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Thread: Measuring a Delta connected load with a wattmeter

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    Measuring a Delta connected load with a wattmeter

    How are the calculations done in a digital power meter to calculate a delta connected load?

    I know for a wye connected load there is a CT and PT on each phase as well as a PT lead on ground/neutral and it simply takes the L-N voltage on one phase multiples it by the current on each phase and then adds up the three phases for the total cpower (assuming pf is 1 here)

    For a delta connected load however does the powermeter somehow use the L-L voltages since there really isn't a L-N voltage on a delta system? Or is a L-N voltage on each phase calculated from each L-L voltage and the calcualted L-N voltage used in the calculation?

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    You can connect the PTs L-L and then the CT primaries to that same L-L. It would have to be correct. The product of the L-L voltage and L-L current would give you the VAs for that phase. I do not think the L-N voltage is calculated from the L-L voltage. The meter merely measures the voltage and current L-L just as it would L-N.
    Last edited by erickench; 11-23-11 at 03:34 PM.
    Eric Kench, P.E.


    If it's not broken don't fix it

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    Usually on a delta system you would have the PTs measure the line to line voltage, and use 2 CT to measure 2 of the line currents. Blondell's Theorem is used to get the 3 phase power (P=Vab*Ia + Vcb*Ic).

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    Quote Originally Posted by erickench View Post
    You can connect the PTs L-L and then the CT primaries to that same L-L. It would have to be correct. The product of the L-L voltage and L-L current would give you the VAs for that phase. I do not think the L-N voltage is calculated from the L-L voltage. The meter merely measures the voltage and current L-L just as it would L-N.
    What are you refering to when you are talking about L-L currents? Are the line currents before the primary of the delta considered L-L currents, or aren't they simply considered Line currents? Each line current splits inside the delta to the other two phases so how are they considered L-L currents?

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    Quote Originally Posted by david luchini View Post
    Usually on a delta system you would have the PTs measure the line to line voltage, and use 2 CT to measure 2 of the line currents. Blondell's Theorem is used to get the 3 phase power (P=Vab*Ia + Vcb*Ic).
    So with a wye connected load we measure line current and L-N current on each phase where as with a delta load only two of the line currents are measured as you described above?

  6. #6
    T.M.Haja Sahib Guest
    Quote Originally Posted by Pitt123 View Post
    So with a wye connected load we measure line current and L-N current on each phase where as with a delta load only two of the line currents are measured as you described above?
    No matter star or delta,balanced or unbalanced loads with neutral or without neutral,take the three RYB power lines to the loads and connect them to your watt meter per instructions in it.It is calculated in the same way as in two element watt meter.
    Last edited by T.M.Haja Sahib; 11-24-11 at 04:53 AM.

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    Quote Originally Posted by Pitt123 View Post
    What are you refering to when you are talking about L-L currents? Are the line currents before the primary of the delta considered L-L currents, or aren't they simply considered Line currents? Each line current splits inside the delta to the other two phases so how are they considered L-L currents?
    I'm referring to the currents that flow from A to B.
    Eric Kench, P.E.


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    Quote Originally Posted by Pitt123 View Post
    So with a wye connected load we measure line current and L-N current on each phase where as with a delta load only two of the line currents are measured as you described above?
    I think you mean L-N voltage above. But yes, in a 4 wire wye system a meter would measure three line currents and three L-N voltages to get the total power: P= Van*Ia + Vbn*Ib + Vcn*Ic.

    And in a 3 wire delta system a meter would measure two line currents and the L-L voltages to get the total power: P=Vab*Ia + Vcb*Ic.

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    So lets say that the 3-phase meter was set up to measure a wye load however the load being measured was actually a delta load and there were unbalanced currents?

    I've seen this before and the current readings were taken so I'm curious how this works in this case?

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    Quote Originally Posted by Pitt123 View Post
    So lets say that the 3-phase meter was set up to measure a wye load however the load being measured was actually a delta load and there were unbalanced currents?

    I've seen this before and the current readings were taken so I'm curious how this works in this case?
    It doesn't matter if the load is delta or wye, it matters if the system is 3 wire or 4 wire. In a 3 wire system, you need two current measurements, in a 4 wire system, you need three current measurements.

    Imagine if you had a 120/208V wye system, with three wye connected loads (we'll assume resistive loads) so that the load currents were 17.333A, 19.067A, & 20.800A. You can see that the line currents are the same as the load currents. So solving for total power:

    P=(17.333*120)+(19.067*120)*(20.800*120)=6864W total.

    Now imagine a 120/208V wye system where you had three (resistive) loads of 2080W, 2288W, & 2496W connected in a delta connection. You can see that your load currents would be Iab=10A<30, Ibc=11A<-90 and Ica=12A<-210. The meter, however, is measuring the line current, not the load current, so the meter would see currents of:

    Ia=Iab-Ica=19.078<-3.005
    Ib=Ibc-Iab=18.193<-118.424
    Ic=Ica-Ibc=19.925<-238.563

    So the meter multiplies these three line currents times the three line voltages (Van=120<0, Vbn=120<-120, Vcn=120<-240) and sums them to get the total power.

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