# Lower Heating value calculation from chemcial combustion equation of Butane

QUESTION Butene, C4H8 is burned in an engine with a fuel-rich air-fuel ratio. Dry analysis of the exhaust gives the following volume percents: $\ce{CO2} = 14.95%$, $\ce{O2} = 0%$, $\ce{CO} = 0%$, $\ce{H2} = 0%$, $\ce{C4H8} = 0.75%$, with the rest being $\ce{N2}$. Higher heating value of this fuel is 46.9 MJ/kg. Write the balanced chemical equation for one mole of this fuel at these conditions. Calculate the air-fuel ratio, equivalence ratio, lower heating value of the fuel, and energy released when one kg of this fuel is burned in the engine with a combustion efficiency of 98%.

My attempt at solution: I write the stoichiometric equation like so: $$\ce{C4H8 + \frac{1}{\phi}(4+(8/4))(O2 +3.773N_2)\rightarrow n_1 C4H8 +n_2 CO2 +n_3 H2O}$$ where, $\phi$ is the equivalence ratio(a engineering term used to describe the ratio of $(F/A)_{\text{actual}}$ to $(F/A)_{\text{stoichiometric}}$.

So,now I have 3 unknowns, $n_1$, $n_2$ and $\phi$.

I unfortunately get $n_2=0$ after doing the mole balance of carbons,hydrogen and nitrogen.

I understand that the percentages given by volume are the mass fractions which should later form a separate equation. But, I still get $n_2=0$. Where have I gone wrong?(Should I still consider putting in $\ce{O2}$ and $\ce{CO}$ into the above equation even though their %by volume are zero?)

• I have edited your question to make your chemical formulas equations look prettier. When using MathJax, enclosed chemistry in \ce{...} to make it look like chemistry. – Ben Norris Jul 7 '13 at 18:42

I see three flaws in your reasoning.

First, butane is $\ce{C4H10}$. The formula $\ce{C4H8}$ refers to several isomers of butene. I will keep using your formula, but you should specify whether you mean butane or $\ce{C4H8}$.

Second, you know the relationship between your variables $n_1$ and $n_2$ because you have the volume percents of $\ce{C4H8}$ and $\ce{CO2}$ in the exhaust. Thus, you have fewer variables.

$$\dfrac{n_1}{n_2} = \dfrac{0.75}{14.95}$$

Third, I am not sure why $\ce{N2}$ is in the equation. It does not participate in the reaction, so it should not be there. Likewise, you do not need to add $\ce{CO}$ and $\ce{H2}$ because you know that none is formed. You can balance the equation and determine the stoichiometric coefficients of $\ce{CO2}$ and $\ce{H2O}$ from the conservation of mass:

$$\ce{C4H8 + 6O2 -> 4CO2 + 4H2O}$$

To include one of your variables, we assume that the reaction does not run to completion, and that some amount $n_1$ of $\ce{C4H8}$ remains unreacted. Since each 1 equivalent of $\ce{C4H8}$ consumes 6 equivalents of $\ce{O2}$, if $n_1$ of $\ce{C4H8}$ is not consumed, then $6n_1$ of $\ce{O2}$ is not consumed. Likewise, $4n_1$ of $\ce{CO2}$ and $\ce{H2O}$ were not produced.

$$\ce{C4H8 + (6-6n_1)O2 -> n_1 C4H8 + (4-4n_1)CO2 + (4-4n_1)H2O}$$

Now, I have rendered your problem down to one variable: $n_1$. Your other variables can be written as equations involving $n_1$ by comparing your chemical equation to mine (they are equivalent). You have enough equations in $n_1$ and $n_2$ to solve for $n_1$, $n_2$, and eventually $\phi$, which then leads to the values you really want.