Calculate power consumed by pressure drop

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1. Jul 28, 2015

NotionCommotion

Hello all! I am interested in how much energy a HVAC VAV box consumes. Can anyone help me come up with the following equation? I don't need to know the exact formula, and can assume an uncompressible fluid and/or ideal gas. Been about 20 years since I have done this and am a bit rusty. Thank you

How many watts are utilized by 1,000 CFM of 55 F air at 2" water pressure above atmosphere going through a restriction that results in a static pressure drop of 0.2" wp over a duration of 1 year?

2. Jul 28, 2015

NotionCommotion

Actually, looks like not watts but watt hours, right?

Do I have this correct?

p=0.2" wc = 50 Pa = 50 n/m^2
Q=1,000 CFM = 0.5 m^3/s
P=p x Q = 25 n-m/s = 25 j/s
For 365 days, E = 788x10^6 j = 220 kw-h

Thanks

Last edited: Jul 28, 2015
3. Jul 28, 2015

That looks reasonable to me.

4. Jul 28, 2015

NotionCommotion

Thanks

PS. I miss doing this stuff!

5. Jul 28, 2015

Staff: Mentor

Starting in English units, it is:
Dp*CFM/6356=hp

Conversion: 746 watts/hp.

I get 206 kWh. Note also, this doesn't include fan efficiency. For a well operating commercial fan, assume 65%. That puts you up to 317 kWh.

6. Aug 2, 2015

NotionCommotion

Good point about fan efficiency. What about the motor efficiency? A couple percent?

7. Aug 2, 2015

Staff: Mentor

Larger motors running near full speed get upwards of 95%. A VFD is about 98% efficient. The pulley-drive on a fan, 90-95% efficient.