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watt-hour meter: finding revolutions per minute?

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  1. Bradfoxthelegend is offline Junior Member Pro Subscriber
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    watt-hour meter: finding revolutions per minute?

    This is the question from the test:

    If a watt-hour meter has a disk constant of 3.6 is used with 600:5 ampere current transformers and 20:1 potential transformers, the primary Kh is?

    The answers are in revolutions per minute. How do I get to revolutions per minute without knowing any of the output values?

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    Watthour Meter

    Quote Originally Posted by Bradfoxthelegend View Post
    This is the question from the test:

    If a watt-hour meter has a disk constant of 3.6 is used with 600:5 ampere current transformers and 20:1 potential transformers, the primary Kh is?

    The answers are in revolutions per minute. How do I get to revolutions per minute without knowing any of the output values?
    =3.6*(600/5)*(20/1)
    =8640

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    Thank you

    Quote Originally Posted by Skyehaven40 View Post
    =3.6*(600/5)*(20/1)
    =8640
    I have been looking for a way to answer that question for quite a bit. So simple and straight forward, Thank you

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  7. mcmahanrf is offline Junior Member Pro Subscriber
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    another way to think about it

    Simplify your current and voltage ratios:

    Current 600:5 = 120:1
    Voltage 20:1 = 20:1

    Now assume a supply voltage and current, pick something that makes your math easy

    Voltage:200V
    Current:120A

    Now calculate the watts for those current and voltage values for both sides of the instrument transformers:

    Primary: 200V * 120A = 24,000 Watts*hr
    Secondary: (200/20)V * (120/120)A = 10 Watts*hr

    Now using your secondary disk constant that was given (3.6, this is the number of watt*hr per revolution of the disk) we can determine the number of turns per hr:

    (10 watts*hr)/(3.6 watt*hr/turn) = 2.77777 turns (in one hour)

    Now working backwards with that turn rate we can find the primary Kh:

    (24,000 watts*hr)/(2.77777 turns) = 8640 watt*hr/turn or 8640 Kh


    Quote Originally Posted by jflan13 View Post
    I have been looking for a way to answer that question for quite a bit. So simple and straight forward, Thank you

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