Cram for Exam Vol 4 Question 10 - 4-Phase Systems and Polyphase Systems with more than 3 phases

[akyip]

Hi guys,

I'm working on Cram for Exam Vol 4, and one of the problems is Question 10, regarding a balanced four-phase system.

Unlike basic 3-phase systems, I am not quite as familiar with polyphase systems above 3 phases.

This question has a balanced 4-phase source feeding a balanced 4-phase load where each load is connected line to neutral.

I understand that the load connected line to neutral is pretty much analogous to a 3-phase Y load. Line current = Phase current. And from what I think I understand, each phase is separated by 90 degrees, because 360 divided by number of phases = 360 degrees / 4 phases = 90 degrees per phase.

But what I'm not quite grasping is how to solve for the line to line voltage in terms of the phase or line-to-neutral voltage for systems with over 3 phases.

In this particular question (question 10 from Cram for Exam Vol 4), V phase = V line / √2. How are these voltages derived? The only thing I understand so far is that V phase = V LN, since the loads are connected line to neutral.

And is there a specific pattern for systems over 3 phases? E.g. each phase spans from 360 degrees divided by number of phases, etc?

Thanks for any insight on this!

 

[akyip]

I think I just figured out the answer to my question about line voltage for a 4-phase system.

Using basic phasor addition, for the line-to-line voltage between two adjacent voltage phases:

V LL = V LN <0 + V LN <90 = 1.414 V LN <45

Magnitude only: V LN = V phase = V LL / 1.414 = V LL / √2

Is that correct? And I should be able to apply the same basic phasor addition logic for higher-phase systems (e.g. 4-phase, 5-phase), right? 

 

[BebeshKing PE]

Hi guys,

I'm working on Cram for Exam Vol 4, and one of the problems is Question 10, regarding a balanced four-phase system.

Unlike basic 3-phase systems, I am not quite as familiar with polyphase systems above 3 phases.

This question has a balanced 4-phase source feeding a balanced 4-phase load where each load is connected line to neutral.

I understand that the load connected line to neutral is pretty much analogous to a 3-phase Y load. Line current = Phase current. And from what I think I understand, each phase is separated by 90 degrees, because 360 divided by number of phases = 360 degrees / 4 phases = 90 degrees per phase.

But what I'm not quite grasping is how to solve for the line to line voltage in terms of the phase or line-to-neutral voltage for systems with over 3 phases.

In this particular question (question 10 from Cram for Exam Vol 4), V phase = V line / √2. How are these voltages derived? The only thing I understand so far is that V phase = V LN, since the loads are connected line to neutral.

And is there a specific pattern for systems over 3 phases? E.g. each phase spans from 360 degrees divided by number of phases, etc?

Thanks for any insight on this!

I haven't seen the Cram Vol.4  yet but this is my personal explanation.

If you know where the square root of three and 30 degree phase shift came from in a 3 phase balanced system, you will be able to derive any polyphase balanced system.

As you mentioned above, each phase is separated by 90 degrees since its 360/4 = 90 degrees. So, if you need to find the line voltage, let's say Vab. So,.

Vab=Van-Vbn = (V<0) - (V<-90) = V(square root of 2 <45)  , where V is the per phase voltage magnitude.

Therefore ,the the line voltage for a 4 phase balanced system is square root of 2 larger than the phase voltage with 45 degrees phase shift.

 

[Cram For The PE]

I think I just figured out the answer to my question about line voltage for a 4-phase system.

Using basic phasor addition, for the line-to-line voltage between two adjacent voltage phases:

V LL = V LN <0 + V LN <90 = 1.414 V LN <45

Magnitude only: V LN = V phase = V LL / 1.414 = V LL / √2

Is that correct? And I should be able to apply the same basic phasor addition logic for higher-phase systems (e.g. 4-phase, 5-phase), right? 

Yes that is correct. But if you just recognize that with four phases you have 90 degree separation you just could use pythagoras theorem. With both sides of a right triangle equal to each other the hypotenuse will be sqrt[2] larger than one side. 

Assume each voltage vector is magnitude of 1 we would have:
hypotenuse=sqrt(1^2+1^2)  

 

[Dude99]

V ph LL = 1/0 - 1/ 360/ph

V3 = 1/0 - 1/120

V4 = 1/0 - 1/90