mathrm{CH}_{4}=809 n x \mathrm{~kJ}$
Now, $2878 \mathrm{~kJ}$ energy is evolved from 1 mole $\left(\frac{1}{n} \mathrm{~L}\right) \mathrm{C}{4} \mathrm{H}{10}$.
$\Rightarrow 809 n x \mathrm{~kJ}$ energy will be evolved from $\frac{809 n x}{2878 n} \mathrm{~L}$ of $\mathrm{C}{4} \mathrm{H}{10}$
$$ =0.28 x \mathrm{~L} \text { of } \mathrm{C}{4} \mathrm{H}{10} $$
Also, the combustion reaction of butane is
$$ \begin{aligned} \mathrm{C}{4} \mathrm{H}{10}+\frac{13}{2} \mathrm{O}{2} & \longrightarrow 4 \mathrm{CO}{2}+5 \mathrm{H}_{2} \mathrm{O} \ \Rightarrow \text { Rate of supply of oxygen } & =\frac{13}{2} \times 0.28 x \times 3 \ & =5.46 x \mathrm{~L} / \mathrm{h} \end{aligned} $$
