Status
Not open for further replies.

#### Caligula

##### Member
Having a debate in the office. This is kind of like when you know you've spelled a word correctly, but when you look at it, it looks wrong.

Say we have a 208V/120 3phase panelboard with a main breaker. Say we connect to it 6 heating loads, each 3kva and 208V single phase. Say these loads are ALL across phase A and B.

This is where the argument begins. What is the minimum size the MCB can be?

One person says you have your 6 loads. For each, 1500va is on phase A and phase B. Add this up. You have 18kva total, but because of the imbalance, you need to look at the current in legs A and B. Each Leg: 1500*6=9000VA. 9000/120 = 75A. Thus 75 is the minimum you can have.

The other says that because we are dealing with 208V 1ph loads, both phases will be experiencing the same current at the same time, because there is no neutral. So for EACH load, you have 3000/208=14.4A. Thus, on the panel, you have 14.4*6=86.5A on EACH leg, and thus 87A is the smallest it can be.

(Obviously this ignores breaker sizing issues like continuous load, etc)

Which is correct?

#### Npstewart

##### Senior Member
Perhaps I dont understand the question, but if there are (6) heaters, 3KW ea, the service to the panel is 208-3 and the loads are single phase, 3KW @208 = 50A. So your MCB for the panel is required to be 50A minimum. The MCB is protecting the feeder and the panel, the individual breakers for the heaters will be protecting the branch circuits within the panel.

#### augie47

##### Moderator
Staff member
If they are all across A&B, my thoughts would go along with the 86A

#### defears

##### Senior Member
Code wise you are supposed to calculate everything in VA, not amps. And you are supposed to balance everything. If you only add load on 2 phases, then I guess just don't x 1.73.

#### skeshesh

##### Senior Member
The second person is correct, but I don't really see what he means by "because there is no neutral". It's just KCL, lack of a neutral does not effect or offer an explanation.

#### infinity

##### Moderator
Staff member
If they are all across A&B, my thoughts would go along with the 86A

I agree.

3KW * 6 heaters /208 volts = 86.5 amps

What would be correct for the 6 heaters across all three phases?

#### Caligula

##### Member
here's the kicker. In my head I agree with #2, just as you guys have. But if you take just about any panelboard schedule spreadsheet or program out there and you input this scenario, the phase and total load will match explanation #1, not #2. I feel like if #2 is correct, this issue would have come up and been built in to everyone's schedules.

I've seen excel and proprietary schedules used by many companies, ranging from very basic all the way up to where you input load type, continuous or noncontinuous, etc., and nowhere do they compensate for this quirk of 208 1ph. Typically a designer will enter 1500W on phase A and 1500W on phase B for each heater circuit and move on.

Does anyone go back manually to compensate for this in their designs?

#### Caligula

##### Member
I agree.

3KW * 6 heaters /208 volts = 86.5 amps

What would be correct for the 6 heaters across all three phases?
This issue grew from connecting an ODD number of heaters across a three phase circuit:
1: A B
2: A C
3: B C
4: A B
5: A C

The same issue arises because one phase has more load than the other. Depending on if you follow the answer #1 mentality or the #2 mentality it affects how you size the breaker of the 3phase circuit.

#### infinity

##### Moderator
Staff member
This issue grew from connecting an ODD number of heaters across a three phase circuit:
1: A B
2: A C
3: B C
4: A B
5: A C

The same issue arises because one phase has more load than the other. Depending on if you follow the answer #1 mentality or the #2 mentality it affects how you size the breaker of the 3phase circuit.

Where did you come up with only 5 heaters? The OP mentions 6 in this example.

#### LarryFine

##### Master Electrician Electric Contractor Richmond VA
Where did you come up with only 5 heaters? The OP mentions 6 in this example.
One fell out of the back of the truck.

#### skeshesh

##### Senior Member
What you have to remember is KVAs are KVAs. When you size the breaker remember that both wires in a L-L load conduct the same current, but the load that your system sees is the KVA. In other words your breaker will have to be sized to 86.5A and your load is in 3000KVA.

