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

Question

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!

Answer 1

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.

Answer 2

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.