Calculate 3-phase current sum of single-phase loads

[RyreInc]

Hello,

I need to calculate the total RMS bus current, for each phase, for an arbitrary combination of single-phase loads. I have an array of heaters on a 3-phase bus, evenly distributed around the phases (line-to-line). I have current transducers on the bus phases, and have run each heater individually to measure it's current. I have a table that looks something like this (values have been normalized):

htr #	I1	I2	I3
1	1	1	0
2	1	0	1
3	0	1	1
4	1	1	0
5	1	0	1

​

If I run say heaters 1 and 2, the simple sum would result in currents, of 2, 1, 1. But because it's three-phase it's really 1.73, 1, 1. (1.73 being √3). Et cetera.

I tried converting to complex coordinates and adding them together, accounting for phase, but of course I get zero when the load is balanced. I also tried squaring the current, and taking the real and imaginary components of that, but am still getting nonsensical results.

Please help!

 

[Gr8blu]

As seen by the upstream power source, KW plus KW equals KW, no matter whether the loads are single phase or three phase.

Current does not behave the same way. You need to first add all KW, divide that result by the phase-to-phase voltage, and divide again by the square root of three, in order to calculate current. That is because current leaving on one phase will return to the source on another phase. The two don't add, as they are the same current.

So: (kw1 + kw2 + kw3 + kw4 +kw5) / (Vph * 1.732) = Iph (this assumes the loads have the same offset between current and voltage in each phase)

By the way - the connection of five identical loads across three phases does NOT result in a balanced condition. To be balanced, each load would require a specific rating.

 

[RyreInc]

Gr8blu, thanks for the reply. Your technique would only work if all my loads are three phase, but they're all single phase

 

[Gr8blu]

Ryre - read the first line of the post again.
As seen by the upstream power source, KW plus KW equals KW, no matter whether the loads are single phase or three phase.

The method works REGARDLESS whether the loads are single or three phase. Let's look at your layout again.

Four of the five heaters are connected to PH A, three to B and three to C.
Current in Ph A = (kw1 + kW2 + kw4 + kw5) / (V * 1.732)
Current in PH B = (kw1 + kw3 + kw4) / (V * 1.732)
Current in PH C = (kw2 + kw3 + kw5) / (V * 1.732)

 

[waross]

Current in Ph A = (kw1 + kW2 + kw4 + kw5) / (V * 1.732)

I am wondering where does the current from kw5 return?

 

[stevenal]

Phase C.

 

[waross]

A phase to B phase, 1 A,
Plus A phase to C phase. 1 A
A phase current = 1.73 amps.
Now add 1 Amp, A phase to C phase.
Now what is the current in A phase?

 

[RyreInc]

Thanks for your patience Gr8blu. I'm still not getting it. To be clear, I am assuming your 'kw' and 'kW' is the power magnitude - no phase info.

For my learning purposes I have normalized everything to one: 1V, 1A → 1W. In the simplest case of just one heater, your calculation results in phase currents of 0.58 A, when it should be 1. Then for more heaters, the calculation does not match what I have gather from SPICE simulations.

Columns C-E show the SPICE results as a reference. N-P show individual heater currents. Q-S is the sum for each phase (power = current here). And T-V is the final current calculation, which is simply columns Q-S divided by sqrt3.

What am I missing?

Thanks again!

 

[waross]

The angle of a single phase load is not the same as the angle of the resultant current of two single phase loads on the shared phase.
1 Amp in phase with A-B, combined with (Vector sum) 1 Amp in phase with B-C = 1.73 Amps on B phase.
Two heaters on phase A-B = 2 Amps in phase with A-B.
Combine (vector sum) with one heater on phase B-C, You have 2 Amps at phase angle A-B and 1 Amp at phase angle B-C.
You do the math. (Vector sum or directed sum)