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Calculate power consumed by pressure drop

  1. Jul 28, 2015 #1
    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. jcsd
  3. Jul 28, 2015 #2
    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
  4. Jul 28, 2015 #3

    boneh3ad

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    That looks reasonable to me.
     
  5. Jul 28, 2015 #4
    Thanks

    PS. I miss doing this stuff!
     
  6. Jul 28, 2015 #5

    russ_watters

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    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.
     
  7. Aug 2, 2015 #6
    Good point about fan efficiency. What about the motor efficiency? A couple percent?
     
  8. Aug 2, 2015 #7

    russ_watters

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    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.
     
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