Q: 3 phase star voltage

Please help a mechanical engineer with an electrical problem.
If 3 equal heating elements are connected in star to a 400 V 3 phase supply
then each element has 230 V across it. In this condition the star point will be at the neutral voltage,
If one element blows each remaining element will have 200 V across it , and the star point will have a 100 V difference from neutral.
I have a system where each leg consists of two heating elements in parallel.
If a single element blows the remaining element on that leg will be subject to a voltage which must be less than 330 V, but more than 230 V. The other two legs will be between 200 V and 230 V each. The star point must have a voltage to neutral of less than 100 V.
Can anyone tell me what the actual voltages will be, and how they are calculated... or a good reference.
--
Jonathan

Barnes's theorem; for every foolproof device
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A phase'-worth (pair) of open elements will "neutral shift" from the normal six-element wye case to four series-parallel elements. The shift is sqrt(3)/6, or 0.288 per-unit - voltage, or 115.5V.
One element open in six would be about half that shift, placing about 115.5/2+230 or about 287.8V on a single 230V element [five of six operating.]
s falke
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wrote:
| Please help a mechanical engineer with an electrical problem. | | If 3 equal heating elements are connected in star to a 400 V 3 phase supply | then each element has 230 V across it. | In this condition the star point will be at the neutral voltage, | | If one element blows each remaining element will have 200 V across it , and | the star point will have a 100 V difference from neutral. | | I have a system where each leg consists of two heating elements in parallel. | | If a single element blows the remaining element on that leg will be subject | to a voltage which must be less than 330 V, but more than 230 V. | The other two legs will be between 200 V and 230 V each. | The star point must have a voltage to neutral of less than 100 V. | | Can anyone tell me what the actual voltages will be, and how they are | calculated... or a good reference.
Based on your voltages and email address, you are obviously in Europe.
In the USA, we have a split single phase system, and as a result, we have extensive experience with a concept called an "open neutral". This is when the neutral wire of the power being provided is disconnected or broken. The results are that imbalance in the phases (2 of them at 180 degrees) causes the common point to shift in the "voltage gradient" between the phases. The two 120 volt circuits then get a different split of the 240 volts, resulting in possibly much higher voltage on half of it.
Europe's single phase system is not like that, and thus the only time you have an "open neutral" is with three phase (400/230). It sounds like what you are describing is just such a situation, possibly intentionally wired that way. If your common midpoint is not connected to the neutral wire, this is why you have this situation. Without the neutral connection, then your power is effectively a "delta", even if the secondary of the transformer is a "star". If you had wired this up with 400 volt elements connected in a delta arrangement, you would not have this problem.
My whole point is that is it wrong to leave out the neutral when the circuit has a common connection.
When you have just 3 elements in a star and one goes out, it's just like one phase goes out. The remaining 2 elements are in series across 400 volts. But with each leg being 2 elements in parallel, your problem is when just one goes out. If the circuit were completely open, the voltage between the common point and the unconnected phase would now be 346.41 volts (based on 400.00 volts phase to phase, and 230.94 volts phase to ground). Your figure of 330 appears to be based on the system being 381.05 volts phase to phase and 220.00 volts phase to ground (which much of Europe still has).
One factor that could complicate this is the temperature of the elements. If the paralleled elements are in close proximity, then the opening of one element can reduce the temperature of another and effect its resistance, and the whole balance.
I have never done this before, but I suspect one possible way to calculate this is to work out the voltage drops across each element once for each phase being disconnected. That is, figure it out with A being disconnected and 400 volts from B to C. Then again with B being disconnected and 400 volts from A to C. Then finally with C being disconnected. Include the phase angles of these voltages. Then add up the 2 vectors for each element (each element will in one case have zero current and thus zero volts when it's phase leg is the disconnected one).
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wrote:

supply
and
parallel.
subject
Yes, England. We have just sycronised with the europeans on 400 / 230 V power from 415 / 240 V

Two phases of 110 to ground at 180 phase angle.

