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Network, 2 loads, find complex power

  1. Jan 4, 2014 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/a/img819/9791/sh5k.jpg [Broken]

    Given the network [above], find the complex power supplied by the source, the power factor of the source, and the voltage Vs(t). The frequency is 60 Hz.


    2. Relevant equations

    ω=1/f

    Pf = cosø

    P = IV

    PL = (Vrms)*(Irms)*(Pf)



    3. The attempt at a solution

    Any hints on where to start would be helpful... not too sure about how to get & distinguish the different types of power

    ω = 1/f = 0.0166

    Apparent power = 12kVA ?

    Z = (0.05 + 0.2j)Ω

    cos-10.7 = 45.5° for the first load and
    cos-10.9 = 25.8° for the other load

    not sure what to do from here
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 5, 2014 #2

    tiny-tim

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    Hi Color_of_Cyan! :smile:
    Apparent power = S = Vr.m.sIr.m.s

    Average power = Scosφ

    Reactive power = Ssinφ

    Complex power = Se = S(cosφ + jsinφ) :wink:
     
  4. Jan 6, 2014 #3
    Ok, if i have

    PL = (Vrms)*(Irms)*(Pf)

    and the power factor is given for the load, can I find the current through that load first then?

    (PL)/(Pf*Vrms) = (Irms)

    = 20000/(0.7*240)

    is the current (119 ∠ 0)A for this load?

    (and isn't VA the same as watts?)


    and your sig is missing the '∠ ' symbol
     
  5. Jan 6, 2014 #4

    tiny-tim

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    no, 119 A is Irms

    I is Imaxcos(ωt + φ)

    (φ is the phase, cosφ is the power factor)

    Irms = Imax/√2 … so knowing Irms tells you nothing about φ :wink:
    yes
    ooh, that's a good idea, i'll add it! :smile:
     
  6. Jan 6, 2014 #5

    rude man

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    No. VA is volts times amps. W is the product of volts times amps times the power factor.

    Anyway, I don't understand your diagram. What's the "240V < 0 deg rms" mean? Is that the voltage on the right-hand side?
     
  7. Jan 7, 2014 #6
    Yes that's what it looks like (even though I could have copied it over better); I think it's the drop across both of the "loads" if they are in parallel like that.

    Thanks. I don't see what would be good about knowing that angle though. I am trying to find the total current first and then the "impedances" of the loads... wouldn't that help find Vs?


    So instead of (PL)/(Pf*Vrms) = (Irms)

    is it just (PL)/(Vrms) = (Irms)?
     
  8. Jan 7, 2014 #7

    tiny-tim

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    "that angle" is the phase, the amount of lead or lag

    the total complex current (is that what you mean by total current"? :confused:) is:
    Imax φ​

    if you want impedance, you need to know phase
     
  9. Jan 7, 2014 #8

    rude man

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    I would start by detrmining the actual components represented by the two blocks, then do a KVL or whatever to determine what the source has to be.
     
  10. Jan 8, 2014 #9
    So would there be separate phases for each of the "blocks" then? I had cos-10.7 = 45.5° and cos-10.9 = 25.8° for the blocks, would these be the phases for them (be it for impedance & current)? Not that good with phases here (yet).
     
  11. Jan 8, 2014 #10

    tiny-tim

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    yes (as the diagram shows)
    yes (except i'm not familiar with this decimal notation: i don't know whether it's cos-1 or sin-1)

    and of course + for leading and - for lagging
     
  12. Jan 8, 2014 #11

    gneill

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    If I may offer a suggestion... Look up "Power Triangle". It's a way to graphically represent the various power components (Real, Reactive, and Apparent). It also specifies their relationships to each other and the power factor.

    This problem specifies two loads: one via real power and power factor, the other via apparent power (kVA rather than Watts) and power factor. You can drop the known values onto a power triangle and determine the related values via Pythagoras.
     
