Calculating Load for a 3 Phase Service

[WireRunner]

I need help. I have a proposed total load of 88,270VA in my building. The load is non-continuous. The service will be 3Φ 208Y/120v. All breakers will be single pole and double pole (single phase loads only). If I balance my loads, what will be my amp load per phase? I just can't seem to figure out how to calculate service load in a 3 phase service. Is it as easy as 88,270VA/208/1.73 = 245A/Leg, and then size service based on that?

I just don't understand the 1.73 in the calculations. If a 3Φ service is loaded with single phase loads...it just gets confusing to me. Any clarity would be helpful.

Thanks!

 

[One-eyed Jack]

I need help. I have a proposed total load of 88,270VA in my building. The load is non-continuous. The service will be 3Φ 208Y/120v. All breakers will be single pole and double pole (single phase loads only). If I balance my loads, what will be my amp load per phase? I just can't seem to figure out how to calculate service load in a 3 phase service. Is it as easy as 88,270VA/208/1.73 = 245A/Leg, and then size service based on that?

I just don't understand the 1.73 in the calculations. If a 3Φ service is loaded with single phase loads...it just gets confusing to me. Any clarity would be helpful.

Thanks!

Yes and NO. The 245 can be the amount per leg if the loads are values that can be placed on opposite phases and equal each other. I don't have the computer savvy to draw and post what I am saying. Some of the long term members can whip something out in a hurry. Hopefully they will chime in.

 

[iwire]

I will go with a simple yes with the loads balanced.

I would not go with a 250 amp service but I would feel fine going with a 300 amp or larger service.

 

[Dennis Alwon]

I need help. I have a proposed total load of 88,270VA in my building. The load is non-continuous. The service will be 3Φ 208Y/120v. All breakers will be single pole and double pole (single phase loads only). If I balance my loads, what will be my amp load per phase? I just can't seem to figure out how to calculate service load in a 3 phase service. Is it as easy as 88,270VA/208/1.73 = 245A/Leg, and then size service based on that?

I just don't understand the 1.73 in the calculations. If a 3Φ service is loaded with single phase loads...it just gets confusing to me. Any clarity would be helpful.

Thanks!

Not an engineer but 1.73= square root of 3. This number is used in 3 phase calculations. So when you calculate loads for 3 phase you would take wattage/208/1.73= amperage. Another way to see this is Wattage/ 360= amps because 208/1.73= 360.

 

[Smart $]

...Another way to see this is Wattage/ 360= amps because 208/1.73= 360.

Umm... that should be 208 ? 1.73 = 360 (= 120 ? 3)


...as in his equation 88,270VA/208/1.73 = 245A/Leg = 88,270VA/(208 ? 1.73)

 

[hardworkingstiff]

I never had any formal training in electrical work. When I took my 1st "master" exam in Norfolk, VA it was after having the minimum J-man experience (classroom training was not required at that time). The State did not have any license program, it was all local. Anyway, all of my electrical experience was residential, no commercial or industrial at all.

One of the inspectors said they (the inspectors in the City would grade the tests) got a good chuckle grading my exam because I did 3-phase calculations the "old fashioned way". I did not understand 1.732, and all my calculations for 3-phase services were done by breaking everything down to 120-volt loads then recombining them times 3. It made sense to me at the time, and I did pass. I was glad I could make them laugh :roll:.

 

[Dennis Alwon]

Umm... that should be 208 ? 1.73 = 360 (= 120 ? 3)


...as in his equation 88,270VA/208/1.73 = 245A/Leg = 88,270VA/(208 ? 1.73)

You are correct-- I took liberties too far on that one. Trying to show that some number/208/1.73 is the same as some number over 360 because 208*1.73= 360

 

[char]

I agree - round up to 300A. However I have never seen a building with an entire non-continuous load. Lighting, refrigeration, most process loads are continuous. Would be curious as to the type of building this is.

 

[WireRunner]

Still confused...

Still confused...

I appreciate your help so far, but I am still a little confused!

Can you check my logic in an example? Here are some equipment specs. Consider all loads non-continuous in this example, I'm just trying to understand the 3 phase calc...

Equipment Calculated Wattage

Refrigerator 13A, 120v, 1Φ 1560w
Freezer 9.3A, 208v, 1Φ 1935w
Meat Slicer 3A, 120v, 1Φ 360w
Dishwasher 33A, 208v, 1Φ 6864w
Glasswasher 14.1A, 208v, 1Φ 2933w
Bottle Cooler 10A, 120v, 1Φ 1200w
Elec Fireplace 13.3A, 120v, 1Φ 1596w

TOTAL WATTAGE 16,448w

So the load per leg would be: 16448/208/1.73 = 45.7A per leg

Is that correct?




To further explain the thought behind my confusion, lets use the refrigerator and the freezer in another example....

Lets say I place the refrigerator(1560w) on Phase A, and the freezer(1935w) on Phase B and Phase C. That is all that is in the panel.


