# 50 hp Power Cube Calculations

Prof. Fling;

I've given some thought to the prior calculations - they arrive at a displacement of ~1.17 CIPR, but if I make a Power Cube with this size pump, it won't produce suitable flow. So, I've decided to use a "rule of thumb" to size the pump, basing it on prior experience with engines and the OSE LifeTrac tractor. It goes like this: In the past, the preferred power for the LifeTrac has employed ~54 horsepower @ 3600 RPM to drive two pumps, each about 0.92 cu in displacement. This has proven satisfactory and is the rule against which the new Power Cube will be measured. The total pump displacement was 1.84 CIPR (Cubic Inches Per Revolution). This gives a ratio of HP:Flow of 54:(1.84 * 3600).

For this new Power Cube, the engine will be producing between 50 and 60 hp @ 2700 RPM (the RPM yielding max torque). So, using the above ratio to derive the new displacement as follows for 50 and 60 HP: 54:(1.84 * 3600) = 50:(R * 2700) -or- R = 50 * 1.84 * 3600 / (54 * 2700) = 2.271 CIPR

54:(1.84 * 3600) = 60:(R * 2700) -or- R = 60 * 1.84 * 3600 / (54 * 2700) = 2.726 CIPR So, this "rule of thumb" measurement puts the displacement at about 2.5 CIPR and 28.3 GPM flow: 2.5 * 0.97 * 2700 / 231 = 28.34 I know this doesn't match industry numbers, but I think it is a realistic estimate for this application.