Voltage drop, .... again

[hardworkingstiff]

In order to keep away from a disucssion of the "correct" voltage drop formula, for the sake of this topic, can we just assume that VD=IR?

On a 120-volt circuit, a 50-amp load 100' one way on a #4 wire should be 3.08-volts. ((100*2)*(.308/1000)*50). This is a 2.57% VD.

Make this 2 circuits that each pull 50-amps and wire them to a shared neutral conductor (single-phase 120/240-volt) and the voltage drop will still be 3.08-volts, but since it is shared across the 2 circuits which have an applied voltage of 240-volts, the voltage drop would be 1.28% instead of 2.57%.

Now let's say this same MWBC is connected to a 120/208 3-phase panel. If only one circuit is loaded, then the voltage drop would be the same as the 1st scenario (2.57%). Now, when you load up both circuits, the voltage drop will not be halved (as a %) since the neutral is carrying the same current as the other two ungrounded conductors.

I know I asked this before, but I'm not convinced the answer was ever really given. What is the VD on the second scenario when each (2) circuit is pulling 50-amps? How did you calculate that?

Thanks,
Lou

 

[winnie]

http://www.mikeholt.com/code_forum/showthread.php?t=81792
Lou,

Your understanding of the situation is correct. In the 120/240V single phase situation, there is no current on the neutral conductor and thus no voltage drop in the neutral conductor. In the 120/208V three phase wye situation, you have full current flow on the neutral conductor, and thus see the full 1.54V dropped on that conductor.

However to figure the voltage drop seen by the load, you need to include the phase angle of the drop on the neutral conductor. You need to figure the phase angle of the current flow in the neutral, which will give the phase angle of the voltage dropped in the conductor, and then do vector subtraction to figure the voltage left for the load.

I 'cheated' and did this graphically with a CAD package, and get a 1.91% drop in the voltage seen by the 120V loads, consisting of a 1.28% drop in phase with the load, and a 1.28% drop that is 60 degrees out of phase with the load.

-Jon

 

[hardworkingstiff]

I guess my problem is I really don't understand phase angles and how they solve this problem.

:-(

 

[winnie]

When I try to explain this in detail, I get myself into trouble because I get stuck trying to describe all of the end cases. Feel free to keep asking questions.

As you know, AC power is characterized by continuously changing values of voltage and current. What we call '120V' or '208V' is really constantly changing, from positive to negative and from 0 to a peak greater than the single number that we use to represent it.

The only 'true and correct' way to add or subtract AC voltages or currents is to look at the ever changing _instantaneous_ value, and to do the addition or subtraction on this continuous time series, to get a resultant time series. For example, if you had two separate circuits sharing a single conductor, the only 'true and correct' way to calculate the current in that shared conductor is to look at the instantaneous current in the separate circuits, and then to add those time series together to get the resultant current time series.

Doing this would be impossible (though it can be approximated), and the result generally means that you are stuck not seeing the forest for the trees. It is far better to say that a circuit is carrying 50A RMS then to try to describe the constantly changing value. Everything having to do with RMS voltage, power factor, phase angle, harmonic content, etc. is a technique to represent the continuously changing value as a small set of numbers that tells you the critical information.

If you assume that everything is sinusoidal, and that you system is operating at a single frequency, then there happens to be a particularly useful simplification that allows you to do the math to figure the voltages and currents in a system. In these conditions you can _represent_ a constantly changing AC voltage or current with only _two_ numbers, the _magnitude_ of that voltage or current, and the _phase_ of that voltage or current. If you imagine a graph of the voltage (or current) versus _time_, then the _magnitude_ tells you how tall the graph is, and the _phase_ tells you the relative offset of the peaks and troughs of the graph.

That's it. The two values, magnitude and phase, tell you all that you need to know about the sine wave (that is, if you make the assumption of constant frequency and pure sine waves).

Here is where the mathematical utility comes it. It turns out that you can treat magnitude and phase as the _length_ and _direction_ of a vector on a plane. This does not mean that the voltage is the length of a line, nor does it mean that the phase angle is the direction of that line. It simply means that you can _represent_ the magnitude and phase of a sinusoidal voltage by using a line of given length and direction.

Once you have this interpretation of the voltages and currents that you are dealing with, it turns out that the rules for the geometry of these lines can be used to represent how voltages or currents will add up. In other words, if 'vector A' represents the current in one half of a MWBC, and 'vector B' represents the current in the other half, then I can simply use the rules of geometry to 'add' these two vectors together, and determine the current in the neutral of the MWBC.

A line of length 50, direction 0, added to a line of length 50, direction 180, will give a net of 0.

A line of length 50, direction 0 degrees, added to a line of length 50, direction 120 degrees, will give a net of length 50, direction 60 degrees.

-Jon

 

[bcorbin]

Winnie...on the off-chance that you fumbled the calc...are you sure you got 1.91%? I get 2.22% when I add 1.28 @ 0 and 1.28 @ 60.

 

[hardworkingstiff]

A line of length 50, direction 0 degrees, added to a line of length 50, direction 120 degrees, will give a net of length 50, direction 60 degrees.

-Jon

Ok, I cranked up my CAD program and drew a line length 50 by 0 deg. and another one 50 by 120 deg. If I drew two circles (50 length radius) from the end of the the lines, they intersect at the start of both lines and at 50 length by 60 deg. Is this the methodolgy that is to be used?

Edit: I just discovered this is wrong.

