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*To*: quattro <quattro@coimbra.ans.net>*Subject*: McMath*From*: "David G. Lawson" <dbldmnd@compuserve.com>*Date*: Fri, 20 Jun 1997 02:51:31 -0400*Sender*: owner-quattro@coimbra.ans.net

McMath by Dave Lawson, 1997 I know there is some interest in learning about turbos and intercoolers and the effect they have on our cars, so I spent some time "doin' the math", which I hope will build a technical foundation in which we can discuss various upgrades. A while back I set off to educate myself on what my car is doing, so I could make informed decisions on expanding the performance envelope. A few months ago I did a post, turbo.edu which listed a number of sights where those with inquiring minds could go and learn about turbos and such. Here I have literally gone through an example of 1 operating condition of an MC engine, explaining the various inputs which enter into turbocharger and intercooler analysis, along with handing out the actual math to do the calculations. If your are interested in why your car is slower driving through the arizona desert in august or why your car will drive differently cruzzin' up pikes peak to meet with Ms. Mouton, get out your calculators. If a technical analysis bores you, hit the D key. Enjoy. Take our much discussed 2.2l MC engine, the amount of air the engine consumes can be calculated using the following equation: w * D Fi = ------- (1728 * 2) where Fi = ideal air flow [ft^3/min] w = engine speed [rpm] D = engine displacement [in^3] Letting D = 135.8 and w = 6000 the ideal air flow of the MC engine is: Fi = 235.8 ft^3/min The problem here is our engine isn't 100% efficient, so we need to factor in the volumetric efficiency and calculate the actual air flow: Fa = Fi * VE Assuming the MC is 95% efficient, we get Fa = 224.0 I know this VE might be a bit high for the MC, but have never heard an actual number mentioned. Remember the ideal gas law which Bob Meyers described a few days ago. Look closely and you will see some of what Bob was describing in the math below. PV = nRT where P = absolute pressure [psig] V = volume n = number of moles R = constant = not used directly in this example T = absolute temperature [ degrees Rankin] = [deg F] + 460 Now lets throw some other conditions into the example Ta = ambient air temp [deg F] = 80 a nice summer day Pb = boost pressure [psi] = 13.2 ==> .9bar close to those IA stage II mods floating around Pi = turbo inlet pressure [psi] = 12.2 let altitude = 5000ft (because thats where I live), this effects the turbo inlet pressure, which sets Pi = 12.2 psi instead of the 14.7 psi for sea level Ei = intercooler efficiency = .7 Another number whose actual value can be debated, remember it is 0 when sitting still and is less efficient when it gets the heat soak treatment Now lets calculate the ideal temperature gain which occurs when the turbo compressor does its job and, ta-da, compresses the air Tgi = (Pr^0.283 * (Ta + 460)) - (Ta + 460) where Pr = pressure ratio = (Pi + Pb) / Pi = 2.29 <=== note this is above the upper limit which QSHIPQ has mentioned previously, 2.0, and shows why you need think in terms of pressure ratio, especially those living @ altitude, or when you are out touring in the hills Using the above described conditions we get Tgi = 142.6 [deg F] Now we need to include the turbo efficiency... Et = turbo charger efficiency = 65% This number comes from those rare compressor maps which KKK doesn't release. It's based in the design of the turbo and includes both the hot and cold side characteristics. This actual number is based on the pressure ratio and the actual air flow calculated above. See, those maps really do have a use. Like the engine, the turbo isn't 100% efficient and this effects the temperature gain. Tg = Tgi / Et = 219.5 deg F So now the actual turbo outlet temp is To = Ta + Tg = 299.5 deg F Now lets calculate the air density ratio based on the compressed air Rd = ((Ta+460)/(To+460)) * Pr = 1.63 Now let this air flow into the intercooler and calculate how much the intercooler reduces the air temperature: Tir = Tg * Ei = 153.6 deg F Now calculate the actual air temp at the intercooler outlet or intake manifold inlet: Tm = To - Tir = 145.9 deg F So the net rise in the air temp after being compressed and cooled is Tr = Tm - Ta = 65.9 deg F With this information we can calculate the intake air tract overall efficiency Eintake = (Ta+460)/(Tm+460) = 0.89 And the net air density ratio after intercooling Rm = Pr * Eintake = 2.04 And now we can calculate the effective air flow of our turbocharged, intercooled MC engines Fe = Fa * Rm = 457.0 ft^3/min By using the turbocharger/intercooler we have more than doubled the air flow our engine would produce if it was normally aspirated. Remember this example is for 1 engine speed at 1 ambient air temp and 1 boost pressure using 1 point off the K26 turbo compressor map. I leave it as an excercise to the reader to calculate the results for cold days, an engine coming off idle(why does my engine come alive @ 3000 rpm?), driving at altitude, a more efficient intercooler, etc. I have developed a spreadsheet to do all these calcs and even keep tabs on the competition. How much air does that 944 turbo S consume or what is the air flow for those 12 sec talons? We can take this further and start calculating the fuel required to mix with this compressed air or how the pressure drop across an intercooler effects the system, but... lets save some subjects to discuss next week. I do hope that owners of the 1.8L turbos have been following the discussions, as these concepts are just as applicable. I haven't yet ran the numbers for these cars, but I would be interested in the results if others have. I would also like to hear about the design of the 1.8T systems, intercooler sizes, diameters of the pipes, measured boost pressures and air temps, etc. - Dave Lawson dbldmnd@compuserve.com 83 ur-q

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