As we have already seen in the basics of power factor in electrical distribution system, most Industrial loads require both Real power and Reactive power to produce useful work. Typically, inductive loads (motors, transformers etc) require two kinds of power to operate:
(1) Active Power (KW) � This actually performs the useful work
(2) Reactive Power(KVAR) � This helps to maintain the electromagnetic field.
The vector sum of the active power and the reactive power gives total power often referred to as apparent power in KVA:
KVA = KW + KVA (vector sum)
Low power factor in an electrical system often occur when inductive loads are operated below their full load capacity especially motors. Consistently operating electrical loads at low power factor will result in higher utility bills because of the poor utilization of electrical energy. In fact, a higher power factor means less KVA and KVAR components and a more efficient utilization of electrical energy while a low power factor implies the presence of more KVA and reactive (KVAR) power components and less efficient electrical energy utilization:
KVA = KW + KVA (vector sum)
Low power factor in an electrical system often occur when inductive loads are operated below their full load capacity especially motors. Consistently operating electrical loads at low power factor will result in higher utility bills because of the poor utilization of electrical energy. In fact, a higher power factor means less KVA and KVAR components and a more efficient utilization of electrical energy while a low power factor implies the presence of more KVA and reactive (KVAR) power components and less efficient electrical energy utilization:
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| Power Factor Concept |
Power Factor Formula
We have already defined power factor in the basics of power factor in electrical distribution system. