Calculate power consumed by pressure drop

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SUMMARY

This discussion focuses on calculating the energy consumption of a HVAC Variable Air Volume (VAV) box, specifically how much energy is utilized by 1,000 CFM of 55°F air at a static pressure drop of 0.2" water column (wc). The calculations provided indicate that the energy consumed over one year is approximately 220 kWh, considering a fan efficiency of 65%, which raises the total to about 317 kWh. The discussion also highlights the importance of accounting for motor efficiency, which can reach up to 95% for larger motors and 98% for Variable Frequency Drives (VFDs).

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  • Basic knowledge of fluid dynamics and pressure drop calculations
  • Familiarity with energy consumption metrics, including watt-hours and kilowatt-hours
  • Knowledge of efficiency ratings for fans and motors in HVAC applications
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  • Explore pressure drop calculations in duct systems for HVAC applications
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HVAC engineers, energy analysts, and anyone involved in optimizing the performance and efficiency of heating, ventilation, and air conditioning systems.

NotionCommotion
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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?
 
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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:
That looks reasonable to me.
 
Thanks

PS. I miss doing this stuff!
 
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.
 
Good point about fan efficiency. What about the motor efficiency? A couple percent?
 
NotionCommotion said:
Good point about fan efficiency. What about the motor efficiency? A couple percent?
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.
 

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