A specific example of the effect of powerfactor correction in reducing electricity consumption.
Why is Power Factor important? The phase relationship between the acpowerline voltage and current is related by a term called power factor. For a purely resistive load on the power line, voltage and current are in phase and the power factor is 1.0.
However, when an ac motor is placed on the power line, the voltagecurrent phase difference increases and power factor decreases because of the motor's inductive load. When this occurs it reduces motor/transmissionline system efficiency and generates harmonics that can cause problems for other systems connected to the same power line.
Powerfactor correction modifies the relationship between powerline voltage and current by making them closer to being in phase. This improves the power factor of the transmission line/motor as a system, and possibly reduces the harmonics and improves the system's efficiency.
BREAKING DOWN THE EXAMPLE ANALYSIS
For convenience, this analysis assumes the use of a 1hp motor. Many applications using a lowhorsepower electric motor will be fed by a #12gauge cable and protected at the load center (main panel) by a 20A circuit breaker. For this analysis, the average twoconductor cable lengths from the load center to an appliance containing an electric motor is assumed to be between 25 ft to 50 ft from the main panel to the appliance, for a total length of 50 ft to 100 ft. Also, we're assuming that motordriven appliances in the home may have a 1/3 to 1hp motor with an 85% efficiency and a lagging power factor of 0.75. The cost of electricity in the U.S. may be anywhere from $0.10/kWh and higher. For this example, the present (May 2009) cost of electricity in the Annapolis, MD area ($0.118/kWh) was used.
Powerfactor analysis of the power delivered to a singlephase, 1hp electric motor fed by a 120V electric circuit requires a knowledge of motor and cable characteristics. The feed line is assumed to be a 50ftlong, #12gauge Romex Cable.
The first task is to determine the resistance of 100 feet of cable (resistance of both the hot and neutral wires). The resistance of #12gauge wire is 1.588 Ω/1,000 feet, so:
Rcable = 1.588 Ω/1,000 ft × 100 feet = 0.1588 Ω
The electrical equivalent of an electric motor can be symbolized as an inductive reactance in series with a resistance. The inductive reactance is due to the stator inductance and reflected inductance of the rotor.
The resistance is caused by wire resistance (both stator and reflected resistance of the rotor) combined with losses due to hysteresis and eddy currents, mechanical resistances such as bearing losses, and windage.
Also, the mechanical load on the motor may be depicted as a resistance loss. If the electrical parameters depicted in Fig. 1 are assumed, a reasonable electrical approximation of a 1hp motor will be obtained.
Therefore:
(See Equations)
pf_{MOTOR} = 0.8
Then:
(See Eq. 2)
(See Eq. 3)
pf_{SYSTEM} = 0.8056
Due to cable resistance, the full 120 V would not be applied to the motor. By the voltage divider rule:
(See Eq. 4)
The power delivered to the system is:
(See Eq. 5)
The power delivered to the motor is:
(See Eq. 6)
Assuming a 75% motor efficiency (for a 1hp motor, 75% to 85% efficiency is common):
(See Eq. 7)
(See Eq. 8)
Applying a powerfactorcorrection capacitor to the system, note that the powerfactorcorrection capacitor is connected to the motor input terminals. It does not affect the power factor of the motor.
It only corrects the power factor that the cable plus load and capacitor pair presents to the source. Changing the series impedance of this motor to an equivalent parallel impedance:
(See Eq. 9)
Or:
(See Eq. 10)
Or:
(See Eq. 11)
Then:
By selecting a capacitor of capacitive reactance equal to 16.6667 Ω, we have the circuit in Fig. 2.
Performing the above computations but with the system load only represented by a resistance:
(See Eq. 12)
Then:
(See Eq. 13)
Then:
Calculating the value of the electrical current feeding the transmission line where XCP cancels XLP:
(See Eq. 14)
pfSYSTEM = 1.0
Due to the cable resistance, the full 120 V would not be applied to the motor. By the voltage divider rule:
(See Eq. 15)
The power delivered to the system is:
(See Eq. 16)
The power delivered to the motor is:
(See Eq. 17)
Assuming a 75% motor efficiency (for a 1hp motor, 75 to 85% efficiency is common):
(See Eq. 18)
(See Eq. 19)
The powerfactorcorrection capacitor only affects the transmissionline losses (the pf of the motor is an inherent characteristic of the motor), so we can compare the savings accrued with the addition of the pf correction capacitor.
Without capacitor:
(See Eq. 20)
With capacitor:
(See Eq. 21)
Thus, the savings due to the pf device is P_{SAVING} = 22.3 W 14.27 W = 8 W.
If a cost assessed by the electrical utility is $0.118/kWH, then the savings per hour is:
(See Eq. 22)
(See Eq. 23)
The required capacitor would be:
(See Eq. 24)
C = 159 µF
The difference in power delivered to the system is 8 W, and the difference in power dissipated in the cable is 8 W. Therefore, the system power savings resides in reducing cable losses
Although an AWG #14gauge cable would have been within the limits prescribed by the National Electrical Code (15A rated), an AWG #12gauge wire was selected to minimize losses (20A rated).
As an afterthefact observation, in an actual application it would have been advisable to consider the use of a 240V rather than 120V motor. This would have reduced line currents by nearly onehalf, and would have allowed the use of an AWG #14gauge cable.
Note: If you want a copy of the program for power factor of a motor, please send a selfaddressed, stamped ($0.70) and an approximately 6 × 8in. envelope to William Rynone, Rynone Engineering, Inc., P.O. Box 4445, Annapolis, MD 21403.
REFERENCES

Introductory Circuit Analysis, Robert Boylestad