Panel schedule add in amps

[hhsting]

Attached panel schedule shows calculation in amps per phase then add amps each phase and then turn then in kva single phase. So ckt #1 total load is 40A but its 208V single phase so what the person did is put 40A on each leg. I thought each leg would see 20A not 40A.

I have always did kvA per phase and then add kvA. I am not sure if the below method of amps per phase is correct or incorrect?

 

[augie47]

40 amp 208 would be 40 amps each phase

 

[hhsting]

40 amp 208 would be 40 amps each phase

Really? 208V single 40A has 8.32 kva. However if it does have 40A each phase then 40x.12x2=9.6kva which is not equal to 8.32kva?


Also they are adding phase A current column, B current column, C current column and then multiply them with 120V to get to connected load. Somehow this does not work that way.

Currents are certain degree out if phase. Just wondering how can one do this right way aside from redoing to kva?

 

[DrSparks]

Really? 208V single 40A has 8.32 kva. However if it does have 40A each phase then 40x.12x2=9.6kva which is not equal to 8.32kva?


Also they are adding phase A current column, B current column, C current column and then multiply them with 120V to get to connected load. Somehow this does not work that way.

Currents are certain degree out if phase. Just wondering how can one do this right way aside from redoing to kva?

When making 3-phase VA calculations, you have to multiply VA by the square root of 3 (approx. 1.732). So 208 3-phase 40 Amps is approx 14.4 kVA.

You only do this for actual 3-phase loads. For 208 single phase, you don't use the factor.

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[hhsting]

When making 3-phase VA calculations, you have to multiply VA by the square root of 3 (approx. 1.732). So 208 3-phase 40 Amps is approx 14.4 kVA.

You only do this for actual 3-phase loads. For 208 single phase, you don't use the factor.

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If you take a look at the load they are using amps per phase not kva in thd A, b, c phase.

The load talking about is 208V single phase


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[infinity]

It says that the load is 1Ø, 40 amps. In a 2-wire circuit the current is the same in each of the phase condcutors.

 

[hhsting]

It says that the load is 1Ø, 40 amps. In a 2-wire circuit the current is the same in each of the phase condcutors.

Look at the total load cal for phase column, they are adding currents and then multiply by 120V

 

[infinity]

Look at the total load cal for phase column, they are adding currents and then multiply by 120V

Your initial question asked about 40 amps single phase that is what my response was clarifying. I don't see where there is multilplying by 120.. At the bottom is just the total of each column.

 

[hhsting]

Your initial question asked about 40 amps single phase that is what my response was clarifying. I don't see where there is multilplying by 120.. At the bottom is just the total of each column.

Lets see:

Phase A current bottom left: 103A
Phase B current bottom left: 69.2
Phase C current bottom left: 74.7

Phase A current bottom right: 217A
Phase B current bottom right: 282.9
Phase C current bottom right: 263.2A

So total of all the above: 1010A (calc not shown panel schedule)

Now 1010Ax.12kV=121.2kVA (calc not shown on panel schedule)

Bottom left you see where it says panel schedule connected kVA it says 121.2kva so then do you see what I mean?

 

[charlie b]

I have always did kvA per phase and then add kvA. I am not sure if the below method of amps per phase is correct or incorrect?

You are right, and this panel schedule is wrong.

If all loads were single pole, or if all loads were three pole, then you could add amps, multiple by voltage, and get the correct total KVA. But this schedule is treating the three pole load on circuit 1-3-5, the two pole load on circuit 13-15, and the single pole load on circuit 19 as though they were similar. They are not. The three 4.8 amps on circuit 1-3-5 need to be multiplied by 208 and then multiplied by the square root of 3. The two 11.7 amps on circuit 13-15 get multiplied by 208, and not by the square root of 3. The 20 amps on circuit 19 gets multiplied only by 120. So the three types of loads are weighted differently in the calculation.

 

[hhsting]

You are right, and this panel schedule is wrong.

If all loads were single pole, or if all loads were three pole, then you could add amps, multiple by voltage, and get the correct total KVA. But this schedule is treating the three pole load on circuit 1-3-5, the two pole load on circuit 13-15, and the single pole load on circuit 19 as though they were similar. They are not. The three 4.8 amps on circuit 1-3-5 need to be multiplied by 208 and then multiplied by the square root of 3. The two 11.7 amps on circuit 13-15 get multiplied by 208, and not by the square root of 3. The 20 amps on circuit 19 gets multiplied only by 120. So the three types of loads are weighted differently in the calculation.

Circuit #13 and #15 are 120V single phase 1 pole breaker so not sure where 208V is coming from?

Circuit #4,#6 is 208V single phase load. You are saying multiply 40x.208V? Or 80x.208?
Same thing with 1,3,5?

