Three phase power meaurement questions

[TheRedDevil18]

I have a few questions regarding three phase power measurement

Three Watt-meter method:

Balanced load:
What would happen if the neutral where removed ?, my guess is that it wouldn't make a difference because the neutral doesn't carry any current

Unbalanced load:
Same question as above

Two Watt-meter method:

What would happen if the neutral where removed for a balanced/unbalanced load ?

 

[cnh1995]

Balanced load:
What would happen if the neutral where removed ?, my guess is that it wouldn't make a difference because the neutral doesn't carry any current

Right.

Unbalanced load:
Same question as above

Two Watt-meter method:

What would happen if the neutral where removed for a balanced/unbalanced load ?

For balanced load, removing the neutral should not make any difference. But without the neutral, phase difference between the currents for unbalanced load will no longer be 120°.

 

[TheRedDevil18]

Right.

For balanced load, removing the neutral should not make any difference. But without the neutral, phase difference between the currents for unbalanced load will no longer be 120°.

So for an unbalanced load, if the neutral where removed, would the wattmeter reading be higher than if it where not removed ?

Like, for the three wattmeter circuit if the neutral is removed, each wattmeter would have a different reading ?

 

[cnh1995]

Like, for the three wattmeter circuit if the neutral is removed, each wattmeter

Yes, in case of unbalanced load.

So for an unbalanced load, if the neutral where removed, would the wattmeter reading be higher than if it where not removed ?

I'm not sure about that. But since the currents are no longer 120° apart, the usual equations W1=VIcos(30+Φ) and W2=VIcos(30-Φ) are not applicable. So my guess is it's not possible to formulate the total power in terms of W1 and W2.

 

[TheRedDevil18]

Yes, in case of unbalanced load.

I'm not sure about that. But since the currents are no longer 120° apart, the usual equations W1=VIcos(30+Φ) and W2=VIcos(30-Φ) are not applicable. So my guess is it's not possible to formulate the total power in terms of W1 and W2.

So basically if you have unbalanced loads, the correct way would be to measure with the neutral ?

Also, when the neutral is in place with an unbalanced load then the power and currents would be balanced for each load ?

 

[informatory]

So,for the second picture it would be it,generally speaking neutral current is used for protecting electric circuits and taking the sufficient current also in this case when you remove neutral phase there is going to happen on some phases (b,y or r) higher or lower voltage ,in first case with neutral powers of each phase you can calculate separately but it would be so called (in my language) literally phase voltages rhat counts for finding powers of each phase.(Also all of these volltages are "lined" because there is no neutral as you call it) Idk if have made the right discussion and satisfied you with answer,so ask again if you have a need
All the best [emoji16]

 

[cnh1995]

So basically if you have unbalanced loads, the correct way would be to measure with the neutral ?

Three wattmeter method will work for both balanced and unbalanced load with or without the neutral.

Also, when the neutral is in place with an unbalanced load then the power and currents would be balanced for each load ?

No. The neutral maintains 120° phase difference between the currents in case of unbalanced load. The power consumption in each phase will be different.

 

[Windadct]

For N conductors - you need N-1 Wattmeters... or current meter for that matter. If you have a neutral point "in" the system but no neutral conductor leaving the system, then you only need 2 Meters for a 3 wire system.

As for the neutral phase angle - huh. I would not say the neutral maintains anything...with respect to current. Consider A-B have a Resistive load ( no neutral current impact)and then C-N have a purely capacititive load--- the neutral current will be 90Deg leading the C-N Voltage - and then in a real system the neutral typically gets a lot of the non-linear elements. From a watt meter standpoint you have to still account for them.

Edit - Spelling level EE