panel size

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Alwayslearningelec

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hello have 2x 3 phase 60a/208v and 4x20a 120v load that need to be fed from panel.
all continuous loads.
1.how can I calculate main breaker size in panel?

thanks
 
Since you don't list full load amps, we will over calculate by using 80% of each load.
2 X 60 X 208 X 1.73 X 80% =34.54 KVA
4 X 20 X 120 X 80% = 7.68 KVA
7.68 KVA + 34.54 KVA = 42.22 KVA
42.22 KVA =Amps X 208 X 1.73
Amps = 117 amps
117 amps X 1.25 = 146 amps.
I would go with a standard 200 amp three phase 208/120 Volt panel to allow for expansion. 200 amp Main circuit breaker.
 
powerpete69 said:
Since you don't list full load amps, we will over calculate by using 80% of each load.
2 X 60 X 208 X 1.73 X 80% =34.54 KVA
4 X 20 X 120 X 80% = 7.68 KVA
7.68 KVA + 34.54 KVA = 42.22 KVA
42.22 KVA =Amps X 208 X 1.73
Amps = 117 amps
117 amps X 1.25 = 146 amps.
I would go with a standard 200 amp three phase 208/120 Volt panel to allow for expansion. 200 amp Main circuit breaker.
Click to expand...
how did you get the 117A? ty

Sent from my SM-N960U using Tapatalk
 
powerpete69 said:
Since you don't list full load amps, we will over calculate by using 80% of each load.
2 X 60 X 208 X 1.73 X 80% =34.54 KVA
4 X 20 X 120 X 80% = 7.68 KVA
7.68 KVA + 34.54 KVA = 42.22 KVA
42.22 KVA =Amps X 208 X 1.73
Amps = 117 amps
117 amps X 1.25 = 146 amps.
I would go with a standard 200 amp three phase 208/120 Volt panel to allow for expansion. 200 amp Main circuit breaker.
Click to expand...
 
powerpete69 said:
Since you don't list full load amps, we will over calculate by using 80% of each load.
2 X 60 X 208 X 1.73 X 80% =34.54 KVA
4 X 20 X 120 X 80% = 7.68 KVA
7.68 KVA + 34.54 KVA = 42.22 KVA
42.22 KVA =Amps X 208 X 1.73
Amps = 117 amps
117 amps X 1.25 = 146 amps.
I would go with a standard 200 amp three phase 208/120 Volt panel to allow for expansion. 200 amp Main circuit breaker.
Click to expand...

I dont follow the part in red.
 
horsegoer said:
hello have 2x 3 phase 60a/208v and 4x20a 120v load that need to be fed from panel.
all continuous loads.
1.how can I calculate main breaker size in panel?

thanks
Click to expand...
I come up with 183 amps, so either way a 200A panel.
Load totals 52.8 kVA
I add the 13.2 kVA 25% for continuous = 66 kVA

Those numbers sound wrong for a proper calculation though, like perhaps minimum circuit ampacity specs, not the kind of data I would use to size a panel per article 220.
Can you tell us more about the loads? Whats the largest motor?
Do you have nameplate info ?
Are they HVAC, motors,commercial cooking equipment , receptacles, lighting?

Even if just the 120V loads are continuous (which I suspect) you would be at 153 amps, so still a 200A panel.
 
horsegoer said:
how did you get the 117A? ty
Click to expand...
Simple algebra:
42.22 KVA =Amps X 208 X 1.73
42,220 = I X 208 X 1.73
I or Amps is your algebraic variable.
42,220 / (208 X 1.73) = I
I = 117 amps.
I would normally calc off the fold load amps but since you didn't have them, I just assumed 80% of each breaker.
I calculated the full load breakers then took 80% to find amps. This calculates everything being on at the same time, continuous load.
With information missing, this should be fine. a 200 amp panel with a 200 amp main breaker works all day long and twice on Tuesday.
 
Adding all loads together and then multiplying by 1.732 does not give an accurate number when the loads are not balanced. You need to calculate each phase separately and size by the highest load. In the OP's case 1 phase is going to have 20 amps more load than the other 2 phases.

Example:
"A" will have 60/60/20/20
"B" will have 60/60/20
"C" will have 60/60/20
 
curt swartz said:
Adding all loads together and then multiplying by 1.732 does not give an accurate number when the loads are not balanced. You need to calculate each phase separately and size by the highest load. In the OP's case 1 phase is going to have 20 amps more load than the other 2 phases.

Example:
"A" will have 60/60/20/20
"B" will have 60/60/20
"C" will have 60/60/20
Click to expand...
I agree with Curt.
Assuming that all the 3-phase and 120V loads have unity power factor (PF = 1.0), then we can add the currents from the 3-phase and single-phase L-N loads directly because they will add in-phase at each bus A, B, and C.