 

[wwhitney]

Just wondering how can one do this right way aside from redoing to kva?

kVA is best. But it would be equivalent to simply identify the 2 pole breakers feeding 2-wire 208V loads, and multiply those load currents by 0.5*sqrt(3). That takes care of the 8.32 vs 9.6 kVA discrepancy you noted.

Cheers, Wayne

 

[infinity]

So if 121 is incorrect what is the actual kva load?

 

[wwhitney]

PS Winnie's comments in the other thread made me realize the following:

1) If the panel has only L-N loads and L-L-L 3 phase loads, the two methods (currents vs kVA) will agree.
2) If the panel has the 2-wire L-L loads all on one particular pair of ungrounded conductors, then the currents method (as in OP) is exact.
3) If the panel has the 2-wire L-L loads evenly balanced on all pairs of ungrounded conductors, then the kVA method is exact.

In the panel schedule in the OP, the L-L loads are distributed around the three pairs of ungrounded conductors, but they are different sizes, so they are not perfectly balanced. So the actual line currents will be somewhere between the 2nd and 3rd case above.

For an exact answer, cases (2) and (3) can be mixed by summing the line currents from the L-L loads separately from the L-N and L-L-L loads; then the largest balanced set of L-L currents (determined by the smallest of the L-L load current sums) can be subtracted from the L-L load current triple, multiplied by 0.5 * sqrt (3) , and added to the L-N / L-L-L load currents (which treats them as in case 3 above). The remainder of the L-L load currents then gets added in without the 0.5 * sqrt(3) factor (as in case 2 above).

This all assumes power factor 1.

Cheers, Wayne

 

[hhsting]

PS Winnie's comments in the other thread made me realize the following:

1) If the panel has only L-N loads and L-L-L 3 phase loads, the two methods (currents vs kVA) will agree.
2) If the panel has the 2-wire L-L loads all on one particular pair of ungrounded conductors, then the currents method (as in OP) is exact.
3) If the panel has the 2-wire L-L loads evenly balanced on all pairs of ungrounded conductors, then the kVA method is exact.

In the panel schedule in the OP, the L-L loads are distributed around the three pairs of ungrounded conductors, but they are different sizes, so they are not perfectly balanced. So the actual line currents will be somewhere between the 2nd and 3rd case above.

For an exact answer, cases (2) and (3) can be mixed by summing the line currents from the L-L loads separately from the L-N and L-L-L loads; then the largest balanced set of L-L currents (determined by the smallest of the L-L load current sums) can be subtracted from the L-L load current triple, multiplied by 0.5 * sqrt (3) , and added to the L-N / L-L-L load currents (which treats them as in case 3 above). The remainder of the L-L load currents then gets added in without the 0.5 * sqrt(3) factor (as in case 2 above).

This all assumes power factor 1.

Cheers, Wayne

Huh You lost me. An example or the panel schedule calculation itself please?

 

[wwhitney]

Called the 3 phases A, B, C.

L-L-L load (A-B-C) 27 A
L-N load (A) 12A
L-N load (B) 5A
L-N load (C) 12A
L-L load (A-B) 25A
L-L load (B-C) 30A
L-L load (C-A) 32A

Simple current total for each phase: (96A, 87A, 101A). [Assuming no demand factors, a 100A panel bus would not be sufficient, but this is an upper bound.]

"kA" current total for each phase (since I haven't specified the voltage, multiply by voltage to get kVA), each L-L load gets multiplied by 0.5*sqrt(3): (88.4A, 79.6A, 92.7A) [A 100A panel bus would be sufficient, but this is a lower bound.]

Mixed method: Current totals for L-L-L and L-N loads only: (39A, 32A, 39A). For L-L loads only: (57A, 55A, 62A). The largest balanced portion for the L-L loads is (55A, 55A, 55A) with a residual of (2A, 0A, 7A). So 55A * 0.5 * sqrt(3) = 47.6A. The final answer is (39,32,39) + (47.6, 47.6, 47.6) + (2, 0, 7) = (88.6A, 79.6A, 93.6A). [This should be exact if all loads have power factor 1. A 100A panel bus would suffice.]

Cheers, Wayne

 

[don_resqcapt19]

Why are you getting into this detail on plan review? The worst case is the ~352 amps on B phase and the panel has a 400 amp main. Who cares if the math is not technically correct?

 

[wwhitney]

Who cares if the math is not technically correct?

It's important to know that the approximation is conservative rather than non-conservative. But it is in this case, so that's enough for plan review.

Cheers, Wayne

 

[hhsting]

Okay so their is also other way to do it:

Total AC current is 1010A. (1010/3)*.208*1.732= 121.28A
So then that would work

 

[don_resqcapt19]

It's important to know that the approximation is conservative rather than non-conservative. But it is in this case, so that's enough for plan review.

Cheers, Wayne

With a load calculation based on Article 220, I don't really care.