Star Delta Transformation

In summary: 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.Adding 0 into the equation means reversing the order of the sources and the loads. Can you please elaborate on what you meant by "special case deduction about the nature of the central nodes"?Can you please elaborate on what you meant by "special case deduction about the nature of the central nodes"?I meant that you would have to use a different approach to combine the impedances in parallel if the nodes of the Y's weren't all at the same distance from the node of the
  • #1

M P

88
1

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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  • #2
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.
 
  • #3
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 ?
 
  • #4
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.
 
  • #5
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
 
  • #6
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?
 
  • #7
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.
 
  • #8
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?
 
  • #9
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.
 

1. What is Star Delta Transformation?

Star Delta Transformation is a method used to simplify and analyze complex electrical networks. It involves converting a circuit with a star (or Y) configuration to an equivalent delta (or Δ) configuration, or vice versa.

2. Why is Star Delta Transformation used?

Star Delta Transformation is used to make calculations and analysis of electrical circuits easier and more efficient. It allows for the conversion of a circuit with mixed resistive and reactive elements to an equivalent circuit with only resistive elements, making it easier to apply Ohm's Law and other circuit laws.

3. When is Star Delta Transformation applicable?

Star Delta Transformation is applicable to circuits with three branches and a common point, where one branch is connected between each pair of elements. This type of circuit is commonly found in three-phase electrical systems.

4. How is Star Delta Transformation performed?

To perform Star Delta Transformation, the values of resistors and impedances in the star configuration are first identified. Then, using the transformation equations, these values are converted to an equivalent delta configuration. The same process can be reversed to convert a delta configuration to a star configuration.

5. What are the advantages of using Star Delta Transformation?

The main advantage of using Star Delta Transformation is that it simplifies circuit analysis and calculations, making it easier to determine voltage, current, and power in a circuit. It also allows for the reduction of complex circuits to simpler ones, making it easier to troubleshoot and identify potential issues.

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