208V 2 Pole Load

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W@ttson

W@ttson

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USA
Hello All,

This topic has been gone over a few times on this forum but I just wanted to display the question in a different form:

For a 10KVA 2 pole single phase load on a 208V/120V system, what is the per phase current?

When doing a panel schedule I would split the 10KVA between the two phases so I would have 5KVA on each phase, then I would do the same for all other loads, distributing them as required whether 120V loads, 208V Single phase loads, or 3 phase 208V loads. I then get the per phase current by taking the column for each phases summed up KVA and dividing by 120V.

However, how do we get around the discrepancy when looking at one load:

208V, 2 Pole load, 10KVA:

10KVA/208V = 48.08A

5KVA/120V = 41.67A

Its close, but not close enough to be sufficiently the same. Is it just a limitation of the assumptions/simplifications we make when trying to do a panel schedule and utilizing loads that should be 100% balanced but are not?
 
spraymax6 said:
Its close, but not close enough to be sufficiently the same. Is it just a limitation of the assumptions/simplifications we make when trying to do a panel schedule and utilizing loads that should be 100% balanced but are not?
Click to expand...

Exactly. When panel schedules are normally done the approximation is that everything is 100% balanced and the same power factor. Since one generally makes the attempt to balance the panel and loads are roughly similar, this approximation is 'good enough' for most applications.

But if you had a situation where you needed to be precise, you would have to treat each load a vector, and then do the vector sum of all the loads on a given bus to determine the actual load.

Lets take the example you gave and extend it a bit. Lets consider a panel and add one 10KVA load at a time, and calculate the current on the bus.

1) Connect a 10KVA resistive load from phase A to phase B. We get 48A on bus A, 48A on bus B, and 0A on bus C.

2) Now add a corresponding load from phase B to phase C. We get 48A on buses A and C, but 83A on bus B.

3) Finally add a third load from C to A. Now we see 83A on all three busses.

4) Keep adding loads and you will see that the simple approximation (assign 1/2 the KVA to each bus and presume balance) gets better as you go along.

-Jon
 
winnie said:
Exactly. When panel schedules are normally done the approximation is that everything is 100% balanced and the same power factor. Since one generally makes the attempt to balance the panel and loads are roughly similar, this approximation is 'good enough' for most applications.

But if you had a situation where you needed to be precise, you would have to treat each load a vector, and then do the vector sum of all the loads on a given bus to determine the actual load.

Lets take the example you gave and extend it a bit. Lets consider a panel and add one 10KVA load at a time, and calculate the current on the bus.

1) Connect a 10KVA resistive load from phase A to phase B. We get 48A on bus A, 48A on bus B, and 0A on bus C.

2) Now add a corresponding load from phase B to phase C. We get 48A on buses A and C, but 83A on bus B.

3) Finally add a third load from C to A. Now we see 83A on all three busses.

4) Keep adding loads and you will see that the simple approximation (assign 1/2 the KVA to each bus and presume balance) gets better as you go along.

-Jon
Click to expand...
What a perfectly clear response with a very good example. Thank you.
 
spraymax6 said:
Hello All,

This topic has been gone over a few times on this forum but I just wanted to display the question in a different form:

For a 10KVA 2 pole single phase load on a 208V/120V system, what is the per phase current?

When doing a panel schedule I would split the 10KVA between the two phases so I would have 5KVA on each phase, then I would do the same for all other loads, distributing them as required whether 120V loads, 208V Single phase loads, or 3 phase 208V loads. I then get the per phase current by taking the column for each phases summed up KVA and dividing by 120V.

However, how do we get around the discrepancy when looking at one load:

208V, 2 Pole load, 10KVA:

10KVA/208V = 48.08A

5KVA/120V = 41.67A

Its close, but not close enough to be sufficiently the same. Is it just a limitation of the assumptions/simplifications we make when trying to do a panel schedule and utilizing loads that should be 100% balanced but are not?
Click to expand...
I'll go with a slightly different angle, (and dodge the question in the process 😆). Why not just add up the KVA of everything instead of doing it "per phase"? I have never understood the purpose of loads on a panel schedule.
 
electrofelon said:
I'll go with a slightly different angle, (and dodge the question in the process 😆). Why not just add up the KVA of everything instead of doing it "per phase"? I have never understood the purpose of loads on a panel schedule.
Click to expand...
The thing was someone off the cuff asked me what is the current on a 10KVA single phase load at 208V. There was a little bit of back and forth and realized that they were doing a panel schedule. I still felt weird telling them that they shouldn't look at it as 48.08A but as a 5KVA load on the one phase.
 
electrofelon said:
I'll go with a slightly different angle, (and dodge the question in the process 😆). Why not just add up the KVA of everything instead of doing it "per phase"? I have never understood the purpose of loads on a panel schedule.
Click to expand...

Because that calculation is only valid in the event that all of your loads are perfectly balanced.

One trick you can do as a conservative way to calculate it, is assume hypothetical loads are added where needed to balance the panelboard. This will get you an upper limit to the current per phase. Maybe one of the phases will have this current, or maybe they all will be less than it.

As an example, consider a situation where we have 8 qty 10 kVA loads connected across 2-pole breakers on the 120/208V 3-phase grid. 8 of these loads obviously cannot be uniformly distributed on a 3-phase panel, so let's suppose that we have 9 of these loads. Add up the kVA and we get 90 kVA. The corresponding kVA per phase is 30 kVa, and divide by the line-to-neutral voltage. The current per phase would be 250A, by this calculation.

Now to properly calculate this, you'd use the square root formula that accounts for finding the line currents given the phase-to-phase currents. Here's what you get instead. You get a total of 144A between phases A&B and between phases B&C, and the runt CA gets 96.2A. Then transform these numbers thru the square root formula, and you get 209A on phase A, 250A on phase B, and 209A on phase C.
 
spraymax6 said:
The thing was someone off the cuff asked me what is the current on a 10KVA single phase load at 208V. There was a little bit of back and forth and realized that they were doing a panel schedule. I still felt weird telling them that they shouldn't look at it as 48.08A but as a 5KVA load on the one phase.
Click to expand...

When current mixes between multiple phase-to-phase loads, and the phase-to-neutral loads, it doesn't simply "add". The magnitude of current alone doesn't give the complete information that enables you to combine it with other currents on other phases. There is vector math involved in the theory behind how these combine, to keep track of the relative timing of the waveforms that we call phase.

An individual 10kVA load across a 2-pole breaker, will yield 48.08A drawn across them, and 48.08A on each of the two conductors associated with the phases of that breaker. But when you mix it with a second otherwise-identical 10 kVA load, that is staggered to the next pair of phases, the current adds up as vectors on the shared phase to become 83.3A, and the 48.08A remains alone on the two lone phases. Notice that 83.3A = sqrt(3)*48.08A.

Given a mixture of 1-pole, 2-pole, and 3-pole loads, here is the procedure to add them up. Unity power factor is assumed.

Step 1: Suppose all the 2-pole loads are shut off, and all the remaining loads are on at their full current. Add up all the loads that are connected to each of the three phases. Call these values Ia0, Ib0, and Ic0. These are the currents on each of the phases, prior to considering the 2-pole loads. Single pole loads only apply to the corresponding phase, while balanced 3-phase loads apply their current to all three phases. An unbalanced 3-phase load, like a 3-phase feeder to a subpanel will have three different currents to apply to this calculation.

Step 2: Add up all the AB loads, and likewise all the BC loads, and then all the CA loads. Call these values, Iab, Ibc, and Ica.

Step 3: Construct the following formula to find current on phase conductor A:
Ia = Ia0 + sqrt(Iab^2 + Ica^2 + Iab*Ica)

Any of the 2-pole loads that touch phase A, are inside the square root equation. We square each of them, and we find a mixed product. Add up these three terms, and take the square root. Then add to the phase conductor current from step 1.

Step 4: Recognize the pattern of the formula in step 3, and repeat it for phase conductors B and C:
Ib = Ib0 + sqrt(Iab^2 + Ibc^2 + Iab*Ibc)
Ic = Ic0 + sqrt(Ibc^2 + Ica^2 + Ibc*Ica)
 
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Carultch said:
Construct the following formula to find current on phase conductor A:
Ia = Ia0 + sqrt(Iab^2 + Ica^2 + Iab*Ica)
Click to expand...
I think that's only exact when Iab = Ica. As the sqrt is the magnitude of the vector sum of Iba and Ica, but if they are of unequal magnitude, that sum won't be parallel to Ia0. For example, if Iab = 0, then the above formula reduces to Ia = Ia0 + Ica, but Ia0 and Ica are 30 degrees apart.