Correct,
Just so, the heating elements are all rated at 230 V, and when all is working as it should the voltage between the star point and nutral is negligable. When a leg fails the voltage to nutral is 116V, which lights a warning lamp conected between the star point and nutral, what I need to know is what the voltage is when only 1 element fails and I have " half a leg " in the curcuit, ( will it be enough to make the lamp glow ? )
If your common midpoint is not connected to the neutral

connection,
Check , 116 to nutral ( not 100 as in origonal post ) + 230 gives 346 V

Complicated enough without worrying about tempreture :-)

disconnected
element
The problem is that I don't know the phase angle .... it's dependant on the star point.

--
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wrote:

supply
and
parallel.
subject
Yes, England. We have just sycronised with the europeans on 400 / 230 V power from 415 / 240 V

Two phases of 110 to ground at 180 phase angle.

Correct,
Just so, the heating elements are all rated at 230 V, and when all is working as it should the voltage between the star point and nutral is negligable. When a leg fails the voltage to nutral is 116V, which lights a warning lamp conected between the star point and nutral, what I need to know is what the voltage is when only 1 element fails and I have " half a leg " in the curcuit, ( will it be enough to make the lamp glow ? )
If your common midpoint is not connected to the neutral

connection,
Check , 116 to nutral ( not 100 as in origonal post ) + 230 gives 346 V

Complicated enough without worrying about tempreture :-)

disconnected
element
The problem is that I don't know the phase angle .... it's dependant on the star point.

--
> | Phil Howard KA9WGN | http://linuxhomepage.com /
http://ham.org/ |
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wrote:

supply
and
parallel.
subject
Yes, England. We have just sycronised with the europeans on 400 / 230 V power from 415 / 240 V

Two phases of 110 to ground at 180 phase angle.

Correct,
Just so, the heating elements are all rated at 230 V, and when all is working as it should the voltage between the star point and nutral is negligable. When a leg fails the voltage to nutral is 116V, which lights a warning lamp conected between the star point and nutral, what I need to know is what the voltage is when only 1 element fails and I have " half a leg " in the curcuit, ( will it be enough to make the lamp glow ? )
If your common midpoint is not connected to the neutral

connection,
Check , 116 to nutral ( not 100 as in origonal post ) + 230 gives 346 V

Complicated enough without worrying about tempreture :-)

disconnected
element
The problem is that I don't know the phase angle .... it's dependant on the star point.

--
> | Phil Howard KA9WGN | http://linuxhomepage.com /
http://ham.org/ |
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wrote:
|> I have never done this before, but I suspect one possible way to calculate |> this is to work out the voltage drops across each element once for each |> phase being disconnected. That is, figure it out with A being | disconnected |> and 400 volts from B to C. Then again with B being disconnected and 400 |> volts from A to C. Then finally with C being disconnected. Include the |> phase angles of these voltages. Then add up the 2 vectors for each | element |> (each element will in one case have zero current and thus zero volts when |> it's phase leg is the disconnected one). | | The problem is that I don't know the phase angle .... it's dependant on the | star point.
I don't know if it's the right way, but what I was suggesting was to figure it as three separate partial networks based on each of the three voltages phases. What you will get for each calculation is a current that would flow when only that one voltage (400V between 2 of the phases) is present. That voltage is then a vector at the appropriate phase angle. Do this for all 3 phase angles, which will give you 2 non-zero vectors for each element. Add those vectors separately for each. It should also show you what your star point voltage relative to neutral would be by taking that voltage and its phase angle in each of the 3 separate scenarios and adding their vectors.
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wrote:
| Just so, the heating elements are all rated at 230 V, and when all is | working as it should the voltage between the star point and nutral is | negligable. | When a leg fails the voltage to nutral is 116V, which lights a warning lamp | conected between the star point and nutral, what I need to know is what the | voltage is when only 1 element fails and I have " half a leg " in the | curcuit, ( will it be enough to make the lamp glow ? )
Can you not connect the neutral to the star point? If you can connect a warning lamp, you do have at least something. If you can connect it with the full current capacity you get with the worst imbalance (which is 2 phases die), you would at least get a consistent voltage until you can shut it down for maintenance. These being heating elements, there should be no harmonic issues that would force a larger neutral.
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Yes I could connect the star point to neutral.... but I would then not get a warning when an element blows....
I know I could install current detection on the neutral line, but using the star imbalance is cheap.
I just rewired the machine, and had to spend 143 on a reel of pyro cable, I had to re use the old elements, and I've no idea how long they will last.
I can't help it if the factory manager does not want to spend a penny more than he absolutely has to.
wrote:

