karlosg Posted October 18, 2005 Posted October 18, 2005 Does anyone have any flow rate charts of the various side mount intercoolers available ?? Quote
Paul Stevens Posted October 18, 2005 Posted October 18, 2005 Does anyone have any flow rate charts of the various side mount intercoolers available ?? http://www.tdm-works.com/ hope thats some use to you , click on intercooler info when page comes up scroll down a bit then youl see loads of info and good comparison of makes of intercoolers Quote
BlackMagic Posted October 18, 2005 Posted October 18, 2005 was this of any use to anyone ??? Interesting read :) Quote
TopLess Posted October 18, 2005 Posted October 18, 2005 Interesting but does it just give info for smic or fmic or both? Quote
karlosg Posted October 18, 2005 Author Posted October 18, 2005 yeah cheers almost solved my query Quote
karlosg Posted October 18, 2005 Author Posted October 18, 2005 Interesting but does it just give info for smic or fmic or both? yep has both Quote
TopLess Posted October 18, 2005 Posted October 18, 2005 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? Quote
karlosg Posted October 18, 2005 Author Posted October 18, 2005 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 Quote
HeXaKing Posted October 18, 2005 Posted October 18, 2005 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. Quote
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