calculation ?

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Mike01

Senior Member
Electrical calculations:
Question regarding electrical fourmlas when you have a 5000w load at 480v-3ph you would (5000/(480*sqrt(3))) this would give you approx 6amps.

For 208v single phase I have herd two different opinions what would be correct:
a. (5000/2) = 2500 (2500/120)=20.8A
b. (5000/208)=24.0A
is a or b correct? For a single phase system?

spsnyder

Senior Member
b.

For single phase P=IV.

bth0mas20

Senior Member
i will also agree with (b.)

charlie b

Moderator
Staff member
Also concur with "b." The notion of dividing the 5000 by 2 sounds like a confusion between "2-pole" and "2 phase." This is a single phase load, and its voltage is 208, not 120.

bsh

Senior Member
A120/208 wye transformer has the phases 120 degrees apart but since you are connecting your load single phase then you would divide by 208. Your "a" answer looks like single phase 120/240. The "b" is correct.
It is interesting to watch the faces of people who "know" about electrical "stuff" when they talk about 480 or 208 volt equipment and you ask them if the load is single or three phase.

kingpb

Senior Member
It's also funny to see the look on peoples face when you tell them that a 120/208wye service is a single phase, 3 wire service and a 208Y/120V is the three phase service. Quite different arn't they!

infinity

Moderator
Staff member
bsh said:
Wouldn't that be single phase 240? Where does the 120 factor in?

rattus

Senior Member
infinity said:
Wouldn't that be single phase 240? Where does the 120 factor in?
Apparently 120/208 means two phases of a wye--a 3-wire service. These two phases are separated by 120 degrees. 2 transformers. Some call this an open wye.

Then 120Y/208 means all 3 phases are provided--a 4-wire service. 3 transformers.

120/240 is a 1-phase services where the voltages on Line 1 and Line 2 are separated by 180 degrees with the neutral being the common reference. This is still a single phase service however. 1 transformer.

Dennis Alwon

Moderator
Staff member
Mike01 said:
Electrical calculations:
For 208v single phase I have herd two different opinions what would be correct:
a. (5000/2) = 2500 (2500/120)=20.8A
b. (5000/208)=24.0A
is a or b correct? For a single phase system?
Since multiplication is communicative "a" would work if you use 104 volts instead of 120. By dividing by 120 and 2 you are assuming a 240 volts sytem not 208. 2500 watts/ 104= 24 amps. This method confuses the issue.

Single phase divide thw wattage by the voltage being supplied to the load.

Smart \$

Esteemed Member
rattus said:
Then 120Y/208 means all 3 phases are provided--a 4-wire service. 3 transformers.
Umm... correctly written (and stated), it is a 208Y/120V 3? 4W (or 4-wire) system or service. The "Y" is optional. The "3? 4W" portion is quite often dropped in trade discussions.

Three phase systems list the higher voltage first. Single phase systems and services list the lower voltage first.

rattus

Senior Member
Smart \$ said:
Umm... correctly written (and stated), it is a 208Y/120V 3? 4W (or 4-wire) system or service. The "Y" is optional. The "3? 4W" portion is quite often dropped in trade discussions.

Three phase systems list the higher voltage first. Single phase systems and services list the lower voltage first.
I should have known that! Now I do.

Mike01

Senior Member
??

??

Question about ?a? would this be the correct way to do it tho because you need to reference the phase to neutral voltages (2-hot legs each leg measures 120v-phase to neutral not 104)? 120v phase to neutral not 208v?

kingpb

Senior Member
Smart \$ said:
Umm... correctly written (and stated), it is a 208Y/120V 3? 4W (or 4-wire) system or service. The "Y" is optional. The "3? 4W" portion is quite often dropped in trade discussions.

Three phase systems list the higher voltage first. Single phase systems and services list the lower voltage first.
Actually, the "Y" is not optional. The 3ph, 4wire is optional, as it is redundant, since 208Y/120V as written means 3ph, 4W.

ddubbs103

Member
Just wondering if any one had any diagrams of services wye and delta etc? thanks!

Mike01

Senior Member
Still Confused??

Still Confused??

I understand that different people indicate the voltage different ways and I believe the IEEE has a way to to this also. But in a 3 phase panel 208Y/120V-3ph, 4w with single phase 208 loads connected, the proper way to calculate the apmacity from the given load would be as follows (correct?):
5000w = 5000w/2 = 2500w/120v = 20.8A per phase.

Correct because the load is calculated as it relates to single phase that is 120v phase to ground. Correct?

Dennis Alwon

Moderator
Staff member
Mike01 said:
I understand that different people indicate the voltage different ways and I believe the IEEE has a way to to this also. But in a 3 phase panel 208Y/120V-3ph, 4w with single phase 208 loads connected, the proper way to calculate the apmacity from the given load would be as follows (correct?):
5000w = 5000w/2 = 2500w/120v = 20.8A per phase.

Correct because the load is calculated as it relates to single phase that is 120v phase to ground. Correct?
Mike read the other post again. Single phase does not mean 120 volts. Single phase can be 120 volts or 208 volts in your example. You stated that the loads are single phase 208 loads thus you would need to divide the 5000 by 208 not 240.

In your example "a" when you divide by 2 and then again by 120 you are essentially dividing by 240 volts. The system you have is 208. So-- just divide the load (5000 watts by 208) . Answer "b" is the correct answer.

rattus

Senior Member
Mike01 said:
I understand that different people indicate the voltage different ways and I believe the IEEE has a way to to this also. But in a 3 phase panel 208Y/120V-3ph, 4w with single phase 208 loads connected, the proper way to calculate the apmacity from the given load would be as follows (correct?):
5000w = 5000w/2 = 2500w/120v = 20.8A per phase.

Correct because the load is calculated as it relates to single phase that is 120v phase to ground. Correct?
Mike,

You are thinking of a different load configuration:

For a single resistive line to line load, one must use the line to line voltage, that is,

I = 5000W/208V

One should note that there is a +/- 30 degree phase angle between the phase voltages and phase currents in this case. The neutral current would be zero. This is the basis for the infamous Oregon Fudge Factor.

For two equal resistive loads on a MWBC, the current would be,

Iphase = 2500W/120V = Ineutral

The phase currents and voltages would be in phase.

Mike01

Senior Member
??

??

Ok but the load is connected to say phase ?A? and phase ?B? the line to neutral voltage of each bus is 120V so would you not divide your load by two to split the load over the two busses and then divide by the line to neutral voltage of the panelboard to see the correct ampacity of the load?

Smart \$

Esteemed Member
ddubbs103 said:
Just wondering if any one had any diagrams of services wye and delta etc? thanks!
What kind of diagrams?

There are some transformer connection diagrams in this pdf.

kingpb

Senior Member
Mike01 said:
Ok but the load is connected to say phase ?A? and phase ?B? the line to neutral voltage of each bus is 120V so would you not divide your load by two to split the load over the two busses and then divide by the line to neutral voltage of the panelboard to see the correct ampacity of the load?
Nope, you should show the total load on one of the phases with the other slot crossed out or referenced as a 2-pole etc. I have seen many ways to accomplish this, but it is not appropriate to divide by 2, and split it between the two lines.

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