Motor Control Center

[New EE]

I have an existing Motor Control Center that I am trying to get a general idea of the existing load. The GE MCC is a 600A, 480V, 3PH, 3W, 8000 line/spectra Motor Control Center with a 3P400A main breaker.
My question is whether I am doing my calculations correctly just to get a general idea of the existing load. I put a power logger on for about 30 minutes and took the highest load readings during this timeframe.

Phase A: 241.7 x .480 = 116.0
Phase B: 244.3 x .480 = 117.3
Phase C: 243.4 x .480 = 116.8

116.0 + 117.3 + 116.8 = 350.1

350.1 / .831 = 421.3 Amps

Is this the correct way just to get an idea of the total existing load on the MCC? Note that the system is 480V, 3PH, 3W (not neutral)

Thanks.

 

[augie47]

are the 241, 244, 243 readings the ampere readings on each phase ?

 

[New EE]

Yes. I used a Fluke clamp on and set it to amps when taking the reading.

 

[iceworm]

... My question is whether I am doing my calculations correctly just to get a general idea of the existing load. I put a power logger on for about 30 minutes and took the highest load readings during this timeframe.

Phase A: 241.7 x .480 = 116.0
Phase B: 244.3 x .480 = 117.3
Phase C: 243.4 x .480 = 116.8

116.0 + 117.3 + 116.8 = 350.1

350.1 / .831 = 421.3 Amps

Is this the correct way just to get an idea of the total existing load on the MCC? Note that the system is 480V, 3PH, 3W (not neutral).

No.

The line currents are vectors. The phase to phase voltages are vectors. Each one has a magnitude and a phase angle. Never add the line currents together - unless you are looking for Summation(I) = 0. And that is a vector calculation

With no neutral, there are no single phase line to neutral loads. With the data you have, The loading is just the currents you listed. And that is the way you write it down:

Phase A = 241.7 Amps
Phase B = 244.3A
Phase C = 243.4A

If all the loads are 3 phase, and, since the magnitudes are fairly close, you might average the currents and say, "The current is 243A". All will understand you to mean the load is balanced and the current in each line is nominally 243A

The CB main is 400A. You might say there is 160A of headroom that you could add load. However, to say that with confidence, it will take more than a 30 minute data test.

ice

 

[New EE]

Thank you. I was obviously way off...

 

[GoldDigger]

Yes. I used a Fluke clamp on and set it to amps when taking the reading.

Meaning that the readings were on each of the three ungrounded lines? Or on the individual phase load legs attached to those lines? Your answer does not make that clear.

 

[petersonra]

I have an existing Motor Control Center that I am trying to get a general idea of the existing load. The GE MCC is a 600A, 480V, 3PH, 3W, 8000 line/spectra Motor Control Center with a 3P400A main breaker.
My question is whether I am doing my calculations correctly just to get a general idea of the existing load. I put a power logger on for about 30 minutes and took the highest load readings during this timeframe.

Phase A: 241.7 x .480 = 116.0
Phase B: 244.3 x .480 = 117.3
Phase C: 243.4 x .480 = 116.8

116.0 + 117.3 + 116.8 = 350.1

350.1 / .831 = 421.3 Amps

Is this the correct way just to get an idea of the total existing load on the MCC? Note that the system is 480V, 3PH, 3W (not neutral)

Thanks.

i am not sure what this calculation is supposed to be. Your load in amps is the average of the load in each of the three phases, or about 243 A. This gives you about 202 kVA, if you want the load in kVA.

I don't think what you measured serves a whole lot of purpose though. If you are trying to figure head room on the MCC you are generally required to use the load calculations from the code, and not what you measured for a short time. I think there is some provision to replace a calculated load with a measured one, but off hand I don't recall just how you go about doing it, and under what circumstances it is considered acceptable.

 

[New EE]

At first I put the data logger on each of the three ungrounded lines and left the data logger on for about 30 minutes.
After that, I took a clamp on meter and took a reading on individual phase load legs to get a reading.
The readings Phase A = 241.7 Amps Phase B = 244.3A Phase C = 243.4A are taken from the clamp on reading taken
on each individual phase load legs.

 

[New EE]

Bob - I'm obviously confused and need to go back to my supervisor for clarification. Sorry about this. I'm still in the learning phase. It was previously explained to me to take a clamp on reading and do the calculation as I had shown to get a general idea of the load. Except I used .480 instead of .277 when doing the calc because it was a 480V, 3PH, 3W system.

 

[petersonra]

At first I put the data logger on each of the three ungrounded lines and left the data logger on for about 30 minutes.
After that, I took a clamp on meter and took a reading on individual phase load legs to get a reading.
The readings Phase A = 241.7 Amps Phase B = 244.3A Phase C = 243.4A are taken from the clamp on reading taken
on each individual phase load legs.

did you wear voltage rated gloves and arc flash rated clothing when you did this?

 

[GoldDigger]

Bob - I'm obviously confused and need to go back to my supervisor for clarification. Sorry about this. I'm still in the learning phase. It was previously explained to me to take a clamp on reading and do the calculation as I had shown to get a general idea of the load. Except I used .480 instead of .277 when doing the calc because it was a 480V, 3PH, 3W system.

Hi New EE,
You need to multiply the line current measurements by the line to neutral voltage (277) assuming you have a close to balanced load with PF near 1.
You multiply the phase (delta) currents by the line to line voltage (480), and whether the PF is 1 or lower you will get the load VA and if you know the PF you can calculate the watts.
If your data logger is recording current, voltage, and phase angle for the three lines it can give you the complete VA and wattage information from that.