Help to Resolve 3 Phase Power Calculation

[Fearless Fred]

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

 

[Fearless Fred]

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

 

[Fearless Fred]

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

 

[Fred]

Hey thanks daestrom .... very good explaination ... also never did consider
that that by definition, X/sqrt(X) = sqrt(X).

Best regards

Fred

 

[Don Kelly]

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

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 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

 

[Fearless Fred]

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


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

 

[Don Kelly]

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

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

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 [email protected]
remove the X to answer

 

[Fearless Fred]

Hi Don!

Thanks for taking the time ...black box and resistor analogy works for me.

Thanks again!

Fred