Turbo Compressor Map Point Plotting
05-11-04 | 02:03 AM
#1
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Joined: Jul 2003
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From: Contra Costa County, CA
Turbo Compressor Map Point Plotting
This is the mathematics required to compressor point plotting, this is from a stock 13B motor plotted @ 7000 RPMs. Enjoy and good luck plotting your own motor. All you have to do here is switch around some numbers and you'll be all set. Sorry if it is kind of cryptic.
Absolute Pressure out of the Compressor (Pco)
Intercooler Pressure Drop is Approx 1.5PSI
Pco = Boost + Atmospheric Pressure + Intercooler Pressure Drop
Pco = 10PSI + 14.7PSI + 1.5PSI = 26.2 PSI
Pressure Ratio = Pr
Pr = Pco =26.2PSI= 1.78231
Atmospheric Pressure 14.7PSI
Pr = 1.7823129251706680272108843537415
R = 53.3 (R is from ideal gas law PV = nRT)
Di = Boost Pressure + Atmospheric Pressure
R x 12 x (460 + Post-Intercooler Temp)
Di=10PSI + 14.7PSI = 24.7PSI = 6.5454 x 10 ^ -5
53.3 x 12 x (460° + 130°) 377364
Di = 6.545404437095218409811216756 x 10 ^ -5
Mf = Di x Displacement in cubic inches x RPM
2 x Volumetric Efficiency
Mf = 6.545x 10 ^ -5 x 80 x 7000 RPM = 20.3625 lbs/min
2 x .90
Cmf = Corrected Mass Flow According to the Compressor Map
545°R = 85°F
CMf = Mf x (Square Root of 545°R (Compressor Inlet Temp in R°)/545°R)
Atmospheric Pressure
Compressor Inlet Pressure
CMf = 20.3635 Lb/min x 1 = 19.32456LB/min
. 14.7PSI
. 13.95PSI
CMf = 19.32456 lb/min
After all that mathematics you plot the points onto the compressor map. The map I used was a Garrett GT28RS Turbo Compressor Map.
Absolute Pressure out of the Compressor (Pco)
Intercooler Pressure Drop is Approx 1.5PSI
Pco = Boost + Atmospheric Pressure + Intercooler Pressure Drop
Pco = 10PSI + 14.7PSI + 1.5PSI = 26.2 PSI
Pressure Ratio = Pr
Pr = Pco =26.2PSI= 1.78231
Atmospheric Pressure 14.7PSI
Pr = 1.7823129251706680272108843537415
R = 53.3 (R is from ideal gas law PV = nRT)
Di = Boost Pressure + Atmospheric Pressure
R x 12 x (460 + Post-Intercooler Temp)
Di=10PSI + 14.7PSI = 24.7PSI = 6.5454 x 10 ^ -5
53.3 x 12 x (460° + 130°) 377364
Di = 6.545404437095218409811216756 x 10 ^ -5
Mf = Di x Displacement in cubic inches x RPM
2 x Volumetric Efficiency
Mf = 6.545x 10 ^ -5 x 80 x 7000 RPM = 20.3625 lbs/min
2 x .90
Cmf = Corrected Mass Flow According to the Compressor Map
545°R = 85°F
CMf = Mf x (Square Root of 545°R (Compressor Inlet Temp in R°)/545°R)
Atmospheric Pressure
Compressor Inlet Pressure
CMf = 20.3635 Lb/min x 1 = 19.32456LB/min
. 14.7PSI
. 13.95PSI
CMf = 19.32456 lb/min
After all that mathematics you plot the points onto the compressor map. The map I used was a Garrett GT28RS Turbo Compressor Map.
05-11-04 | 04:57 PM
#2
Rotary Enthusiast
Joined: Dec 2001
Posts: 1,209
Likes: 5
From: Delaware
although you calculated some variables out to 20 significant digits, actual mass flow is about 50% more than your values!
I used same 10 psi boost at 7000 rpm, 85/14.7 ambient, 1.5 IC drop, 90% ve, 130F at manifold, and 13.95/85 garrett ref's.
My spreadsheet yields 31.01 lb/m corrected flow.
In your Mf equation, the VE goes up top. The 2 below is for boingers, so double the rotary displacemet or use a 1 to reflect rotary displacement is based on one rev at 100% ve.
I used same 10 psi boost at 7000 rpm, 85/14.7 ambient, 1.5 IC drop, 90% ve, 130F at manifold, and 13.95/85 garrett ref's.
My spreadsheet yields 31.01 lb/m corrected flow.
In your Mf equation, the VE goes up top. The 2 below is for boingers, so double the rotary displacemet or use a 1 to reflect rotary displacement is based on one rev at 100% ve.
05-11-04 | 08:22 PM
#3
Thread Starter
Looking for a project...
Joined: Jul 2003
Posts: 138
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From: Contra Costa County, CA
Ok, so the general equation wont' work when specified towards a rotary motor. Just a piston. Could you write out the equation for me then so I could rework and replot. Thanks.
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