| 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).