# 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
17,457
4,701
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
2,498
507
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
17,457
4,701
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
507
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
17,457
4,701
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
2,498
507
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