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Overview
• Power Factor Definition
• Leading and Lagging Power Factor
• Power Factor Correction
04-Power Factor
ECEGR 450
Electromechanical Energy Conversion
2
Dr. Louie
Questions
Power Factor
• What is power factor?
• Recall Power Factor:
PF
• What are the real-world consequences of low
power factor?
cos
v
i
• Power Factor gives an indication of how much
apparent power S is used for real work, P
cos
P
P2
Q2
P
|S|
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Power Factor
• Power factor is non-negative
• cos( ) = cos(– )
• Need to distinguish between
S
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Power Factor
• For example let
• Case 1: i = 30o
v
= 0o
 Capacitive circuit
 PF = 0.866
and –
• Case 2:
Q
i
= -30o
Same power factor
 Inductive circuit
 PF = 0.866
P
-Q
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Leading/Lagging Power Factor
Identify as Leading, Lagging or Unity
Must describe the PF value along with whether
the current leads or lags voltage
(a)
(b)
(c)
 Lagging: current lags voltage (inductive)
 Leading: current leads voltage (capacitive)
 Useful mnemonic: ELI the ICE man
V
1
0o
V
0o
I
1
I
0.5
0.5
2
(b)
V
0
V
I
0.5
0
V
o
1
V
120 0 V
I
11.22
0o
0.5
30o
2
leading
I
2
unity
30o
2
20.7 A
0
P | S | cos
Q | S | sin
o
2
2
2
0.5
• For this circuit
(c)
S (120 0 )(11.22 20.7 )
| S | 1346VA
1
I
Power Factor Example
• Then
o
30o
2
Identify as Leading, Lagging or Unity
1
0o
2
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(a)
1
V
0o
2
2
I
0.5
30o
2
lagging
1346 20.7
1346 cos(20.7 )
0.476 kVAR
I
1.26 kW
L = 0.01
R=10
Vs =120
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Power Factor Example
Leading/Lagging Power Factor
• Note: current lags the voltage (by 20.7 degrees)
V
120 0 V
I
11.22
10
• Inductive circuits have a lagging power factor
 Consumes reactive power
 Q is positive
20.7 A
• Capacitive circuits have a leading power factor
• Power factor is lagging
• PF = cos(20.7o) = 0.94 (lagging)
I
 Supplies reactive power
 Q is negative
L = 0.01
R=10
Vs =120
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Power Factor Correction
Power Factor Correction
• Low power factor requires more current to do the
same amount of work
• Higher current increases losses in transmission of
the electricity (|I|2R)
• Consider a load that consumes 1kW of power at a
power factor of 1.0 and operates at 115 V (rms).
The load is connected to a voltage source by a
wire with a resistance of 0.1.
• We want to know the power supplied by the
source
Rw =0.1
Vs
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Sload
Sload
I
1000
Rw =0.1
0j
VloadI
1000 0
115 0
8.695 0A
115 0
0.1(8.695 0)
Vs
I
I
Load
115.8695 0 V
• Now find Ps
Ps | 115.8695 || 8.695 | cos(0)
1007.5 W
• Load consumed 1kW, source supplied 1.0075 kW
• 7.5W of loss, caused by the power consumed by
the wire resistance
• By conservation of energy, we could have
computed the power loss in the wire and added it
to the power consumed in the load to arrive at
1.0075 kW
PLoss | I |2 R w
7.5 W
Ps
1007.5 W
1000
7.5
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Power Factor Correction
• Consider a load that consumes 1kW of power at a
power factor of 0.8 lagging and operates at 115 V
(rms). The load is connected to a voltage source
by a wire with a resistance of 0.1.
• What is the power supplied by the source?
I
• First compute I
cos 1(0.8)
| Sload | 0.8
Load
Dr. Louie
Pload
1000
Sload
1250 36.87 VA
Sload
VloadI
I
17
Rw =0.1
36.87
Vs
| Sload | 1250VA
I
Vs
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Power Factor Correction
Rw =0.1
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Power Factor Correction
• Now find Vs
Vs
Load
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Power Factor Correction
• First compute I
I
I
Load
1250 36.87
10.869 36.87 A
115 0
10.869
36.87 A
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Power Factor Correction
Power Factor
Completing the calculation
I
10.869
36.87 A
0.1(8.695
8.695
j6.521)
j6.521 A
Vs
115
Ps
(115.87)(10.869) cos(36.55 )
115.8695
j0.652
115.87
0.32 V
1011.7 W
Losses increased to 11.7W
To operate efficiently, unity PF (PF=1) is desired
Typically, PF ranges from 0.8 to 0.95
Most loads are inductive (reactive power
consumers)
• To increase PF, capacitors (reactive power
suppliers) can be placed near the loads
•
•
•
•
 Known as power factor correction
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Power Factor Correction Example
Power Factor Correction Example
A load draws 10 A of current at a power factor of
0.5 lagging. What size capacitor should be added
to the load in parallel to increase the power factor
to 0.8 lagging? Assume the voltage at the load is
120 V at 60 Hz.
• First compute S at the load
cos 1(0.5)
S
VI*
60
120 0
(10
60 )
1200 60 VA
Lagging ( is positive), so
i
is negative
• What are P and Q?
I
Find the size of Capacitor
Vs
S
Load
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Power Factor Correction Example
1200 60
P
jQ
600
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Power Factor Correction Example
• What are P and Q?
S
Q
P
• We need to reduce the amount of Q consumed in
order to make the PF equal to 0.8
• The real power will remain at 600 W (since the
capacitor is placed in parallel)
• At a PF of 0.8:
j1039.2 VA
cos 1(0.8) 36.9 (target )
tan 1(Q )
P
Q
tan
P
Q P tan
450.5 VAR (target Q)
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Power Factor Correction Example
Summary
• We then need to find a capacitor that supplies
1039.2-450.5 = 588.7 VAR when placed in
parallel with the load
Q
| V |2
jXc
| V |2
jXc
j24.44
Q
1
C
108.53 F
2 fXC
•
Power factor does not distinguish between inductive and
capacitive circuits
•
ELI the ICE man
•
Low power factor increases loss in the system, and can
mitigated by adding inductors/capacitors to the load
• See text Example 1.3 for another way of solving
this problem
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