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HP Calculation for AHU Motor

12/03/2009 8:06 AM

Hello, how to calculate the motor HP for chilled water air handling unit (AHU) central air conditioning system. Is that any formula or rule of thumb we can calculate from cfm and external static pressure to get the elect hp. Tq.

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Guru
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#1

Re: Electrical HP for Air Handling Unit Motor

12/03/2009 9:22 AM

The power requirement for any fluid mover is the pressure rise multiplied by the volumetric flowrate divided by the efficiency.

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Power-User
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#2

Re: Electrical HP for Air Handling Unit Motor

12/03/2009 11:22 AM

Fan bhp = (CFM x TSP)(1 + Drive Loss %) / (6356 x static eff)

Note that you need to use TSP not ESP (in WC), so you will need a rough idea of the AHU internal losses (~2" WC?). Use about 3% drive loss for belt drive. You will also need to know the static efficiency of the fan, which will vary significantly depending on the type of fan.

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Anonymous Poster
#4
In reply to #2

Re: Electrical HP for Air Handling Unit Motor

12/04/2009 8:48 AM

Tqvm bro, for the guidence.

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Guru

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#3

Re: HP Calculation for AHU Motor

12/04/2009 12:26 AM

Expanding on DAC1267-

Just to be sure you know terms-

TSP = Total Static Pressure (discharge pressure minus inlet pressure (negative) plus internal losses in coils, filters, housing)

ESP = External Static Pressure (discharge pressure minus inlet pressure (negative)) only

One other factor that MIGHT be an issue is air density- are you at or near (+/- 500 feet) of sea level with "normal" (less than 100F) inlet air and normal (above 50F) outlet air? Then no problem. If you are more than 500 feet above seal level or higher than 100F inlet air, then your operating HP will be lower because you will be pumping less dense (lower mass) air. If you are discharging colder than 50F air, then the HP will be higher due to higher density (more mass per unit volume) air.

You can easily make the pressure adjustment by multiplying the formula results by the ratio of actual air pressure (inches mercury) divided by 30 (air pressure are sea level).

The temperature adjustment is the ratio of actual air temperature (F) plus 460 (the temperature difference between "0" F and absolute "0") divided by 528 (460 plus 68F (the temperature of "standard air")).

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