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.

 

[waross]

If you have measured the power in each phase correctly the total power will be the sum of the power in each phase. However if you are asking a question like this there is a strong possibility that you may not have measured the individual phase Wattages correctly.

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

 

[racobb]

Or if you are asking about kVA (total apparent power) then it will be volts X amps, or kW/pf.

If it is only total kW you want, then as Bill says just add the 3 measured values together.

Alan

 

[Gadol]

Bill, Alan thank you so much for replaying.
The data that I get from my monitoring board are: A, V(LL), KVA,PF and KW with accumaltive Kwhr. the reason for my question is that my calculated power,
the sum of (A*V(LL)*PF*3)/1000 per phase does not match the KW from my board.
In addition can I calculate Kw from the Kwhr?
Thanks,
Gadol.

 

[Skogsgurra]

Use sqrt(3) instead of 3. Or use V(LN) instead of V(LL).

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[rbulsara]

Replace the 3 with square root of 3 (or 1.732) in your formula. This is assuming the amps being more or less equal in all three phases.


For single phase KW calcs you need to use per phase voltage which is V(LL)/SQRT(3). This is useful when the currents are not balanced.

Now you if take that single phase value and multiply with 3, you will get the total 3 phase power. This should match what I said in my first paragraph.


Rafiq Bulsara

http://www.srengineersct.com

 

[rcw retired EE]

"can I calculate Kw from the Kwhr?"

(I assume by Kwhr you mean kilowatt-hours and not kw-hours reactive. Kiloatt hours is usually abbreviated kwh or kWH).

Kwh = average kw x time.

If your load is relatively constant and you know the length of time that the meter was measuring the load, you can calculate the average KW = KWH/(time in hours).

Read the kwh meter at the beginning of a time period, then read it the end of the period. The difference in the readings divided by the time is the average kw.

Think of your cars odometer and speedometer. The speedometer tells how fast you are going (that's kw). The odometer tells how far you have gone (that's kwh). Using kwh to calculate kw is like using a trip odometer and stop watch to calculate average speed.

 

[Gadol]

Thank you so much, it is very helpful info.

When I add the 3 calculated power for each phase I get unrealistic value of total power, but when I avgerage the 3 power values per phase it's matchs my device power rating?

Do I do the right calculation by averaging the power for each phase?
Thanks,
Gadol.

 

[racobb]

Maybe your monitor is not configured properly?

Alan

 

[rbulsara]

Did you calculate the values using the "correct" formula as stated in my earlier post and that of Skogsgurra?

Post your calculations and measurements.

Rafiq Bulsara

http://www.srengineersct.com

 

[Gadol]

Yes I have used the correct formula.
For each phase:
Pph=(Aph*Vph(LN)*PF*3)/1000
Pph1=32.77kw Pph2=36.42kw Pph3=36.83kw
the Power monitoring board gives P=38.6kw
So avereging the phases is more like it, is it correct?


Thanks,
Gadol

 

[rbulsara]

Still not clear:

Delete 3 from the Pph formula.

Power per phase or Pph=(Aph*Vph(LN)*PF)/1000
Total 3 phase power = 3*Pph

Is the P that meter reads is total power?

What is the Amperage for each phase A, B and C?

Rafiq Bulsara

http://www.srengineersct.com

 

[Gadol]

Hi,
the monitor gives me total power in Kw.
Current per phase: 44.2, 48.3, 49.2 Amps
pf=~1
Thanks,
Gadol.

 

[davidbeach]

The 3 in Pph=(Aph*Vph(LN)*PF*3)/1000 is definitely wrong. Assuming LN voltage of 277V, your powers are:
A=12.24kW
B=13.38kW
C=13.63kW
Total=39.25kW

Also, it is kW, NOT Kw or kw.

 

[Gadol]

David and all,
thank you so much, now it is making sense.
Happy Holidays,
Gadol

 

[rockman7892]

davidbeach

can you please explain how you came up with the kW values in your last post based on the currents given?

Also I thought that with a Delta connected load the only way you could sum up the individual kW values is is you were measuring them on the phase currents of the delta being inside the delta. I wansn't sure that you could simpy add the kW values gathered by the the Line currents and L-N voltages for the line currents on a delta configuration as you could with a wye connected load.

 

[waross]

Hi rockman;

Gadol (Petroleum)
21 Dec 09 18:25
Hi,
the monitor gives me total power in Kw.
Current per phase: 44.2, 48.3, 49.2 Amps
pf=~1

Use 277V


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

 

[rockman7892]

Waross

O.k. I see what I was doing wrong, I was still throwing in a 1.73 when using the 277 L-N voltage. I get the correct answers now.

So from this I take it that it does not matter weather the load is connected in wye or delta, when using line currents and L-N voltages to come up with individual line kW's and then add them together for the total.

 

[rockman7892]

The only reason I ask about using the individual L-N voltages and Line currents on a detla connected load is because I have a fluke power meter that can be set up to measure a wye or delta network. Since it will give the same power reading no matter how the load is connected I was wondering what the advantage was for using one over the other. I guess with using the wye configuration of the meter you can see the individual line kW's for a delta connected load for example?

 

[Gadol]

Hi,
now I'm confused, Delta connection doesn't have N leg, so how come we still have the V(LN) we should only have V(LL)..?
davidbeach, what is the correct answer?
so given the current for each phase, which voltage should we used to calculate the power? and if the total power in the sum of individual phase in Delta connection?

Thanks,
Gadol.