Current in Unbalanced Load 2

[roydm]

I have been asked to monitor the current in each phase of some delta connected heaters (unity PF) using current transducers in just two phases
I can't remember the formula for calculating current in third phase.
If someone could supply this I would be most gratefull.
Too long out of school!
Thanks in advance
Roy

 

[jghrist]

David is correct. If the only load is the normally non-varying heater load, you only need to know if there is a change. See Waross' 9/25 post for a little more detail. The extra CT will make it easier to troubleshoot, however and would be worth the $80.

 

[roydm]

Davidbeach,
Yes you're right I guess I got hung up on the question of measuring 3rd phase without a transducer for a few reasons.
1/ My control system is Opto22 with two inputs per module.
2/ I have often seen the 2 Wattmeter circuit for measuring 3 phase Watts so it seemed reasonable that 2 CTs would be able to measure 3 phase current.
3/ At some point finding the formula became the challenge
As I said several times math is not my strong point so I posted the question thinking it would be a simple task for someone with an engineers skills.
I got to thinking about your "Kirchhoff's current law" and although it seems to be applied to DC if you think about a 3 phase circuit at any instant in time with the phases at 0, 120 & 240 degrees it should still apply.
Jghrist,
"Maybe there is a formula, but I haven't had time to work it out" I searched high and low but I couldn't find one, several posters hinted at one but someone else would find a flaw.

If I think about a circuit with CTs on A & B, if both currents are equal you don't know if the load is between A-B or A-C & C-B or some combination therof
but if the A & B aren't equal then you know there's current in C but how much, not A-B obviously because of the phase angle.

It seemed like a simple question at the time, I'm sorry I wasted everyones time.
As I said I will use 3 transducers but I won't stop wondering!
Kind regards
Roy

 

[waross]

The formula is just a+b.
Each watt meter in the two watt meter method measures not only the power of the phase that it is connected to but 50% of the power in the third phase.
If you use two watt meters to measure the power draw of three 10 kW heaters in delta, each meter will indicate 15 kW.
If you are measuring a single phase circuit with a watt meter the voltage is in phase with the real current. When you add a second load in delta, the phase angle of the current changes. The current is the vector sum of both currents. If you extract the horizontal component of the current vector, you will find that it equals the sum of the in phase current plus 50% of the second phase current.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[davidbeach]

Kirchhoff's Current Law applies to AC just as well as it applies to DC. When working with AC it is imperative that you use the AC current, not just the magnitude of the AC current; the phase angle matters.

 

[waross]

I agree David and thank you.
My post was not clear. What I meant by a+b was the indicated wattage. Each watt meter will be measuring its share of the total load already due to the phase angles of the current as seen by the watt meters.
I should have said that the total watts is the sum of the indications of each of the two watt meters.
My bad.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[stevenal]

Suggestion:
OP mentions "current transducers." Not sure what the nature of the transducer is, but this method should work for CTs that are well matched, using single phase ammeters. Connect A phase CT polarity to the A phase meter polarity terminal. Likewise with B phase. Common the two non-polarity meter terminals together and route to the C phase meter non-polarity terminal. C phase meter polarity terminal is connected to the two CTs non-polarity terminals. As Slavig said, C= -(A+B), and summing the currents by parallel connection will include the phase relationship. All three currents are being monitored with two transducers as required by OP. No watt-meters (voltage connections required), or further math needed.

 

[waross]

Another solution may be the broken delta normally use with PTs. Any unbalance in the load will show up as a voltage across the corner resistor.
3 CTs a resistor and a voltage relay.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[slavag]

Stevenal is totally right.
Is standard solution for ptotection in case of two CT's.
Regards.
Slava

 

[LionelHutz]

Stevenal hit on the easy answer.

If you're still not sure on this, look up a GE Multilin 369 manual in the back it should be section 7.6.7 (2-phase CT Configuration) for an example of this wiring configuration. Hey, it was the first manual I had to look up.

This likely won't work though with 4-20mA current sensors. They typically convert the measured current to an RMS output (which doesn't have the phase info any more). You'd have to install plain old CT's on the leads and then install 3 x 4-20mA current transducers connected to the CT's.

 

[slavag]

Hi.
Please back to OP.
current transducers
next:
Opto22 with two inputs per module
it's 4-20mA for my pinion.
price is 80$,
If Roy have convential CT's, evry very simple power meter or 3-ph ampermeter are suitble.
Regards.
Slava