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?

 

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.

 

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?

And the hardest question of all, what does it take to get TAKYMOUNIR to show some work!

 

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!

 

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!

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

 

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.

 

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.

 

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.

 

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.

true, no phase angle just RMS V times RMS A

 

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.