My calculation of ((1560w+1935w)/208/1.73) = 9.71A per phase.


My logical thought says I have 13A on Phase A, 9.3A on Phase B, and 9.3A on Phase C. Isn't that what my amp meter would read if I checked the phases, like in a single phase panel? How could it read 9.71A on each phase?


I think I just don't understand the 3 Phase waveform and how the 1.73 reates to it and it's calculations.

Any help is appreciated! Thanks so much!

 

[brother]

I appreciate your help so far, but I am still a little confused!

Can you check my logic in an example? Here are some equipment specs. Consider all loads non-continuous in this example, I'm just trying to understand the 3 phase calc...

Equipment Calculated Wattage

Refrigerator 13A, 120v, 1Φ 1560w
Freezer 9.3A, 208v, 1Φ 1935w
Meat Slicer 3A, 120v, 1Φ 360w
Dishwasher 33A, 208v, 1Φ 6864w
Glasswasher 14.1A, 208v, 1Φ 2933w
Bottle Cooler 10A, 120v, 1Φ 1200w
Elec Fireplace 13.3A, 120v, 1Φ 1596w

TOTAL WATTAGE 16,448w

So the load per leg would be: 16448/208/1.73 = 45.7A per leg

Is that correct?




To further explain the thought behind my confusion, lets use the refrigerator and the freezer in another example....

Lets say I place the refrigerator(1560w) on Phase A, and the freezer(1935w) on Phase B and Phase C. That is all that is in the panel.


My calculation of ((1560w+1935w)/208/1.73) = 9.71A per phase.


My logical thought says I have 13A on Phase A, 9.3A on Phase B, and 9.3A on Phase C. Isn't that what my amp meter would read if I checked the phases, like in a single phase panel? How could it read 9.71A on each phase?


I think I just don't understand the 3 Phase waveform and how the 1.73 reates to it and it's calculations.

Any help is appreciated! Thanks so much!

Good question. Im not sure If i have this right and others are welcome to chime in . I think since it would only be those 2 loads and they are single phase, it would be single phase calc. so it would be (1560+1935)/208=16.8 amps. So just for those 2 you would have to size your feeders to that size of 16.8 amps ( this is NOT accounting for demand factor).

 

[charlie b]

The thing you are missing, and the thing that confuses many people with regard to three phase systems, is that the current that leaves the source via any one phase will return to the source via the other two phases. The three currents do not act independently of each other. When the loads are not balanced on the three phases, the mathematical technique for calculating the current on each wire becomes very complicated. It is simplest to assume, or pretend, that the loads are balanced, and use the formula that has already been discussed. As long as we do our best to spread the loads evenly across the three phases, the formula?s result will be close enough.

The value of 1.732 (the square root of 3) will come into play on any calculation you perform on a three phase system. To prove that would require the use of trigonometry. I have written up a derivation of the formula for power in a three phase system, and it shows the origin of that square root of three. I have sent it to forum members in the past, if they ask for it by PM, and give me an email address. I think I will figure out how to post it as a FAQ, so I won?t have to send out email messages anymore.

In the mean time, try this experiment: Draw two lines, starting at the same point. Make them each 1 inch long, and draw them at an angle of 120 degrees apart from each other. Then measure the distance between the tips of the lines. You will get a measurement of about 1.732 inches. If you recall any trigonometry from high school, in particular the fact that in a 30-60-90 triangle the hypotenuse is always twice the lengths of the shortest side, you can work out for yourself (using the Pythagorean Theorem) the fact that the third side will be 1.732 times longer than the shortest side.

 

[charlie b]

I have written up a derivation of the formula for power in a three phase system, and it shows the origin of that square root of three. I have sent it to forum members in the past, if they ask for it by PM, and give me an email address. I think I will figure out how to post it as a FAQ, so I won?t have to send out email messages anymore.

I tried to post it, but was told (by the forum police) that it was too long. So if anyone wants to see it, please send me a PM, and give me an email address.

 

[brother]

I tried to post it, but was told (by the forum police) that it was too long. So if anyone wants to see it, please send me a PM, and give me an email address.

I would like to have that info. So was I wrong in my post about the single phase loads that he posted before?? I want to be sure I understand this correctly as well.

 

[charlie b]

So was I wrong in my post about the single phase loads that he posted before??

If you refer to your post 10, then I would say yes. But let us be clear about the situation.

Lets say I place the refrigerator(1560w) on Phase A, and the freezer(1935w) on Phase B and Phase C. That is all that is in the panel.

Here is how I would predict the current readings. It matters very much if we are talking about the branch circuit currents or the total currents seen by the panel.

