# Power calculation in an unbalanced 3 phase heater

A

#### antonslark

Hi there I have measured the following line currents in a 3-phase heating set-up:

A - 13A
B - 11A
C - 5A

The set-up consists of 3 delta connected heating elements.

The usual power equation for a balanced load would be W = LV x LA x 1.732 x PF, but how do I do this in an unbalanced load as above? Just taking the average of the three values seems too simple?
Any help much appreciated!

#### PhilCorso

Replying to Atonslark's 26-Apr-10 (08:30) query... the total power taken by the heater is:

Total P (Watts)= Vab x Iab + Vbc x Ibc + Vca x Ica, where:

o Vab, Vbc, and Vca, are the line-voltages, and

o Iab, Ibc, and Ica, are the delta-currents.

o Heater PF is assumed to equal 1.0.

If the currents you presented are wye or line-currents, Ia, Ib, and Ic, then formula you presented is incorrect if line and/or phase-voltages are unbalanced.

For solution, contact me.

Phil Corso (cepsicon [at] aol [dot] com)

P

#### -Pat

If this is a Delta system my question is do you have a ground in the system causing the unbalance??

You said three Delta connected heaters. Is that a total of three or nine elements?

I do agree that you need to get the I and E of each element and then total them up individually to get the correct load of the system. This is assuming pure restive load.

#### PhilCorso

Antonslark... have you given up?

If not, can you provide the nameplate rating of the original heater, that is 3-ph kW, Line-Volts, and Line-Amperes?

Regards, Phil Corso (Cepsicon [at] aol [dot] com)

J

#### Jim

I have a heater setup that is unbalanced. They are wired in a delta configuration. The line to line voltage is 480. I know the resistive load between each line. I can then calculate the phase current, but since it is unbalanced how do I calculate the line current?

#### PhilCorso

Jim... contact me off-line.

Phil Corso (cepscon [at] aol [dot] com)

M

#### mukesh shah

> If this is a Delta system my question is do you have a ground in the system causing the unbalance??

no

> You said three Delta connected heaters. Is that a total of three or nine elements?

three element connected parallel to two phase. the watt is 3000 +3000 + 2000 with each phase.

<b>moderator's note: this is not the original poster, but it looks as if he has the same problem.</b>

P

#### Phil Corso

Mukesh... I have interpreted your question to mean that you have a Delta-Connected supply serving an unbalanced-load!

The load consists of two (2) 8,000 Watt loads: one connected across, say Phase R-S; the other across Phase S-T (or Ph T-R)!

If you want the magnitude of each of the line-currents, Ir, Is, and It, then please provide the phase-to-phase voltage!

Or, do you want the formula(s) to calculate the currents?

Regards Phil Corso

E

#### Eric Lemmon

Coincidentally, I too have a similar problem. I have a 32.4 kW three-phase 480 VAC heater in which there are nine 3.6 kW elements connected in delta, with three of these elements in parallel between each pair of phases. When all heater elements are functional, the line current is 39 amperes in each phase.

My question is, what will be the respective line currents if one of the heating elements opens? For this question, let's assume that one of the three elements between phases A and B is open. I know that the total heat load will drop to 28.8 kW, but I can't seem to find the right formula to make the calculation of resulting line currents.

P

#### Phil Corso

Eric... for the loss of one element in Phase A-B then the line-currents A, B, C, are 37.2A, 37.2A, 39.0A, respectively.

If you would like to know the formulas, with just an exchange of name, job, and location, just contact me at:

Cepsicon[at]aol[dot]com

E

#### Eric Lemmon

Phil, Is it possible that you meant to write 32.7A, 32.7A, 39.0A?

> for the loss of one element in Phase A-B then the line-currents A, B,
> C, are 37.2A, 37.2A, 39.0A, respectively.