unbalanced 3-phase loads (WYE,3-wire)

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banquo

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Hi: 1st post, so please be kind if it's a stupid one... I did try a simple search first, and did not see this topic answered.

If we have a 3-phase circuit connected in 3-wire WYE (star) with line voltages equal in each phase, but the loads are not balanced in each phase, then clearly the voltages will not be balanced either. From memory, the phase with the highest load resistance will see the highest voltage, and despite having the highest resistance, will also carry the highest power....:-?
So, for example, if we have (for the sake of argument) 100V between each pair of lines, then the phase voltages would be 100/sqrt3 = 57.735V, but only if the resistance in each phase is balanced. Otherwise, the 'star-point' will 'float' and voltages per phase will be dependent on what the resistance in each of the three phases actually is.
Where the resistances are not balanced, what is the easiest way to work out what the voltage in each phase will be....?
I remember doing this on a drawing board (but that was 35 years ago, and I can't remember how to do it :roll: ) and I've seen a calculation that uses imaginary numbers to work out the voltages, currents and powers in each phase.

I would prefer a mathematical solution to this, especially if it could be built into a spreadsheet, and used on a day to day basis....

Tall order?

The knowns are the line to line voltages (same between each phase), and the phase resistances (different in each phase).
The unknowns are the phase voltages, phase currents and phase powers.

Answers on either mathematical or graphical resolution would be appreciated, although like I said, a mathematical resolution would be ideal!
 
banquo said:
If we have a 3-phase circuit connected in 3-wire WYE (star) with line voltages equal in each phase, but the loads are not balanced in each phase, then clearly the voltages will not be balanced either. From memory, the phase with the highest load resistance will see the highest voltage, and despite having the highest resistance, will also carry the highest power....:-?
Click to expand...
I'm not following your question. When you say
From memory, the phase with the highest load resistance will see the highest voltage, and despite having the highest resistance, will also carry the highest power.
Click to expand...
are you referring to the resistance of the conductor?
 
A three phase load will be by definition a balanced load. A three phase 100 amp load will place 100 amps on each phase. The only way to have an unbalanced load is to have different amperage line to neutral loads on each phase. Since they are connected in parallel, not series, each load will see the 100 volts for which it was designed (in your example). I think you are trying to over-complicate the concept.
 
banquo said:
The knowns are the line to line voltages (same between each phase), and the phase resistances (different in each phase). The unknowns are the phase voltages, phase currents and phase powers.
Click to expand...
That cannot happen if the center point of the wye (what you called the ?star point?) is connected to planet Earth, as it almost always is for US installations. Are your systems not grounded (or should I call it ?earthed??) in Scotland? :-?

The mathematics of analyzing an unbalanced system are not simple. You can do it with vectors, but I have long since forgotten how. You can also do it with ?symmetrical components.? I have forgotten how to do that also, but I forgot it more recently. ;)
 
haskindm said:
A three phase load will be by definition a balanced load.
Click to expand...
Not necessarily. You might be looking at the total loads on a panel that has a mix of single-phase and three-phase loads. That panel could easily have an unbalanced total load. But even one three-phase load by itself has the potential to be unbalanced. I could construct a three-phase heater, and use different resistance-heating elements on the three legs. I can?t say why anyone should choose do such a thing, but if they do, the device will still be properly called a three-phase load, despite having unbalanced currents in each leg.
 
It sounds to me like the original poster is describing a 'three terminal' load which internally has a wye connection, but that the 'star point' of the load does not have any connection to the source neutral. Consider, for example, a 'lost neutral' situation from a wye source to a wye load.

The supply is some arbitrary 'perfect' three phase source. The load is resistive. What happens to the voltage of the 'start point' of the load?

Looks like a 'solution of simultaneous equations', except that the variables are all phasors.

-Jon
 
Wouldn't the different sized heaters on different legs be connected in series? This would cause higher voltage and lower voltage across the heaters. If the were connected as L to N the voltage would be stabilized. I am trying to picture why someone would do that. Not arguing, just trying to learn.
 
What I had intended for my example (probably no relation to the original question), was to build a new heater as follows. Take three elements of the same voltage rating by of different resistance values. Wire the device internally such that Phase A hits the top of one element, Phase B hits the top of another, Phase C hits the top of the third, and the neutral wire (from the source) connects to the other side of all three elements. This makes for a single heater that has a three-phase WYE connection, and that will not appear to the power system as a balanced load.

