Engineering Delta and star transformation of AC circuits

[gneill]

Hi, regarding question d. This is how I expressed my working out to get my answer (looking at the previous posts I now realize my answer is wrong but would like to know if my train of thought is correct)
Total impedance = 45+j45

In what way is that the "total impedance"? Total impedance of what?

 

[Spongecake]

For part C would the circuit look like this (see picture) before I make it into 1 single Y load.

 

[gneill]

For part C would the circuit look like this (see picture) before I make it into 1 single Y load.

Part C also specifies: "...and with the star-points of the two Y-sub-circuits connected together," What do you suppose is meant by that?

 

[Spongecake]

Would it be like this?

 

[gneill]

Would it be like this?

That would be my interpretation, yes.

 

[Spongecake]

ok brilliant i will give it a go. Thank you for your help gneill

 

[Spongecake]

So i have worked out that each of the impedance's would be 7.5+j7.5
Is this correct?

 

[gneill]

So i have worked out that each of the impedance's would be 7.5+j7.5
Is this correct?

Yes. But you could have verified this by checking the earlier parts of this thread.

 

[Jason-Li]

Yes. But you could have verified this by checking the earlier parts of this thread.

Hi Gneill,
Reading through the thread, I can't quite figure out why part (c) is the same as part (b). I have labelled up his drawing to reflect these calcs:
So Zy1=45+j45Ω
Zy2=ZΔ / 3 = 15+j15

Would each Y1 impedance not be in parallel with a Y2 impedance hence the calculation for part (c) would then be:
Zy combined = ((45+j45)(15+j15)) / ((45+j45)+(15+j15)) = j1350 / (60+j60)

j1350 / (60+j60) multiplied by conjugate = (81000+j81000) / 7200 = 11.25+j11.25 Ω

 

[gneill]

Reading through the thread, I can't quite figure out why part (c) is the same as part (b). I have labelled up his drawing to reflect these calcs:
So Zy1=45+j45Ω
Zy2=ZΔ / 3 = 15+j15

Zy1 is not 45+j45Ω. It's the original given value of (15 + j15)Ω

 

[Jason-Li]

Zy1 is not 45+j45Ω. It's the original given value of (15 + j15)Ω

Sorry got a bit lost there, I follow you now and have gotten 7.5Ω.

One other query regarding part (d), I have calculated out the same procedure as mentioned earlier in this thread using P = 3VI for total power consumed, which gives 11481.667-j11481.667. Could I assume that the 'imaginary' number here is effectively Reactive power Q? So my final answer should be 11481.667W or 11.48kW?

I know you mentioned that real power is just P = VI but should that not be S = VI, hence using the two numbers I get for S = VI would then be reactive & real power rather than using P = VI which would be the wrong formula technically?

 

[gneill]

One other query regarding part (d), I have calculated out the same procedure as mentioned earlier in this thread using P = 3VI for total power consumed, which gives 11481.667-j11481.667. Could I assume that the 'imaginary' number here is effectively Reactive power Q? So my final answer should be 11481.667W or 11.48kW?

The imaginary part is the reactive power, yes. I think you might have forgotten to take the complex conjugate of the current when you calculated $p = V I$, as the sign of the imaginary component is suspect (to me).

The final real power consumed is 11.48 kW as you've stated.

I know you mentioned that real power is just P = VI but should that not be S = VI, hence using the two numbers I get for S = VI would then be reactive & real power rather than using P = VI which would be the wrong formula technically?

I'm not sure where this question comes from as it's been a long time since the thread has been active and I haven't re-read through all the posts.

You can calculate the complex power by $S = V I^*$, where $I^*$ is the complex conjugate of the current $I$. $P = VI$ wouldn't make sense if V and I are both complex numbers, as $P$ must be strictly real.

 

[Jason-Li]

You can calculate the complex power by S=VI∗S=VI∗S = V I^*, where I∗I∗I^* is the complex conjugate of the current III. P=VIP=VIP = VI wouldn't make sense if V and I are both complex numbers, as PPP must be strictly real.

I see, that does answer why it should be S = VI* thanks. Why is it you multiply by the conjugate? Would the negative Q value not mean it was leading?

 

[gneill]

Why is it you multiply by the conjugate? Would the negative Q value not mean it was leading?

It's related to the fact for finding the power you want the phase difference between the current and voltage. Say that the voltage phasor $V$ has some phase angle $\phi_V$ and the current phasor $I$ some phase angle $\phi_I$. Multiplying these two phasors would yield $|V||I|$ with a phase angle of $\phi_V + \phi_I$. But you want the phase difference, $\phi_V - \phi_I$. Taking the complex conjugate of the current gives you this.

You should be able to find the mathematical derivation of the complex power in a textbook, or via google.