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Can somebody explain relation between combustion efficiency and pressure in lean mixtures using any kind of chart or mathematical equations?

At first glance I thought, combustion efficiency increases with pressure initially and then gets constant. But not sure by how much amount it varies based upon pressure for different fuels.

Then I tried to derive a formula to calculate or estimate Combustion Efficiency using Collision theory. As per my understanding, more collisions need to happen between reactants for better combustion efficiency. But collision theory alone is not sufficient to calculate or estimate Combustion Efficiency.

Also does Combustion efficiency varies between simple fuels like Hydrogen and methane under similar conditions? If so, why and at what margin?

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In an engine, higher pressures lead to a higher brake power. Why? Check the Otto and Diesel diagrams:

https://www.mechanicalbooster.com/2017/10/diesel-cycle.html https://www.mechanicalbooster.com/2017/10/otto-cycle.html

If you increase the combustion pressure, then you will maximize the area contained among the lines and the work output will increase. As you can see in the pictures, the expansion is considered to be isentropic, so these are the formulae you can use:

$T_3/T_4=(p_3/p_4)^{(\gamma-1)/\gamma}$

Where $\gamma$ is the adiabatic constant for the combustion gases, roughly 1.4.

If you increase the pressure ratio you increase the temperature ratio too and the output work of this process can be calculated this way:

$\sum_{p_3}pv^{\gamma}dv$

Where v is the specific volume or the volume if you prefer. This is an integral between $p_3$ and $p_4$ (sorry for the wrong format).

It's a politropic isentropic process, that's why the politropy index equals $\gamma$. And remember that $pv^{\gamma}=constant$, that's the curve you must integrate. So $p_3 v_3 ^{\gamma}=p_4v_4^{\gamma}$. You can use $v=R*T/p$ in both occasions, rembering that the gas can be modeled as ideal and R has an approximate value of 287 J/kgK.

In conclusion, as that pressure ratio increases, the integral will get bigger. That's why efficiency will increase, because the output work is increasing higher than the input (compression) energy (heat+work) and the losses.

$\eta=W_o/E_i$ or $\eta=1-Losses (heat+work)/E_i$.

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  • $\begingroup$ But there is a limit to how much you can increase the compression ratio... $\endgroup$ – Solar Mike Jun 18 at 8:40
  • $\begingroup$ Not in theory, but yes in practice. $\endgroup$ – user20096 Jun 18 at 8:41
  • $\begingroup$ even in theory, if the compressed air T is above the self-ignition point of the fuel then you have a problem... $\endgroup$ – Solar Mike Jun 18 at 8:44
  • $\begingroup$ That doesn't happen in Diesel engines. $\endgroup$ – user20096 Jun 18 at 8:47
  • $\begingroup$ So, you changed fuel... Are we not working with hydrogen and methane? $\endgroup$ – Solar Mike Jun 18 at 8:49

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