PSERC. Another way to calculate power factor: “displacement” power factor. (3.6
ms / 16.7 ms) x 360 degrees = 77 degrees current lags voltage by 77 degrees.
Power Factor and Reactive Power Ward Jewell Wichita State University Power Systems Engineering Research Center (pserc.org)
PSERC
Energy to lift a 5 pound weight 2 feet high: 2 ft x 5 lb = 10 ft-lb = 0.0000038 kWh = 0.0033 “calories” (which are actually kcal)
Value at 10.3 cents per kWh: (average residential US price, summer 2006)
0.000039 cents PSERC
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As dragline bucket lowers, motors generate, return electricity to source
PSERC
Induction motor with no load 800
energy to motor
735.249
power (watts)
600
400
200
p ( t)
0 energy from motor
0
200
400
− 465.196 600
0 0
0.002
0.004
0.006
0.008
0.01
0.012
t
0.014
time (seconds)
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0.016
0.018 0.017
Induction motor 800 735.249
power (watts)
600
average power: 130 watts
400
200
p ( t) 0
200
400
− 465.196 600
0
0.002
0
0.004
0.006
0.008
0.01
0.012
0.014
0.016
t
0.018 0.017
time (seconds)
PSERC
Incandescent lights 350 306.8
power (watts)
300
250
average power: 150 watts
200
p ( t) 150
100
50 0 0
0 0
0.002
0.004
0.006
0.008
0.01
0.012
t
0.014
time (seconds)
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0.016
0.018 0.017
0
Incandescent Lights
PSERC
Induction motor with no load
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Lights and Motor Power
Current
Voltage
Incandescent lights
0.15 kW
1.3 A
118.0 V
Induction motor with no load
0.13 kW
5.1 A
117.7 V
PSERC
Why do the Volts and Amps matter?
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Motors and Resistance Heat: 100 MW Customer voltage
Power lost in wires
Resistance Heat
12.3 kV
1.0 MW
Motors
11.7 kV
2.3 MW
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Incandescent Lights
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Incandescent lights power: Power = 118 V x 1.3 A = 153 W = 0.15 kW = power measured by meter
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Incandescent Lights
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Induction motor with no load
PSERC
Induction motor power: 117.7 V x 5.1 A = 600 W? = 0.6 kW? NOT the power measured by meter
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Induction motor with no load
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Define some new values: Apparent power = volts x amps For the motor: 117.7 V x 5.1 A = 600 VA = 0.6 kVA VA: volt-ampere PSERC
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Define some new values: Power Factor = Average (“real”) (kW) power Apparent (kVA) power For the motor: pf = 0.13 kW / 0.60 kVA pf = 0.22
VI2 – average power2 2
2
( 0.60kVA) − ( 0.13kW) = 0.59 kVAR
0.58 kVAR
reactive power = 0.58 kVAR
reactive power =
Appa rent p ower =
Define some new values: the power triangle for the motor:
0.60 kVA
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VAR: volt-ampere reactive real power = 0.13 kW PSERC
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Induction motor with no load
PSERC
Lights and Motor Real Reactive Apparent Power Current Voltage power factor power Power Incandescent lights
0.15 kW
0 kVAR
0.15 kVA
1.0
1.3 A
118.0 V
Induction motor with no load
0.13 kW
0.58 kVAR
0.60 kVA
0.22
5.1 A
117.7 V
Note: the motor’s reactive power will stay near its no-load value of 0.58 kVAR as its load and real power (and thus apparent power and power factor) vary from no load to full load. PSERC
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Power factor and reactive power are indicators of
power losses in wires voltage drop between supply and load
PSERC
Typical Power Factors Induction motor
0.7-0.8
Resistance heat
1.0
Incandescent lights
1.0
Fluorescent lights
0.6-1.0
Battery Chargers
0.6-1.0
Computers
0.5-1.0
Variable Speed Motor Drives
0.5-1.0
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Power factor: lagging or leading?
Most loads with lower power factor are inductive. Current lags voltage. Power factor is “lagging.”
