Something easy but dont know

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Timboe

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Hi,
To find the total amps on a 480volt three phase system. How do you find the total amps? Do you add the amps on the three legs together to come up with a total sum. Say i have a 200 amp breaker and put a meter on leg 1 and got 50 amps and the same on leg two and three would the total amps be 150? How many amps can i have on each leg saying all are equal to max out this 200 amp breaker.
thanks timboe:confused:
 
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27hillcrest

Senior Member
Omes law will give you all the answers. The breaker will trip when any phase of the breaker reaches 200 amps. You should only load a breaker up to 80%
 
roger

roger

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Timboe, you will be more correct to think of total watts.

Three phase 480 volt system = 480 x 1.732 = 831 volts


200A x 831V = 166,200 watts

166,200w / 277v = 600 amps phase to neutral or 200 amps per phase
166,200w / 480v = 346.25a / 1.732 = 200 amps phase to phase

Roger
 
Dennis Alwon

Dennis Alwon

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27hillcrest said:
You should only load a breaker up to 80%
Click to expand...
It may be good advice for future addition but it is not code. If you calculate the load of a 200 amp service at 80% then we would only be able to load the breaker to 160 amp calculate load. This is not the case. If I do a calculated load on a new house at 190 amp then a 200 amp service is sufficient. As other have said 80% is for continuous load.
 
iwire

iwire

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Dennis Alwon said:
If I do a calculated load on a new house at 190 amp then a 200 amp service is sufficient.
Click to expand...

I agree but I would like to point out that the load calculations have already added 25% to the continuous loads.

In other words that 190 amps is basically a fictional number and will never be seen as a real load at the service.
 
George Stolz

George Stolz

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This issue (breaker loading) has actually come up for me recently, here was the setup:

  • Ceiling-mounted heater in vestibule.
  • 50A breaker spec'd for unit on the prints
  • #6 conductors spec'd
  • Unit's overall amp rating is 49.98
  • The motor is a .12 HP, 1.9A blower motor.
  • 75 degree conductors required by unit (by this I assume 75* terminals)
  • The wiring method was THHN in EMT.
When I suggested that a 60A breaker would last longer, I got a funny look. Was I wrong?

My thought is, a breaker has a thermal and a magnetic trip mechanism, right? If the breaker is loaded to 100% of it's rating for two hours at a time consistently, it seems to me it would fail (over time) faster than a 60A breaker under the same conditions, due to operating the breaker so close to it's rating.

Edit to add: I assume the thermal portion of the breaker would wear quicker.
 
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George Stolz

George Stolz

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iwire said:
Was that before or after applying 25% for space heating loads?
Click to expand...

I have just banged my head against the wall repeatedly. Shame on me for not looking at the codebook before throwing an opinion out! :mad: :mad:

No, I did not look and see 424.3(B) calling it a continuous load, I assumed it was just like an A/C and considered undersized if it ran for three hours or more. I'll send the foreman a link to this thread, I'm not on that job as of Monday.

Thanks for the catch, Bob. :cool:
 
George Stolz

George Stolz

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georgestolz said:
  • Ceiling-mounted heater in vestibule.
  • 50A breaker spec'd for unit on the prints
  • #6 conductors spec'd
  • Unit's overall amp rating is 49.98
  • The motor is a .12 HP, 1.9A blower motor.
  • 75 degree conductors required by unit (by this I assume 75* terminals)
  • The wiring method was THHN in EMT.
Click to expand...
There was one other detail I left out: The unit says "maximum fuse size 65A".

65A is not a standard fuse size, and 49.98A x 125% = 62.5A.

So the code minimum is a #6 THHN on a 70A breaker (240.4(B)), correct?

Bear in mind, the nameplate on this unit is a hand-written fill-in-the-blank style, I'm not completely sure that the only OCPD allowed for this appliance is a fuse. It has a pair of GE two-pole breakers installed as disconnecting means for it.
 
iwire

iwire

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georgestolz said:
There was one other detail I left out: The unit says "maximum fuse size 65A".

65A is not a standard fuse size, and 49.98A x 125% = 62.5A.

So the code minimum is a #6 THHN on a 70A breaker (240.4(B)), correct?
Click to expand...

No, 110.3(B)

Also we do not have to use 240.4(B) and we do not have to use standard sizes.

IMO this unit will need an OCPD between 62.5 and 65 amps.

The cheapest way to get there is a fusible disconnecting means with 65 amp fuses.
 
growler

growler

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Timboe said:
Say i have a 200 amp breaker and put a meter on leg 1 and got 50 amps and the same on leg two and three would the total amps be 150? How many amps can i have on each leg saying all are equal to max out this 200 amp breaker.
thanks timboe:confused:
Click to expand...

If you can balance the load then you can have 200 Amps on each leg. It's a 200 amp breaker.

But you can't use a meter to check the load because this will only tell you what the load is at the time you check it. The minute that another motor starts or another light is switched on the load will increase. What if you check in winter and they have gas heat but sooner or later summer comes around and then the A/C load kicks in. You must do a load calculation and this is not always easy because the equipment may not have even been purchased and changes are made. I don't hold with the 80% idea but I wouldn't max out the panel either. Someone mentioned 190 Amps. , that's about as close as I would wish to go before up gradeing.
 
LarryFine

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Henrico County, VA
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Timboe said:
Say i have a 200 amp breaker and put a meter on leg 1 and got 50 amps and the same on leg two and three would the total amps be 150?
Click to expand...
No, you'd have the same current, but because power is equal to volts times amps, you'd have three times the power that the same current on one phase would provide (ignoring 3-phase fudge-factors).
 
R

rattus

Senior Member
Not really:

Not really:

roger said:
Timboe, you will be more correct to think of total watts.

Three phase 480 volt system = 480 x 1.732 = 831 volts

200A x 831V = 166,200 watts

166,200w / 277v = 600 amps phase to neutral or 200 amps per phase
166,200w / 480v = 346.25a / 1.732 = 200 amps phase to phase

Roger
Click to expand...

480Vrms is the maximum voltage seen in this system. 831V cannot be measured anywhere. This is misleading.

Apparent power (balanced load) in KVA is given by,

Pa = 277V x 200A x 3
= (277V x 1.732) x 200A x 1.732
= 480V x 200A x 1.732 = 166.3KVA

Iline = 200A (given)
Iline to line = 200A/1.732 = 115.5A
 
Smart $

Smart $

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rattus said:
Iline = 200A (given)
Iline to line = 200A/1.732 = 115.5A
Click to expand...
Umm... I believe your second line is a little misleading, too. It is accurate when all three lines are conducting equally...
I_A-B = I_B-C = I_C-A = 115.5A​


However, in a maximally unbalanced Line-to-Line situation...
I_L-L,max = I_L,max = 200A​
...where the Line-to-Line load(s) are across only one combination of three lines, say A-B, and may conduct up to 200A. Any additional load to either line in a maximally unbalanced state will exceed the trip rating. The other Line, in this case Line C, can still conduct 200A but no part of that 200A can be conducted through Line A or Line B. Therefore, the current has to return to the system through another conductor. In this case, that would be a neutral in a wye-configurated system...
I_A-B,max = I_C-N,max = 200A​

What it all amounts to is, the vectorial sum of currents cannot exceed the trip rating on any one Line.
 
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