How to Analyze a Balanced Three Phase Star Load Diagram?

[DunceKirchhoff]

http://i59.tinypic.com/727ole.jpg

My answers


http://i59.tinypic.com/mkzvqp.jpg

Can some see if I am going in the right direction, thanks.

 

[DunceKirchhoff]

Anyone??

 

[gneill]

It's a bit of a pain having to follow links to question bodies. Consider loading your pics to PF rather than some other image hosting site, and including them in the post body (via the "Go Advanced" editing frame where icons on the frame give you the ability).

The only issue I'd have with your results as shown is that the question asks you to use V_AB as the reference phasor for the calculations, which should, I think, yield a "defined" angle for V_AB of zero degrees rather than 30°. The other voltages would then similarly be shifted by -30° for part (a), and would carry through to the voltages calculated for part (b) and currents for part (c).

 

[DunceKirchhoff]

Hi Neil

Sorry will do that in the future.

Yeah usually the ref is 0 degrees agreed.
But in this question the line voltage is not given but the phase voltage Vao.
So given that information I have tried to work it out.
Also the line voltage leads the phase voltage by 30 degrees.

Could be wrong

 

[DunceKirchhoff]

I am wrong sorry, yes it has asked me to use vab as a reference.
So vab=0 degrees
Vbc -120 degrees
Vca 120 degrees
Whats the phase voltage?

 

[gneill]

I am wrong sorry, yes it has asked me to use vab as a reference.
So vab=0 degrees
Vbc -120 degrees
Vca 120 degrees
Whats the phase voltage?

For V_AO', V_BO' and V_CO' you can just shift their normal values by the same -30°. Confirm by re-calculating V_AB etc., from them to see that the resulting angles match what you got for part (a).

 

[DunceKirchhoff]

So basically I have put vab, vbc and vca where vao, vbo and vco should be? And the phase angle is -30 instead of 30?

 

[gneill]

So basically I have put vab, vbc and vca where vao, vbo and vco should be? And the phase angle is -30 instead of 30?

Previously you had:
$V_{AO'} = 100\ ∠\ 0°$
$V_{BO'} = 100\ ∠ -120°$
$V_{CO'} = 100\ ∠ \ 120°$

which I believe is a "standard" set of angles for the lines of a star configuration. Now you want to have the voltages V_AB, etc., come out having the desired set of angles where V_AB has an angle of zero degrees. Those angles were determined in part (a).

Since all the various relative angles are fixed on a phasor diagram, if you had to shift V_AB by -30° to make it the zero degree reference then ALL the angles will shift by the same amount. It's equivalent to rotating the whole phasor diagram by 30°. So if you were to subtract 30° from each of the above you should find that if you recalculate V_AB etc. from V_AO' V_BO' and V_CO', the angles will turn out to match the ones from part (a).

 

[DunceKirchhoff]

Ok thanks I got you. Well explained.