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Thread: Converting Kwh to KVA

  1. #1
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    Converting Kwh to KVA

    I'm wanting to find out if I'm misunderstanding something in this particular conversion process. I believe that if you know the energy usage and the time span it was used, it is a fairly simple matter (using the / to denote division and * to denote multiplication):

    kwh/time=kw

    Using dimensional analysis, as long as you use the proper unit conversions and they cancel this conversion will yield kw (i.e. if time is in days, you need to multiply the number of days by 24 to obtain the resultant hours which will then cancel with the hour in kwh). Then, using an assumed power factor:

    kw/pf=kva

    and

    kva/(voltage*1.73)=three phase current

    I've always understood that kwh is actually a combination of units from two different unit systems, the SI unit is watt and the standard English unit is hour and there was no direct conversion from kwh to kva that can occur.

    But the above analysis follows every mathematical convention I was ever taught in school. Can anyone help me to see if there is flawed logic here? Thanks.

  2. #2
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    Lightbulb

    Lets go with time elapses, 1st.
    The 6 second, differential
    Last edited by cadpoint; 06-29-07 at 08:10 PM.
    If your even thirsty, your two quarts low.

  3. #3
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    Quote Originally Posted by cadpoint
    Lets go with time elapses, 1st.
    The 6 second, differanctial
    One more time in English please.

  4. #4
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    KWH/time = average KW for that time period. Using an assumed average power factor will get you average KVA for that time period.

    Using average values with a varying load can cause problems in some calculations.

    Otherwise, I don't see an issue with your logic.
    Bob Wilson

  5. #5
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    Quote Originally Posted by vw55
    I've always understood that kwh is actually a combination of units from two different unit systems, the SI unit is watt and the standard English unit is hour . . . .

    A “watt” is indeed an SI unit (i.e., metric). Its fundamental units are:
    1 watt = 1 (kilogram) (meter squared) / (second cubed).

    The basic unit of measure for time is the second, both in SI Units and in English Engineering Units.

    Quote Originally Posted by vw55
    . . . and there was no direct conversion from kwh to kva that can occur.

    There cannot be a “direct conversion” between KWH and KVA. They are different parameters. It is like trying to get a direct conversion between MPH (miles per hour) and feet.

    Your process of deriving KVA from KWH, a process that requires knowing two other parameters (time and power factor) is valid.
    Charles E. Beck, P.E., Seattle
    Comments based on 2014 NEC unless otherwise noted.

  6. #6
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    dimensional analysis

    Using dimensional analysis,

    Its well layed out

    http://www.physics.uoguelph.ca/tutorials/dimanaly/

    Quote Originally Posted by vw55
    ...
    as long as you use the proper unit conversions and they cancel this conversion will yield kw (i.e. if time is in days, you need to multiply the number of days by 24 to obtain the resultant hours which will then cancel with the hour in kwh). Then, using an assumed power factor:

    kw/pf=kva

    and

    kva/(voltage*1.73)=three phase current

    I've always understood that kwh is actually a combination of units from two different unit systems, the SI unit is watt and the standard English unit is hour and there was no direct conversion from kwh to kva that can occur.
    from the Web Site

    "Most physical quantities can be expressed in terms of combinations of five basic dimensions. These are mass (M), length (L), time (T), electrical current (I), and temperature, represented by the Greek letter theta (q). These five dimensions have been chosen as being basic because they are easy to measure in experiments. Dimensions aren't the same as units."

    Quote Originally Posted by vw55
    But the above analysis follows every mathematical convention I was ever taught in school. Can anyone help me to see if there is flawed logic here? Thanks.
    from the Web Site

    "An equation in which each term has the same dimensions is said to be dimensionally correct. All equations used in any science should be dimensionally correct. The only time you'll encounter one which isn't is if there is an error in the equation. So dimensional analysis is a valuable tool in helping you to detect an equation in which you made an error in algebra,.... "

    Be advised that I didn't mean to be short with you ...
    If your even thirsty, your two quarts low.

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