unbalanced three phase load 1

[ossie]

this is not my area of expertise? but i was asked the question and i am now curious.

There are 3 small reactors. Each one is a standalone ultra violet lamp operating at 2000V - fed via a 400/3500V transformer. Each lamp is rated at 27kW (not important) One. two or three lamps may be on together -
each is fed from its own panel and transformer. Each panel has a separate 3-phase supply from an MCC.

Each transformer primary takes 107A at 400V connected across two phases.

This brings us to the question:
Say lamp 1 is connected across phases L1 and L2
and lamp 2 is connected across phases L2 and L3
When two lamps are on, phase L2 is carring a bigger load but what isthat load?

The loads are not balanced 3-phase loads so it will not be 2 x 107A

I think we have to go back to bacics and consider phase angles and sine waves, etc. to work it out. But I have forgotten all that.

Because the lamp is across 2 phases, we are looking at line voltage 400V and line current 107A - not phase values. It makes sense if you sketch out the star of the phase vectors at 120 degrees and show 2 lamps connected.

Voltage is ok
V line = V phase x square root of 3 (1.732)
i.e. 400 = 230 x 1.732

But I have forgoten how to convert line current to phase current.

I reckon phase current is approx 2/3 of 2 x 107A = 142A
It is probably a factor like 1.732 divided by 3

Connecting 3 lamps will be the same as for 2 lamps - except all phases will then be balanced.

Would you know how to calculate the phase current? Or know anyone who knows? Have you ever dealt with anything like this?

 

[Shortstub]

If you want to learn how then I suggest the following:

Transform the balanced delta into a balanced wye, then the problem becomes one of two current-loop fed from two voltages. One is E/sqrt3 @ 0deg. The other is E/Sqrt3 @ (+) or (-) 120deg. The unbalanced case can be solved by eliminating one of the 3 loads!

This was done in another thread but I can't find it.

 

[busbar]

For 2-wire {1©ª} 107-ampere loads, one would see ~185 amperes line current on the common phase to two lamps, and the same ~185 amperes on each phase for three lamps. In both cases, ¡î3 is the current multiplier with balanced voltages.

 

[Shortstub]

Ossie,

Earlier thread I mentioned was:

Thread238-61660, "Unbalanced Resistive Loading of Three Phase System"

 

[rbulsara]

Firstly there will not be an unbalanced system!!!

By connecting only two transformers you are creating a Open delta system with equal load on both the connected transformers hence creating a 'Balanced' 3 phase load on an open delta system.

In such a system, if L2 is common to the two xfmrs, the current in L2 (I2) will be the vector sum of I1 and I3. I1 and I3 are 120 degree apart.

Since I1 and I3 are equal say it is I. To find curent L2

I/_0 + I /_120 = I /_-120. =I in L2

I + I cos120 +j(I.sin120)

I - 0.5I +j(.866I)
0.5I +j 0.866I
I /_-120

Scalar value comes back to I.

Meaning the current in L2 will be same as L1 and L3 but 120 from behind current in L1.

You are more than confused. 27KW at 400 is 62A, unless it has a 0.58 pf!

Assuming a unity PF 27 kW load, 107A will be the Line current when all 3 transformers are connected and loaded, per your description.

When all three xfrms are connected (closed delta) you will have 62A in each phase (transformer winding) and 1.732 times 62=107A in each line.