Voltage drop on 3? Lighting Circuit

[winnie]

In the case of a perfectly balanced set of 277V loads, there would be no current flow on the neutral, and you would get the correct answer by using the 480V three phase calculation, and then converting back to single phase, as George did in post #5. IMHO the only 'error' in that post was the use of 0.00193 ohms per foot for the wire; IMHO the wire is lightly loaded and thus cool, and thus a lower resistance would apply.

The the case of the _imbalanced_ load, Rattus has the technique right; figure out the current on the neutral, which will be the vector sum of the various phase currents. Apply this current to the resistance of the neutral conductor, which gives a _vector_ voltage for the neutral voltage drop. Depending upon the phase angle of this voltage relative to the phase voltages, this may increase or reduce the voltage that reaches the load.

However it is probably simpler and faster to use the approach of applying the 277V equation to each individual load, assuming no benefit from the shared neutral. This will over-estimate the voltage drop; so if you are good with the simplistic approach, then no need to do a more refined calculation.

Assuming/approximating the following:
0.5 ohm for 300 feet of single conductor
loads carry full load current (as if they were getting full voltage)
loads are all unity power factor
supply is 277V line-neutral.

I get:

276V phase A-neutral at the load
272V phase B-neutral at the load
267V phase C-neutral at the load

For what its worth, I estimate 6A flowing on the neutral, with a phase angle of 210 degrees.

-Jon

P.S. just saw rattus' last post. I used the spreadsheet ATAN2 function, which takes an X and Y input to give an angle, rather than Y/X to give a value. Eliminates the ambiguity of atan. I got -150 degrees, which I reported as 210 degrees. I then did my current calculations graphically with a CAD package. Oh, my estimate of 6A was really 6.25, so we agree there.

 

[hardworkingstiff]

Assuming/approximating the following:
0.5 ohm for 300 feet of single conductor
loads carry full load current (as if they were getting full voltage)
loads are all unity power factor
supply is 277V line-neutral.

I get:

276V phase A-neutral at the load
272V phase B-neutral at the load
267V phase C-neutral at the load

I don't understand these numbers.

Taking just the A-phase load (2000 VA), I get 7.22 amps (2000/277 = 7.22)

Using E=IR, I get 7.22 * .5 = 3.61 volts needed to push the amps through 300' of wire.

277-3.61 = 273.4-volts left at the load, having 3-phase would bring it up to 276-volts?

Edit: based on this post, I retract my earlier post on the voltage at the loads.

 

[George Stolz]

Thanks so much for all the replies.

I'm glad this wasn't easy, because I was having trouble coming to the right answer. (I hadn't come to the right answer, but you know what I mean. )

What I did for a ballpark figure (which is about all I had use for) was making a new sheet on my Vd spreadsheet on my PDA, using wattage instead of current (to speed using the numbers on a typical panel schedule).

When I realized unbalanced loads would behave differently, I simply took a ratio between a phase's contribution to the overall wattage of the circuit (in this case, Phase A is 22%, B is 33%, C is 44% of 9000 overall) and took the overall voltage drop (figuring the circuit as a balanced three-phase circuit) and split it between them proportionately.

This ended up showing A=274.7, B=273.5, C=272.3.

This didn't match Jon's results too well, so I think a little bit better rule of thumb than I've got. (I've read through the posts and realized that my proportion-method is never going to be right on the money, but if it's reasonably close that would suit me. )

However it is probably simpler and faster to use the approach of applying the 277V equation to each individual load, assuming no benefit from the shared neutral. This will over-estimate the voltage drop; so if you are good with the simplistic approach, then no need to do a more refined calculation.

As in, do a single-phase 277V calculation for each?

 

[winnie]

Using E=IR, I get 7.22 * .5 = 3.61 volts needed to push the amps through 300' of wire.

277-3.61 = 273.4-volts left at the load, having 3-phase would bring it up to 276-volts?

Exactly. The numbers that I got suggest that for the A phase there is a 3.6V drop in the 'hot' feeding the load, which is nearly balanced by a 3.1V 'boost' on the neutral side. This is because the larger B and C loads are causing a voltage drop in the neutral, but the phasing of this neutral voltage drop means that it _adds_ to the A phase voltage.

