How to convert (A) current to Power (KW, KVA, etc)

[adindas]

Dear All

I have been given the following data taken from ampere meter:
Line 1 (Red)
Maximum current: 384.3 A
Minimum current: 65.6
Average: 132 A

Line 2 (Yellow)
Maximum current: 372.9 A
Minimum current: 49.2
Average: 122.5

Line 3 (Blue)
Maximum current: 277.3 A
Minimum current: 75.5
Average: 135.7A

How could I know the maximum demand power requirement based on the above data. Assuming voltage (V) is 240/415 Volt?


Thank you
Adindas

 

[charlie b]

Much depends on the reason you are trying to calculate the power. For example, if you are planning to add more load to this system, and need to know if there is available spare capacity, that is different than trying to find the actual peak power that occurred at any moment in time throughout the measurement period. Neither of these two things can be determined from the data you presented, by the way.

Just as a way of getting an idea of peak power, I would ignore the apparent imbalance in the loading, and use the highest value of current as the basis of my calculation. I would take 384.3 amps, multiply by 415 volts, multiply again by the square root of three, and give a maximum value of 276.2 KVA. All I can say for certain is that this number is a ?bounding condition,? meaning that it was never exceeded during the measurement period.

 

[mivey]

I like charlie b's method for determining equipment size.

I guess if you want a historic theoretical peak for something else, like for a billing peak kVA, you could calculate the power separately. Take the sum of the peak amps (1034.5) times 415/sqrt(3) or 239.6 volts (assuming balanced voltage) to get the VA for a total of 247.9 kVA. This would not be good for sizing 3-phase equipment.

[edit: this would give you the highest theoretical billing kVA but I suspect the actual would be less]

 

[charlie b]

Do you realize that you just calculated the average power value from the three peak currents?

 

[mivey]

Do you realize that you just calculated the average power value from the three peak currents?

I would have thought the average kVA over this time period was 93.5 kVA, unless I'm having a blank moment.

[edit: If the 3 peak currents happened at the same time, this 247.9 kVA is what would show up as the peak 3-phase kVA, correct?]
[edit: I guess assuming the phase angles had a 120 degree displacement? Am I missing something (it would not be the first time)?]

 

[charlie b]

I would have thought the average kVA over this time period was 93.5 kVA, unless I'm having a blank moment.

That is correct, but it is not the same thing I had said. You added the three peak currents, and did some math with that number. If instead you averaged those three peak currents, and said the "average peak" was 1034.4/3, or 344.8, and then calculated the corresponding power value by multiplying 344.8 x 415 x 1.732, you would have come up with the same answer of 247.9 KVA.

 

[mivey]

That is correct, but it is not the same thing I had said. You added the three peak currents, and did some math with that number. If instead you averaged those three peak currents, and said the "average peak" was 1034.4/3, or 344.8, and then calculated the corresponding power value by multiplying 344.8 x 415 x 1.732, you would have come up with the same answer of 247.9 KVA.

Gotcha (I think). Your number was the upper bound of what could happen on a per phase basis. Mine was the upper bound of what happens on a system basis. And yes, the 247.9 can be calculated either way.

You can also get your number by converting the 415 to a phase voltage of 239.6 and multiplying it by the phase current of 384.3 to get the worst case for a phase of 92.08 kVA. Multiplying by 3 gets your system boundry of 276.2 kVA (actually, you got KVA:grin.

[edit: I started out looking at the three phases separately and was going to sum V1A1+V2A2+V3A3 to get the system peak (the worst case based on all readings instead of the worst case based on one reading). Then I realized/assumed V=V1=V2=V3 so just used (A1+A2+A3)*V. Did not even think of the average route until you pointed it out]

 

[adindas]

Thank you

Thank you

I also have the same opinion to determine the maximum demand is better to get the result in KVA and/or and measurement using KW/KVA meter. But this is all the result I have got


Thank you to all who spend time to answer my question. Thank you also for correction. Indeed it is 240/415 V (an old British installation in the UK)

I also have the same opinion to determine the maximum demand is better to get the result in KVA and/or and measurement using KW/KVA meter. But this is all the result I have got and I was asked o predict the maximum demand.

I just got the ne information that the spike/maximum current on each phase did not occurred at the same time:


The first spike current (maximum) occurred on 19/03/08 and 20/03/08occurred in Blue phase (L3)
At this moment
Red (L1) = 240A (approximately. It can not be precisely determined from the graphics)
Yellow (L2) = 210A (approximately. It can not be precisely determined from the graphics)
Blue (L3) = 277.3 A
The demand at this moment = 240 x (240+210+277.3) = 174.55 KVA

The second spike current (maximum) occurred on 20/03/08 and 21/03/08 occurred in Yellow phase (L2)
At this moment
Red (L1) = 220A (approximately. It can not be precisely determined from the graphics)
Yellow (L2) = 372.9A
Blue (L3) = 190A (approximately. It can not be precisely determined from the graphics)
The demand at this moment = 240 x (220+372.9+190) = 187.89 KVA

The third spike/maximum current occurred on 22/03/08 occurred in Red phase (L3)
At this moment
Red (L1) = 384.5 A
Yellow (L2) = 270 A
Blue (L3) = 170A (approximately. It can not be precisely determined from the graphics)
The demand at this moment = 240 x (384.5 + 270 + 170) = 197.88 KVA

Focusing on three phase current (excluding the spike/maximum current)

It is apparently that the maximum current is about 270A
Based on this The Maximum demand will be= 1.73 x 415x 270 = 193.85 KVA

Conclusion:
Based on the above figure and assumption it is reasonable to say that the maximum demand will be:
About 200 KVA

Dis you see any loophole in my comclusion ?

