what is the kva

Discussion in 'Homework Help' started by TAKYMOUNIR, Aug 27, 2013.

  1. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
    351
    1
    If V (t) = 679 sin(377 xt) and I (t) = 1414 sin(377xt + Φ), where Φ is a phase angle. What is the KVA produced by this voltage and current? What is the power delivered by this voltage and current as a function of Φ? What is the power factor as a function of Φ?
     
  2. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    What this problem is trying to teach you is the difference between Watts (true power) and Volt-Amps (VA's) which is apparrent power. And those two things are related by the phase angle.
     
  3. WBahn

    Moderator

    Mar 31, 2012
    17,788
    4,808
    And the hardest question of all, what does it take to get TAKYMOUNIR to show some work!
     
    DerStrom8 and TAKYMOUNIR like this.
  4. Alan brad

    New Member

    Aug 29, 2013
    9
    0
    Hey what doubt do you have ? i don't understand your question...any way KVA means Kilo volt ampere..and its delivered power may be reactive power or Apparent power.and phi means cosine of the angle between voltage and current!
     
  5. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    Volt-amps is apparrent power. Reactive power is VARs which stands for volt-amps-reactive. Real power is Watts. Both VA and VAR have power factor in the calculation.

    http://en.wikipedia.org/wiki/AC_power
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    I think "apparent power" is just the product of the effective voltage and the effective current without regard to the power factor. It is the magnitude of "complex power". The power factor is taken into account as being the cosine of the angle of the complex power. "Real power" and "reactive power" are the real and imaginary parts of complex power, and hence are essentially apparent power adjusted appropriately for power factor.
     
    Last edited: Aug 30, 2013
  7. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
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    I think that real power (Watts) is obtained by multiplying apparrent power (V-A which is the product of the VAC and current) times the cosine of the phase angle. To get reactive power you multiply the App Pwr times the sine of the phase angle, or the cosine of the complement of the phase angle.
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    I agree. My point was taking issue with the statement that, "Both VA and VAR have power factor in the calculation.". VA does NOT have power factor in the calculation -- that is what makes it only "apparent" power.
     
  9. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    true, no phase angle just RMS V times RMS A
     
  10. donpetru

    Active Member

    Nov 14, 2008
    186
    25
    Next, consider the problem (give the following information):
    V (t) = 679 sin (377 xt)
    and
    I (t) = 1 414 sin (377xt + Φ)

    377 means w = 2*pi*f (ie, pulsation), we obtain f = 377/2 * 3.14 = 60 Hz.
    Now, the presence of Φ in the mathematical relationship of I (t) and the "+" in front of it, means that the current is before voltage and the consumer is capacitive. If the sign was negative, this means that the consumer was inductive (but we don't have this case).

    To solve the problem we need to know the value of Φ because:
    S [kVA] = sqrt (P*P + Q*Q); (1)

    where: P [kW] = V * I * cosΦ (2) and Q = V * I * sinΦ (3)

    On the other hand, we have the following values:

    679
    = 1.41 * Vrms

    and

    1414
    = 1.41 * Irms

    Then we can calculate simply:

    S [kVA] = Vrms * Irms = (679 / 1.41 ) * ( 1414 / 1.41) = 481 [Vrms] * 1002.83 [A] = 482364.5 [VA] = 482.3 [kVA].

    This is the result of theoretical but practical, to find out exactly, be applied to equation (1) above.
     
    Last edited: Sep 1, 2013
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