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Delta star conversion help please

  1. Sep 15, 2013 #1
    Hello, I would like someone to give me some help or clarification relating to the conversion from a delta to star source. I understand that in star, Line Voltage is sqrt(3) times the phase voltage, and line voltage also leads phase by 30°. I also understand that Line and Phase currents are the same too. In Delta, I understand that Line and Phase Voltages are the same, and that Line Currents are sqrt(3) times the phase Currents, but LAG Phase currents by 30°. My issue seems to be relating to the time when converting a delta source to a star source is required. What I am not sure about is what to take as reference? If you start with a Delta Source and convert to Star, do you take the Delta Voltage as reference, and then divide by sqrt(3) and subtract 30° to get the equivalent star phase voltage, or do you say that the delta voltage is the same as the star phase voltage (in terms of angle and magnitude) and then multiply by sqrt(3) and add 30° to get the equivalent star line voltage.

    Also, when converting from delta to star, do you make the initial delta voltages to be at 30°, so that when you divide by sqrt(3) and subtract 30° to get the equivalent star phase voltage this will be at 0 degrees?, or do you take the voltages in the initial delta source to be at 0°?

    E.g., should the conversion be undertaken as below if the delta voltages are 415V, as there are several ways of doing this, depending on what is taken as the reference (0° angle) (Vp = Phase Voltage and Vl = Line Voltage)

    If Delta VL = 415V@30° → Star Vp = 415/√3 @ 0° & Star VL = 415V@30°


    Delta VL = 415V@0° → Star Vp = 415/√3 @ -30 & Star VL = 415V@ 0°


    Delta VL = 415V@0° → Star Vp = 415V@0° & Star VL = √3 * 415V @ 30°

    I'm really sorry if this has confused anyone but I would appreciate some help as this should be so simple and is really bugging me. I'm not sure whether I'm making a mountain out of a molehill here or whether I have missed something really obvious, its the fact that our lecturer said it doesn't matter what is taken as reference, although surely this would knock everything out by 30° if true?

  2. jcsd
  3. Sep 17, 2013 #2
    I think you are making it harder than it really is :)

    First of all, from your language ("line/phase voltage, line/phase currents"), I assume you are talking about Three-phase Circuits. In three-phase circuits, there are some differences with respect to a simple Δ-Y conversion (you should have been a bit more specific in defining the context). Moreover, you are referring to balanced three-phase circuits (you must specify it!), otherwise nothing of what you said holds true (and complications arise). From your post, it seems to me that, although you have a general picture, you are very confused on the details. I don't know if the terminology is confusing you, but it is important to be very precise here.

    Let's take a step back. There are four kinds of balanced three-phase circuits. They are classified in reference to the source-load configuration: 1) Y-Y; 2) Y-Δ; 3) Δ-Δ; 4) Δ-Y. The question is: which cases are the problematic ones? If (as I think) you are referring to the Δ-source, Y-load case (4), apart from the fact that you should have specified it :) , then, assuming an abc sequence (this is important!), you can say that:

    a) The phase voltages, denoted by [itex]\textbf{V}_{ab}, \textbf{V}_{bc}, \textbf{V}_{ca}[/itex], can be written in the following way:
    [itex]\textbf{V}_{ab} = V_p\angle{0°} \ \ \ \textbf{V}_{bc}=V_p\angle{-120°} \ \ \ \textbf{V}_{ca}=V_p\angle{+120°}[/itex]
    b) It does not matter what you take as a reference, although it is convenient (for the notation) to take [itex]\textbf{V}_{ab}[/itex].
    c) If you want to convert the Δ-source to the equivalent Y-source, then, denoting the equivalent phase voltages with [itex]\textbf{V}_{an}, \textbf{V}_{bn}, \textbf{V}_{cn}[/itex], you must use the following formula:
    [itex]\textbf{V}_{an} = \frac{V_p}{\sqrt{3}}\angle{-30°} \ \ \ \textbf{V}_{bn}=\frac{V_p}{\sqrt{3}}\angle{-150°} \ \ \ \textbf{V}_{cn}=\frac{V_p}{\sqrt{3}}\angle{+90°}[/itex]
    c.bis) If the Δ-source has an internal impedance [itex]\textbf{Z}_S[/itex] for each phase, then the equivalent Y-source will have an internal impedance of [itex]\frac{\textbf{Z}_S}{3}[/itex] for each phase.

    I hope this clarifies things. There are a lot of imprecisions and terminology abuses in your post, please be very careful in this context, since slipping a single word can change the meaning of the whole sentence and generate confusion. If you still find it problematic, remember to always specify context in great detail. Questions such as: "is it a balanced three-phase circuit?", "if so, what configuration is it in?", "Is the source in positive or negative sequence?" are the first steps you should take to avoid confusion. Hope this helps.
  4. Sep 17, 2013 #3
    Hi, Thanks for your help. Sorry for being a bit vague in my description. From what you have explained in part 'c' of your answer where you have converted a delta source to an equivalent star source, am I correct in thinking that you have taken the voltages of the original delta source as the 0° reference, as your equivalent star 'a' phase voltage (Van) is now at -30°
  5. Sep 18, 2013 #4
    In doing the conversion, I took [itex]\textbf{V}_{ab}[/itex] (and only this) as a reference, since this is the standard notation. The rest of the formula is simply a direct application of the properties of the abc sequence. Let me clarify: this is just for theoretical convenience, it does not mean that, in an actual exercise, the phase of [itex]\textbf{V}_{an}[/itex] will always be [itex]-30°[/itex]. In fact, sometimes the text of the exercise will tell you explicitly which phase voltage to take as a reference; often, you do not need a reference at all. Try to solve the following:

    Exercise. Given a balanced Δ-Y circuit, knowing that [itex]\textbf{V}_{ab} = 100\sqrt3 \angle{17°}[/itex] and that the source is in positive sequence, calculate the equivalent Y-source if the internal load for each phase is [itex]\textbf{Z}_S = 30 + j12 \ \Omega[/itex].
  6. Sep 23, 2013 #5
    Hi, Sorry for the delay in replying, been moving back to uni so been quite busy. Given the exercise shown above, for the equivalent Y source, I believe it would be as follows;

    Van = (100√3)/√3 ∠(17°-30°) = 100V∠-13°

    Vbn = (100√3)/√3 ∠-13°-120° = 100V∠-133°

    Vcn = (100√3)/√3 ∠-13°-240° = 100V∠-253° (or +107°)

    Zs for Equivalent Star = Zs (Delta) / 3

    Zs (Equivalent Star) = (30+j12)/3 = 10+j4 Ω per phase

    I think that This is correct. Thank you for your help microgravity
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