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*To*: "David G. Lawson" <dbldmnd@compuserve.com>*Subject*: Re: McMath*From*: Dave Weiss <dzweiss@worldnet.att.net>*Date*: Sat, 21 Jun 1997 11:47:22 -0700*CC*: Quattro List <quattro@coimbra.ans.net>*Sender*: owner-quattro@coimbra.ans.net

"David G. Lawson" <dbldmnd@compuserve.com> wrote: > 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. Good job bringing this to the light of day! I haven't had a chance to go through it all in the detail it deserves, but would like to build on one part that jumps right out: compressor efficiency is usually taken as (ideal work / actual work) rather than (ideal temperature gain / actual temperature gain). The MacInnes book does treat efficiency as a ratio of temperature gain, perhaps because that's a safe approximation; the rigorous method yields a lower discharge temp. Maybe one of the gurus can tell us for sure. First calculate the ideal outlet temp (as you did) for an adiabatic compression from inlet T,P to outlet P, then calculate the ideal work, and use efficiency and heat capacity to figure actual outlet temp. The notation is changed slightly to protect the guilty, and to conform to my versions of Perry et al. Let P1 = inlet pressure, psia T1 = inlet temperature, deg R = deg F + 460 P2 = outlet pressure, psia T2 = outlet temperature, deg R T2i= ideal (adiabatic) outlet temperature, deg R = T1 * r^[(k-1)/k] R = gas constant = 1544 lbforce/ (deg R lbmole) = 10.73 psi ft3 / (deg R lbmole) = many other values in different units Wi = ideal work to compress (adiabatically), ft lbf/lbmole = k/(k-1) * R * T1 * {r^[(k-1)/k]} - 1} Wa = actual work to compress, ft lbf/lbmole WL = lost work = Wa - Wi, ft lbf/lbmole (cap L so it doesn't look like a number) nu = turbo efficiency = Wi / Wa, unitless cp = heat capacity at constant pressure, ft lbf/ deg R lbmole = about 5453 ft lbf/ deg R lbmole for air k = heat capacity ratio cp/cv, unitless, about 1.395 for air r = pressure ratio P2 / P1, unitless Assume 80 F, 5000 ft above sea-level, 1.9 bar absolute manifold pressure, 65% turbo efficiency, ideal gas, ignore humidity, ignore effect of temperature on heat capacities: T1 = 540 deg R P1 = 12.2 psia (didn't check this) P2 = 1.9 * 14.504 = 27.6 psia r = 27.6 / 12.2 = 2.26 k/(k-1) = 1.395 / 0.395 = 3.53 (k-1)/k = 0.395 / 1.395 = 0.283 T2i = T1 * r^[(k-1)/k] = 540 * (2.26^0.283) = 680.2 R = 220 F --so far, so good! Wi = k/(k-1) * R * T1 * {r^[(k-1)/k] - 1} --ideal work = 3.53 * 1544 * 540 * (2.26^0.283 - 1) = 763877 ft lbf / lbmole Wa = Wi / nu --actual work = 763877 / 0.65 = 1175195 ft lbf / lbmole WL = Wa - Wi --lost work = 1175195 - 763877 = 411318 ft lbf / lbmole T2 = T2i + (WL / cp) --actual outlet T = 647.6 + 411318 / 5453 = 723.0 R = 263 F still hot stuff! Dave Weiss '91 V8 5-spd (getting ready for Blackhawk Farms next weekend) '93 90 CS UnQ

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