turbo and intercooler math

[CBX]

I posted this over in the doo turbo forum, But this is one of the most active turbo forums and I think most would learn alot.

I found a great article on autospeed.com with equations showing how to calculate pressure drop from air temperature increases, intercooled and non intercooled efficency, and Non turbo pressure ratio (a number that would represent how much the turbocharged engine was boosted as if it were pure atmosphere pressure (like a labratory) and un-assisted. by a turbo or supercharger.) It explains perfectly why when the intake pressure is doubled, horsepower isnt doubled because of efficency.

Heres the link to the article.

http://autospeed.com/cms/title_Intercooler-Maths/A_109789/article.html

I tried to copy the article here, but no matter what i did the math formula's looked messed up no matter how I edited it. For some reason the forum software defaults on align left for text. It won't let me manually move text around to make the formula's look normal. Even if i align center the formula's.

If someone can figure this out i'd be glad to put the entire article up on here for easy reference. Thanks.

Enjoy!

 

[pernikm]

Excellent post! This is more useful information than originally thought. This gives us the ability to scale dyno results without actually requiring charge tube temperatures.

I also want to point out a couple of things. First off the pressure ratio calculation he gives is for sea level since 1 bar is equal atmospheric pressure at sea level. If anyone wants to use this equation at altitude it still works. Simply add the boost gauge presure to the the local atmospheric pressure and divide by the local atmospheric pressure. If you don't know the atmospheric pressure just google atmospheric vs. altitude. There is a real nice table available. I posted that on another thread, but I got kicked off the thread by the OP because he did not like math. You can work in bar or psi it does not matter, you are just converting to an absolute pressure system.

The other thing I want to point out is that the T2=T1(P2/P1)^.286 is not a magic formula. It is the equation for Isentropic compression of any gas based on the ratio of specific heats. The general form for for this equation is:

T1/T2 = (P2/P1)^((k-1)/k) where k is the ratio of specific heats for the gas. Since we are compressing air, use k = 1.4

The derivation is algebraic and actually quite simple.

 

[CBX]

atmoshpere values by elevation and barometric pressure.