typical post turbo charge air temps
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Ahh, no I just misread it (didn't notice P / T swap), thanks for the clarification. (to clarify neither is solved for anything in that form, but yes)
Last edited by bn880; Feb 16, 2023 at 10:42 AM.
(P2/P1) = (T2/T1)^(k/(k-1))
Where the pressures and temperatures are in absolute values, and k = ratio of specific heats = 1.4 for air. After you solve for T2, divide (T2-T1) by the efficiency of compressor, so about 0.70. Then add that value to T1. That should get you close.
So for 1.4 bar of boost, (we'll say 1 bar = 1 atm for simplicity), P2 = 2.4 bar, P1 = 1bar, T1 = 295K (~71F).
T2 = ~379K. T2 - T1 = 84. 84/0.70 = 120. So the final temp is about 120+295 = 415K or 288F.
Where the pressures and temperatures are in absolute values, and k = ratio of specific heats = 1.4 for air. After you solve for T2, divide (T2-T1) by the efficiency of compressor, so about 0.70. Then add that value to T1. That should get you close.
So for 1.4 bar of boost, (we'll say 1 bar = 1 atm for simplicity), P2 = 2.4 bar, P1 = 1bar, T1 = 295K (~71F).
T2 = ~379K. T2 - T1 = 84. 84/0.70 = 120. So the final temp is about 120+295 = 415K or 288F.
T2=T1*(P2/P1)^(k/(k-1))
where
T1 = 71F converted to Kelvin => 295K
T2 = Unknown
P2 = 2.4 bar
P1 = 1.0 bar
k or gamma where gamma = cp / cv or otherwise denoted as the polytropic coefficient = 1.4
T2 = 295*(2.4)^(1.4/(1.4-1))
T2 = 295*(2.4)^(.285)
T2 = 378.6 ~ 379K
Delta T = 379-295 = 84
Efiiciency of Compressor at P2/P1 = 70%
84/.7 = 120 K
T2 = T1 + 120K = 415K converted to Farenheit = 287.3 F ~ 288F
Equations were likely supplied by a Turbo engineer because they are mirrored here :
https://motoiq.com/compressor-efficiency-and-more/
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