Voltage drop on 3? Lighting Circuit

[George Stolz]

Suppose I had a 480Y/277v MWBC with 3000W on each leg, feeding L-N lighting on each leg.

Would I use the 3? voltage drop calculation with 277v or 480v?

(Might seem like a simple question, but I'd like to be sure.)

Thanks,

 

[Dennis Alwon]

I would use 277 volts-- the line to neutral load.

One reason I say this is because the chances are the distance will be different for each MWBC-- so what distance would you use. I think you would have to find the VD for each circuit and then I wuld think you would use single phase cal.--- but what do I know

 

[charlie b]

Sorry, Dennis, but I must beg to disagree. You have a total of 9,000 watts on a 480 volt circuit. That gives you a current of 10.8 amps. That is the current you use to calculate the voltage drop. You then use 480 volts as the basis of calculating the percent voltage drop.

 

[Dennis Alwon]

Sorry, Dennis, but I must beg to disagree. You have a total of 9,000 watts on a 480 volt circuit. That gives you a current of 10.8 amps. That is the current you use to calculate the voltage drop. You then use 480 volts as the basis of calculating the percent voltage drop.

If I ran a single circuit-- one neutral and one hot on the same system what would I use-- 3phase or single phase caluclation and 277 or 480 volts???

 

[George Stolz]

Thanks for the replies.

So, using the formula I use, for #12 at 300 feet, this would be:

Vd = 1.732 x .00193 x 10.8 x 300
Vd = 10.83v

480 - 10.83 = 469.2v
469.2 / 1.73 = 271.2v

So, at a light fixture 300 feet away a voltage of 271.2 (a 2% drop) would be measured, right?

 

[mdshunk]

Lighting, particularly parking lot lighting, is weird. You've got a load at various points along the way to the furtherest load. You sorta need to calculate the Vd at each stop to get the final realized voltage at the last stop.

 

[coulter]

george -

I'm assuming you are looking for the voltage drop at the light fixture. If so, I'm not sure what charlie is calculating. Dennis has it right.

It is a single phase, two wire circuit, fed from 277V. There is voltage drop on the conductor going out to the lamp and on the neutral conductor coming back. You have the round trip distance, wire size, and the current. Nothing three phase about it.

carl

 

[charlie b]

First of all, ?voltage? does not enter into the equation for ?voltage drop.? The value of voltage only comes in when you want to determine the percent of voltage that was dropped. But you might need to know the voltage, in order to convert the given value of Power (in watts or in VA) to the corresponding value of current.

If you have a single circuit with a single 277 volt load, then you use 277 volts to calculate current. If (as in this case) you have a multi-wire branch circuit, with 277 volt single phase loads balanced among the three phases of a 480/277 volt power source, then you must use 480 volts, and the 1.732 factor, to calculate current from power.

The difference is that the current flowing out of one phase will return to the source on both the other two phases. This is a three phase system, supplying a group of single phase loads. You need to treat it as a three phase calculation.

 

[infinity]

If your three equal loads were at the end of the MWBC you would in effect end up with a 3 phase load. The neutral current would end up being zero.

 

[George Stolz]

Okay, so now let's say I have:

2000W on Phase A.
3000W on Phase B.
4000W on Phase C.

All three phases travel the same distance, and are all 12 gauge conductors.

What is the voltage at A, B, and C loads?

 

[mdshunk]

You have a slight contraction in terms when you use the term "phase", but I assume you're still talking about a 277V load?

That's a single phase load. No matter that it's supplied by a MWBC. Single phase calc for each load if you're talking about 277.

 

[rattus]

You have a slight contraction in terms when you use the term "phase", but I assume you're still talking about a 277V load?

That's a single phase load. No matter that it's supplied by a MWBC. Single phase calc for each load if you're talking about 277.

For the unbalanced case that George presents, there is neutral current in the MWBC which will be less than the phase currents but should still be considered.

