Control.com - Nerds in Control

I was asked how to calculate the power to a 3 phase delta connected heater if the load is not balanced. We all know the formula W = LV x LA x 1.732 x PF for 3 phase power but I can't remember how to do the same for an un-balanced load. My intuition tells me that using the average current for LA would be close but that seems too simple.

Anyone remember the formula?

Thanks,
Roy

Responding to Roy Matson's 21-Dec (00:34) question... yes, you are correct. Averaging is not the answer. But first, please clarify your need:

Given the three line currents, Ia, Ib, and Ic, as well as the three line-to-line voltages, Vab, Vbc, and Vca, you want to determine:

1) The heater's total power consumption in Watts?

2) The power consumption of the three heating element?

The answer, like the your previous qustion, can be found in the Control List archives. If a search proves unfruitful, then contact me... on or off list!

Regards, Phil Corso (cepsicon@aol.com)

Roy, have you resolved the calculation question asked in your 21-Dec-07 (00:34) post?

Phil Corso (cepsicon@aol.com)

Phil,

Not exactly, I asked several engineer buddies, they couldn't remember either however I concluded that 2 out of 6 elements were toast.

Thanks,
Roy

Roy, if you still want the formula, please provide (if available) the three ph-ph voltages, and the three line currents measured before discovering that 2 elements were gone!

Regards, Phil Corso (cepsicon@aol.com)

Roy, the unsymmetrical loss of elements can be easily explained if the load was wye-connected, with two-paralleled elements per phase.

If the above connection has no neutral wire the remaining resistances are exposed to voltages that could be much higher than expected line-neutral voltage!

Regards, Phil Corso (cepsicon@aol.com)

Phil,

Yes, even I could calculate a Y configuration but the elements were in delta. When site finally got back to me the phase currents were as follows.

Working element
3.3, 3.3, 3.3

Faulty element
1.9, 1.9, 0

So I concluded there must be 2 elements out. I am still curious as to how to calculate unbalance kW.
It seems from the above results if you subtract the lowest reading 3.3 and calculate that
3.3 x 600 x 1.732 = 3429
then add
1.9 x 600 = 1140, end result 4569 will be correct.
But that's easy, what if all 3 are different?
I thought there would be a simple formula.
It's not that important now, so don't stress over it.

Regards,

Roy

Roy, it depends on what parameter you have chosen for current!

1) The general formula for a balanced delta-connected resistive-load is:

a) If 3.3A repesents line-current, Ia, then total power is:

Pt = Sqrt(3)(Vab)(Ia).

b) If 3.3A represents phase-current thru an element of the delta-connected load, Iab, then total power is:

Pt = (3)x(Vab)x(Iab).

2) The general formula for an unbalanced delta-
connected resistive load is:

a) If one has access to the individual elements then total power is:

Pt = (Vab)(Iab)+(Vbc)(Ib)+(Vca)(Ic).

b) For the case you cite, it would be helpful to know what parameter the two 1.9A measurements represent!

I apologize for misunderstanding your original question. I thought you wanted to determine the total power for a three-phase load (delta or wye) given the 3 phase-phase voltage magnitudes, |Vab|, |Vbc|, & |Vca|, as well as the 3-line currents magnitudes, |Ia|, |Ib|, & |Ic|.

The above problem requires the use of a mathematical approach known as Symmetrical Components!

Regards, Phil

Roy, I forgot to mention in my earlier post that the 1.9A measurement is indicative of one blown element!

That is, current magnitude, 1.9A, for the damaged case, is 57.7% of the current magnitude, 3.3A, representing the undamaged case. An additional clue is the fact that two of the line-currents are equal, and the 3rd is zero!

Regards, Phil Corso

In keeping with one of Control.com's goals, i.e., education of List members (admittedly some acrimoniously), the following Table compares three cases illustrating current distribution and power in a 3-phase, delta-connected system! They are: 1) the balanced case with equal resistive branch-elements; 2) open-circuiting of one branch-element; and 3) open-circuiting of two branch-elements:

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

TABLE A:  Per-Unit A & Pwr, D-Load

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

Unit    Balanced 1-Res Opn  2-Res Opn

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

Ia      1.00        1.00       0.00

Ib      1.00        0.58       0.58

Ic      1.00        0.58       0.58

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

Iab     0.58        0.58       0.00

Ibc     0.58        0.00       0.58

Ica     0.58        0.58       0.00

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

Pwr     1.00        0.67       0.33

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

The above was derived from the fact that ANY three-phase power problem, configured as wye or delta, balanced or unbalanced, can be represented as a two-loop circuit consisting of just two sources of voltage, and three resistive or complex loads! If additional detail is required, contact me!

Phil Corso (cepsicon@aol.com)