This is much simpler if you consider a single phase center tapped system. Imagine some sort of 277/554V center tapped single phase system, with two 'hot' legs and a neutral. Call the hot legs A and C. Use 300' of 12ga conductor to feed a MWBC with loads of 2000VA and 4000VA respectively on the A and C side (I'm making this up to match the 3phase example).

On the 'A' side hot conductor you will have 3.6V of drop. On the 'C' side hot conductor you will have 7.2V of drop. On the neutral you will have a net drop of 3.6V, but 'pulled' toward the heavier load. This 'drop' on the C side is seen as a 'boost' on the A side.

As in, do a single-phase 277V calculation for each?

Exactly. You know that the result that you get with this calculation will be wrong, and higher than what will happen in reality. But if the voltage drop given by this simplistic approach is 'good enough', then you know that the performance of the real system will be better.

You will not get an exact answer from any of these calculations; the best you can hope for is to put more effort in and get a _better_ approximation. As soon as the approximation is good enough (by whatever metric of 'good enough') then why bother going further.

When I suggest doing a 277V calculation as if the system does not have a shared neutral, I mean: if conductor size is not being set by voltage drop, check the voltage drop using a conservative approximation. If the voltage drop is acceptable, then move on. If the calculated voltage drop is not acceptable, then do a more refined calculation and check again. If the voltage drop is still not acceptable, then adjust the design and calculate again.

-Jon

 

[charlie b]

Charlie - My copy of Stevenson has likely been packed away for thirty + years, but I have a copy of Grainger and Stevenson (1994) here. Tell me what sections you need and I'll scan them and send them to you.

That's OK. I'm at work now, and so is my copy of Stevenson. If I can take some time at lunch, I'll give it a look-through.

For those who don't know that book by the last name of its author, we are talking about "Elements of Power System Analysis," by William D. Stevenson, Jr. My copy is a fourth edition, copyright 1982. It is a common textbook for an introductory course in the analysis of electrical power systems. It's usually a senior or graduate level course.

 

[rattus]

Phasors again:

Phasors again:

I don't understand these numbers.

Taking just the A-phase load (2000 VA), I get 7.22 amps (2000/277 = 7.22)

Using E=IR, I get 7.22 * .5 = 3.61 volts needed to push the amps through 300' of wire.

277-3.61 = 273.4-volts left at the load, having 3-phase would bring it up to 276-volts?

Edit: based on this post, I retract my earlier post on the voltage at the loads.

Stiff, we have a neutral current which is not in phase with Va. In the strange world of phasors, the drop in the neutral reduces the drop on line A, but it may increase the drop on the others.

If the loads were balanced, there would be no neutral current, and we could ignore it.

I would assume that only the 4KW load is on and compute the drop for a single phase circuit (this could happen). That is, include the neutral resistance. Then your total drop is,

Vdrop = 1 Ohm x 14.4 A =14.4 V.

That is 5% worst case

Switching on the other loads will reduce this drop. Believe it or not!

 

[kingpb]

I think rattus has made an excellent point that for sizing the conductor you must assume the two lighter loaded circuits are off (since you cannot guarantee they won't be in normal operation), and use the single circuit model for the calculation. This will guarantee an adequate design and the situation can only improve as one or both of the other circuits is energized.

 

[rattus]

Fun Experiment:

Fun Experiment:

If you have access to a 120V, 3-phase MWBC, do this:
Connect a 100W load, e.g., light bulb to each line. Now measure the neutral current with 1, 2, and 3 loads drawing current. Results won't be perfect unless everything is perfectly matched, but you will get the point.

 

[rattus]

Winnie, thanks for the tip on Atan2(x). I didn't know how to do that with a spreadsheet.

 

[Smart $]

I think rattus has made an excellent point that for sizing the conductor you must assume the two lighter loaded circuits are off (since you cannot guarantee they won't be in normal operation), and use the single circuit model for the calculation. This will guarantee an adequate design and the situation can only improve as one or both of the other circuits is energized.

While I agree one must consider that only one circuit may be conducting, IMO it should not be an absolute rule. If these circuits are, as suggested, for parking-lot lighting, most likely they will all be operating concurrently, with the only exception being a defective load or circuit. Thus the voltage drop will be substantially less than anticipated while in "normal" operation.