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Watts Calculation Method for 240V Circuits

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  • Cybatrex
    replied
    Originally posted by AtTheKeyboard View Post
    I might be over-answering on this one, please forgive me if so.

    Besoeker is correct - this is in units of VA, not watts, based on the inputs. Watts would actually be equal or less, but isn't relevant to this problem.



    Method 1 is good, but Method 2 is toast.

    With a 120/240 split-style transformer, each Line-to-Neutral leg is 120V, so it's 34.1 Amps (added across 120V legs), times 120V, not 240V, which is why the result doubled. Another way to look at it is - The average current on each side of the transformer - at (17.1+17.0)/2 - times the line-to-line voltage, 240V. 17.05A * 240V = 4,092VA

    Revised Method 2 - 34.1A * 120V = 4,092VA
    Thats exactly what I needed to see. Perfect, thank you so much.

    Leave a comment:


  • Cybatrex
    replied
    You all have been tremendously helpful. Thank you so much!

    Leave a comment:


  • ggunn
    replied
    Originally posted by Besoeker View Post
    A point in passing. I usually use the letter or symbol for the unit - W rather that Watts.
    This stems from losing a mark on an exam question for spelling out the unit in the answer.
    I was told that spelling it out was the incorrect thing to do. IDK whether that's true or not. But it stuck with me.
    IMO you had a jerk for an instructor who warped you for life.

    Leave a comment:


  • ggunn
    replied
    Originally posted by kwired View Post
    I would say that is "more accurate" if there is a neutral conductor involved.

    If it is just a two wire circuit - then it is simply V x A. The fact he measured .1 amps difference on one line vs the other is an accuracy issue with measuring method - the current has to be the same on both lines as there is no where for any differential current to go to, outside of potential fault current issues, but if those are present we have an unintended multiwire circuit and current can be flowing in more then just a single loop.
    Kirchoff lives!

    Leave a comment:


  • kwired
    replied
    Originally posted by Besoeker View Post
    If it is 120-0-120V then I suppose L1*120 + L2*120 would be the more accurate. But the 0.1A would be in the neutral but probably not significant. If is is actually there.
    I would say that is "more accurate" if there is a neutral conductor involved.

    If it is just a two wire circuit - then it is simply V x A. The fact he measured .1 amps difference on one line vs the other is an accuracy issue with measuring method - the current has to be the same on both lines as there is no where for any differential current to go to, outside of potential fault current issues, but if those are present we have an unintended multiwire circuit and current can be flowing in more then just a single loop.

    Leave a comment:


  • AtTheKeyboard
    replied
    I might be over-answering on this one, please forgive me if so.

    Besoeker is correct - this is in units of VA, not watts, based on the inputs. Watts would actually be equal or less, but isn't relevant to this problem.

    Originally posted by Cybatrex View Post
    Method 1:
    L1: 17.1A x 120V = 2,052 Watts
    L2: 17.0A x 120V = 2,040 Watts

    Total Watts1: 4,092 Watts

    Method 2:

    L1: 17.1 Amps
    L2: 17.0 Amps

    Lt = 34.1 Amps x 240V = 8,184 Watts

    Total Watts2: 8,184 Watts
    Method 1 is good, but Method 2 is toast.

    With a 120/240 split-style transformer, each Line-to-Neutral leg is 120V, so it's 34.1 Amps (added across 120V legs), times 120V, not 240V, which is why the result doubled. Another way to look at it is - The average current on each side of the transformer - at (17.1+17.0)/2 - times the line-to-line voltage, 240V. 17.05A * 240V = 4,092VA

    Revised Method 2 - 34.1A * 120V = 4,092VA

    Leave a comment:


  • Besoeker
    replied
    Originally posted by Ingenieur View Post
    that is more accurate
    in this case the currents happened to be balanced
    not always (or even usually) the case
    Yes. And Gar's point about it being VA rather than W should be noted.

    A point in passing. I usually use the letter or symbol for the unit - W rather that Watts.
    This stems from losing a mark on an exam question for spelling out the unit in the answer.
    I was told that spelling it out was the incorrect thing to do. IDK whether that's true or not. But it stuck with me.

    Mods, forgive the digression.

    Leave a comment:


  • kwired
    replied
    Originally posted by Cybatrex View Post
    I need assistance with determining my Watts correctly with Ohm's law using the amperage reading from my Fluke clamp meter. I welcome anyone who can correct the errors of my ways and offer a brief explanation so I can learn correctly.

