208V-1ph calculation

[dabatman]

I think this has been discussed before but I can't locate the thread to answer my question so I thought I would post this.

Essentially I have a power supply that can run off 208V-1ph. My load is 3200 Watts. Initial instinct would be to say 3200/208V=15.4A on each leg. I agree this is what you should see if you measure with an amp clamp on the conductor on say the A phase at the load.

My question comes from when you actually figure a panel schedule. I have 10 of these in a system and so looking at choosing a service, I believe with I've seen before that actual load on the system is calculated differently. From above, you actually have 15.4*2*120=3696 Watts. That's why I think I've seen before that you figure the load in your panel schedule as

Amperage =Watts / (2*208V/sqrt(3)) to get 13.32 A

which essentially gets you 3200/240.

So do I end up with 3200 Watts but 3696 VA in the panel schedule? I usually would show VA on my panel schedule but this problem has me a little bit confused on what the correct answer actually is. I would also welcome anybody's suggestion on an actual text that would show this calculation to see it more in depth.

 

[Hameedulla-Ekhlas]

From above, you actually have 15.4*2*120=3696 Watts. That's why I think I've seen before that you figure the load in your panel schedule as

Amperage =Watts / (2*208V/sqrt(3)) to get 13.32 A

which essentially gets you 3200/240.

.

these are new formula for me too.

I know P = VI for single phase but in your equation where 2 comes from
I = Watts/(208*sqrt3) for three phase again in your equation where 2 comes from

 

[charlie b]

Presuming that the overall power source is a three phase system, here is your mistake:

From above, you actually have 15.4*2*120=3696 Watts.

Why did you switch to 120 volts? You have two legs of a single phase 208 volt circuit, and the power is known to be 3200 watts. So on the panel schedule, you put 1600 watts on each phase.

 

[charlie b]

. . . but in your equation where 2 comes from

He was taking the calculated amps, 15.4, and saying that was the same current in each of two legs, and saying the voltage for a single phase was 120 volts, and that is why he multiplied 15.4 times 2 and then times 120. He was working the power numbers backwards.

 

[dabatman]

Switch to 120V

Switch to 120V

What I meant to show was that if you balance the loads across the phases evenly then you end up with 15.4 A on each leg. If you calculate the watts based on that, then it ends up being 15.4*120V*2 legs. Back-calculating if you will.

From Hameedulla-Ekhlas's question, I was thinking of it as:

Assuming Unity PF

Watts = 3*(V*I) 3phase
Watts = V*I 1-phase

It seems like I've seen before then to figure your amps then you get

Watts = 2*Vphase*Iphase which eventually comes out to the formula I have above.

I may be wrong but that's what I'm asking the question for.

 

[Hameedulla-Ekhlas]

What I meant to show was that if you balance the loads across the phases evenly then you end up with 15.4 A on each leg. If you calculate the watts based on that, then it ends up being 15.4*120V*2 legs. Back-calculating if you will.

From Hameedulla-Ekhlas's question, I was thinking of it as:

Assuming Unity PF

Watts = 3*(V*I) 3phase
Watts = V*I 1-phase

It seems like I've seen before then to figure your amps then you get

Watts = 2*Vphase*Iphase which eventually comes out to the formula I have above.

I may be wrong but that's what I'm asking the question for.

I was just wondering regarding 2 in three phase and single phase. Charli has cleared it for me. thanks

 

[charlie b]

What I meant to show was that if you balance the loads across the phases evenly then you end up with 15.4 A on each leg. If you calculate the watts based on that, then it ends up being 15.4*120V*2 legs.

No it doesn't. That was my point. It is a basic, and common, error. The fundamental truth is that watts plus watts equals watts, but amps plus amps does not equal amps. What I mean is that the current in only leg cannot be added to the current in another leg, without also taking into accout the fact that they are offset from each other by a phase angle of 120 degrees.

Watts, however, do not have an angle associated with them. So you need to do all your math in units of power. Convert to amps only at the end. In your case, you began by converting the 208 volt, 3200 watt load into units of amps. Don't do that. Stick with power. You assign 1600 watts to each leg, and move to the other loads. When you have all the loads accounted for, and balanced as best you can, then and only then do you figure out how many amps are on each phase of the panelboard.

 

[charlie b]

Watts = 3*(V*I) 3phase

No. For three phase systems, with unity power factor, Watts = SQRT(3)*(V*I). The factor of the square root of three comes into play because the three phases are separated from each other by 120 degrees of arc.

 

[skeshesh]

What I meant to show was that if you balance the loads across the phases evenly then you end up with 15.4 A on each leg. If you calculate the watts based on that, then it ends up being 15.4*120V*2 legs. Back-calculating if you will.

From Hameedulla-Ekhlas's question, I was thinking of it as:

Assuming Unity PF

Watts = 3*(V*I) 3phase
Watts = V*I 1-phase

It seems like I've seen before then to figure your amps then you get

Watts = 2*Vphase*Iphase which eventually comes out to the formula I have above.

I may be wrong but that's what I'm asking the question for.

