Required flowbench blower drive horsepower

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Postby 84-1074663779 » Sun Nov 13, 2005 7:08 pm

Assuming an 100% efficient blower (impossible) the horsepower required to compress air thermodynamically. Temperature and barometric pressure will affect this, but for simplicity let's assume standard air conditions.

Hp = CFM x inches of water / 6339

By dividing that figure by blower efficiency you get the required total shaft drive power of the blower rotor(s). If the actual blower efficiency is not known, assume 40% as that is a fairly typical conservative figure for most vacuum motors and superchargers. You may actually do a little better than that, but don't count on it at the design stage.

Motor output horsepower can then be converted to watts by multiplying by 745.7

Motor input power = Motor output watts / motor efficiency

If the motor efficiency is not known, assume 85%. A more accurate figure can be worked out from the motor rating plate, if it has one.

Total Amps = Motor input watts / mains supply voltage




Here is a worked through example:

Hp = 500 CFM x 45 inches of water / 6339
Hp = 3.549

Shaft power = 3.549 / blower efficiency
Shaft power = 3.549 / 0.4
Shaft power = 8.8725 Hp

8.8728 Hp x 745.7 = 6,616 watts

Motor input power = 6,616 watts / motor efficiency
Motor input power = 6,616 watts / .85
Motor input power = 7,783 watts

Total Amps = Motor input watts / mains voltage
Total Amps = 7,783 / 240v
Total Amps = 32.43 A

So you can see a decent sized flow bench can be very power hungry !
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