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Q: 3 phase star voltage

  • Thread starter Jonathan Barnes
  • Start date
J

Jonathan Barnes

Jan 1, 1970
0
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
there is a fool greater than the proof.

To reply remove AT
 
J

Jonathan Barnes

Jan 1, 1970
0
| 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.

Yes, England. We have just sycronised with the europeans on 400 / 230 V
power from 415 / 240 V
In the USA, we have a split single phase system,

Two phases of 110 to ground at 180 phase angle.
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).
Correct,

what you are describing is just such a situation, possibly intentionally
wired that way.

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
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).

Check , 116 to nutral ( not 100 as in origonal post ) + 230 gives 346 V
One factor that could complicate this is the temperature of the elements.

Complicated enough without worrying about tempreture :)
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.http://ka9wgn.ham.org/ |
 
J

Jonathan Barnes

Jan 1, 1970
0
| 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.

Yes, England. We have just sycronised with the europeans on 400 / 230 V
power from 415 / 240 V
In the USA, we have a split single phase system,

Two phases of 110 to ground at 180 phase angle.
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).
Correct,

what you are describing is just such a situation, possibly intentionally
wired that way.

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
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).

Check , 116 to nutral ( not 100 as in origonal post ) + 230 gives 346 V
One factor that could complicate this is the temperature of the elements.

Complicated enough without worrying about tempreture :)
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.http://ka9wgn.ham.org/ |
 
J

Jonathan Barnes

Jan 1, 1970
0
| 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.

Yes, England. We have just sycronised with the europeans on 400 / 230 V
power from 415 / 240 V
In the USA, we have a split single phase system,

Two phases of 110 to ground at 180 phase angle.
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).
Correct,

what you are describing is just such a situation, possibly intentionally
wired that way.

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
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).

Check , 116 to nutral ( not 100 as in origonal post ) + 230 gives 346 V
One factor that could complicate this is the temperature of the elements.

Complicated enough without worrying about tempreture :)
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.http://ka9wgn.ham.org/ |
 
D

Don Kelly

Jan 1, 1970
0
Jonathan Barnes said:
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
there is a fool greater than the proof.

To reply remove AT
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
[email protected]
remove the urine to answer


..
 
J

Jonathan Barnes

Jan 1, 1970
0
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.

| 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.
http://ka9wgn.ham.org/ |
--------------------------------------------------------------------------
---
 
D

Don Kelly

Jan 1, 1970
0
Jonathan Barnes said:
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.
--------
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
[email protected]
remove the urine to answer
 
D

daestrom

Jan 1, 1970
0
Complicated enough without worrying about tempreture :)


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

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 (R=
10ohms) 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
 
J

Jonathan Barnes

Jan 1, 1970
0
| 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.

If I can return the favour I will. ( post to sci.engr.mech )


--
Jonathan

Barnes's theorem; for every foolproof device
there is a fool greater than the proof.

To reply remove AT
 
D

daestrom

Jan 1, 1970
0
daestrom said:
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 (R= 10ohms) 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.

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
 
D

daestrom

Jan 1, 1970
0
Jonathan Barnes said:
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 :)

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.

daestrom
 
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