• If you measure the current on Phase A of the branch circuit feeding the fridge, you will get a current of 1560/120, or 13 amps.
• If you measure the current on Phase B or Phase C of the branch circuit feeding the freezer, you will get a current of 1935/240, or 8 amps.
• If you measure the current on either Phase A, B, or C of the feeder to the panel, you will get a current of approximately (1560+1935)/(208*1.732), or about 9.7 amps. However, the load will not be exactly balanced, so the three currents you measure will not be the same, and neither will be exactly 9.7 amps. To get an accurate prediction for a non-perfectly balanced loading would require the mathematical tool called, "symmetrical components." I no longer remember how to use that tool. So let's just pretend things are balanced, and accept the 9.7 amp predicted reading.

I would like to have that info.

Send me an email address via PM, and I will send you two Word files. They provide the derivation of the forumulas for power in single phase and three phase systems.

 

[Smart $]

...

Here is how I would predict the current readings. It matters very much if we are talking about the branch circuit currents or the total currents seen by the panel.

• If you measure the current on Phase A of the branch circuit feeding the fridge, you will get a current of 1560/120, or 13 amps.
• If you measure the current on Phase B or Phase C of the branch circuit feeding the freezer, you will get a current of 1935/240, or 8 amps.
• If you measure the current on either Phase A, B, or C of the feeder to the panel, you will get a current of approximately (1560+1935)/(208*1.732), or about 9.7 amps. However, the load will not be exactly balanced, so the three currents you measure will not be the same, and neither will be exactly 9.7 amps. To get an accurate prediction for a non-perfectly balanced loading would require the mathematical tool called, "symmetrical components." I no longer remember how to use that tool. So let's just pretend things are balanced, and accept the 9.7 amp predicted reading.

I have to disagree with you. With only these two loads powered by the panel, the feeder conductors carry the same current as the branch circuits, as do the panel's busbars. There are no currents from these two loads combining at the panel bus bars.

 

[brother]

[/LIST]I have to disagree with you. With only these two loads powered by the panel, the feeder conductors carry the same current as the branch circuits, as do the panel's busbars. There are no currents from these two loads combining at the panel bus bars.

Ok, so is my previous post correct on this?? since the feeder conductors will see the same current as the branch circuit conductors (when the frig and freezer are the only loads) then the feeders would have to sized to 16.8 amps correct??

 

[Smart $]

Ok, so is my previous post correct on this?? since the feeder conductors will see the same current as the branch circuit conductors (when the frig and freezer are the only loads) then the feeders would have to sized to 16.8 amps correct??

Sorry, but no... 16.8A is incorrect.

There would be 13A on Line A, and 8A on Lines B and C.

Conductors would have to be sized for these current values as a minimum. Though the NEC doesn't make it blatantly obvious, we are supposed to account for imbalance of current on circuits of more than two conductors.

 

[charlie b]

[/list]I have to disagree with you. With only these two loads powered by the panel, the feeder conductors carry the same current as the branch circuits, as do the panel's busbars. There are no currents from these two loads combining at the panel bus bars.

I disagree with me also. Good catch.

 

[winnie]

To further explain the thought behind my confusion, lets use the refrigerator and the freezer in another example....

Lets say I place the refrigerator(1560w) on Phase A, and the freezer(1935w) on Phase B and Phase C. That is all that is in the panel.


My calculation of ((1560w+1935w)/208/1.73) = 9.71A per phase.


My logical thought says I have 13A on Phase A, 9.3A on Phase B, and 9.3A on Phase C. Isn't that what my amp meter would read if I checked the phases, like in a single phase panel? How could it read 9.71A on each phase?

The answer is that your analysis is exactly correct, but that you are in some sense 'overthinking' the problem.

In the panel that you describe, you would have 13A, 9.3A, 9.3A. There is no way for the current to 'jump around' to average things out.

You are overthinking only in that when these calculations are done, the _assumption_ is that the loads will be evenly distributed, and will average out, and that the entire set of loads will look like a single balanced three phase load.

Of course, in reality, the load will _not_ be perfectly balanced, but as others have said, this is generally not a big deal.

Take this as a caution to not create an intentionally unbalanced situation, say by putting all 120V loads on phase A, and all 208V loads between phases B and C. Distribute stuff so it looks like a balanced 3 phase load.

Additionally, if you have a situation that you know must be unbalanced, for example a high leg delta service, then you know that the conventional three phase balanced equation is not sufficient.

As an aside, I often fall back on Lou's approach as a way to break down complex situations, particularly when I am working with 'high phase' systems. Find the line-neutral voltage, multiply by the line current, adjust for power factor, and then multiply by the number of phases. Lou's approach also makes it clear where the 1.732 comes from; 120 * I * 3 = 208 *I * X, and X falls out as the square root of three.

-Jon

 

[LarryFine]

Take this as a caution to not create an intentionally unbalanced situation, say by putting all 120V loads on phase A, and all 208V loads between phases B and C. Distribute stuff so it looks like a balanced 3 phase load.

My favorite way of doing this is to stack the 2-poles on one side and the 1-poles on the other, breaker count permitting, of course.