I've been given the following to memorize; But my head is so crammed with many others that I need to find another way; Are these formulas Empirical? I was hoping that if I derive it from a much more primitive form; I maybe able to remember it.

$${ (\frac{A}{F})_t = W_{ta} = \frac{Q_h \frac{kJ}{kg} }{3117} \frac{kg_{air}}{kg_{fuel}} }$$

$${ (\frac{A}{F})_t = W_{ta} = \frac{Q_h \frac{kCal}{kg} }{745} \frac{kg_{air}}{kg_{fuel}} }$$

$${ (\frac{A}{F})_t = W_{ta} = \frac{Q_h \frac{Btu}{lb_m} }{1340} \frac{kg_{air}}{kg_{fuel}} }$$

I know if I'm provided the answer to one of these; I can easily derive the others because I've memorized the conversions. What I don't get is how ${\frac{kJ}{kg}}$ got to ${ \frac{kg_{air}}{kg_{fuel}} }$. I cant figure out where the 3117 came from.

${Q_h}$ is Higher Heating Value, ${ \frac{A}{F}}$ is Air to Fuel ratio

  • 2
    $\begingroup$ What are $Q_h$, $A$, and $F$? As they are written, these equations are a rather bizarre mix of units and variables. Are the units of $Q_h$ $\tfrac{\text{kJ}}{\text{kg}}$ in the first equation? $\endgroup$ – Chris Mueller Aug 28 '15 at 2:11
  • $\begingroup$ @ChrisMueller Qh is Higher Heating Value, A/F is Air to Fuel ratio; yes they are the units of Q_h. $\endgroup$ – james Aug 28 '15 at 4:28
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    $\begingroup$ @james rather than adding clarifications to your question in a comment, please always edit them into the question body itself. Then you can add a comment notifying whoever asked for clarification, that the question has been updated. Comments are ephemeral and disposable, so all the useful information needs to be in the question body itself. $\endgroup$ – 410 gone Aug 28 '15 at 6:09
  • $\begingroup$ You can't just program the formulas into your calculator for future reference? Not that I used to do that for tests in college, or anything... $\endgroup$ – grfrazee Aug 28 '15 at 12:49

Are these formulas Empirical? I was hoping that if I derive it from a much more primitive form

This is probably an approximation, let's try to calculate stoichiometric air to fuel ratio and heat of combustion for liquid octane as an example:

$$ C_8H_{18} + \lambda(O_2 + 3.76N_2) \rightarrow 8CO_2 + 9H_2O + 3.76\lambda N_2$$

First we calculate heating value, Cengel (2005) defines the heating value as :

the amount of heat released when a fuel is burned completely in a steady-flow process and the products are returned to the state of the reactants. In other words, the heating value of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel.

Making energy balance we can get $Q_h$ around $-47,891 \ \text{kJ/kg fuel}$

Second, we calculate stoichiometric air to fuel $(\lambda = 8 + \frac{18}{4} = 12.5)$: $$ (\frac{A}{F})_{stoich.} = \frac{12.5(32 + (3.76*28)}{(8*12) + 18} = 15.052 $$

Substituting in your formula: $$ (\frac{A}{F})_{stoich.} = \frac{|Q_h|}{3117} = \frac{47891}{3117}= 15.36$$

which is slightly higher than the calculated theoretical value. Of course there is no direct relation between heating value of fuel and stoichiometric air to fuel ratio as there are many parameters that could affect that relation, you can derive the air to fuel ratio by making energy balance and substitute with elements' enthalpy and heat of formation but that won't be any easier than simply memorizing the formula you were given.


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