I've been confused by this sometimes as well. This is how I think about it: you need to first think about the system you have, from what I gather from your question you have 208V phase-to-phase (p-p) and 120V phase-to-neutral (p-n). When you connect a single-phase, 2-POLE, 208V load, your voltage is still 208V phase to phase, hence P=V*I.
There are other times when you have a single phase system (or at least single phase subpanel in a system), where the voltage is 120/240v, 1ph, 3wire. In this case you have 240V phase-to-phase and 120V phase-to-neutral, so you can look at it as V(p-p)=2*V(p-n).
Now to get back to your original question if you have a have a 208V, 1-phase, 2-pole load in a 120V/208Y system, you find the load on each leg by using:
Load on each phase = P(VA)/208V(p-p)/2
Like Charlie pointed out is I think where you're making a mistake is when you get to the panel you assume you have to use 120V, which this is incorrect because with the 2-Pole, 1phase load your voltage is the Phase-to-phase voltage not the phase-to-neutral.

 

[Hameedulla-Ekhlas]

Assuming Unity PF

Watts = 3*(V*I) 3phase
.

Incase of Unity PF

Watts = 3 * (V*I) 3phase it is also ok but the voltage is not line to line it is line to phase.

Watts = sqrt 3 *V*I it means the voltage is line to line.

 

[dabatman]

Mistake

Mistake

Sorry I should have added more notes

Pφ = Vφ* Iφ* cos(θ) and so PLL = 3 * (Vφ* Iφ* cos(θ) )

so same idea. I think I have what my problem is.

I was trying to figure out it out and show amps on each leg. If I show watts say of 1600 on each leg then that solves my problem.

You end up with 1600 VA on each leg but if you divide by 120V then you get 13.32 A. My calculation was a goofy way (possibly wrong way?) to get to the same conclusion.

The long and short is that I need to just show the panel schedule with VA instead of trying to look at amps. Thanks for the help.

 

[Hameedulla-Ekhlas]

Sorry I should have added more notes

Pφ = Vφ* Iφ* cos(θ) and so PLL = 3 * (Vφ* Iφ* cos(θ) )

.

Dont put with Power phase and line to line signs. Just call single phase power and three phase power.

 

[LarryFine]

Batty, nobody has asked this, but is your source three phase, or just two lines? If it's the former, here's what I'd do:

With 10 loads, the distribution would be three between each of two pairs of lines, and four between the third pair, so I'd figure it as if there were 12 loads, as far as sizing the feeder and such.

 

[kingpb]

I believe this seems reminiscent of the "Oregon Fudge Factor", i..e the magical 1.15.

 

[rattus]

Batty, nobody has asked this, but is your source three phase, or just two lines? If it's the former, here's what I'd do:

With 10 loads, the distribution would be three between each of two pairs of lines, and four between the third pair, so I'd figure it as if there were 12 loads, as far as sizing the feeder and such.

Good answer Larry, now who wants to figure the line currents for this scenario?

 

[kwired]

I think this has been discussed before but I can't locate the thread to answer my question so I thought I would post this.

Essentially I have a power supply that can run off 208V-1ph. My load is 3200 Watts. Initial instinct would be to say 3200/208V=15.4A on each leg. I agree this is what you should see if you measure with an amp clamp on the conductor on say the A phase at the load.

My question comes from when you actually figure a panel schedule. I have 10 of these in a system and so looking at choosing a service, I believe with I've seen before that actual load on the system is calculated differently. From above, you actually have 15.4*2*120=3696 Watts. That's why I think I've seen before that you figure the load in your panel schedule as

Amperage =Watts / (2*208V/sqrt(3)) to get 13.32 A

which essentially gets you 3200/240.

So do I end up with 3200 Watts but 3696 VA in the panel schedule? I usually would show VA on my panel schedule but this problem has me a little bit confused on what the correct answer actually is. I would also welcome anybody's suggestion on an actual text that would show this calculation to see it more in depth.

watts and VA are the same if unity power factor.

Is your load 3200 watts @ 208 volt or 3200 watts @ 240 volt and you will be using it on 208 volts. This will make a difference in your calculations If resistive load your total will be less if 240 equipment operating at 208. If motors they will draw more current @ 208 than the 240 rating but will still use roughly the same power.

You need to figure total VA first and then divide by voltage to calculate the amperage, and factor in the 1.73 if balanced across three phases.

I come up with 3200 VA @ 208 volts = 15.4 amps.

If 3200 VA @ 208 is balanced across 3 phases = 8.9 amps.

If you have 10 of these you can balance 9 of them:

9x3200/208/1.732 = 79.9 amps plus one more unit drawing 15.4 across two phases you pick which two you want unbalanced.

 

[Smart $]

Good answer Larry, now who wants to figure the line currents for this scenario?

...

I come up with 3200 VA @ 208 volts = 15.4 amps.

If 3200 VA @ 208 is balanced across 3 phases = 8.9 amps.

If you have 10 of these you can balance 9 of them:

9x3200/208/1.732 = 79.9 amps plus one more unit drawing 15.4 across two phases you pick which two you want unbalanced.