Thank you Adindas

 

[mivey]

It is apparently that the maximum current is about 270A
Based on this The Maximum demand will be= 1.73 x 415x 270 = 193.85 KVA

Conclusion:
Based on the above figure and assumption it is reasonable to say that the maximum demand will be:
About 200 KVA

Dis you see any loophole in my comclusion ?

Thank you Adindas

I can't even guess if this is reasonable because I have no idea how the conclusion is being used. What do you want to do with the conclusion you make (size equipment, determine available capacity, verify a bill, etc)?

Using just that data alone, you might say that the max is 200 kVA but that may not be correct. It would be better to find the maximum of the sum of the 3 currents. There may be a time when no individual current reaches the historic max, but the sum of all three is higher than the sums found at the max points.

Also, I'm not sure how you conclude that 270A is the maximum when you have evidence that the currents have been higher than this.

I see you have 3 choices, depending on what you are trying to do with the information:

#1) If you are trying to size equipment, use charlie b's method. This takes the highest current of any phase and uses it to establish a system kVA (267.4 kVA). This would be sure your equipment size or available capacity is not phase dependent. This assumes all phases have the highest current historically recorded in any phase. It is truly a system worst case based on historic data.

#2) If you just want to know the highest billing demand possible based on what happened in each individual phase, use the sum of the peak current for each phase (247.9 kVA). This is a worst case looking at each phase individually. You normally don't size equipment differently for each phase so this method would not be good for that purpose.

#3) If you want to approximate the actual billing demand, you will need to calculate the kVA for each reading. This will not match the billing demand exactly because the billing meter demand is averaged over some time period (5 min, 15 min, 30 min, 60 min, etc)

 

[adindas]

Dear mivey

Thank you for your time looking at my figure.
The purpose of my coclusion is that to see the maximum demeand. The Switchboard could only take up to 200 KVA and I wanted to see whether the load could still be added. I am aware that it is better to use KW/KVA meter for this purposes but this is the only data that I have got and I need to make conclusion based on this. I am also aware that evry phase current might have different angle so algebraic sum will not be accurate,

It is apparently that the maximum demand will not higher than 200KVA as the Incoming breaker did not trip at the time tha data was taken or at the time the maximum current of 384.5 A occured.

I concluded that 270A is the maximum was based on symmetrcail current, current is same at all phase. I did not take the maximum current as the maximum current (the spike) is single phase.

For instance at the time
Red (L1) = 384.5 A, the Yellow (L2) was just = 270 A and the Blue (L3) = was just 170A.

So if I took 1.73 x 415V x 384.5 = 276.051 the maximum demand will be too high then it should be (as other phase was just 270, and 170A). Also the maximum demand will not higher than 200KVA as the Incoming breaker did not trip at the time the L1 384.5A occured.

I raised this issue because I am just affraid of fallacy, loophole in my assumption and I know this forum is a good place to ask as this is the forum of electrical engineer.

Thank you
Adindas

I can't even guess if this is reasonable because I have no idea how the conclusion is being used. What do you want to do with the conclusion you make (size equipment, determine available capacity, verify a bill, etc)?

Using just that data alone, you might say that the max is 200 kVA but that may not be correct. It would be better to find the maximum of the sum of the 3 currents. There may be a time when no individual current reaches the historic max, but the sum of all three is higher than the sums found at the max points.

Also, I'm not sure how you conclude that 270A is the maximum when you have evidence that the currents have been higher than this.

I see you have 3 choices, depending on what you are trying to do with the information:

#1) If you are trying to size equipment, use charlie b's method. This takes the highest current of any phase and uses it to establish a system kVA (267.4 kVA). This would be sure your equipment size or available capacity is not phase dependent. This assumes all phases have the highest current historically recorded in any phase. It is truly a system worst case based on historic data.

#2) If you just want to know the highest billing demand possible based on what happened in each individual phase, use the sum of the peak current for each phase (247.9 kVA). This is a worst case looking at each phase individually. You normally don't size equipment differently for each phase so this method would not be good for that purpose.

#3) If you want to approximate the actual billing demand, you will need to calculate the kVA for each reading. This will not match the billing demand exactly because the billing meter demand is averaged over some time period (5 min, 15 min, 30 min, 60 min, etc)