Transcript DC Motors - Anne Arundel Community College

Power Factor and Power
Factor Correction
Learning Objectives

Define power factor.

Define unity, leading and lagging power factors.

Define power factor correction and unity power
factor correction.

Calculate the inductor or capacitor value required to
correct AC series parallel networks to the desired
apparent power.

Compare currents, voltages, and power in AC
series parallel networks before and after power
factor correction.
Review
AC Power to a Resistive Load
2
P  V RM S I RM S  I RM S R 
2
V RM S
(w atts)
R
AC Power to a Inductive Load
2
QL  I
2
RMS
XL 
V RMS
(V AR )
XL
AC Power to a Capacitive Load
2
QC  I
2
RM S
XC 
V RM S
XC
(V AR C AP )
Review
Apparent Power


For a load with voltage V and current I, the
power that “appears to flow” to the load is VI
where V and I are rms values.
S = VI (VA)
S is called the apparent power and has units of
volt-amperes (VA).
Review
Real and Reactive Power

The power triangle also shows that we can find
real (P) and reactive (Q) power.
P  V I cos   S cos 
(W )
Q  V I sin   S sin 
(V A R )
P

S
QL

P
QC
S
Power Factor


Power factor (FP) tells us what portion of the
apparent power (S) is actually real power (P).
Power factor is a ratio given by
FP = P / S

Power factor is expressed as a number
between 0 to 1.0 (or as a percent from 0% to
100%)
Power Factor

From the power triangle it can be seen that
FP = P / S = cos 

Power factor angle is thus given
 = cos-1(P / S)



For a pure resistance,  = 0º
For a pure inductance,  = 90º
For a pure capacitance,  = -90º
S
Q
NOTE: Ө is the phase angle of ZT, not the
current or voltage.

P
Example Problem 1
Determine the power factor for this circuit. Is it
inductive or capacitive?
Example Problem 2
For the circuit below, determine PT, QT, ST, FP and
draw the power triangle.
If E is 120VRMS, determine the supply current IRMS.
Unity power factor (FP = 1)




Implies that all of a load’s apparent power is
real power (S = P).
If FP = 1, then  = 0º.
It could also be said that the load looks purely
resistive.
Load current and voltage are in phase.

P,S
Q=0
Lagging power factor ( > 0º)


The load current lags load voltage
Implies that the load looks inductive.
S
Q

P
VARind
Leading power factor ( < 0º)


The load current leads load voltage
Implies that the load looks capacitive.
P

Q
S
VARcap
Why is Power Factor Important?


Consider the following example: A generator is
rated at 600 V and supplies one of two
possible loads.
Load 1: P = 120 kW, FP = 1
Load 2: P = 120 kW, FP = 0.6
How much current (I) is the generator required
to supply in each case?
I
+
Load
600 V
120 kW
-
Why is Power Factor Important?

For the load with Fp = 0.6, the generator
had to supply 133 more amperes in
order to do the same work (P)!

Larger current means larger equipment
(wires, transformers, generators) which
cost more.

Larger current also means larger
transmission losses (think I2R).
Why is Power Factor Important?


Because of the wide variation in possible
current requirements due to power factor, most
large electrical equipment is rated using
apparent power (S) in volt-amperes (VA)
instead of real power (P) in watts (W).
Is it possible to change the power factor of the
load?
Power Factor Correction

In order to cancel the reactive component of
power, we must add reactance of the opposite
type. This is called power factor correction.
Power Factor Correction


In practice, almost all loads (commercial,
industrial and residential) look inductive (due
to motors, fluorescent lamp ballasts, etc.).
Hence, almost all power factor correction
consists of adding capacitance.
Example Problem 3
The 600 V, 120 KVA generator is connected to a load with
P = 120 kW and QL = 160 kVAR.
a. Determine the power factor of the load.
b. Determine the size (in VAR) of capacitive load (QC)
required to correct the power factor to unity.
Power Factor Correction


Transmission lines and generators must be
sized to handle the larger current requirements
of an unbalanced load.
Industrial customers are frequently fined by the
utility if their power factor deviates from the
prescribed value established by the utility.
Example Problem 4
a.
Determine S, PT, QT, and FP.
b.
Determine the value of the capacitance (in F) required to bring
the power factor up to unity (freq of 60 Hz).
c.
Determine generator current before and after correction.