Q:

I am trying to calculate the increase in temperature in a combustion engine due to compression. 250cc volume, 12 to 1 compression, initial pressure 14#/sqin, compression pressure 135#/sqin. Beginning temperature 30 degrees. Using ideal gas laws I come up with 4300 degrees, that cannot be right. Air is not ideal gas, but it should not be that far off.

- darrel kruger (age 62)

cottonwood, AZ, USA

- darrel kruger (age 62)

cottonwood, AZ, USA

A:

The numbers you give aren't quite consistent. This is close to an adiabatic compression, meaning that it happens too rapidly for there to be much heat flow out of the gas. For an ideal gas with a temperature-independent heat capacity, we have that pV^{γ }stays constant, where p is pressure and V is volume. Those conditions are pretty well obeyed by air, with γ =1.4. So for a factor of 12 compression you expect p to go up a factor of 12^{1.4}=32.4. Yet you seem to say that p only goes up by 135/14= 9.6.

At any rate, for an ideal gas the*absolute* temperature T is proportional to pV, or in our ideal case V^{1- γ} or p^{(γ-1)/γ}. If we use your x12 compression in V, we'd get a factor of 2.7 for the increase in T. If we use the x9.6 in p, we'd get a factor of 1.9 increase in T.

When you say"Beginning temperature 30 degrees" I assume that's in °C, or T=303K. So you end up with T of 2.7 *303 K or 1.9*303K, depending on whether your V or p numbers for compression were right. You can do the calculations more carefully, but that's in the range of about 300°C to 540°C.

Mike W.

At any rate, for an ideal gas the

When you say"Beginning temperature 30 degrees" I assume that's in °C, or T=303K. So you end up with T of 2.7 *303 K or 1.9*303K, depending on whether your V or p numbers for compression were right. You can do the calculations more carefully, but that's in the range of about 300°C to 540°C.

Mike W.

*(published on 01/07/2013)*