# Help to Resolve 3 Phase Power Calculation

• posted
Hi All!
I'm having trouble resolving for three phase power in a 600V 3 Phase delta
system:
Using the Formula for total power in a three phase system with a field
measured 600V phase to phase and a field measured 30A per phase using a
clip-on ammeter.
Pt = 3Pa = SqRt3 x IL x VL x PF where Pa is power in one of the arms of the
Using the above circuit measurments
Pt = 3Pa (3 x 30 x 600 = 54000VA)
why does Pa not equal using the ohter formula
Pt = SqRt3 x IL x VL x PF (1.73 x 30 x 600 x 1 = 31140VA) ??
Do I have to substitute 347 for 600 in the formula 3Pa ??
TIA
Fred
• posted
You've got the formulas correct. For the "single phase" load the line volts are as you state 600/1.73, the VA are then this figure × your stated phase current (30 Amps) × PF × Efficiency..Multiply this single phase VA by 3 to give the total power. This assumes a balanced three phase load here. Jaymack
• posted
Thanks John
Still perplexed by the different product results of the formula, using 600V and 30A measured on a delta system, with the
Pt=3Pa formula resulting in 54000VA and
SqRt3 x IL x VL x PF resulting in 31140VA??
Fred
• posted
Suppose that the Pt=3Pa formula should be 3 x VL / 1.73 x IL to solve this. Or 3 xVL x IL/1.73 depending on wye or delta system??
Fred
• posted
If you were to 'break up' your three-phase into three single-phase circuits, there are two common ways to do it. One is to assume the three single-phase connections where tied together to make the three-phase circuit in a wye formation, and the other is to assume the three single phase connections were tied together in a delta formation.
If you break it up assuming the wye formation, then the single phase current is equal to the line current. But the line voltage you get when you connect three single-phase circuits together in wye formation is sqrt(3) times higher than the single phase voltage. So....
Pt = 3Pa = 3*(Vline/sqrt(3) *(Iline) = sqrt(3)*Vline*Iline
(the not quite obvious step here is that 3/sqrt(3) = sqrt(3) )
If you instead assume that the three single-phase circuits were connected in delta, then the phase voltage equals the line voltage. But the current of two of the single phase circuits combine to form line current. Because of the 120 degree phase shift, the two currents don't add up directly and you find (with a little geometry) that Iline = sqrt(3)*Iphase. So...
Pt = 3Pa = 3*
(Vline)*(Iline/sqrt(3) = sqrt(3)*Vline*Iline
Hope this helps,
daestrom
• posted
Yep. But notice that by definition, X/sqrt(X) = sqrt(X)
So 3 x Vl x Il/1.73 = 1.73 x Vl x Il
daestrom
• posted
Hey thanks daestrom .... very good explaination ... also never did consider that that by definition, X/sqrt(X) = sqrt(X).
Best regards
Fred
• posted
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You have indicated that you have measured the phase to phase voltage which is the "line" voltage as well as the "phase" voltage for a delta. What is unclear is where you measured the current. If you measured the current in one of the 3 incoming leads-you have measured the "line" current. If you have measured the current within one of the branches of the delta- you have measured phase current. Using line measurements P=root(3)*Vl*IL*pf is correct Using phase measurements P=3*Vp*Ip*pf These expressions are true for both Y and delta
• posted
Hi Don!
The current was measured in the line using a clip on ammeter.
I'm still not sure if this current measurement with the clip-on meter on a phase conductor is the "line" or "phase" measurement.
Fred
• posted
If the measurement was taken on one of the three main supply leads, this is line current that you are measuring, irrespective of whether the load is connected in star or delta. Study the different formulae for three phase and single phase calculations for your problem in understanding. It may assist if you were able to draw vector diagrams; (which I doubt since you are struggling with the concepts); or study a reference book for such calculations. Regards
• posted
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Think of the load in a black box with 3 wires going into the box from outside -Any voltage or current measured outside the box is a line quantity and you can't tell from measurements whether the load is Y or delta. Inside the box the quantities measured in the elements will be phase quantities (think of three resistors -phase voltage is that across a resistor and phase current is the current in the resistor). These will depend on the connection
In a delta, the incoming line current is split between two resistors so for a balanced connection the line current is 1.73 times the phase current but the line voltage and phase voltage are the same. In a Y the resistors are connected line to neutral and there is only one resistor connected to each incoming line so line current and phase current are the same. However the line voltage (as before phase to phase voltage is 1.73 times the phase voltage.
Line values outside the box, phase inside. The given equations work for both Y and delta.
One caution- power factor is the angle between phase current and voltage- NOT that between line current and voltage (+/- 30 degree difference).
If you wish to contact me directly, I can send diagrams. However, I can't do this between the 15 th and 25 of October.
Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer
• posted
Hi Don!
Thanks for taking the time ...black box and resistor analogy works for me.
Thanks again!
Fred

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