Understanding Power in Star Connection

[buell23]

Hello

I looked for a good forum which I can use for my specific questions.
I hope I am right here.
English is unfortunately not my first language, but I am working on improving it. So please have understanding for it.

I need help in three phase drive technology. I read in my book that power is defined by

p1 = u*i = u*sin(wt+phi)*i*sin(wt)

Now I ask myself why the first term of sin() contains a +phi and the second term does not.
I mean in the star connection for example u and i have an angular phase shift. So why is also in the second term not a +phi?

 

[NascentOxygen]

Hi buell23! Welcome to the famous Physics Forums.

In power calculations we are interested only in the relative phase difference, and in the case you cite between voltage and current the phase difference is ɸ.

The most general picture would be with v(t)=V·sin(wt+β) and i(t)=I·sin(wt+β+ɸ) but that is a pointless complication most of the time. Power engineers are very pragmatic specialists.

 

[buell23]

Hey
Thank you for your help

Ok, does that mean, that v(t) itself is angular phase shifted by u*sin(wt+beta).
Also the same with i(t) --> i*sin(wt+beta)
And the addition of phi signalizes that there is an angular difference between u and i?

Therefore I can write it more simple as p=u*i*sin(wt)*sin(wt+phi)?
Am I right?

 

[NascentOxygen]

Hey
Thank you for your help

Ok, does that mean, that v(t) itself is angular phase shifted by u*sin(wt+beta).
Also the same with i(t) --> i*sin(wt+beta)
And the addition of phi signalizes that there is an angular difference between u and i?

Therefore I can write it more simple as p=u*i*sin(wt)*sin(wt+phi)?
Am I right?

That's the right idea.

In the rare case, if you were asked to show instantaneous power equations for all 3 phases together on one graph, then that β would be (assumedly) 0° for one phase, 120° for another, and -120° for the third.

 

[buell23]

hey NascentOxygen

Thank you very much, you helped a lot.