    Amperage Readings
    L1: 17.1 Amps
    L2: 17.0 Amps

    Ohms Law
    P(w) = V x I


    Method 1: Calculated ohms law for each leg of the circuit, calculated them individually, then added the wattage together.
    L1: 17.1A x 120V = 2,052 Watts
    L2: 17.0A x 120V = 2,040 Watts

    Total Watts1: 4,092 Watts


    Method 2: Combined amperage for both legs and used 240V by combined average to calculate watts which is double Method 1.
    L1: 17.1 Amps
    L2: 17.0 Amps

    Lt = 34.1 Amps x 240V = 8,184 Watts

    Total Watts2: 8,184 Watts





    Any help with this would be greatly appreciated.
    If you only have a two wire circuit you can't have a different current on one conductor then you have on the other, there is only one current path and current is same everywhere in the circuit. I'm sure accuracy of the measuring method is what resulted in the minor difference you did measure. That said and as others have been pointing out - you have ~17 amps at 240 volts and not ~17 amps at 120 volts - times two.

    Add a third point in the system it gets more complicated, especially if you have 208/120 system. That situation you could possibly have 17 amps at 120 volts - times two, or even three. But that never can happen when there is only a two wire circuit.

    Leave a comment:


  • Ingenieur
    replied
    Originally posted by Besoeker View Post
    If it is 120-0-120V then I suppose L1*120 + L2*120 would be the more accurate. But the 0.1A would be in the neutral but probably not significant. If is is actually there.
    that is more accurate
    in this case the currents happened to be balanced
    not always (or even usually) the case

    Leave a comment:


  • Besoeker
    replied
    Originally posted by Cybatrex View Post
    I need assistance with determining my Watts correctly with Ohm's law using the amperage reading from my Fluke clamp meter. I welcome anyone who can correct the errors of my ways and offer a brief explanation so I can learn correctly.

    Amperage Readings
    L1: 17.1 Amps
    L2: 17.0 Amps

    Ohms Law
    P(w) = V x I


    Method 1: Calculated ohms law for each leg of the circuit, calculated them individually, then added the wattage together.
    L1: 17.1A x 120V = 2,052 Watts
    L2: 17.0A x 120V = 2,040 Watts

    Total Watts1: 4,092 Watts


    Method 2: Combined amperage for both legs and used 240V by combined average to calculate watts which is double Method 1.
    L1: 17.1 Amps
    L2: 17.0 Amps

    Lt = 34.1 Amps x 240V = 8,184 Watts

    Total Watts2: 8,184 Watts





    Any help with this would be greatly appreciated.
    If it is 120-0-120V then I suppose L1*120 + L2*120 would be the more accurate. But the 0.1A would be in the neutral but probably not significant. If is is actually there.

    Leave a comment:


  • ggunn
    replied
    A common rookie mistake is to think that a 240V 100A 2 pole breaker delivers 200A @240V. 100A on L1 + 100A on L2.

    Leave a comment:


  • gar
    replied
    180526-0821 EDT

    Cybatrex:

    Change the word watts to volt-amperes. It may be watts, but only if the load is pure resistance like an incandescent bulb, or resistive heater,

    .

    Leave a comment:


  • GoldDigger
    replied
    Originally posted by Cybatrex View Post
    Thank you so much. To clarify,

    I would take the amperage to either L1 or L2 and then multiply by 120V or 240V?

    L1 X 240V = CORRECT 1?

    (L1+L2) X 120V = CORRECT 2?
    1. Yes.
    2. Yes.
    For 240 you can use one L current value if the two are identical
    If they are not identical, you can take the average and multiply it by 240. Algebraically this is 100% identical to the formula in 2.

    Sent from my XT1585 using Tapatalk

    Leave a comment:


  • Cybatrex
    replied
    Originally posted by GoldDigger View Post
    The problem with your second method is that the same current (approximately) is flowing through the same series circuit between L1 and L2. You only count that current once! That is where your extra factor of two came from.

    Sent from my XT1585 using Tapatalk
    Thank you so much. To clarify,

    I would take the amperage to either L1 or L2 and then multiply by 120V or 240V?

    L1 X 240V = CORRECT 1?

    (L1+L2) X 120V = CORRECT 2?

    Leave a comment:


  • GoldDigger
    replied
    Originally posted by Cybatrex View Post
    I need assistance with determining my Watts correctly with Ohm's law using the amperage reading from my Fluke clamp meter. I welcome anyone who can correct the errors of my ways and offer a brief explanation so I can learn correctly.

    Amperage Readings
    L1: 17.1 Amps
    L2: 17.0 Amps

    Ohms Law
    P(w) = V x I


    Method 1: Calculated ohms law for each leg of the circuit, calculated them individually, then added the wattage together.
    L1: 17.1A x 120V = 2,052 Watts
    L2: 17.0A x 120V = 2,040 Watts

    Total Watts1: 4,092 Watts


    Method 2: Combined amperage for both legs and used 240V by combined average to calculate watts which is double Method 1.
    L1: 17.1 Amps
    L2: 17.0 Amps

    Lt = 34.1 Amps x 240V = 8,184 Watts

    Total Watts2: 8,184 Watts





    Any help with this would be greatly appreciated.
    The problem with your second method is that the same current (approximately) is flowing through the same series circuit between L1 and L2. You only count that current once! That is where your extra factor of two came from.

    Sent from my XT1585 using Tapatalk

    Leave a comment:

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