Displacement Power Factor•Source voltage waveform is assumed to be an undistorted sinusoid with zero phase angle.•Due to reactance of the load, the current waveform may exhibit a “constant” phase shift , with respect to the voltage waveform.

1 12 cos 2 cosAV RMS RMS

T T

P v t i t dt V t I t dtT T

2cos cos cos sin sinRMS RMS

T

V It t t dt

T

2cos cos cos cos sin sinRMS RMS

T T

V It t dt t t dt

T

22cos cos sin cos sinRMS RMS

T T

V It dt t t dt

T

= 0

Orthogonal

= T/2

cos (F) is defined as the “Displacement Power Factor” (DPF).

The cosine of the phase angle by which the current wave is “displaced” from the voltage wave.Note that DPF > 0 for leading and lagging phase: || < /2

The power company is supplying a currrent equal to I(RMS), but they’re only getting paid for I(RMS)cos(). The additional current causes unrecoverable losses due to series resistance in the transmission lines. Customers are required to correct their Power Factor to as close to unity as possible.

cosAV RMS RMSP V I

2cos

2RMS RMS

AV

V I TP

T

“Apparent Power”

The Problem: the customer’s load presents an impedance ZL (or an admittance YL = 1/ZL) at its utility connection.

The customer is required to place a compensating reactance XC (or a susceptance BC = -1/XC) in parallel with the power input lines to correct the power factor to unity.

ZL = 1/YL

YL = GL + jBL

XC = -1/BC

= 1/BL

BC = -BL

+

VRMS

_

Yin = GL + j0Zin = 1/ GL

RTX

Factory

Example:

A factory is supplied with 440 VAC, 60 Hz from the utility. It draws 650 amps RMS with a phase lag of 15 degrees (/12 rad) Determine the Apparent Power, Average (real) Power, required DPF correction circuitry, and RMS utility current after correction.

650 121.48 cos 12 sin 12

440 0L

LL

IY j

V

1.43 0.38

0.38L L

C C

G jB j

B C

0.38

0.001 1000120

CC

BC F F

440 1.43 630 AU U U U LI V Y V G

440 650 286000 WAPPP cos 12 276254 WAV APPP P

20 AmperesUI