In panel schedules what is usually done is showing showing the load in KVA per phase and adding those into phase subtotals which will ultimately be added as a total panel load (and also taking into account largest motor load, continuous loads and such). In this way, if you divide the KVA value by 2 and place that value in the relevant phases and proceed as described, you will get a correct average panel amps by dividing the total KVA by L-L voltage and 1.732. Meanwhile the circuit breaker has to be sized per the correct current value (86.5A). Also, don't forget to do a sanity check if you have a lot of L-L loads to make sure one phase is not overloaded above its rating. If you take the phase subtotal in a case like that and divide by 120V it will not give you the correct current value of the largest leg.

#### kingpb

##### Senior Member
Don't confuse Branch Circuit and Feeder Circuit.

The only reason you divide the load, 3kVA, by 208V is to determine the proper breaker size for each load, i.e. branch circuit. In this case 1.25 x 14.4A = 18.0A, you would use a 20A, 2P breaker on each load.

The Feeder circuit sizing is based on the fact you have a 3ph panel. The minimum breaker size will be based on either the sum of the phases or 3 times the largest total phase load, using KVA, whichever makes it greater.

So, in this case you have 6 x 1.5KVA on phase A, 6 x 1.5KVA on phase B, and the big goose egg on Phase C. The sum of the three phases is 18KVA. Divide 18KVA/208V/1.732 = 50A, but because, the load on each of two phases is 9KVA/120V = 75A, you have a problem.

In this case the sum of the phases is not correct, that only works for balanced 3-phase panels, otherwise you actually have to base the Feeder size on 3 times the largest phase because you have to be able to deliver 9KVA to each of the two phases. So, it would be 3 x 9KVA = 27KVA
now divide 27KVA/208/1.732 = 75A. Which you'll notice is the same as the 9KVA/120V = 75A.

Multiplying by 1.25 the minimum Feeder breaker you need is 100A, 3-pole. and you will need at least a 27KVA, 3-phase transformer or source ahead of the panel.

If you were to balance the panel you could potentially save money by reducing the Feeder. Now each phase would have 6KVA, and 3 x 6KVA = 18KVA, and 18KVA/208/1.732 = 50A, so you could use a 70A or 75A 3-pole breaker, and your 3 phase transformer or source ahead of the panel would only need to be capable of 18KVA, not 27KVA.

#### cdcengineer

##### Senior Member
I say approx. 86.5... Final answer

#### Besoeker

##### Senior Member
In other words your breaker will have to be sized to 86.5A and your load is in 3000KVA.
Hmmm...Interesting concept.
A 208V supply, a 3000kVA load.....and an 86.5A breaker...

#### skeshesh

##### Senior Member
Hmmm...Interesting concept.
A 208V supply, a 3000kVA load.....and an 86.5A breaker...
This is what happens after I post right after the drive hope and dont proof read... sigh. I was going to say good catch, but its such a gross mistake I'm surprised someone didn't set me straight earlier.

I meant 6 x 3000VA.

#### infinity

##### Moderator
Staff member
So what would the load be if all six heaters were distributed evenly across all three phases?

#### kwired

##### Electron manager
So what would the load be if all six heaters were distributed evenly across all three phases?
Balanced

#### Little Bill

##### Moderator
Staff member
So what would the load be if all six heaters were distributed evenly across all three phases?
At the risk of being corrected----- 50A (49.96A)

#### infinity

##### Moderator
Staff member
At the risk of being corrected----- 50A (49.96A)

Have any math to prove that?

#### Dennis Alwon

##### Moderator
Staff member
Have any math to prove that?
Perhaps I am wrong but I agree with Bill

If 6 heaters are distributed evenly then that means each phase has 4- 1500 watt loads. 4 * 1500 = 6000 . 6000/120= 50 AMPS. Also 18000/360= 50

AB
BC
CA
AB
BC
CA

Status
Not open for further replies.