93.67A on the two lines having seven connected, and 80.02A on the one with six connected. I used exactly 15.4A per load. The OP mentioned the loads are power supplies, so this is of course calculating as though each load is exhibiting an identical power factor, and conducting at full rated load.

 

[mull982]

93.67A on the two lines having seven connected, and 80.02A on the one with six connected. I used exactly 15.4A per load. The OP mentioned the loads are power supplies, so this is of course calculating as though each load is exhibiting an identical power factor, and conducting at full rated load.

This is the answer I come up with as well.

However I have this type of problem has always confused me. We know that a 3200kW load @208V will conduct 15.4A which can be seen on each leg of the 208V. If we were just putting this single load across two phases in a panel schedule I believe that we would use 15.4A on each phase of the panel schedule.

Now there was some discussion above about dealing with kW values as opposed to amps when dividing up multiple loads on a panel schedule. To do this with our 3200kW load we divide it in half to get 1600kW per phase and dividing this by the phase voltage of 120V we see that there is 13.3A on each leg. This 13.3A vs the 15.4A shown above is a factor of 1.15 which I believe is the Oregon fudge factor.

So I guess my question is, if we have just a single 3200A load on two phases can we use the 15.4A in the panel schedule. Is it only when we have multipile loads spread across different phases that we have to start working in kW values when dividing up loads? Is the 1.15 difference and the requirement to work with kW units when using multiple loads due to the same results we see when we add L-L loads up on a particular phase using a vector summation? For instance if we add up all the L-L loads connected to phase A, we must use the vector magnitudes and angles of all loads connected to this phase and sum these up to come up with the total connected Amps on this phase. This would be different, then simply adding the current magnitudes from each load without considering angles. Is this the same concept?

 

[Smart $]

...

However I have this type of problem has always confused me. We know that a 3200kW load @208V will conduct 15.4A which can be seen on each leg of the 208V. If we were just putting this single load across two phases in a panel schedule I believe that we would use 15.4A on each phase of the panel schedule.

That is correct. If this load were the only load on the panel, you would indeed have 15.4A on each connected line of the panel's feeder.

Now there was some discussion above about dealing with kW values as opposed to amps when dividing up multiple loads on a panel schedule. To do this with our 3200kW load we divide it in half to get 1600kW per phase and dividing this by the phase voltage of 120V we see that there is 13.3A on each leg. This 13.3A vs the 15.4A shown above is a factor of 1.15 which I believe is the Oregon fudge factor.

You again are correct, in essence. What you need to correct, though, is your use of terminology. In panel schedules, it is only proper to use VA or kVA as the unit of measure. We are not concerned directly with power (W), only amperes... to protect conductors and equipment. Because typical systems have at least two voltage classes of loads, working in VA, kVA units makes it easy to calculate total load on the panel.

So I guess my question is, if we have just a single 3200[V]A load on two phases [Lines] can we use the 15.4A in the panel schedule.

Since we do not use Amperes as schedule units, we would not use 15.4A in the schedule. I believe we would still half the VA amount for each connected Line column, so it would be correct should future loads be added. I guess what I'm saying is we can't simply change the way it is done for one oddball situation. Additionally, while columns are totalled at the bottom of the schedule, sizing of feeders is first based on total connected load (column totals added together) and then adjusted for the difference in columns (an unbalanced condition) as necessary.

That said, a qualified person should be the only one doing panel schedules to begin with, and only qualified persons should be making any determinations thereof. Such qualified people should know the limitations of accuracy and application of such a panel schedule. All in all, a panel schedule should only be used as a rough guide to panel loading. The actual calculation should be done per Article 220 and an unbalanced condition should be handled as best determined (i.e. means of resolution is not stipulated in the NEC).

Is it only when we have multipile loads spread across different phases that we have to start working in kW [kVA] values when dividing up loads?

As noted previously, it is mostly to combine and work with two voltage classes of loads.

Is the 1.15 difference and the requirement to work with kW [kVA] units when using multiple loads due to the same results we see when we add L-L loads up on a particular phase using a vector summation?

Yes. The 1.15 "Oregon Fudge Factor" is a number based on 1? Line-to-Line loads being balanced.

For instance if we add up all the L-L loads connected to phase A, we must use the vector magnitudes and angles of all loads connected to this phase and sum these up to come up with the total connected Amps on this phase. This would be different, then simply adding the current magnitudes from each load without considering angles. Is this the same concept?

Exactly... though Article 220 calculations have no provision to do vector summation. The prescribed calculation rely on basic math to make it easier on those that either aren't aware of the concept of or reason for vector summation, don't know how, or do not have the resources at hand to do a vector summation.

And through all of the above, I should note that it is of my opinion... nothing more

 

[kingpb]

A word of caution to the unsuspecting, is that you still need to be aware when using the panel total KVA to size the service or transformer ahead of the service.

The reason being that a panel with a unbalanced phase could result in a total KVA (sum of all three phases) that when divided by three (transformer power per phase) is less KVA than the largest unbalanced phase KVA. Hopefully, after applying the NEC safety margin of 125%, this will rectify itself, but not always.

Due to my conservative nature, I will typically size the transformer and panel based on at least 3X the most heavily loaded phase. This will guarantee some flexibility as panel loads change.