lamp
the
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a
the
cable,
last.
-------- I gave you a way to calculate the voltages - however I didn't discuss the problem with using a lamp to indicate trouble. One point that you were concerned about- too low a voltage to give a bright enough lamp. 46V on a 115V lamp will be dull and a coloured lens might obscure it (is it on or not?). Another is that people tend to ignore lamps if they do not come on when they are not actually looking at it unless quite bright and obvious. A third is that when one element blows- the remaining element on that leg is stressed at higher than normal voltage (276V vs 230) so heating will be about 44% higher than normal in this element- accelerating failure. It could be that some voltage sensing relay (rather than current sensing) driving a lamp and light would actually save money by saving the second element. Such a device could operate if the voltage exceeds a set limit. If your objective is to shut the heaters down in case of a failure- then such a relay would be easier and cheaper than hoping an operator will notice. Just an opinion. "Penny wise, pound foolish". Work on the plant managers.
--
Don Kelly
snipped-for-privacy@peeshaw.ca
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wrote:
| Yes I could connect the star point to neutral.... but I would then not get a | warning when an element blows.... | | I know I could install current detection on the neutral line, but using the | star imbalance is cheap.
So you'd rather risk blowing the other element just to get a warning?
How much of a warning do you need? Will a small light do?
You could wire up a resistive shunt that would give you maybe a 3 volt drop and put a small light across it. Since the neutral current would be zero under normal conditions, the shunt should stay cool and not affect things. Then when an element does blow, you'll get current across the neutral and shunt, and voltage across the light.
A current transformer could also be used, but if it were me, I'd configure it so the current ratio produces no more than 4 milliamps, better if even less. A 1 watt light at 230v uses 4.348 milliamps, and could be wired right into that. It would be dimmer at lesser currents. A 5000:1 current transformer would give you 4 milliamps at 20 amps current. Just be sure to handle these things safely since the voltage will rise to get that 4 milliamps through.
| I just rewired the machine, and had to spend ? 143 on a reel of pyro cable, | I had to re use the old elements, and I've no idea how long they will last. | | I can't help it if the factory manager does not want to spend a penny more | than he absolutely has to.
What is the cost of replacing a blown element? Include the labor cost.
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<snip>
It seems to me (although I haven't sat down and done this calc), that you could convert the unbalanced wye resistances into an unbalanced delta configuration using the well known wye-delta conversions. Then you could figure the phase current in each delta leg (one would be significantly different than the others because of the unbalanced wye). Sum the phase currents into line currents (the line directly opposite from the 'odd' delta leg would be of interest). (each phase current is in-phase with its line-line applied voltage)
With line currents known, apply them to the original wye configuration and figure the IR drop in each leg of the wye. From this, you can readily find the relation between lines and neutral (i.e. where the neutral has shifted).
Oh, WTH. Here's a go at it. Suppose each resistor is 10 ohms and you have six of them arranged in a two-parallel, wye configuration. So the resistance in each leg is 5 ohms to start with. Then one resistor opens so that two legs are still 5 ohms, but the third leg is now 10 ohms, and line-line voltage is 100V. Using the wye-delta transformations (say the 'C' phase has the faulty resistor)....
Zab = (Zoa*Zob + Zog*Zoc + Zoc*Zoa) / Zoc Zab = (5*5 + 5*10 + 10*5)/10 = 12.5 ohm Zbc = (5*5 + 5*10 + 10*5)/5 = 25 ohm Zca = (5*5 + 5*10 + 10*5)/5 = 25 ohm
So the current through Zab = 100/12.5 = 8 amps, through Zbc = 4 amps and through Zca = 4 amps. Now, 'by inspection' since the currents through Zbc and Zca are equal and inphase with their respective line-line voltages, they are the easiest to sum. But the others can be properly summed as well, just be sure to consider the angles involved. I'll leave the vector addition as 'an excercise', but I believe you will find each of the two delta legs attached to the C line (Zbc and Zca) will contribute cos(30)*4 amps to the line current so the total line current is cos(30)*8 = ~6.93 amps. (for those interested, sin(30)*4 = 2 amps of the current in Zbc will 'cancel' sin(30)*4 = 2 amps of the current in Zca. And notice that sqrt(3)*4 ~6.93)
Going back to the wye configuration, and knowing that Cline current is ~6.93 amps, the voltage drop across the one remaining resistor in the C leg (R10ohms) is ~69.3 volts. Before the fault, when it's balanced, the voltage drop was ~57.7, so the neutral has shifted 11.6 volts (~20%), directly toward the midpoint between the A and B lines.
daestrom
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Mind you, the 11.6 volts is for a the assumed 100V line-line voltage. So if you had 400 V line-line, you would have a corresponding shift of 46.4 volts. (other voltages are ratio-able from this example) The same ~20% increase. And since the remaining heater in that leg would have 120% of normal voltage applied to it, it would disipate (120%)^2 power or 1.44 times its normal power.
daestrom
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My thanks to the group.
I have received answers to my problem of ; 57.8 V, 46 V, and 11.6 V.
I think my favourite theory is the one giving 46 V :-)
I am about to log off this group ( hit and run ) but when an element blows I will take a measurement and report back.
If I can return the favour I will. ( post to sci.engr.mech )
--
Jonathan

