# Flow rate of the fuel consisting of ethanol and water burning in air of a given flow rate

I hope that someone could give me some hints or tips on how to approach this question, even though these types of questions are usually closed off as homework.

A burner is run by a fuel mixture of $$\pu{90 mol.\%}$$ ethanol and $$\pu{10 mol.\%}$$ water. Assume a complete combustion with air ($$\pu{79 mol.\%}$$ $$\ce{N2}$$, $$\pu{21 mol.\%}$$ $$\ce{O2}$$). The whole process takes place at $$\pu{100 kPa}$$.

Calculate the inlet flow rate of the fuel given that the flow rate of air is $$\pu{100 mol s-1}$$ and that the excess of air is $$20\%$$.

The answer key only gives the answer $$\pu{6.48 mol s-1}$$ of the fuel without explaining the calculations.

My attempt at the question was:

$$\ce{C2H5OH + H2O + 3O2 -> 2CO2 + 4H2O}$$

Knowing that we have $$20\%$$ excess in air, we could calculate the theoretical moles needed.

$${\text{fractional excess} = (n_\mathrm{feed} - n_\mathrm{theo})/n_\mathrm{theo}}$$

$$\to n_\mathrm{theo} = \pu{83.3 mol}$$

I then used the molar ratio from the reaction to calculate how much ethanol and water was needed. I just don't see how the answer becomes $$\pu{6.48 mol s-1}$$. Any help is greatly appreciated! • A burner combusts fuel. What takes part in such reactions? – Eashaan Godbole Jan 11 '19 at 14:54
• Also, the question looks like just stoichiometry. Why did you include a heat exchanger etc? – Eashaan Godbole Jan 11 '19 at 15:01
• @EashaanGodbole I added the water in the reaction since they stated in the problem that the fuel consisted of both ethanol and water, although I know that a combustion reaction does not include water in the reactants. The diagram of the system was given, there are other questions which involve the heat exchanger specifically. – Johan Jan 11 '19 at 15:16
• The inlet water should not be included in the balanced chemical reaction formula, and you balanced the formula incorrectly. Your first step should be to get that right. – Chet Miller Jan 11 '19 at 17:12

I think you overcomplicate the problem. There are many variables you don't need and many components that are duds (nitrogen, water, pressure). What important is that we burn ethanol:

$$\ce{C2H5OH + 3 O2 -> 2 CO2 + 3 H2O}$$

so that the ratio between ethanol and oxygen is $$1:3$$. I have taken chemistry engineering classes long time ago and I don't remember exact notations, so I'd rather denote molar flow rate with an n-dot notation:

$$\dot{n} = \frac{\mathrm{d}n}{\mathrm{d}t}$$

where $$n$$ – amount; $$t$$ – time. Since fuel is only 90% alcohol, then

$$\dot{n}(\text{fuel}) = \frac{\dot{n}(\ce{C2H5OH})}{x(\ce{C2H5OH})}$$

where $$x$$ denotes molar fraction. From the combustion reaction

$$\dot{n}(\ce{C2H5OH}) = \frac{\dot{n}(\ce{O2})}{3}$$

and for the air the following is fulfilled:

$$\dot{n}(\ce{O2}) = x(\ce{O2})\frac{\dot{n}(\text{air})}{1 + α}$$

where $$α$$ is the partial excess of air. Now we can put everything together and plug in the numbers:

\begin{align} \dot{n}(\text{fuel}) &= \frac{x(\ce{O2})\dot{n}(\text{air})}{3x(\ce{C2H5OH})(1 + α)}\\ &= \frac{0.21\cdot\pu{100 mol s-1}}{3\cdot 0.9\cdot (1 + 0.2)}\\ &= \pu{6.48 mol s-1} \end{align}

One hell of a burner.

You calculated 83.3 moles of air reacting with the ethanol. Since air is 21% oxygen, this means that 83.3 x 0.21 = 17.5 moles of oxygen react with the ethanol. Since each mole of ethanol reacts with 3 moles of oxygen, the number of moles of ethanol that react is 17.5/3=5.833 moles ethanol. Since the fuel mixture is 90% ethanol, there are 10 moles of fuel mixture for every 9 moles of ethanol. So the number of moles of fuel mixture are 5.833 x 10/9=6.48 moles of fuel mixture.