# How An Intercooler Makes You Horsepower

I took the required formulas, found some actual values from a real car, and put it together so it’s easy for you guys to see how intake manifold air temp. relates to power output.

Lets start with a 70 F day because I know most people feel that when it’s not real hot out, their cheap IC works fine. The car used in this example is an STi making 525 crank hp on a TR70 turbo kit with an APS D/R 725 intercooler. That doesn’t really matter since it’s just an example, but that’s where the numbers came from as supplied by APS.

Here are the numbers:
crank hp = 525
ambient = 70 F
compressor outlet = 415 F
intake manifold (post IC) = 105 F
Pressure Drop Across Core and all Ducting (psi) = 1.5 psi

Now the math:
Figuring out the efficiency of this intercooler:
Compressor outlet temp minus ambient temp:
415 – 70 = 345 F {this is the temperature gain from ambient to what comes out of the turbo}

Temp gain from turbo – temp dropped by intercooling + ambient temp = intake manifold temp.
345 – X + 70 = 105
X = 310 {this is the temperature dropped by intercooling}

Now that you have these numbers you can calculate the efficiency of the intercooler under these conditions:
Temp gain from turbocharging * cooling efficiency = temp drop from intercooling
345X=310
X=.898 = 89.8 % efficient
At almost 90% efficient this is working nicely.

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How does this affect power? The density charge affects power and density is related to temperature.
Dc = Density change
Dt = Compressor outlet temp
It = Intercooler outlet temp
Dc = [ (Dt + 460) / (It + 460) ] – 1
Dc = [ (415 + 460) / (105 + 460) ] – 1
Dc = (875/565) – 1
Dc = .549 (rounded to 3 places)

This means the intercooler made a 54.9% increase in density over the non intercooled charge. Some of the power increase this denser air causes is negated by pressure drop. Flow restrictions in the core and piping do play a role as well. That said, this IC only has a rated drop of 1.5 psi across the core and piping under these conditions (26 psi, above listed temps, etc.).

For anyone wondering where the number 460 came from, it’s used to convert our Fahrenheit temps to Rankine for this equation. More info here:
http://en.wikipedia.org/wiki/Rankine_scale

NOW THE BIG QUESTION!!! How much power am I making because of this intercooler?
HPr = Rise in HP
Dc = density change from the intercooler
Ap = Ambient Pressure (I’m using 14.7 for sea level)
Bm = Boost pressure at the manifold (26.1 psi)
Bc = Boost pressure at the compressor (27.6 psi)

HPr = Dc + 1 – [ (Ap + Bc) / (Ap + Bm) ]
HPr = .549 + 1 – [ (14.7 + 27.6) / (14.7 + 26.1) ]
HPr = 1.549 – (42.3 / 40.8)
HPr = 1.549 – 1.037
HPr = .512

That’s a 51.2% increase in power which basically means you’d have around 350 crank hp if you removed the intercooler, but had similar flow restriction and pressure drop. Of course that’s not realistic, but you’d still be under 400 hp without the intercooler. Sound too good to be true? It’s not…

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Now lets say this intercooler is a less efficient ebay unit, or that it’s a good unit, but a top mount which is heat soaked a bit. Again I’m going to simplify here and say the post IC intake temp is all that changes from this. Instead of a post IC temp of 105, lets say the core is at hotter than before and it’s not quite as efficient, so it only cools the air to 200 F.

Compressor outlet temp minus ambient temp: (THIS STAYS THE SAME)
415 – 70 = 345 F {this is the temperature gain from ambient to what comes out of the turbo}

Temp gain from turbo – temp dropped by intercooling + ambient temp = intake manifold temp.
345 – X + 70 = 200
X = 215 {this is the temperature dropped by intercooling}

Now that you have these numbers you can calculate the efficiency of the intercooler under these conditions:
Temp gain from turbocharging * cooling efficiency = temp drop from intercooling
345X=200
X=.58 = 58 % efficient

The density charge affects power and density is related to temperature.
Dc = Density change
Dt = Compressor outlet temp
It = Intercooler outlet temp
Dc = [ (Dt + 460) / (It + 460) ] – 1
Dc = [ (415 + 460) / (200 + 460) ] – 1
Dc = (875/660) – 1
Dc = .326 (rounded to 3 places)

In this example, the intercooler only bumped the density charge up 32.6 %

The power difference?
HPr = Rise in HP
Dc = density change from the intercooler
Ap = Ambient Pressure (I’m using 14.7 for sea level)
Bm = Boost pressure at the manifold (26.1 psi)
Bc = Boost pressure at the compressor (27.6 psi)

HPr = Dc + 1 – [ (Ap + Bc) / (Ap + Bm) ]
HPr = .326 + 1 – [ (14.7 + 27.6) / (14.7 + 26.1) ]
HPr = 1.326 – (42.3 / 40.8)
HPr = 1.326 – 1.037
HPr = .288

Now the intercooler is only making 28.8% more power than having no cooling at all (again assuming other things being equal).

The difference between the intercooler adding 51.2% power or 28.8% power is sizable in this case…about 75 hp.

I’ve kept these numbers realistic for a common situation many of you face. Now imagine it’s 80 degrees out with a track temp of 120 F and you’re at an autoX or drag event where you’re hot lapping. Think how enormous your losses can be when that intercooler gets hotter!