Barnes's theorem; for every foolproof device
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wrote:

My 11.6 volt result is if you have 100V line to line in my example. If you have 400V line-line, then 4X11.6 to get 46.4V. Same answer as Don.
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wrote: |> On Sun, 5 Sep 2004 12:17:02 +0100 Jonathan Barnes
|> |> | Please help a mechanical engineer with an electrical problem. |> | | My thanks to the group. | | I have received answers to my problem of ; 57.8 V, 46 V, and 11.6 V. | | I think my favourite theory is the one giving 46 V :-) | | I am about to log off this group ( hit and run ) but when an element blows I | will take a measurement and report back.
You can avoid blowing additional elements by just connecting the neutral, and using a small voltage drop across a shunt in that neutral to see if the first element blew.
Hey, if conditions (either manufacturing or operation) can let one blow, surely the 2nd one will blow if it has 46.4 more volts now. Not good unless you have an instantaneous opening of the power in this condition.
OK, here's yet another idea:
Instead of paralleling two elements on each phase, how about paralleling two WYEs. That is, wire up three elements to one star point, and wire up three more elements to the 2nd star point kept isolated from the first.
I'm just trying to save you from blowing TWO elements together.
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supply
and
parallel.
subject
Here's an approach: Assume that the resistor connected between a and the star point (s) is Ra, ditto for b and c so we have a star Ra, Rb, Rc a-----Ra-------| b-----Rb-------|s c-----Rc-------|
Taking a as a reference point and converting to current sources at a and b leads to Vba/Rb+Vca/Rc =Vsa(1/Ra +1/Rb +1/Rc) and the voltage at s with respect to the system neutral is Vsa -Vna Vca=-345+j200 Vba=-345-j200 Vna= -230
If Ra=Rb=Rc then -690 =3Vsc so Vsa = -230V (rounded off to 3 figures which is all that one needs or should use). Vna =-230V so the voltage to neutral is 0.
If Ra is infinite (open) then -690 =2Vsa and Vsa=-345V so Vsn = - 115V
If Ra=2Rc=2Rb then -690 =2.5Vsa and Vsa = - 276V and Vsn = -46V
In the above cases, the star point floats along a line through a and n bisecting the bc line o the voltage triangle
This can be done for any combination of Ra,Rb, Rc but the star point will shift to anywhere within the line voltage triangle.
--
Don Kelly
snipped-for-privacy@peeshaw.ca
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