Neutral current and associated losses

[Larah]

Hello all I have used this forum for some time and have found the advice very useful, however this is my first post on this forum.

My request concerns a 4 wire low voltage network (230v phase to neutral) connected to a delta wye transformer.

I understand that when the load across the 3 phases is not equal current flows in the neutral conductor and the greater the unbalance the greater the current flowing.

What I would like explained is in regard to losses. If the neutral current was say 10amps does this mean the feeder cable is drawing an additional 10amps (measured at the substation)? So the losses over the day would be 10x24 = 240 Ah @ 250V = approx 60kWh.

Or is the losses simply calculated from IxIxR?

If so can someone explain the current flowing in the neutral conductor please.

Thanks in advance





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[GoldDigger]

Hello all I have used this forum for some time and have found the advice very useful, however this is my first post on this forum.

My request concerns a 4 wire low voltage network (230v phase to neutral) connected to a delta wye transformer.

I understand that when the load across the 3 phases is not equal current flows in the neutral conductor and the greater the unbalance the greater the current flowing.

What I would like explained is in regard to losses. If the neutral current was say 10amps does this mean the feeder cable is drawing an additional 10amps (measured at the substation)? So the losses over the day would be 10x24 = 240 Ah @ 250V = approx 60kWh.

Or is the losses simply calculated from IxIxR?

If so can someone explain the current flowing in the neutral conductor please.

Thanks in advance





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OK.

The power lost as resistive heating in the wires will be exactly the sum of the I²R heating, added up over all of the wires in the system.

If you have a three phase wye, you will look at the current in each phase wire and the resistance of that phase wire and find I²R for that wire.
Do the same for all three wires and add up the results.
Now do the same calculation for the neutral, using the unbalance current only, and add it in too.
For a balanced wye you will have no extra power loss in the neutral since there will be no current.

If instead you run three separate single phase circuits (one for each phase) and use separate neutrals for each one, with no interconnection among the neutrals at the load end, then you will get I²R losses in each neutral equal to the losses in the associated phase wire. So double the losses in the case of a balanced system. This is one of the many advantages of a three phase system with a common neutral wire.

 

[kwired]

Hello all I have used this forum for some time and have found the advice very useful, however this is my first post on this forum.

My request concerns a 4 wire low voltage network (230v phase to neutral) connected to a delta wye transformer.

I understand that when the load across the 3 phases is not equal current flows in the neutral conductor and the greater the unbalance the greater the current flowing.

What I would like explained is in regard to losses. If the neutral current was say 10amps does this mean the feeder cable is drawing an additional 10amps (measured at the substation)? So the losses over the day would be 10x24 = 240 Ah @ 250V = approx 60kWh.

Or is the losses simply calculated from IxIxR?

If so can someone explain the current flowing in the neutral conductor please.

Thanks in advance





Sent from my iPhone using Tapatalk

Remember for the neutral to carry current there must be unbalance current among the phases, so loss wise you are lessening the loss on at least one phase and moving it to the neutral conductor(when there is unbalance). There can be some differences in the losses depending on the circumstances, but for the most part that is minor losses, in general you are still just moving conductor losses to a different conductor when you unbalance the system.

 

[Larah]

Thanks for the quick responses guys.

So to calculate the losses I use:

IxI x R x the circuit length

For each phase + neutral

Then x 24 to get the kWhrs per day and hence calculate the cost.


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[Smart $]

Thanks for the quick responses guys.

So to calculate the losses I use:

IxI x R x the circuit length

For each phase + neutral

Then x 24 to get the kWhrs per day and hence calculate the cost.

First, "R" is specified in ohms per distance and its value varies with ambient temperature and current levels (i.e. affects conductor temperature and thus resistance).

Then there's the question of how steady state the current levels will be.

A lot depends on how precise you need your calculation result to be.