Regardless of whether the power system has a connection from the neutral to planet Earth, the three voltage values from L-N for the three elements will not be the same.

Once again, I cannot think of a reason to build a heater this way. It is just intended as an illustration of my point.
 
Charlie, stick with your example, but rather than an unbalanced heater by design, what about an unbalanced heater by accident.

Consider a 480V heater composed of 9 277V elements, connected as 3 parallel elements per phase, with all phases connected at an internal neutral point.

The internal neutral is not brought out; the heater is simply connected to a 480V supply.

What happens to the neutral voltage if an element burns out?

-Jon
 
haskindm -
I don't know why either.

jon -
I think you got it

Call the voltage at the load Y point = Vn

Resistance in A phase = Ra
Resistance in B phase = Rb
Resistance in C phase = Rc
Current in A, B, C phase = Ia, Ib, Ic
Voltage on A, B, C phase = Va, Vb, Vc

Resitance is known. Va, Vb, Vc is known.

Unknowns are Vn, Ia, Ib, Ic

Equations are:

1) Va - Vn = IaRa
2) Vb - Vn = IbRb
3) Vc - Vn = IcRc
4) Ia + Ib + Ic = 0

Four equations, four unknowns, it's just an algebra problem - only slightly complicated by Va, Vb, Vc, Vn are vectors.

Assuming CW rotation, and A phase is horizontal, pointing to the right, Vphase to neutral = E volts

Va = E + j0
Vb = -.5E + jE(3^.5)/2
Vb = -.5E - jE(3^.5)/2

Solving for:
Vn = (real part) + j(imaginary part)

Vn(real) = (E/Ra - .5E/Rb - .5E/Rc)/(1/Ra + 1/Rb + 1/Rc)

Vn(imaginary) = j(E3^.5/(2Rb) - E3^.5/(2Rc))/(1/Ra + 1/Rb + 1/Rc)

Unless I made an algebra slip somewhere. And I could have cause I didn't spend very long at it and didn't check my work

carl

edited to add voltage definitions
 
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Firstly, thanks a lot for all the responses!
I'm sorry, but I obviously didn't make the scenario as clear as I'd hoped.
The idea was to calculate the 'what if' under a fault condition, and several of you took the time to work that out for yourselves, as I didn't explain it well enough in the first place... (where's the red faced smiley when you're looking for one...)?
Here's the scenario:
Let's say we're in the USA, in an industrial plant with a 480V, 3-wire supply (no neutral). We come across this quite a lot in the US, more so than in other countries. In Europe, we have typical industrial voltage of 400V between lines, and 99% of installations have a neutral which is grounded, and the single phase supplies, both industrial and domestic, are between one line and neutral, which is where you get 230V local domestic supply voltage. We'll design a lot of equipment to operate in 4-wire star (WYE), where each phase is a 230V load, but we build three identical loads and connect to make a balanced 3-phase load. With the centre point connected to neutral/ground, then the voltage in each phase stays the same, whatever the resistance in each phase, so we don't have to worry about a fault condition.
When we supply equipment for the USA, we're often told there's no neutral on site, so we have to work with 480V.
If the appliance in question can't be designed to operate from such a high voltage, one option is to connect in wye, and design the resistors to operate from 277V (480/sqrt 3). The resistors are always single phase items, and, for the sake of argument, let's say each one is designed to operate from 277V, and has a resistance of 10 ohms.
The design power for this resistor will be 277?/10 = 7673W.
Again, for the sake of argument, let's say we have three of these resistors connected in parallel in each phase, with nine in total. The phase resistance is now 3.333 ohms, the power per phase is 23,040W, and we have a balanced 3-phase WYE load of 69.1kW or so.
By summation of vectors, the voltage at the centre of the WYE will be zero, so it acts as a reference point if the loads are balanced.
Now, let's look at the fault condition.
One of the resistors, or the connection to one of the resistors fails in one phase. In the second phase, let's say two resistors fail.
Now we have an unbalanced 3-phase wye, with 10 ohms in one phase, 5 ohms in the second and 3.3333 ohms in the third. The centre of the wye is floating, so the voltage is no longer 277V in any of the phases.
My question was, how to calculate the voltage and current in each phase, given that we know the 480V, we know the resistances are 10, 5 and 3.333 and that's all we know.
The point of doing this is to calculate the worst case scenario, so we can design the resistors, and their cables and connectors, to accept the higher voltage and current (per resistor) that will result from the lack of balance.

Coulter's answer to this question looks familiar, very familiar, and I remember some of the structure from the calculations I've seen in the past.

Algebra never was my strong point.... :wink:
 
Have you considered the use of a 'zig-zag' transformer in your equipment to derive a local neutral? Or a resistive derived neutral for relaying purposes?

This could either be sized to maintain the correct equipment voltages in the event of a device failure, or could be part of a protective relay system to shut down power in the event of a device failure and phase imbalance.

-Jon
 
Yes Winnie, we usually look at these options, and other means of providing a neutral, but the question relates to those cases where (usually for cost reasons) we're stuck with the 3 wires.
 
banquo said:
...Coulter's answer to this question looks familiar, very familiar, and I remember some of the structure from the calculations I've seen in the past. ... Algebra never was my strong point....
Click to expand...

banquo -
I don't recommend using what I wrote with out verifying. I can send the derivation if you want. Get someone to give you a hand. Vector algebra is messy, but not difficult - mostly just an excercise in keeping track of the conjugants and signs.

Just for a check, using your example, for:
Ra = 10
Rb = 5
Rc = 3 1/3 = 10/3
|Va| = E = 277

I get:
Vn = -24.9 - j14.4 = 28.8 angle(210)

V(Ra) = Va-Vb = 277 + j0 - (-24.9 - j14.4) = 302 angle(2.7)

V(Rb) = -138.5 + j239.9 - (-24.9 - j14.4) = -113.6 + j254.3 = 278.5 angle(114.1)

V(Rc) = -138.5 - j239.9 - (-24.9 - j14.4) = -113.6 -j225.5 = 252.5 angle(243.3)

The numbers look about right, Ra has the largest voltage drop, and Rc has the lowest. Phase angles look about right.

I can't stess enough, if you are going to use this, please check the math - I could have holes everywhere.

Good luck.

carl
 
charlie b said:
What I had intended for my example (probably no relation to the original question), was to build a new heater as follows. Take three elements of the same voltage rating by of different resistance values. Wire the device internally such that Phase A hits the top of one element, Phase B hits the top of another, Phase C hits the top of the third, and the neutral wire (from the source) connects to the other side of all three elements. This makes for a single heater that has a three-phase WYE connection, and that will not appear to the power system as a balanced load.

Regardless of whether the power system has a connection from the neutral to planet Earth, the three voltage values from L-N for the three elements will not be the same.

Once again, I cannot think of a reason to build a heater this way. It is just intended as an illustration of my point.
Click to expand...

Reply: NO. I think the voltage between each line and neutral of the supply system will remain the same supply voltage but the current in each element will be different due to its resistance.

GHK
 
Banquo, I would look for a way to connect the load line-to-line, and not in a floating-neutral Y. This is a variation of the shared-neutral MWBC debate; an open neutral replaces imbalance current with imbalance voltage.

It's one thing to design a MWBC that relies on its neutral to maintain voltage balance. It's another thing to intentionally design a circuit where one malfunctioning element can guarantee several more.

As you know, if the heaters receive design voltage, and then you get an open element, the remaining elements of that parallel group will receive over-voltage, which will cascade until the entire group has burned opened.

Then you'll be left with the remaining parallel groups in series across the remaining two phases. I would try to avoid using a floating-neutral Y connection for any loads that can open like the above scenario.

Single-element loads would be entirely different. One opening would leave the other two in series line-to-line as I mentioned above. There would be loss of heating, but no damage beyond the initial burn-out.

If you want the math, look at Carl's post again. I'd rather risk his "holes" than try to come up with the numbers on my own.
 
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Thanks to all for these replies.
This is a great forum! :cool:

We're going to run Carl's calcuklations and see how it pans out. Easy enough to bolt some resistors together to prove we get what we calculate.

I'll post back with what we find.

Larry - be interested in any pointers to the MWBC debate. Also, we always recommend either a delta connection or a 4-wire WYE, and this question relates only to these cases where we have no choice (or the user has already made his own choice, so we end up picking up the pieces).

At the end of this, I want to be able to survey these pieces, and demonstrate mathematically why we don't recommend what they did.
I now feel we're 99% of the way there, and all your contributuions are appreciated.
 
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