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Thread: watts per square foot calculation

  1. #1
    Join Date
    Apr 2008
    Posts
    2

    watts per square foot calculation

    The following information was obtained from the drawings available for review: the main service switchboard is rated for 1,600-ampere, 277/480-volt, 3-phase, 4-wire service with room for another identical main service switchboard. The existing configuration will provide the equivalent of approximately six watts per square foot of gross building area for this mostly open storage warehouse building with day lighting. Industrial Bldg. is 183,894 square feet and includes mezzanine.

    Ohm's Law. The potential difference Vab across a resistance R from a to b with a current I from a to b is given by Ohm's Law, an Empirical law,
    Vab = IR.

    Is there a formula to determine the watts per square foot, including the power factor? It has been too long since I worked with this and cannot make sense of it anymore. I forgot what the PF is as well. Any help is appreciated. Thanks!!

  2. #2
    Join Date
    Feb 2003
    Location
    Seattle, WA
    Posts
    16,737
    Well, first of all the fact that the gear is rated for 1600 amps does not mean that it is served by a transformer capable of supplying 1600 amps. But let’s take for granted, for now anyway, that it is.

    1600 amps times 480 volts times 1.732 (the square root of three, a term that always finds its way into three phase calculations) gives you a total of 1,330,000 VA, or 1,330 KVA. If all the loads were resistive (i.e., no motors, no inductive lighting ballasts, no welders), you could use this value as though it were the same as watts. Thus, in your building of 183,894 square feet, you have 1,330,000/183,894, or about 7.2 watts per square foot.

    If there are inductive loads, and this is likely, you would get watts from VA by multiplying VA times the power factor. Let’s assume a power factor of 0.90, just for discussion.

    1,330,000 VA times 0.9 is 1,197,000 watts. Then, 1,197,000/183,894 is about 6.5 watts per square foot.
    Charles E. Beck, P.E., Seattle
    Comments based on 2008 NEC unless otherwise noted.

  3. #3
    Join Date
    Jul 2007
    Posts
    205
    Yeh but for this task you don't want voltage you want power.

    so do P=VI

    and you can include power factor right in the equation so its P=VI*PF if you so desired although most people I know just use unity (1) power factor and convert the power into KVA.

  4. #4
    Join Date
    Apr 2008
    Posts
    2

    Thanks guys!

    Thanks guys!

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