mikeb
11-13-2006, 01:08 PM
Say I have a turbo with the following compressor map:
http://home.earthlink.net/~mvburress/map.jpg
Looking at the A. Graham Bell book Forced Induction Performance Tuning on selecting a turbo by a compressor map we find the following formula to calculate CFM of airflow:
CFM = (L X rpm x VE x Pr) / 5660
Where:
L = Engine size in litres
rpm = max rpm
VE = volumetric efficiency
Pr = pressure ratio
So lets say we have a 7.15 litre engine, max rpm of 6700, a VE of 93 (?), and pressure ratio of 1.68 (10 lbs of boost).
(pressure ratio is calculated by adding 14.7 to your desired boost and then dividing the result by 14.7)
So 7.15 * 6700 * 93 * 1.68 = 7484677.2
Divide by 5660 and the answer is 1332 cfm.
To convert cfm to lb/min you multiply by .07
.07 * 1332 = 93 lb/min
Map 93 lb/min with a pressure ratio of 1.68 on the compressor map and you are off the right side of the map. Not good.
If you redo the calculations using a 4000 rpm max value you get 55 lbs/min, which does map into the 74% efficiency island at 1.68 pressure ratio. Better.
So what i'm surmising is that this turbo on the hypothetical engine would make good torque down low (or would it? impeller speed seems to be on the slow side and we are way to the right of the surge line) but would begin to run out of steam just over 5000 rpm. In other words it would be a good street turbo on a larger engine.
Am I interpeting this correctly?
http://home.earthlink.net/~mvburress/map.jpg
Looking at the A. Graham Bell book Forced Induction Performance Tuning on selecting a turbo by a compressor map we find the following formula to calculate CFM of airflow:
CFM = (L X rpm x VE x Pr) / 5660
Where:
L = Engine size in litres
rpm = max rpm
VE = volumetric efficiency
Pr = pressure ratio
So lets say we have a 7.15 litre engine, max rpm of 6700, a VE of 93 (?), and pressure ratio of 1.68 (10 lbs of boost).
(pressure ratio is calculated by adding 14.7 to your desired boost and then dividing the result by 14.7)
So 7.15 * 6700 * 93 * 1.68 = 7484677.2
Divide by 5660 and the answer is 1332 cfm.
To convert cfm to lb/min you multiply by .07
.07 * 1332 = 93 lb/min
Map 93 lb/min with a pressure ratio of 1.68 on the compressor map and you are off the right side of the map. Not good.
If you redo the calculations using a 4000 rpm max value you get 55 lbs/min, which does map into the 74% efficiency island at 1.68 pressure ratio. Better.
So what i'm surmising is that this turbo on the hypothetical engine would make good torque down low (or would it? impeller speed seems to be on the slow side and we are way to the right of the surge line) but would begin to run out of steam just over 5000 rpm. In other words it would be a good street turbo on a larger engine.
Am I interpeting this correctly?