Lost neutral connection

[hhhansen]

Hi

Here is a question which is bothering me.

In case one have a normal 3-phase installation with 3 lines and neutral it is well known that the line voltage increases from 230 V to 400 V in case the neutral connection is lost.

But I don´t see how this can be verified matematically. Can anyone assist with a proff of this ?

B.R. HHHHansen.

 

[waross]

You may know that but I don't.
I can't prove your hypothesis but I may be able to disprove it if that will help.
In the special case where one phase is heavily loaded and two phases are very lightly loaded, the voltage on the lightly loaded phases may approach 400 Volts. In the case where the loading on all three phases is well balanced the voltages will stay at or near 230 Volts.
For ease of explanation let's look at a single phase application first:
In our example we will use a 120/240 Volt circuit.
On one 120 Volt circuit we have a 1 Amp load. (120 Ohms).
On the other 120 Volt circuit we have a 100 Amp load. (1.2 Ohms).
Now the neutral is lost. The circuit becomes 121.2 Ohms across 240 Volts.
The current drops to 1.98 Amps. (240 Volts / 121.2 Ohms)
The voltage across the 120 Ohm load increases to 237.6 Volts. (1.98 Amps x 120 Ohms).
Now we have been considering a simple case with fixed resistance.
In the real world things are not that simple.
Many loads exhibit non-linear current/voltage characteristics.
Incandescent lamps:
As the voltage drops the resistance drops dramatically. The hot resistance of an incandescent lamp may be 10 times the cold resistance.
Transformers:
As the voltage increases the transformer may saturate and the current may increase by a factor of 5 or more in the case of full saturation.
Motors:
As the voltage drops the current increases. If the motor stalls the current increases dramatically. As the voltage rises the current may drop slightly and then rise. If the voltage rise is enough to cause saturation the current will rise dramatically.

You may calculate the expected voltages with an open neutral based on the currents of the loads with the proper voltage applied. However when the neutral opens and the voltages shift, the impedances may change dramatically and the actual voltages may not be as calculated.
Now consider the special case where the load on one line is inductive and the load on the other line is capacitive. When the neutral is lost the loads become an inductive, capacitive series circuit and the voltage across one part of the load may exceed the applied voltage.
I think that fairly well covers many of the issues with an open neutral on a single phase circuit.
Suffice to say that the calculations may be a little more difficult for a three phase circuit.

Bill
--------------------
"Why not the best?"
Jimmy Carter