Anyone remember the formula?

Thanks,

Roy

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)

Phil Corso (cepsicon@aol.com)

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

Regards, Phil Corso (cepsicon@aol.com)

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)

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

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

Another point is whether I should consider load as star or delta for power calculations and how it will be affecting calculations, which one will be more practical? I have a facility to measure voltage between electrode-hearth. But due to unbalance load nature I cannot measure the correct voltage. Any suggestions?

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)

I neutral=-(Ia+Ib+Ic)

First, calculate Ia=Vab/Zab(Zab=R for resister)

similarly calculate Ib=Vbc/Zbc and Ic=Vac/Zbc

Total power dissipated=total I^2*R

(Calculate I using vector form.)

It will be easier to see if we draw 3 phase, delta connected, and then calculate Ia, Ib, Ic separately, also calculate real power for each phase separately and then total it up.

Note: Real current value will be calculated by vector form.

I hope this helps.

Son

www.electrical-designer-guide.com

To get idea of the unbalance magnitude you face, please provide:

A) the 3 line-line voltages.

B) the 3 phase-neutral voltages.

C) the 3 line currents.

D) the transformer nameplate data.

Regards, Phil

Could you provide an example calculation of leg current using heater elements arranged 1-1-2 across a three phase delta supply?

I sent him the general solution covering n-elements/phase arranged as 3-phase, 3-wire delta-load

If you want a copy, contact me at cepsicon@aol.com

Regards, Phil Corso

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