How do I calculate the power in a star to delta transformation?

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SUMMARY

The forum discussion centers on calculating power in a star to delta transformation, specifically addressing the conversion of impedances in electrical circuits. Participants discuss the transformation of a Y-connected load to a Δ-connected load, with specific impedance values such as ZΔ = 45 + j45 and ZY = 15 + j15. The conversation emphasizes the importance of understanding how to combine impedances in parallel and the implications of identical loads when driven by the same sources. Key insights include the necessity of justifying impedance combinations and the application of Δ-Y transformation formulas.

PREREQUISITES
  • Understanding of star (Y) and delta (Δ) transformations in electrical circuits
  • Knowledge of complex impedance and its representation (e.g., Z = R + jX)
  • Familiarity with circuit analysis techniques, including parallel and series combinations
  • Ability to apply Δ-Y transformation formulas for impedance calculations
NEXT STEPS
  • Study the principles of star to delta (Y to Δ) transformations in electrical engineering
  • Learn how to calculate equivalent impedances in parallel and series configurations
  • Explore the implications of identical loads in parallel circuits and their behavior
  • Review power calculation methods for Δ-connected loads using voltage and impedance
USEFUL FOR

Electrical engineering students, circuit designers, and professionals involved in power system analysis will benefit from this discussion, particularly those focusing on impedance transformations and circuit analysis techniques.

M P
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Homework Statement



zad n.png

Homework Equations

The Attempt at a Solution



a) the question is to transform star to delta ?
 
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M P said:
a) the question is to transform star to delta ?
The question asks you to find a single Δ-connected load that is equivalent to total of the two loads shown.
 
gneill said:
The question asks you to find a single Δ-connected load that is equivalent to total of the two loads shown.
attempt:
90+90j for single delta ?
 
M P said:
attempt:
90+90j for single delta ?
We won't confirm or deny a guess. You'll have to show how you got there.
 
I was not guessing I just used the T to π converter supplied with course and added result to another ZΔ =(45+j45) doubling them given 90+90j
 
M P said:
I was not guessing I just used the T to π converter supplied with course and added result to another ZΔ =(45+j45) doubling them given 90+90j
Can you justify adding them? How do the voltage sources "see" them when you combine them?
 
gneill said:
Can you justify adding them? How do the voltage sources "see" them when you combine them?

angle is changing ? on the other connection V ∠0 ∠-120 ∠-240 thank you for the interest.
 
Ignore the phase angles of the sources for now. How do the individual impedances of the two loads combine as seen by the sources? You have two Δ loads. What connections are in common? Can you see serial or parallel reduction possibilities?
 
attempt: ZΔ x ZΔ /ZΔ+ZΔ and then x3 ?
 
  • #10
M P said:
attempt: ZΔ x ZΔ /ZΔ+ZΔ and then x3 ?
Can you elaborate?
 
  • #11
gneill said:
Can you elaborate?
I have Y and Δ from the start. I transform Y to Δ and obtain 2 x Δ with ZΔ = 45 + j45 and to get single Δ after your hints I thought I need to combine each 2 impedances in parallel so ZΔ x ZΔ /ZΔ+ZΔ and I need to do that 3x ?
 
  • #12
gneill said:
Can you elaborate?

An thank you for the prompt reply :cool:
 
  • #13
M P said:
I have Y and Δ from the start. I transform Y to Δ and obtain 2 x Δ with ZΔ = 45 + j45 and to get single Δ after your hints I thought I need to combine each 2 impedances in parallel so ZΔ x ZΔ /ZΔ+ZΔ and I need to do that 3x ?
Okay, that is correct. But it is important that you can see why the individual impedances are in parallel by considering the circuit diagram. If you can't see it, be sure to ask.
 
  • #14
I will improve I promise.
 
  • #15
gneill said:
Okay, that is correct. But it is important that you can see why the individual impedances are in parallel by considering the circuit diagram. If you can't see it, be sure to ask.

So similar to b) with 2 x Y at 15+j15 and ZYx ZY/ ZY+ZY
 
  • #16
M P said:
So similar to b) with 2 x Y at 15+j15 and ZYx ZY/ ZY+ZY
How are we to interpret your coded messages? o_O A few words of explanation and justification would help.
 
  • #17
Sorry. after Δ to Y transformation I have 15+j15 for each impedance 2xY and then combine each impedance ZYx ZY/ ZY+ZY as in a)
 
  • #18
M P said:
Sorry. after Δ to Y transformation I have 15+j15 for each impedance 2xY and then combine each impedance ZYx ZY/ ZY+ZY as in a)
Okay, what leads you conclude that you can combine the Y's in that fashion? I'm not saying it's incorrect for this very special case, but you should be able to justify it. That way you won't inadvertently make the error of trying to do the same thing when the scenario is different and combining them in this fashion would be incorrect.
 
  • #19
as per drawing??
 

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  • #20
now I am thinking what happens when I add 0 to it? I wonder if you have any hints...:)
 
  • #21
M P said:
now I am thinking what happens when I add 0 to it? I wonder if you have any hints...:)
What do you mean by "add 0 into it"?

The diagram alone doesn't explain why you can combine the like-labeled impedances in parallel. Note that none of those pairs shares two nodes... So you need to make a special case deduction about the nature of the central nodes of each Y...
 
  • #22
gneill said:
What do you mean by "add 0 into it"?

The diagram alone doesn't explain why you can combine the like-labeled impedances in parallel. Note that none of those pairs shares two nodes... So you need to make a special case deduction about the nature of the central nodes of each Y...

adding 0 I meant c) and connection with star points not sure how can I solve this. Also "special case deduction about the nature of the central nodes" that is another scary thing to me I thought transforming and parallel calculation is enough. I am not very good combinatorist.
 
  • #23
If you read the question statement carefully, for (b) you are expected to perform a Δ-Y transformation of the single Δ load obtained in part (a). That should be a straightforward application of the Δ-Y formulas.

It's part (c) that asks you to combine two identical Y loads. A very important fact to note is that they are identical. When driven in parallel by the same sources you can be sure that they will behave identically as well. That let's you say something about their center nodes.
 
  • #24
regarding d) I am not sure how to get PF? Any hints or webs I can catch up on it? thank you in advance..
 
  • #25
M P said:
regarding d) I am not sure how to get PF? Any hints or webs I can catch up on it? thank you in advance..
With a Δ load you have the voltage across each of the individual load impedances. The voltage and impedance are sufficient to calculate the power.
 

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