It is our turbochargers that allow our little 181 CI (3 L) engines to make 360 CI (6 L) power. However, a penalty comes with that power ... heat. The turbochargers can heat the air hot enough to boil water. To determine how hot, we need to understand adiabatic processes. You can look up "adiabatic" in your physics texts but you won't find the formula that pertains to our particular situation. I'll show you how to derive the expression we need. Then we'll see how hot that air really gets.
An adiabatic process is one in which heat does not enter or leave the system. Though this is not strictly true in a turbocharger, it is a close approximation over a very short time period. An adiabatic temperature or pressure change can be calculated from the volume change using the following two equations (which usually are found in your physics texts):
(T2/T1) = (V1/V2)g-1 and (P2/P1) = (V1/V2)g,
where g is the ratio of air's heat capacity at constant pressure to air's heat capacity at constant temperature; T1, P1, and V1 are the initial temperature, pressure, and volume; and T2, P2, and V2 are the final temperature, pressure, and volume. Taking the log of both of these equations, and knowing that log ab = b log a, the following equation relating adiabatic temperature change to pressure change can be derived:
(T2/T1) = (P2/P1)[(g-1)/g].
Physics texts will tell you that for the diatomic gasses like N2 and O2 in air, g is assigned a value of 1.4. By substituting and rearranging we get
T2 = T1 x (P2/P1)0.286.
Actually, in the above equation the exponent that you usually see is 0.283, meaning that g is closer to 1.395. You can use either number, the difference it makes in the calculations is about 0.25%. We need to note that temperature in all of these equations must be absolute temperature in either Kelvin (Centigrade scale) or Rankine (Fahrenheit scale). If you want to work in 0C then 0K = 0C + 273.15. If you are more comfortable with 0F then 0R = 0F + 459.69. Also note that one degree on the Kelvin scale equals one degree on the Centigrade scale, and one degree on the Rankine scale equals one degree on the Fahrenheit scale. I prefer the Fahrenheit scale and will round off 459.69 to just 460 for the examples here. I'll assume that we are at sea level at a standard pressure of 14.7 psi with an ambient air temperature of 700F.
The expression (P2/P1) represents the turbocharger compressor pressure ratio (PR), where P1 is atmospheric pressure on the inlet side and P2 is the absolute pressure on the outlet side. T1 is the air temperature entering the compressor and T2 is the air temperature leaving the compressor.
Many people think that 12 psi in the manifold is a moderate boost level so I'll use that as an example. If we can assume there is a 2 psi pressure loss from the turbo to the manifold, then the PR at the compressor is about 1.95 ( (14 +14.7)/14.7). Let's also assume we manage to keep the intake air temperature down to 800F (or 5400R). The air temperature leaving the compressor, due to adiabatic heating only, would be 1940F which is 6540R = 5400R x (1.95)0.286. This is an increase of 1140F! But the air is actually hotter than that.
Turbochargers compress the air by increasing the velocity of the molecules. When the air molecules don't move in a direction toward the intercooler (IC), they serve only to heat the air. The more the air is heated above that predicted by adiabatic compression, then the less efficient the turbocharger is. Compressor flow maps usually show at various flow amounts and pressure ratios the measured efficiency of a turbocharger. Compressors usually operate in the 55-75% efficiency range depending on engine load, pressure ratio, turbine speed, and compressor size (65-70% is average, 75% or more is where we would like to be).
When we factor in compressor efficiency (CE) the final temperature of the air leaving the compressor is
T3 = T1 + (T2-T1)/CE.
The intercooler of course is supposed to cool the air. Intercooler efficiency, IE, can be calculated by measuring the air temperature going into the intercooler, Tin, the air temperature leaving the intercooler, Tout, and the ambient air temperature, Tamb.
IE = (Tin-Tout)/(Tin-Tamb) or Tout = Tin - IE x (Tin-Tamb).
High efficiency levels are near 90% and lower levels are toward 50-60%. If an 80% efficient intercooler had been used in the above example, then manifold air temperature at 12 psi boost would have dropped to about 1040F = 2420F – 0.8 x (242-70)0F. Intercooler air cooling by the way is not an adiabatic process. Ideally, heat is removed from the air without a corresponding decrease in pressure. If you think about it, you'll see that intercooler efficiency is far more important than the efficiency of the turbocharger.