Cheers, Wayne
 
Carultch said:
Construct the following formula to find current on phase conductor A:
Ia = Ia0 + sqrt(Iab^2 + Ica^2 + Iab*Ica)
Click to expand...
wwhitney said:
I think that's only exact when Iab = Ica. As the sqrt is the magnitude of the vector sum of Iba and Ica, but if they are of unequal magnitude, that sum won't be parallel to Ia0. For example, if Iab = 0, then the above formula reduces to Ia = Ia0 + Ica, but Ia0 and Ica are 30 degrees apart.
Click to expand...
I agree with your statement except that it's also exact if Ia0 = 0. It's obvious you knew that, but I'm just mentioning it so that someone else doesn't misinterpret your first sentence.
 
synchro said:
I agree with your statement except that it's also exact if Ia0 = 0.
Click to expand...
Good point, I was actually reasoning this out for myself, so in that sense I didn't know that yet. I was only considering the case of all currents positive, which I didn't state.

Cheers, Wayne
 
wwhitney said:
I think that's only exact when Iab = Ica. As the sqrt is the magnitude of the vector sum of Iba and Ica, but if they are of unequal magnitude, that sum won't be parallel to Ia0. For example, if Iab = 0, then the above formula reduces to Ia = Ia0 + Ica, but Ia0 and Ica are 30 degrees apart.

Cheers, Wayne
Click to expand...
If Iab equaled Ica, the formula would reduce the square root expression to either Iab*sqrt(3) or Ica*sqrt(3).
 
Carultch said:
If Iab equaled Ica, the formula would reduce the square root expression to either Iab*sqrt(3) or Ica*sqrt(3).
Click to expand...
That's true, but doesn't change the validity of my comments, as far as I can see.

The magnitude of the vector sum of Iba and Ica is correctly given by the sqrt expression, but the direction of the vector sum is parallel to Ia0 only if Iba = Ica or Ia0 = 0. When non-parallel, the magnitude of the final vector sum will be less than the expression you listed.

Cheers, Wayne
 
wwhitney said:
That's true, but doesn't change the validity of my comments, as far as I can see.

The magnitude of the vector sum of Iba and Ica is correctly given by the sqrt expression, but the direction of the vector sum is parallel to Ia0 only if Iba = Ica or Ia0 = 0. When non-parallel, the magnitude of the final vector sum will be less than the expression you listed.

Cheers, Wayne
Click to expand...

So what would be the correct formula? Is there a cosine term on either Ia0, or the sqrt expression or both? And what would be the angle?
 
Well, here's the calculation in coordinates, I'm not sure if there's a more elegant formula and I didn't check if it simplifies to something nice:

We want to add Ina (which you called Ia0), Iba, and Ica. Since you assumed power factor one for everything, we know the latter two are at 30 degree angles to the former, on opposite sides. We can write the vectors in components parallel and perpendicular to Ina:

Ina= Ina * (1,0)
Iba = Iba * (√3,-1)/2
Ica = Ica * (√3,1)/2

So the sum is (Ina + √3 * (Iba + Ica) / 2 , (Ica - Iba) / 2). Now for the magnitude we need to take the square root of the sum of squares of the two components. Which looks messy enough that I won't write it out, although under the square root, the coefficients of Ina2, Iba2, and Ica2 will all be 1, it's just the cross term coefficients that need determining.

Cheers, Wayne
 
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PS I guess the cross terms end up being √3 * (Ina*Iba + Ina*Ica) + Iba * Ica. But you'll have to check my algebra.

Cheers, Wayne
 
wwhitney said:
Well, here's the calculation in coordinates, I'm not sure if there's a more elegant formula and I didn't check if it simplifies to something nice:

We want to add Ina (which you called Ia0), Iba, and Ica. Since you assumed power factor one for everything, we know the latter two are at 30 degree angles to the former, on opposite sides. We can write the vectors in components parallel and perpendicular to Ina:

Ina= Ina * (1,0)
Iba = Iba * (√3,-1)/2
Ica = Ica * (√3,1)/2

So the sum is (Ina + √3 * (Iba + Ica) / 2 , (Ica - Iba) / 2). Now for the magnitude we need to take the square root of the sum of squares of the two components. Which looks messy enough that I won't write it out, although under the square root, the coefficients of Ina2, Iba2, and Ica2 will all be 1, it's just the cross term coefficients that need determining.