 

[charlie b]

That's a single phase load. No matter that it's supplied by a MWBC. Single phase calc for each load if you're talking about 277.

No, Marc, sorry, but it’s not. For the “single phase calculation” to be applicable, the conductor going from the source to the load, the load itself, and the conductor from the load back to the source, must all three be in series. As current goes from the source to the load on one conductor, some voltage is dropped. More voltage (in fact most of it) will be dropped through the load itself. But how can you figure out how much voltage will be dropped on the return trip, if you do not know (1) What conductor that current will be traveling through, and (2) How much current will be flowing in that conductor?

In a MWBC, the phase conductors, the neutral, and the load(s) are not in series. Current flowing from the source to the load has more than one path available for the return trip. In George’s imbalanced loading condition, current flowing towards the Phase A load will have three possible return paths, and it is going to take all three. That is why you are into a three-phase calculation.

As to your new question, George, it is no easy task. It may seem easy, and it is certainly easy to describe the situation, but it is not easy to solve. At my office, I have access to a computer program that does this sort of stuff for far more complicated problems. The manual way to solve it starts with a set of three equations that relate voltages to currents and to known values of impedance. The equations are expressed in matrix format. Then you perform a transformation that converts the three imbalanced equations into nine new equations (three sets of three, with each set of three being balanced). They are called the “positive sequence,” the “negative sequence,” and the “zero sequence.” The new balanced equations are relatively easy to work with. Once they are solved, you do the inverse transformation to get back to the original terms of voltages and currents.

This process is called "symmetrical components." It has been many years since I did such matrix manipulations. In fact, I don’t think I have ever used them in a real world problem, since I learned how in grad school. So I might not have the process described correctly. But I hope to at least give you an idea of why we always take the easy road, and assume that the loads are balanced, when performing voltage drop calculations.

 

[mdshunk]

Symmetrical components only apply when you're talking about unbalanced L-L loads with a 3-phase supply, and not L-N. No?

 

[charlie b]

I don't think so. But my textbooks are at my office, and I can't check it out just now (some lame excuse about having to help with Easter dinner).

Anyway, I hope everyone had a lovely day.

 

[coulter]

Okay, so now let's say I have:

2000W on Phase A.
3000W on Phase B.
4000W on Phase C.

All three phases travel the same distance, and are all 12 gauge conductors.

What is the voltage at A, B, and C loads?

George-

I'm thinking I was not clear on the circuit(s).

I originally assumed you had three 2W, 277V circuits. In retrospect, that was unfounded, I know you would know the answer to that. Don't know what I was thinking

So, just to clarify, you are dealing with a 4W, 480V circuit, three hots, one neutral?

carl

 

[coulter]

First of all, ?voltage? does not enter into the equation for ?voltage drop.? ....

Charlie -

Excellent catch. I completely missed that.

carl

 

[coulter]

...This process is called "symmetrical components." ...

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.

carl

 

[hardworkingstiff]

Okay, so now let's say I have:

2000W on Phase A.
3000W on Phase B.
4000W on Phase C.

All three phases travel the same distance, and are all 12 gauge conductors.

What is the voltage at A, B, and C loads?

No one else has answered, and I really don't know how to correctly calculate this, but using 300' as the distance, my guess would be:

A = 270.8-volts
B = 267.7-volts
C = 264.1-volts

 

[rattus]

A good approximation:

A good approximation:

An approximate solution can be obtained with a calculator or even a slide rule for the technology challenged, but these methods are tedious.

It is easier to use a spreadsheet to compute the neutral current and its phase angle. But, this is iffy due to the ambiguity of the tan^-1 function.

I get,

In = 6.2A @ 150

Then you compute the drops in all conductors and then compute the effect on each phase voltage as a percentage. I haven't done this yet. Kinda hurts my head.

Easier still is the use of a circuit simulator such as PSPICE, but I only have the student version.

Of course, there must be stock formulas which approximate this drop without the use of fancy techniques.