The 1.73 factor is needed when adding currents from L-L single-phase loads because the L-L voltages (and therefore L-L load currents for PF =1.0) are at 30° from the L-N voltages. But in the OP's case there are no L-L single-phase loads.

So in Curt's example for the OP's loads, the line A current is 60A + 60A + 20A + 20A = 160A and the B and C currents are each 60A + 60A + 20A = 140A. With continuous loads that would bring you to a 1.25 x 160A = 200A main breaker requirement.

If the power factor of some loads is less than 1.0 and therefore the load currents don't add exactly in-phase, then the line currents will be reduced but in most cases only by a small amount. So to be conservative this should be ignored.
 
Last edited:
powerpete69 said:
Simple algebra:
42.22 KVA =Amps X 208 X 1.73
42,220 = I X 208 X 1.73
I or Amps is your algebraic variable.
42,220 / (208 X 1.73) = I
I = 117 amps.
I would normally calc off the fold load amps but since you didn't have them, I just assumed 80% of each breaker.
I calculated the full load breakers then took 80% to find amps. This calculates everything being on at the same time, continuous load.
With information missing, this should be fine. a 200 amp panel with a 200 amp main breaker works all day long and twice on Tuesday.
Click to expand...

That's what I'd go with.
 
curt swartz said:
Adding all loads together and then multiplying by 1.732 does not give an accurate number when the loads are not balanced. You need to calculate each phase separately and size by the highest load. In the OP's case 1 phase is going to have 20 amps more load than the other 2 phases.

Example:
"A" will have 60/60/20/20
"B" will have 60/60/20
"C" will have 60/60/20
Click to expand...
I've always wondered why you would assume that the 4-120 volt loads would be 2 on A, 1 on B and 1 on C? They could all end up on the same phase. Is there something in Article 220 that directs us to consider the 120 volt loads the way that you did in your example?
 
synchro said:
So in Curt's example for the OP's loads, the line A current is 60A + 60A + 20A + 20A = 160A and the B and C currents are each 60A + 60A + 20A = 140A. With continuous loads that would bring you to a 1.25 x 160A = 200A main breaker requirement.
Click to expand...

I agree, based on the OP saying 'load' rather than 'protective device' for the 60A and 20A values.
 
infinity said:
I've always wondered why you would assume that the 4-120 volt loads would be 2 on A, 1 on B and 1 on C? They could all end up on the same phase. Is there something in Article 220 that directs us to consider the 120 volt loads the way that you did in your example?
Click to expand...
Yes, they could end up being on the same phase. My example was just a best case scenario of balancing the loads. The previous posts were basing the minimum calc on all 3 legs being perfectly balanced which not possible with (4) 120 volt loads.
 
I tried placing the three-phase loads as 96A (60 X 0.8 X 2 = 96A) per phase and then spread the 4 X 20A, 120V single-phase loads on the three lines to neutral, one phase is loaded to 20 X 0.8 X 2 = 32A while the two remaining phases to neutral loaded at 16A!
I get a maximum line current of 198.28A, while the two lines are loaded 182.28A with a neutral current of 16A! I'd go for a 250A main breaker, IMO.
 
KVA is KVA.
If your loads are unbalanced, then balance them. Engineers try to do it on drawings.
Electricians should try to do in the field as well. This thing will run all day on a 150 amp panel. But I would still install a 200 amp.
If you want to spend some extra money on a 250 amp breaker, then you need a 400 amp bus.
150 amp, 200 amp or 250 amp will all work.
 
powerpete69 said:
KVA is KVA.
If your loads are unbalanced, then balance them. Engineers try to do it on drawings.
Electricians should try to do in the field as well. This thing will run all day on a 150 amp panel. But I would still install a 200 amp.
If you want to spend some extra money on a 250 amp breaker, then you need a 400 amp bus.
150 amp, 200 amp or 250 amp will all work.
Click to expand...

Depends on the panelbaord too as to what in practice would be actually used. For example a Siemens P1 true panelboard is there smallest and it starts out with a 250A bus. for loadcenters, I doubt any of my normal supply houses stock a 150A three phase loadcenter.
 
topgone said:
I tried placing the three-phase loads as 96A (60 X 0.8 X 2 = 96A)...
Click to expand...

The OP said the load was 60A continuous, why are you applying an 80% factor?
I would agree with you if the question was about continuous loads on a 60A breaker or fuse.
 
jim dungar said:
The OP said the load was 60A continuous, why are you applying an 80% factor?
I would agree with you if the question was about continuous loads on a 60A breaker or fuse.
Click to expand...
Yes but he never stated if the 60a was the OCP or the actual load. Therefore I would take 80% of the 60a as well since we don't know the FLA or MCA of what he is feeding.
 
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