3 Phase Power Monitor 2

[Gadol]

Hello,
I Monitor power of 3 Phase 480V (Delta connection), I get the power in Kw for each phase, how can calculate the total power?
Thanks.

 

[davidbeach]

Measured voltages were never provided so some assumptions needed to be made. We've already had to assume a power factor; assuming a point somewhere inside the delta that has an equal potential to all three phases is less of a stretch and doesn't require further knowledge of phase angles.

Given to amount of data missing you could also try:

((I_a+I_b+I_c)/3)/((V_ab+V_bc+V_ca)/sqrt(3))

That will provide W, divide by 1000 for kW. I haven't made that calculation but it will give a number very close to my previous answer. It won't be exact because you can't have a delta connected load, identical phase angles, and different current magnitudes all at the same time. That's also why you wouldn't get the some results if you assumed one of the phases to be the reference voltage for all three phases.

Neutral doesn't have to exist to be useful. Three wire systems are metered L-G all the time. If you had a neutral conductor you'd have to use it as your neutral voltage or measure the current in the conductor.

 

[rockman7892]

So what you are saying is if I have a delta connected load I can use the Line currents and L-N voltages to find watts on each phase and then multiply by 3 for total, or go inside the delta and use the phase currents and L-L voltages to find watts on each L-L connection and then multiply by 3 and I will arrive at roughly the same answer?

With the power meter if I have all voltage probes connected including one on ground/neutral then should I get the same results no matter if I tell the meter I have a wye or delta connection? Only values would be displayed differently.

 

[Gadol]

davidbeach,
Data is not the problem, I get from my power monitoring board more then enough data.
let's go to the basic:
I get voltage and current for each phase:
I1=47.8A I2=59.1A I3=69.7A
V1=244.4V V2=247.3V V3=245.9V
PF=~1
What is the Power for Delta connection?
My board gives a power of 43.07kW, does it match the calculate one?

Thanks,
Gadol

 

[waross]

If we assume a power factor of 99% the KWs will match.

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

 

[davidbeach]

(47.8*0.2444)+(59.1*0.2473)+(69.7*0.2459)=43.44kW. Obviously your PF is merely approximately 1, but not exactly 1. You're still missing half the data as you only provide magnitudes and no phase angles. In AC circuits about all you can do with just magnitudes is compare to set points and meter apparent power (VA). Pretty much everything else requires phase angles.

 

[rbulsara]

gadol:

Calculations using your readings yield 43.44 kW, very close to the meter reading of 43.07.

244.4 *47.8 =11.68
247.3 *59.1 =14.62
245.9 * 69.7 =17.14
Total 43.44

The load or source connection (wye or delta) does not matter when you are measuring the line quantities. You can consider both of them as black boxes connected with three lines.

You can assume either delta or wye configuration for calculations.

If you assume a wye connection, the line voltage is divided by sqrt 3 and the phase current is same as the line current.

If you assume a delta connection, the line current is divided by sqrt 3 and the phase voltage is same as the line current.

The final per phase value of kVA or kW will remain the same.

Also for total 3 phase power formula, when you multiply the phase values for either assumption, it will result in the the same final formula of Sqrt(3)*V(LL)*I (LL) for a balanced system

You can also add the three (3) individual phase kVA (or kW if you include power factor), to get the total 3 phase power.


The formula Sqrt(3)*V(LL)*I (LL) only works for balanced system. For unbalanced one you must calculate individual phase values and add them.

Rafiq Bulsara

http://www.srengineersct.com

 

[rbulsara]

A correction to my last post ( ..blame the cut and paste attempt..)

If you assume a delta connection, the line current is divided by sqrt 3 and the phase voltage is same as the line voltage

Rafiq Bulsara

http://www.srengineersct.com

 

[rockman7892]

rbulsara

Your example with the "black box" makes perfect sense now.

Why does the power meter then give the option of of setting up the measurement for a wye or delta network? Is this just for a matter of wheather you want to look at L-N or L-L readings?

For adding the individual phase KVA's then we dont need to worry about phase angles for anything other than determining kW values when measuring current? Obviously if you were calculating line current from phases currents then I believe you would have to account for phase angles.

 

[rbulsara]

rockman:

I can't comment on on your particular power meter question as there is no accompanying data. It may have to do with how and where in the circuit, you connect the meter. You should consult the meter manufacturer's application engineer.

Think about this, your utility meter does not care if your motor or transformer is wye connected or delta. Does it?

As for the phase angles, do not confuse the phase angle for a power factor of a circuit with the 120 degree phase difference between the three phases of a 3 phase system.

The SQRT(3) quantity is the result of the 120 degree relationship between the phases and you do not need to consider any other phase angle in a 'balanced' three phase system to calculate kVA.

The phase angle relationship for determining the power factor is between the voltage and the current of a circuit. That is, between the phase voltage and the current in a given phase.

You could have a different power factor in each phase, but different load (kw) and yet have the same 'apparent' current. For kVA calculation such a circuit will still be balanced. Not for kW calcs.

To your last line in your last post, when calculating the line current from the phase current (if you are measuring at the right spot), only SQRT(3) factor comes or does not come in play depending up on the load connection as explained earlier. No phase angle.

I would suggest you get a basic book on AC polyphase circuit analysis and spend some time with it.



Rafiq Bulsara

http://www.srengineersct.com

 

[rockman7892]

rbulsara

My last paragraph dealt with the fact that for multiple L-L single phase circuits tapped off of a 3-phase, 3-wire, bus you would need to factor in the phase angles of all the L-L currents to determine the overall line currents on each of the lines feeding this bus.

It looks like you do not have to do the same with a three phase load, even if the different phases of the load were drawing different loads which would esentially be an unbalanced condition.

 

[rbulsara]

Yes, but that still amounts to the factor of sqrt3.
Phase angle is only important for calculating kW (or power factor).

The best is to look at an example:

Consider a 208Y/120V source and (3) loads of 10kW each at 208V. If they are connected to A-B, B-C and C-A lines respectively, they become a balanced circuit, connected in delta. The total load is 30kW.

Current in each 208V circuit is 10000/208=48A. This is the 'phase' current of the delta created by three loads.

You can calculate the 3 phase line amps using the 3 phase formula kVA= SQRT(3)*V(LL)*I(LL)/1000, which will be 83.2 amps (from A=1000*30/(1.732*208)

This is the same as what you will get by multiplying the 48A of 'phase' current, we calculated before, with Sqrt(3).

48A*Sqrt(3) = 83.2A.

If you have only two load circuits on, one of the line currents will be 83.2A and other two 48A each. Total real load 20kW.

If you have only one of the load circuits on, two of lines will have 48A each, the other will have 0A. Total real load 10kW.

So no phase angle is required to calculate kVA or amps.

However for kW, you need to take into account angle between the current and voltage you are using.

You may want to go through the exercises of considering the "black box" approach and calculate kVA and kW per phase. (Hint: L-N voltage is 30 degree out of phase with the line-line voltage).

Rafiq Bulsara

http://www.srengineersct.com

 

[rockman7892]

rbulsara

Great example! Makes perfect sense now.

I understand that the pahse angle on each line will need to be know for determining the kW on each line.

I also understand your hint regarding the phase shift. No matter what the source or load (black boxes) the line current will be in phase with the L-N voltage, and 30deg shifted from the L-L voltage on a given line for three phase loads.

Single phase loads is slighty different for you would need to know weather the current was a L-L or L-N current.