 

[tallgirl]

Winnie...on the off-chance that you fumbled the calc...are you sure you got 1.91%? I get 2.22% when I add 1.28 @ 0 and 1.28 @ 60.

You have to use 120 degrees -- 3 x 120 = 360.

So, it's 1.28v @ 0* and 1.28v @ 120*. The answer should be 1.28v @ 60* (or actually at 240*, I think) and look like an equilateral triangle if drawn as vectors.

 

[bcorbin]

You're misunderstanding the problem, Tallgirl. The phases are indeed 120 degrees apart. However, we were calculating the voltage drop of single phase loads. The voltage drop in the neutral is only 60 degrees out of phase with the voltage drop in the phase itself.

 

[hardworkingstiff]

I 'cheated' and did this graphically with a CAD package, and get a 1.91% drop in the voltage seen by the 120V loads, consisting of a 1.28% drop in phase with the load, and a 1.28% drop that is 60 degrees out of phase with the load.

-Jon

I think I understand this. My thought process though has some more confusion.

1st, back to the single-phase load. Let's say we have (same parameters) A-phase loaded to 50-amps and B-phase loaded to 25-amps. Neutral current would be 25-amps. My thinking is that B-phase would see a voltage drop of .77-volts (since all of it's load is shared through A-phase and no neutral current (hence additional voltage drop) would be applied to B-phase. A-phase would see the neutral voltage drop (.77-volts) plus the A-phase voltage drop (1.54-volts) for a total VD of 2.31-volts.

Now the 3-phase. What would the voltage drops be on A-phase (50-amp load) and B-phase (25-amp load) as stated in the above paragraph?



Thanks,
Lou

 

[tallgirl]

You're misunderstanding the problem, Tallgirl. The phases are indeed 120 degrees apart. However, we were calculating the voltage drop of single phase loads. The voltage drop in the neutral is only 60 degrees out of phase with the voltage drop in the phase itself.

But if it's added as vectors, don't you still use 0* for the first phase angle (just because ...) and 120* for the second (because that's the relationship between the two), then the result is (coincidentally) 240* because the vector "points" back towards the origin, even though it's drawn @ 60*?

Or is the vector from the origin to the end point? In which case, yeah, 1.28v @ 60*.

 

[hardworkingstiff]

Check me please

Check me please

Now the 3-phase. What would the voltage drops be on A-phase (50-amp load) and B-phase (25-amp load) as stated in the above paragraph?

Based on what Jon has posted, my answer to this question would be:

A-phase load would have a 2.75-volt drop (2.29%)

B-phase load would have a 1.48-volt drop (1.23%)

Right/Wrong?

 

[bcorbin]

I got:

2.78V (2.31%)

1.54V (1.28%)

Close enough to be rounding, but far enough apart to indicate one of us did it wrong.

 

[hardworkingstiff]

I got:

2.78V (2.31%)

1.54V (1.28%)

Close enough to be rounding, but far enough apart to indicate one of us did it wrong.

I'm pretty sure I'm wrong.

 

[hardworkingstiff]

Ok, I'm making progress. I understand what I've been doing wrong with the vectors.

On the 120/208 service,

A-phase load is 50-amps
B-phase load is 25-amps
Neutral is 43.3-amps

How does the voltage drop in the neutral get shared between the loads (how much for A-phase load and how much for B-phase load)?

 

[Smart $]

...
How does the voltage drop in the neutral get shared between the loads (how much for A-phase load and how much for B-phase load)?

Both loads get the full amount of the drop, but it is vectorial...

View attachment 257

50.0 A * 100 ft * 0.308 Ω/1000 ft = 1.54 V @ 0?
25.0 A * 100 ft * 0.308 Ω/1000 ft = 0.77 V @ -120?
43.3 A * 100 ft * 0.308 Ω/1000 ft = 1.33 V @ -210?

View attachment 258

Note the total VD for the load on B-phase is 2x the phase conductor VD, the same as if it had a dedicated "neutral". If A-phase load were to be more than 50A, the B-phase load total VD would actually be greater than 2x the phase conductor VD.

 

[tallgirl]

Both loads get the full amount of the drop, but it is vectorial...

(Snip ...)

Did you draw that by hand or do you have a program that does that? I've been thinking about getting Eclipse out and writing some Java programs I can put on my PDA and play with.

 

[hardworkingstiff]

How is the phase angle (-210deg) of the neutral determined?

 

[hardworkingstiff]

Forgive me for being so dense. I have a hard time understanding why the VD on the B-phase load (25-amps) would be the same if it was the only load on the neutral as when it shares the neutral with an A-phase load. In my mind it seems that there should be some reduction in VD since some of the power for the B-phase load would be coming through the A-phase load. I know this is different than the single phase scenario, but it just seems to make sense that there would be some VD reduction by sharing a neutral even if it's only on 2-phases of a 3-phase system.

Thanks everyone for your patience and attempts to educate me.
Lou

 

[Smart $]

Did you draw that by hand or do you have a program that does that?

I draw my diagrams manually CAD-wise.

 

[Smart $]

How is the phase angle (-210deg) of the neutral determined?

...By the direction of the parallelogram's diagonal and relative to the "start" or "zero" end of the vector.

If you happen to draw the vectors end-to-end, there'd be no parallelogram or diagonal. It would look like Julie mentioned earlier...

View attachment 260

Note also if you decide to use positive angle orientation of phasing, the diagrams would appear to be flipped vertically compared to mine (and missing the "?" symbol).