A little "Math for the Boost Guys"
 

Example:
The Inlet Temperature is 70 deg F, the suction pressure is -0.5 psig (a slight vacuum due to Inlet Plumbing), the discharge pressure is 19 psig, and the efficiency is 72%. What is the discharge temperature?

T in= 70 deg F + 460 = 530 deg R

P in= -0.5 psig + 14.7 = 14.2 psia

P out= 19 psig + 14.7 = 33.7 psia

P out/P in = 33.7/14.2 = 2.373 (this is the pressure ratio)

T out = (530 + 530 x (-1+(2.3730) raised to the power of .263 ))) all of this divided by 0.72

= 717.8 deg R - 460 = 257.8 deg F

So the theoretical outlet temperature is 257.8 deg F.

(I sure would like to have an intercooler to cool that hot air down before it goes into my engine.)

So assuming a 70% efficient inter-cooler

257 deg F – 70 deg F ambient temp = 187.886 deg F Temp rise from the turbocharger compressor

So if you have a 70% efficient inter-cooler: 188 deg F times .70 = 131 degrees of temp removal

So the discharge air out of the intercooler is 56 deg f hotter than ambient temp.

The actual Air Temp going into the engine is 126 degrees at 19 psi boost pressure

Tom V.

Most use .283 but it would be closer to .2857. But that would be negligible.
But the .263 is off about 16º


:confused:

Thanks for catching the typo on the .263 vs the .283 commonly used in MacInnis book or the .2857 number that some use today.

Any more suggestions?

Tom V.

Note to members, Constants are constantly being redefined.

Most of the Historical ones like the MOLE were redefined again this year.
https://chemistry.stackexchange.com/...fined-for-2019

So don't worry too much about someone posting a number as being gospel for eternity, it does not work that way in engineering.

2019 redefinition of the SI base units

https://en.wikipedia.org/wiki/2019_r...ase_units#Mole

So take numbers for the most part as a close approximation in formulas.

Good info, John, in my case a typo, in your case you need to do more reading as your number is not absolute either by any means.

Tom V.

Tom,
If anyone tries to work this out the bold line has extra ( )

(530 + 530 * (-1 + (2.3730 ^.263))) / .72 = 923.94619

530 + 530 * (-1 + (2.3730 ^.263)) / .72 = 717.83508

Stan

Thanks Stan, typing is not high on my skill set.

Tom V.