Converting amps on a known three phase breaker to total kW

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Jezo

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I have a three phase breaker and took amp readings on each of the wires. With those readings to do I sum them up for my conversion to kW or do I take the average across all three for my conversion?

=SUM(SQRT(3)*0.95*<EITHER THE SUM OR THE AVERAGE OF THE THREE PHASES GOES HERE>*208/1000)
 
Jezo said:
I have a three phase breaker and took amp readings on each of the wires. With those readings to do I sum them up for my conversion to kW or do I take the average across all three for my conversion?

=SUM(SQRT(3)*0.95*<EITHER THE SUM OR THE AVERAGE OF THE THREE PHASES GOES HERE>*208/1000)
Click to expand...
Average, but you will get kVA, not kW. To get kW you need the power factor. If the load is purely resistive you can assume it is 1.0, so kVA = kW. But if it is motors or electronics, you have to measure or guess. What's the purpose? How accurate do you need to be?
 
Simple Method

Simple Method

My simple minds says take 120V times each current measured and sum the three numbers to get VA. As above, VA is not Watts.
 
Jraef said:
Average, but you will get kVA, not kW. To get kW you need the power factor. If the load is purely resistive you can assume it is 1.0, so kVA = kW. But if it is motors or electronics, you have to measure or guess. What's the purpose? How accurate do you need to be?
Click to expand...

It appears as if Jezo has declared a PF of 0.95 is his equation.
 
Jraef said:
Average, but you will get kVA, not kW. To get kW you need the power factor. If the load is purely resistive you can assume it is 1.0, so kVA = kW. But if it is motors or electronics, you have to measure or guess. What's the purpose? How accurate do you need to be?
Click to expand...

I am using .9 as the PF, I have the actual measurement someplace but this is just the peliminary bits. Its for computer server loads.

They plug in using a three phase pin and sleeve connection, we took amp readings at the breaker for each of the phases feeding each plug.

So for example the one three phase breaker has readings of

A-PhaseB-PhaseC-Phase
4.14.34.2

Now with that, using the average and a .9 PF, I get
1.361808
For the total kW consumption of the entire plug.

The other side of it, it has two power supplies, have basically the same readings, so then total for the entire device would be about 2.6 kW.
 
130328-1131 EDT

Approximate power is:

120 * (4.1 + 4.2 + 4.3) * 0.9 = 1361 W.

This assumes your reference to 208 V was line to line and all voltages are balanced, and phase angles are equal. The 120 is the line to neutral voltage or at least imagined neutral if the source is delta. Note: real power is the line current in phase with the line to neutral voltage times the line to neutral voltage.

Is your power factor really as good as 0.9

.
 
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I'll adjust the powerfactor with the actual one for my final chart.

Thanks all, I got it. The confusion arose when one person said add and I said average, it confused us because we didnt realize one person was doing the calculations on each phase individually as a 110 figure and I was doing it together as a 208 figure.

But its all sorted out now and we both were right in our own way.
 
gar said:
130328-1131 EDT

Approximate power is:

120 * (4.1 + 4.2 + 4.3) * 0.9 = 1361 W.

This assumes your reference to 208 V was line to line and all voltages are balanced, and phase angles are equal. The 120 is the line to neutral voltage or at least imagined neutral if the source is delta. Note: real power is the line current in phase with the line to neutral voltage times the line to neutral voltage.

Is your power factor really as good as 0.9

.
Click to expand...


good point on the PF. Many times, .9 is the "ideal" , but not actual. Here is a good write-up on the topic of "real world" IT power supply loading if you are interested.

http://www.modularpowersolutions.com/wp-content/uploads/2012/10/MPS-White-Paper_Server-Power-Factor1.pdf
 
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