Power factor is the ratio between the Power (Watt) and the Apparent Power measured in VA(voltage-ampere) drawn by an electrical load . It is written as PF= P/S.
Using the Power triangle:
P= VpIpcosθ
S= VpIp
Q=Ssinθ
S= VpIp* w/ angle
We can sum up all the Power and Apparent Power (S).
P = P1 + P2 + P3
S= S1 + S2 + S3
EXAMPLE
LINE VOLTAGE = 34.5 60Hz Find C when the power factor is 0.94 leading
Solve first the P, S and Q using the 0.78 pf lagging.
Using the power triangle:
pf= arcos(0.78) = 38.74
The given is the Apparent Power equal to 24 MVA, we only need to solve the P and Q to complete the triangle.
P = Scos38.74
P = 24cos38.74
P = 18.72 MV
Q= Ssin38.74
Q = 24sin38.74
Q = 15.01 MVAR
These values is for the old.
FOR THE NEW ONE
Using the 0.94 pf leading and the P= 18.72 ( note that the power will just the same to the old one)
θ= arcos(0.94)
θ= 19.94
solve the Q;
using the trigonometric function
Q=Ptanθ
Q= 18.72tan(19.94)
Q= 6.79MVAR
we can disregard the S, since we only need the Q.
the power triangle for the New one is:
Since the Q in the new is in below, the sign is negative
Qc= Qnew – Qold
Qc= -6.79 -15.01
Qc = -21.79
Solving for the Capacitor
C= Q/2πf V^2rms
C= -21.79×10^6/ 2π(60) (34.5×10^3)^2
C= + 48 microFarad (disregard the sign)
FranCircuit 2 