Transcript Power Factor Correction

Lesson 22:
AC Power Factor and
Power Factor Correction
1
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.
2
Power Factor
• Power factor (FP) tells us what portion (or ratio) of
the apparent power (S) is actually real power (P).
• Power factor is a ratio given by:
P
FP  cos 
Vrms * I rms
P
FP 
S
• Power factor is expressed as a number between 0 to
1.0 (or as a percent from 0% to 100%).
− The closer to 1.0 the power factor gets, the more resistive.
− The closer to 0.0 the power factor gets, the more reactive.
3
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.
4

P
Power Factor
Leading or Lagging
• The term leading and lagging are defined in
reference to the current through the load.
− If the current leads the voltage across the load then the
load has a leading power factor.
− If the current lags the voltage across the load then the
load has a lagging power factor.
FP  cos
FP  cos
FP  cos(40  (20))
FP  cos(80  30)
FP  cos60  0.5 leading
FP  cos50  0.64 lagging
5
Unity Power Factor (FP = 1)
• Unity Power Factor 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.

Q=0
P,S
6
Lagging Power Factor ( > 0º)
• The load current lags load voltage:
• Implies that the load looks inductive.
S
Q
VARind

P
7
ELI
Leading Power Factor ( < 0º)
• The load current leads load voltage:
ICE
• Implies that the load looks capacitive.
P

Q
VARcap
S
8
Example Problem 1
a.
Determine P,Q, S and the FP for this circuit.
a.
Draw the power triangle.
b. Is it a leading or lagging power factor?
c. Is the circuit inductive or capacitive?
.
.
In Rectangular:
0.534
9
=> 53.4% real power
Example Problem 2
a. Determine total current, apparent power, and the power
factor for this circuit. Is it a leading or lagging power factor?
b. Determine total current, apparent power, and the power
factor if the capacitor reactance is decreased to 40 ohms.
What kind of power factor does it have?
or: FP 
P 293W

 0.97 lagging
S 303VA
Reducing the capacitor reactance to 40 ohms creates a unity Power Factor because Ѳ now
becomes zero for ZT, which also implies the circuit is now resistive instead of inductive.
10
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
• Determining how much current (I) is required is one
such reason...
I
+
Load
600 V
120 kW
11
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).
12
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?
The answer is yes…through power factor correction…
13
Power-Factor Correction
• In the real world, almost all loads are inductive.
• In order to cancel the reactive component of power, we must
add reactance of the opposite type.
• This is called power factor correction.
Capacitor bank in
shipboard power panel
for FP correction
14
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.
15
Power-Factor Correction
How it changes the power triangle:
FIG. 19.28 Demonstrating the impact of
power-factor correction on the power
triangle of a network.
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Power-Factor Correction
Solution Steps
1. Calculate the reactive power (Q) of the load.
2. Insert a component in parallel of the load that will cancel out
that reactive power.
e.g. If the load has QLD = 512 VAR, insert a capacitor
with QC = -512 VAR.
3. Calculate the reactance (X) that will give this value of Q
Normally the Q=V2/X formula will work.
4. Calculate the component value (F or H) required to provide
that reactance.
17
Power factor correction capacitors
for A, B, and C phases at the Crofton , MD substation
Rating: 230 kV, 360 MVAR
size comparison
Capacitor banks
18
Example Problem 3
a.
b.
Determine the value of the capacitance (in F) required to bring the
power factor up to unity (freq of 60 Hz).
Determine load current before and after correction.
a)
ZT 
1
1
1
*
60 20  j 30
IT 
 25.335.8
ES
120V 0

 4.74 A  35.8
Sign
ZT 25.335.8
Change
ST  Es * ( I )  (1200) * (4.74 A35.8)
*
T
ST  568.7VA35.8  462W  j 332VAR
b) Because the QT = 332 VAR, we can insert a
capacitor with QC = -332 VAR
PT
FP 
19
QT
P
462W

 0.81 or 81%
S 568.7VA
Example Problem 3 cont.
a.
b.
Determine the value of the capacitance (in F) required to bring the
power factor up to unity (freq of 60 Hz).
Determine load current before and after correction.
b) Because the QT = 332 VAR, we can insert a
capacitor with QC = -332 VAR
QTUnity
V2
 332VAR 
XC
V2
(120V ) 2
XC 

 43.3
| QTUnity |
332
Notice that XC ≠ XL!
Notice that XC ≠ XLD !
X CUnity 
c) Adding a unity cap changes ZT:
ZTUnity 
1
1
1
1
*

60 20  j 30  j 43.4
CUnity 
 31.20
1

2 fCUnity
1
1

 61.1 F
2 fX CUnity 2 (60 Hz )(43.3 )
NEW Current:
E
120V 0
IT  S 
 3.85 A0
ZT 31.20
OLD Current:
E
120V 0
IT  S 
 4.74 A  35.8
ZT 25.335.8
20
Notice, there is now less current needed for the load after unity
power factor correction
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.
21
Example Problem 4
a.
b.
c.
Determine S, PT, QT, and FP.
Determine the value of the capacitance (in
F) required to bring the power factor up to
unity (freq of 60 Hz).
Determine generator current before and
after correction.
a)
11738 W
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Example Problem 4 cont.
a.
b.
c.
Determine S, PT, QT, and FP.
Determine the value of the capacitance (in
F) required to bring the power factor up to
unity (freq of 60 Hz).
Determine generator current before and
after correction.
b) Since QLD = 8438 VAR, let’s insert a capacitor with QC = -8438 VAR.
c)
Old
Current
Notice, there is now less
current needed for the
generator after unity
power factor correction
New
Current
23
QUESTIONS?
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