  13. Jan 8, 2014 #12
    It was for power factor & the angle. For the complex current of the leftmost block then I got

    mag = 119 * √2 = 168.2

    then the angle = cos-10.7 = 45.5°

    so would the complex current for it be (168,2∠45.5°)A ?

    for the other block would it be

    I rms = 12000/240

    then 50 * √2 = 70.7

    then cos-10.9 = 25.8

    so would the complex current for the other block be (70.7 ∠ 25.8)A ?
     
  14. Jan 10, 2014 #13

    NascentOxygen

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    Almost. :smile:

    How are you going to designate the angle to show 'lagging'?

    It is probably the convention that voltages and currents are denoted by their RMS values, so there is no need to convert to peak values (since you invariably are going to convert back to RMS values in the end).

    Now that you know both currents, sum them to give the current through the L and R elements representing the transmission line.
     
  15. Jan 10, 2014 #14
    Nice, thanks. I still don't know how to even tell between lead or lag though.

    Converting to rect. form they are:

    (117.9 + 120j)A

    and

    (63.6 + 30.77j)A

    so Itot = 181.5 + 150.77j

    Itot = (236 ∠ 39.7)A


    The impedance of the other 2 elements is (0.05 + 0.2j)Ω = ( 0.206 ∠ 75.96)Ω

    so V = IZ

    = ( 0.206 ∠ 75.96)Ω * (236 ∠ 39.7)A

    = (48.61 ∠ 115.6)V

    = (-21 + 43.8j) - why is the real component negative?

    Adding the two voltages now (240V being the other one) I get:

    Vtotal = 219 + 43.8j

    =(223 ∠ 11.3)V

    Would this be the same as Vs?
     
  16. Jan 10, 2014 #15

    NascentOxygen

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    A lagging power factor indicates the load is inductive. It is a property of an inductor that the current lags the voltage. The angle of current relative to voltage will lie between 0° and -90°.

    The converse applies to a leading power factor, and is an indication of the load being capacitive. The angle of current relative to voltage now lies between 0° and +90°.

    One of the above won't be quite right. You need to treat one as leading and the other lagging. You diagram in the first post indicates this difference.
     
  17. Jan 10, 2014 #16
    But the currents both have positive angles there in the current phasors... how do you know if the angle is negative then (if that is what you were implying) ?
     
  18. Jan 11, 2014 #17

    tiny-tim

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    Because (as NascentOxygen :smile: says) the diagram says that one is leading and the other is lagging …

    it's like a diagram that says that two velocities are 30 mph north and 70 mph south … even though they're both positive, you know that they're opposite because it says so! :wink:
     
  19. Jan 11, 2014 #18
    Okay okay, so I guess the current goes from + to - (on the diagram for the 240V drop) where it "leads" then?
     
  20. Jan 11, 2014 #19

    NascentOxygen

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    You "guess"?! Who guesses when you can google?

    On the phasor diagram, it's the Imaginary component of the current which is drawn pointing in the negative direction to show lagging current. The Real component is not drawn any differently, whether for leading or lagging conditions. It's solely the Imaginary component that determines the lead or lag. http://www.labsanywhere.net/circuit/lectures/lect12/lecture12.php
     
  21. Jan 12, 2014 #20
    Well that helped to see it, so the currents would then be

    0.7 leading pf block: (168,2∠45.5°)A = (117.9 + 120j)A

    and now for the lagging block:

    0.9 lagging pf block: (70.7 ∠ -25.8)A = (63.6 - 30.77j)A



    which would mean

    Itot = (181.5 - 89.23j)A

    Itot = (202 ∠ -26.1°)A


    impedance of the other 2 elements is still (0.05 + 0.2j)Ω = ( 0.206 ∠ 75.96)Ω

    so V = IZ

    = ( 0.206 ∠ 75.96)Ω * (202 ∠ -26.1°)A


    = (41.612 ∠ 49.86)V = (26.82 + 31.8j)V

    so Vtot = Vs(t) = (266.82 + 31.8j)V then?


    Then apparent power is finding Vrms * Irms. Would that just be the magnitudes of Vtot and Itot multiplied out?
     
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