Firing angle in rectifiers and inverters.

  1. Hello everyone I am a newbie to electrical engineering, I have this problem I've been strugling with, no textbook seems to cover this problem..


    A three-phase, 440-V, generator delivers 5400W of active power through two recti-fies, A and B, into a three-phase, 230-V, line as shown in Figure 3.
    The dc current through the inductor is 20A. If power loss in both rectifiers are negligible, calculate:

    (a). Firing angle of the rectifier A;
    (b). Firing angle of the rectifier (inverter) B.

    This is my attempt ;

    V out = 5400/ 20 = 270
    For inductive load

    VOUT = 1.35 × VLINE × cos α

    Where ‘α’ is the firing angle of the rectifier.

    Therefore,

    cos α = V / 1.35 x Vline

    α = 29

    how does this seem? and how is the inverter's firing angle different?
     
  2. jcsd
  3. 1. The problem statement, all variables and given/known data

    Hello everyone I am a newbie to electrical engineering, I have this problem I've been strugling with, no textbook seems to cover this problem..


    A three-phase, 440-V, generator delivers 5400W of active power through two recti-fies, A and B, into a three-phase, 230-V, line as shown in Figure 3.
    The dc current through the inductor is 20A. If power loss in both rectifiers are negligible, calculate:

    (a). Firing angle of the rectifier A;
    (b). Firing angle of the rectifier (inverter) B.



    2. Relevant equations



    3. The attempt at a solutionThis is my attempt ;

    V out = 5400/ 20 = 270
    For inductive load

    VOUT = 1.35 × VLINE × cos α

    Where ‘α’ is the firing angle of the rectifier.

    Therefore,

    cos α = V / 1.35 x Vline

    α = 29

    how does this seem? and how is the inverter's firing angle different?
     
  4. NascentOxygen

    Staff: Mentor

    Where is Fig 3?
     
  5. uart

    uart 2,777
    Science Advisor

    Seems ok. And you can use the same formula (with the relevant line voltage) for the inverter side.

    BTW The constant 1.35 comes from 3 sqrt(2) / pi.
     
  6. but what is the relevent line voltage for the inverter? isn't it the same?
     
  7. Not sure this is the right place to post this (perhaps this is why no one answers, but I don't know I'm new to PF). Here's some help:

    In fact, there are a lot of text books that cover this. For instance see "Power electronics", from Mohan. I can't see your Figure 3, but I assume you have a perfect voltage source connected to 2 full-bridge thyristor converters (1 rectifier, 1 inverter) and some load. There should also be an inductor on the DC bridge. To find the solution for rectifier A (source):

    [tex]P=V_{DC}I_{DC}[/tex]

    where the voltage on the DC bus is given by:

    [tex]V_{DC}=\frac{3\sqrt{2}}{\pi}V_{LL}\cos{\alpha}[/tex]

    where [itex]V_{LL}[/itex] is the line-line voltage of the AC side. The DC bus current is given in the problem, which is [itex]I_{DC}[/itex]. Just isolate these for [itex]\alpha[/itex]...

    For rectifier (inverter) B, same procedure, but by using a negative power and use the 230V line-line voltage. Isolate again for [itex]\alpha[/itex]... This angle should be between 90 and 180 degrees since this converter is operating in inverter mode.

    M.
     
  8. berkeman

    Staff: Mentor

    (Moderator's note -- two threads merged...)
     
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