Cheers, Wayne
Click to expand...

Thanks for clarifying.

Total the x-components and get:
Iax = Ina + Iab*sqrt(3)/2 + Ica*sqrt(3)/2

Total the y-components and get:
Iay = Ica/2 - Iab/2

Combine in Pythagorean theorem:
Ia = sqrt(Iax^2 + Iay^2)
Ia = sqrt((Ina + Iab*sqrt(3)/2 + Ica*sqrt(3)/2)^2 + (Ica/2 - Iab/2)^2)

Simplifying:
Ia = sqrt(Iab^2 + Iab*Ica + Ica^2 + Ina*sqrt(3)*(Iab + Ica) + Ina^2)

I compared this formula to the formula that I had thought was correct, and it turns out that the results are very close to each other. "Ia = Ina + sqrt(Iab^2+Ica^2+Iab*Ica)" is a conservative calculation as we expected, regardless of the specifics. I explored the example of Iab = 100A, and Ica = 50A (that I thought would be extreme), with a broad sweep of Ina from 0 to 200A. The greatest difference I got between these calculations was 1.44A at Ina = 200A. For a more extreme example of Iab = 100A and Ica = 10A, the greatest difference was 6.7A at Ina=200A.

You would have to deliberately go out of your way to intentionally make it unbalanced, for there to be any practical implications of the difference between these calculations. And it is not an advantage to do so, because a balanced configuration makes the best use of the properties of 3-phase.
 
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wwhitney said:
a more elegant formula
Click to expand...
The law of cosines for vector addition can be derived using the inner product <,> with the understanding that for a vector a, the length |a| = sqrt(<x,x>). The inner product is bilinear and symmetric, so for two vectors a and b:

<a+b,a+b> = <a,a> + 2<a,b> + <b,b>

where <a,b> = |a| |b| cos(θab) and θab is the angle between a and b.

This generalizes to an arbitrary number of vectors, the square length of the sum is the sum of the square lengths plus the sum over all unordered unequal pairs of twice the product of their lengths times the cosine of the angle between them.

So in the earlier computation we got coefficients on the mixed terms of 2*(√3 / 2) for the two pairs of currents 30 degrees apart, and of 2*(1/2) for the pair of currents 60 degrees apart.

Cheers, Wayne
 
Carultch said:
An individual 10kVA load across a 2-pole breaker, will yield 48.08A drawn across them, and 48.08A on each of the two conductors associated with the phases of that breaker. But when you mix it with a second otherwise-identical 10 kVA load, that is staggered to the next pair of phases, the current adds up as vectors on the shared phase to become 83.3A, and the 48.08A remains alone on the two lone phases. Notice that 83.3A = sqrt(3)*48.08A.
Click to expand...
If you have two 10kVA loads, each 208 volts single phase, and each is connected to phases A and B on a panel, those currents would just 'add'? So the current on each phase would be 96 amps (48 for each load)? I'm assuming so since the currents for both loads would be coming in at the same phase / time.
 
winnie said:
Exactly. When panel schedules are normally done the approximation is that everything is 100% balanced and the same power factor. Since one generally makes the attempt to balance the panel and loads are roughly similar, this approximation is 'good enough' for most applications.

But if you had a situation where you needed to be precise, you would have to treat each load a vector, and then do the vector sum of all the loads on a given bus to determine the actual load.

Lets take the example you gave and extend it a bit. Lets consider a panel and add one 10KVA load at a time, and calculate the current on the bus.

1) Connect a 10KVA resistive load from phase A to phase B. We get 48A on bus A, 48A on bus B, and 0A on bus C.

2) Now add a corresponding load from phase B to phase C. We get 48A on buses A and C, but 83A on bus B.

3) Finally add a third load from C to A. Now we see 83A on all three busses.

4) Keep adding loads and you will see that the simple approximation (assign 1/2 the KVA to each bus and presume balance) gets better as you go along.

-Jon
Click to expand...


Okay I am lost

A load of 10kw on A and B phases has 10kw load on each phase not 5 kw on each. That is why we get 48 amps on A and B phases. Now we add a load to B and C phase and we still have 48 on Phase A but how do we only have 83 amps on B phase. Why not 96. How do we get 83 amps.

BTW, I am sure you are correct but I am missing it. It seems like you are dividing by 240 instead of 208
 
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