Neutral Point of Heater Bank

AJJewsbury:

Assuming the heating elements are resistive loads, there are three basic single-fault failure modes:

May I add a fourth - or a subdivision of (a) if you prefer - where several individual elements are connected in parallel on each phase. (probably not unlikely for a 52kW unit),  If one element fails open circuit the current on that phase drops and the star point moves towards the other two phase (towards, the point where the elements on the other two phases would see 200V, but not quite), The remaining elements on the phase with the faulty element then see an increased voltage and start to overheat (with probably undesirable consequences). At least I think that was Geoff's example.


  - Andy.

 

Say we only have 2 equal elements on each phase, the L-star on the undamaged phases decreases to about 210 V rms from 230 V rms, and the L-star voltage on the phase with damaged element increases by about 50 V.


If there are more parallel strings, and only one element string goes O/C, the overvoltage and undervoltages are less.


If you want a quick way of doing the maths, for resistive loads, the formula in A722.2 could be used to calculate the voltage shift of the star point (increases on the phase with the highest resistance / lowest current), and then you can do a bit of trigonometry (or cheat and use a drawing or maths package) in the vector representation of the voltages:



 

AJJewsbury:

Assuming the heating elements are resistive loads, there are three basic single-fault failure modes:

May I add a fourth - or a subdivision of (a) if you prefer - where several individual elements are connected in parallel on each phase. (probably not unlikely for a 52kW unit),  If one element fails open circuit the current on that phase drops and the star point moves towards the other two phase (towards, the point where the elements on the other two phases would see 200V, but not quite), The remaining elements on the phase with the faulty element then see an increased voltage and start to overheat (with probably undesirable consequences). At least I think that was Geoff's example.


  - Andy.

 

Say we only have 2 equal elements on each phase, the L-star on the undamaged phases decreases to about 210 V rms from 230 V rms, and the L-star voltage on the phase with damaged element increases by about 50 V.


If there are more parallel strings, and only one element string goes O/C, the overvoltage and undervoltages are less.


If you want a quick way of doing the maths, for resistive loads, the formula in A722.2 could be used to calculate the voltage shift of the star point (increases on the phase with the highest resistance / lowest current), and then you can do a bit of trigonometry (or cheat and use a drawing or maths package) in the vector representation of the voltages: