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
load
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

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

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

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

----------------------------
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

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

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

----------------------------
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

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