simple calculation forgotten

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augie47

augie47

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It's been 4 decades since I have studied calculations and suddenly I have forgotten a simple one.... (old age creeping in more every day)
Lets say I have a 208 3 phase system and I'm adding (7) 5 kw single phase loads.
Obviously two phases are going to carry more load than the other.
How do I calculate the load per phase ?
 
I would say that there area max of 5 place where a phases is connected. I would then base my calculation on that 5 * 5000 = 25000 watt max on the phase. Probably too simple huh....
 
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augie47 said:
It's been 4 decades since I have studied calculations and suddenly I have forgotten a simple one.... (old age creeping in more every day)
Lets say I have a 208 3 phase system and I'm adding (7) 5 kw single phase loads.
Obviously two phases are going to carry more load than the other.
How do I calculate the load per phase ?
Click to expand...
Three wire or four wire? Are the single phase loads phase to neutral?
If the latter is the case, then two phases will have two of the 5kW loads and the third will have three. Phase to neutral voltage is 208/sqrt(3) = 120V as I'm sure you know.
Currents, assuming unity power factor, would be about 42A in each 5kW load (from 5000/120).
Thus two phases would have about 84A and the third about 126A. Bold for those who don't care for the approximation. It would be 125A at nominal voltage.
Don't know if this answers your question.
 
jumper,
Thats the way I looked at it, but something in the back of my mind made me think it was not that simple. :grin:
 
jumper said:
I took the balanced load of 6 pieces at 208v across all 3 phases and then added the 7th.

6x5000W= 30000W
30000W/208V/1.732=83A per phase
5000W/208V=24A

2 phases have 107A and 1 phase has 83A.
Click to expand...


augie47 said:
jumper,
Thats the way I looked at it, but something in the back of my mind made me think it was not that simple. :grin:
Click to expand...

Gus, you're first instinct was correct...it is not that simple.

Assuming that the delta connected loads have the same power factor (are 120deg separate) then 1 phase would have 83.2A and the other two phases would have 104.6A. This is from KCL: Ia=Iab-Ica, etc.
 
david luchini said:
Gus, you're first instinct was correct...it is not that simple.

Assuming that the delta connected loads have the same power factor (are 120deg separate) then 1 phase would have 83.2A and the other two phases would have 104.6A. This is from KCL: Ia=Iab-Ica, etc.
Click to expand...

David,
That stirs the old brain cells... can you elaborate, please.
 
augie47 said:
David,
That stirs the old brain cells... can you elaborate, please.
Click to expand...

Sure, starting with your 5kw, 208V, 1ph load, you can see each load has a current of 24A for each load. Since there are 7 of these loads, lets assign 3 loads between A-B, 2 between B-C and 2 between C-A.

So the phase-to-phase load currents are Iab=72, Ibc=48 & Ica=48. If the load currents are 120 degs separated, we could write Iab=72<0, Ibc=48<-120 and Ica=48<-240.

The current flowing in each phase to supply the load would be determined by KCL: Ia=Iab-Ica, Ib=Ibc-Iab and Ic=Ica-Ibc.

So, Ia=72<0 - 48<-240 in vector form. Converting to complex form to make the subtraction easier we have Ia=(72+j0) - (-24+j41.57) = 96-j41.57. Converting back into vector form, we have Ia=104.6<-23.4, or Ia has a magnitude of 104.6.

Solving for Ib and Ic would show that Ib also has a magnitude of 104.6 and Ic has a magnitude of 83.14.

There may be a simpler way to arrive at the same solution, but this is the method that I remember.
 
So David-- I stated that 15000 amps on phases A & B. 15000/208 = 72 amps

Now isn't that the highest amperage the system will see? I guess I don't understand if that is the draw on the highest two phase why would we need to go any further. Your calculation tells me I need a larger amp. wire than needed for the 72 amps?
 
Dennis Alwon said:
So David-- I stated that 15000 amps on phases A & B. 15000/208 = 72 amps

Now isn't that the highest amperage the system will see? I guess I don't understand if that is the draw on the highest two phase why would we need to go any further. Your calculation tells me I need a larger amp. wire than needed for the 72 amps?
Click to expand...
Haven't read the whole thread verbatim but the thing that sticks in the front of my mind is that Gus said he's adding these 7 loads to the system. Does he want the load per phase for just the 7 or the entire system? It may make a major difference.
 
Dennis Alwon said:
So David-- I stated that 15000 amps on phases A & B. 15000/208 = 72 amps

Now isn't that the highest amperage the system will see? I guess I don't understand if that is the draw on the highest two phase why would we need to go any further. Your calculation tells me I need a larger amp. wire than needed for the 72 amps?
Click to expand...

You have 15000W connected BETWEEN A and B, and 10000 between B and C, and 10000 between C and A. That means you have 72A load between A & B, but phase A also sees current from the load between A & C. So the current on phase A is the load current from A-B plus the load current from A-C.

As jumper noted, a balanced connection of 6 5000W loads would give a balanced current of 83A (30,000/208/1.732), so the largest phase draw couldn't be 72 Amps.
 
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Smart $ said:
Haven't read the whole thread verbatim but the thing that sticks in the front of my mind is that Gus said he's adding these 7 loads to the system. Does he want the load per phase for just the 7 or the entire system? It may make a major difference.
Click to expand...

correct.. actual panel is more than adequate. I was trying to recall the math.
 
augie47 said:
correct.. actual panel is more than adequate. I was trying to recall the math.
Click to expand...
Need to include existing load in the calculation. Also, the existing load may be unbalanced... so adding an unbalanced could actually balance the system better than it had been ;)
 
augie47 said:
correct.. actual panel is more than adequate. I was trying to recall the math.
Click to expand...
There is no prescribed NEC method.

The generally accepted method is the panel schedule method...

3? loads have 1/3 their kVA value entered in each column.
1?, 2-pole loads have 1/2 their kVA value entered in each connected-line columns.
1? L-N loads have their kVA value entered in the connected-line column.

The sum of each column being indicative of per line loading and overall balance. Not exact... but as I said, the generally accepted method.
 
david luchini said:
You have 15000W connected BETWEEN A and B, and 10000 between B and C, and 10000 between C and A. That means you have 72A load between A & B, but phase A also sees current from the load between A & C. So the current on phase A is the load current from A-B plus the load current from A-C.

As jumper noted, a balanced connection of 6 5000W loads would give a balanced current of 83A (30,000/208/1.732), so the largest phase draw couldn't be 72 Amps.
Click to expand...

Gotcha forgot about a to c connection
 
Besoeker said:
But is it a 208V single phase load?
Click to expand...

augie47 said:
Lets say I have a 208 3 phase system and I'm adding (7) 5 kw single phase loads. Obviously two phases are going to carry more load than the other.
Click to expand...

If the loads are 120V single phase and connected 3/ph, 2/ph and 2/ph, then only one phase would carry more than the other two. I think the question was clearly about 208V single phase loads.
 
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