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Posted
well it does and it doesn't seems to only give info mostly on side mounts and then compares them to a few front mounts, not fair really?

 

just looked again , yeah seems a bit one sided , but at least helps if your after a s/m

Posted

Just done some quick calculations, hope they help :rofl: :hyper: :confused:

 

Stock intercooler, stock turbo.

Given 40 lb/min air flow @ 300 deg F and 19 psig from the turbo to make 15 psig boost in the intake manifold; 85 deg F outside temperature; an intercooler outlet temperature of 140 deg F has been measured, as has the cooling air temperature of 160 deg. What is the UA of the stock intercooler?

 

First, calculate Q

Q = m * Cp * DT

Q = 40 lb/min * 0.25 BTU/lb-F * (300-140 F) = 1600 BTU/min

 

Calculate DTlm

DT1 = turbo air temperature in - outside air temperature out = 300 - 160 = 140

DT2 = turbo air temperature out - outside air temperature in = 140 - 85 = 55

P=0.74, R=0.47, F=0.875

 

DTlm = F*(DT1-DT2)/ln(DT1/DT2) = 0.875*(140-55)/ln(140/55) = 74.4/0.934=79.6 F

 

Calculate UA

UA = Q/DTlm = (1600 BTU/min)/79.6 F = 20.1 BTU/min-F

 

What is the cooling air flow?

Q = m * Cp * DT, or Q/(Cp * DT) = m,

m = (1600 BTU/min)/[0.25 BTU/lb-F * (160-85 F)] = 85.33 lb/min of outside cooling air

 

Stock intercooler, big turbo

How will the same stock intercooler perform with a bigger turbo and more boost?

Given 53 lb/min @ 350 deg F and 27 psig from the turbo to make 22 psig in the intake; 85 deg F outside temperature. Cooling air flow is still 85.33 lb/min.

 

This requires some trial and error to solve since we don't know the intercooler outlet temperature. There IS a way to calculate it directly, but that involves some more equations and is a little tedious so I'll skip it and do it the hard way, by assuming an intercooler outlet temperature and then checking to see if it is right. I'll do that by calculating Q for the overall exchanger and then Q for just the turbo air; if they come out the same then my guess was correct.

 

m=53 lb/min, Cp=0.25, U*A=20.1

lets start by assuming that the intercooler outlet temp = 140

Q = m * Cp * DT

Then DT = (350 - 140 ) = 210 and Q = 2782.5 BTU/min

Cooling air flow = 85.33 lb/min

DT for the cooling air = Q/(m*Cp)

DT = 2782.5 BTU/min / (85.33 lb/min * 0.25 BTU/lb-f) = 130.4 F

since DT = T out - T in, then 130.4 = T out - 85 and T out = 215.4 F

 

So the cooling air inlet is 85 F and the outlet is 215.4 F, and the turbo air inlet is 350 F and the outlet is assumed to be 140 F. Now calculate DTlm:

P=0.792, R=0.62, and F=0.75

DT1=134.6, DT2=55

DTlm=(134.6-55)/ln(134.6/55) * 0.75 = 66.7

 

Now calculate a new Q, Q= UA * DTlm

Q=20.1*66.7=1340.7

 

Since this isn't the same Q we got when we assumed an outlet temp of 140 deg, we have to get a new outlet temp and run through all this again.

I'll assume a new intercooler outlet temp of 170.

Q=(m*Cp*DT)=2385

cooling air DT = 2385/(85.33*0.25) = 111.8

Cooling air outlet = 85 + 111.8 = 196.8

P=0.68, R=0.62, F=0.84

DTlm=97.3

Q=1954.7 still not close enough

 

Last try!

T IC out = 182

Q=(m*Cp*DT)=2226

cooling air DT = 2226/(85.33*0.25) = 104.4

Cooling air outlet = 85 + 104.4 = 189.4

P=0.63, R=0.62, F=0.88

DTlm=111.0

Q=2232 close enough

 

Well, this time the Q we guessed at (by guessing the IC outlet temp) and the Q we calculated from the overall equation are pretty close, so we can say we've found the answer. It appears that this intercooler, which worked fine in a basically stock application (cooling the air to the intake manifold to 140 deg F) isn't working as well in this high HP application, being able to cool the air down to only 182 deg!

 

Last example: same turbo and air flow as before, but we have a new intercooler with the same heat transfer coefficient but 50% more area (intercooler and a half). We'll assume that it also flows 1.5 times the cooling air flow.

 

U * A old IC = 20.1

U * 1.5 * A = 1.5 * 20.1 = 30.15 = UA for new intercooler

m turbo air = 53 lb/min, Cp = 0.25 BTU/lb-F, T in = 350 deg

m cooling air = 1.5 * 85.33 = 128 lb/min, Cp = 0.25 BTU/lb-F, T in = 85 F

 

Assume intercooler outlet temp = 140 F

Q = m*Cp*DT = 53 * 0.25 * (350-140) = 2782.5

cooling air DT = 2782.5/(128*0.25) = 87

Cooling air outlet = 85 + 87 = 172

P=0.79, R=0.41, F=0.85

DTlm=89.0

Q=2684 not too bad, we'll try it once more

 

Assume intercooler outlet temp = 142 F

Q = m*Cp*DT = 53 * 0.25 * (350-142) = 2756

cooling air DT = 2782.5/(128*0.25) = 86.1

Cooling air outlet = 85 + 86.1 = 171.1

P=0.78, R=0.41, F=0.86

DTlm=91.7

Q=2763 close enough

 

So this tells us that in this high performance car the intercooler-and-a-half outlet temperature is about the same as the outlet temperature of the stock turbo/stock intercooler car.

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