PSERC
Induction motor with no load
voltage
current
3.6 ms
Current lags voltage by about 3.6 milliseconds
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Another way to calculate power factor
16.7 ms
3.6 ms
One 60 Hz cycle = 1/60 seconds = 16.7 ms PSERC
Another way to calculate power factor: “displacement” power factor (3.6 ms / 16.7 ms) x 360 degrees = 77 degrees current lags voltage by 77 degrees cosine (77 degrees) = 0.22 power factor is 0.22 lagging pf = cos θ θ = angle between voltage and current PSERC
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Incandescent lights
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Current and voltage are “in phase.”
Incandescent lights: displacement power factor: angle between voltage and current = 0 degrees pf = cos(0 degrees) = 1.0
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true power factor: pf = 0.15 kW / 0.15 kVA pf = 1.0
Page 15
If voltage and current are sinusoidal displacement pf (DPF) = true pf (PF)
lights
motor
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Correcting (increasing) power factor
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Page 16
Capacitors to improve power factor: capacitors release energy when inductors consume 1.2 1
Capacitor current
0.5
iL( t) 0 ic ( t)
Inductor current
0.5
1 − 1.2 0
0.002
0.004
0.006
0.008
0
0.01 t
0.012
0.014
0.016
0.018 0.017
PSERC
Induction motor with power factor correction capacitor
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Induction motor with power factor correction capacitor Real Reactive Apparent power power power
Power factor
Current Voltage
Induction motor
0.13 kW
0.58 kVAR
0.60 kVA
0.22
5.1 A
117.7 V
Induction motor with capacitors
0.13 kW
0.11 kVAR
0.18 kVA
0.96
1.5 A
118.4 V
PSERC
Wire losses: motors with capacitors Customer voltage
Power lost in wires
Motors
11.7 kV
2.3 MW
Motors with power factor correction capacitor
12.3 kV
1.0 MW
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Incandescent lights with power factor correction capacitor
PSERC
Incandescent lights with power factor correction capacitor
Incandescent lights Lights with capacitors
Real Reactive Apparent power power power 0.15 0 kVAR 0.15 kVA kW 0.15 kW
0.64 kVAR
0.66 kVA
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Power factor 1.0
0.23 leading
Current Voltage 1.3 A
118.0 V
5.5 A
119.9 V
Wire losses: lights with capacitors Customer voltage
Power lost in wires
Resistance heat
12.3 kV
1.0 MW
Resistance heat with power factor correction capacitors
13.0 kV
2.0 MW
PSERC
Leading power factor Current leads voltage in a capacitor. Too much capacitance causes low leading power factor. (just as bad as low lagging power factor)
Leading power factor causes high voltage and increased wire losses. Use the correct amount of capacitance. (more is not better)
Switch capacitors off when motors are off (just put capacitor on same switch as motor) PSERC
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If voltage and current are sinusoidal displacement pf = true pf
lights
motor PSERC
If waveform is not sinusoidal: PC voltage and current
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If waveform is not sinusoidal: PC voltage and current
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Harmonic distortion
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Low power factor caused by harmonic distortion cannot be corrected by capacitors Harmonic currents are not accompanied by harmonic voltage, so average (real) power in harmonics is almost zero. pf = average power / apparent power decreases PSERC
Common harmonic loads
computers motor drives battery chargers rectifiers induction heaters arc furnaces
To correct low power factor caused by distorted current waveforms, the harmonic currents must be filtered. PSERC
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Capacitors can make harmonic distortion worse:
Lights with power factor correction capacitor
This is rare, but should be considered in the presence of harmonic loads
PSERC
Summary
Induction motors and other inductive equipment load the electric power system differently than incandescent lights and resistive heaters Power Factor and Reactive Power are indicators of power lost in wires and reduced customer voltage Low displacement power factor caused by induction motors (and other inductive loads) can be corrected with power factor correction capacitors Power factor correction capacitors must be sized properly Power factor correction capacitors cost much less than utility power factor charges and will eliminate those charges Power factor correction capacitors should be disconnected when motors are disconnected Low harmonic power factor is corrected with filters, not capacitors. Capacitors may make it worse.
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Ward Jewell 316.978.6340
[email protected] pserc.org (slides are posted under “presentations”)
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