1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Delta and star transformation AC circuits

  1. Jan 12, 2015 #1
    1. The problem statement, all variables and given/known data
    For the balanced three-phase loads shown in FIGURE 3,
    ZY = (15 + j15) Ω and ZΔ = (45 + j45) Ω. Determine:
    Uploaded file C1.png

    (a) the equivalent single Δ-connected load,

    (b) the equivalent single Y-connected load obtained from the Δ-Y transformation of (a) above,

    (c) the equivalent single Y-connected load obtained by transforming the Δ sub-load of FIGURE 3 to a Y and with the star-points of the two Y-sub-circuits connected together,

    (d) the total power consumed in case (a) above if the line voltage of the three-phase supply is 415 V at 50 Hz.

    2. Relevant equations


    3. The attempt at a solution
    For (a) P=Q=R=(15+15i)
    Star "PQR" --->Delta "ABC" equivelant =
    A=PQ+QR+RP/R
    Since the loads QPR are all the same value and same equation form then A=B=C, ((15+15i)*(15+15i)+(15+15i)*(15+15i)+(15+15i)*(15+15i))/(15+15i)=45+i45
    Delta equivelant is A=45+i45, B=45+i45, C=45+i45

    For(b) The reverse of (a) I assume; Delta ---> Star =
    Q=AC/A+B+C
    P=AB/A+B+C
    R=BC/A+B+C

    Q=P=R= 15+i15

    Questions seems deceptively easy for my liking

    (c) Is the diagram C2.png how the transformation and two Y sub-circuit connected star points should look like?
    I need a hint on how to form the equations for this if it is correct(im sure its obvious but im not sure.)
    Any help greatly appreciated.
     

    Attached Files:

    • C1.png
      C1.png
      File size:
      44.4 KB
      Views:
      430
    • C2.png
      C2.png
      File size:
      48 KB
      Views:
      340
  2. jcsd
  3. Jan 12, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    For (a) they want a single Delta load. You've converted the Y to a Delta but now you have two Deltas as loads. Combine them into one.

    For (b) they want you to take the single Delta from above and make a single Y.

    Your diagram for part (c) doesn't make sense. You should have the single Delta that results from part (a) connected appropriately to the source. Having the line voltage and a Delta connection should make finding the power fairly straightforward... note how the individual impedances of the loads are connected to the sources.
     
  4. Jan 12, 2015 #3
    So do i need to combine two deltas using product over the sum rule in (a) and use that answer to make a Y transformation in(b)?
     
  5. Jan 12, 2015 #4

    gneill

    User Avatar

    Staff: Mentor

    That's the idea. I think you'll find that finding the parallel impedances is greatly aided by the values chosen for the problem ;)
     
  6. Jan 14, 2015 #5
    (a)
    ((45 + j45) (45 + j45))/(90 + j90)
    =22.5+j22.5 ohms

    (b)
    ((22.5 + j22.5) (22.5 + j22.5))/((22.5 + j22.5) + (22.5 + j22.5) + (22.5 + j22.5))
    =7.5+j7.5 ohms

    Is question (c) essentially the product over the sum of ZY = (15 + j15) with the two Y three phase loads in parallel which is
    ((15 + j15) (15 + j15))/((15 + j15) + (15 + j15)) = 7.5+j7.5 ohms

    For (d) The single delta load (22.5+j22.5), the power is basically (I^2)*R? It gives the frequency of 50Hz but I can't find a power equation or direct link for this. I know power factor depends on frequency and watts law uses power factor in its equation.
     
  7. Jan 14, 2015 #6

    gneill

    User Avatar

    Staff: Mentor

    For (d) note that you are given the Line voltage. Note how Line voltage presents on the Δ load (there's an individual line voltage across each of the impedances). So you have potential difference and impedance for each. So you can calculate the current in a given impedance, right?

    After that, what's the expression for complex power given the voltage V and current I?
     
  8. Jan 14, 2015 #7
    I=V/Z =9.2222-j9.2222 A , P = VI = 415*(9.2222-j9.2222)= 3827.22-j3827.22 watts.
    Thats just for the single delta load in (a)
     
  9. Jan 14, 2015 #8

    gneill

    User Avatar

    Staff: Mentor

    Okay, one thing. To compute the complex power you need to use the complex conjugate of the current in order make the power's angle work out with the correct sign (not that it's a big issue here where you just need the real component of the complex power). So the equation for the complex power is p = VI*, where I* represents the complex conjugate of I. The complex conjugate just changes the sign of the imaginary component.

    So, you can pick out the real power dissipated by one of the impedances, right? What then is the total real power dissipated?
     
  10. Jan 14, 2015 #9
    The total real power is the sum of the single delta load (3827.22+j3827.22) watts and the three phase ZY load (415*13.833+j13.833= 5740.695+j5740.695i)
     
  11. Jan 14, 2015 #10

    gneill

    User Avatar

    Staff: Mentor

    No, real power is real only. No imaginary part. Imaginary power dissipates no energy, as it's simply exchanged back and forth between the load and the power source over a cycle.

    Now, didn't the single Delta load from part (a) represent the entire load? After all, you took the time to change the Y load to its Delta form and then combine it with the existing Delta, resulting in a "single Delta load", right? If you want to find the total power dissipated, then it makes sense to start with that single Delta load and find the power dissipated by one of its impedances, then scale to three of them...
     
  12. Jan 14, 2015 #11
    the real power dissapated in one of the impedances times 3 (3827.22*3)=11481.66 watts.
     
  13. Jan 14, 2015 #12

    gneill

    User Avatar

    Staff: Mentor

    Yup. You might express it in kW and trim the significant figures a bit :)
     
  14. Jan 14, 2015 #13
    Well that seems simple enough! Thanks for your help once again.:)
     
  15. Mar 3, 2015 #14
    I don't understand the need to times by three again.

    Line voltage = 415v
    Therefore phase voltage = 239.6

    239.6/22.5+j22.5 = 5.33+j5.33A

    P=VI*

    P= 240 x 5.33-j5.33

    P=1279.2 -j1279.2

    x3 because of the three phases.

    3837.6-j3837.6 = total power

    3837.6W is actual power.

    Why is there a need to times by three again?
     
  16. Mar 3, 2015 #15

    gneill

    User Avatar

    Staff: Mentor

    It's the line voltage, not the phase voltage that's impressed across the individual load impedances in the Delta configuration. By converting to phase voltage you've effectively divided by ##\sqrt{3}## twice when calculating the power. That's the same as dividing by three.
     
  17. May 14, 2015 #16
    isn't vl = vp on a delta network?
     
  18. May 14, 2015 #17

    gneill

    User Avatar

    Staff: Mentor

    Yes, if the source is a delta configured source. In this case the source is not specified and we're given only the line-to-line voltage.
     
  19. Jun 4, 2015 #18
    Hie guys I'm still not getting part (c) even after reading this, in my calculation I transformed the delta of fig. 3 to a Y giving me 15 + j15 same as with the other Y.

    'And with star-points of the two Y-sub-circuits connected together'
    Does this part of the question change my calculation above ( well, if I calculated right)
     
  20. Jun 4, 2015 #19

    gneill

    User Avatar

    Staff: Mentor

    No, it doesn't change your calculation. But your calculations aren't finished. You left off at the point where you have two Y-connected loads. You have yet to combine them into a single load.
     
  21. Jun 5, 2015 #20
    Thanks gneill,

    So if I combine them using product over sum I get 5+j5. Is that correct?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Delta and star transformation AC circuits
Loading...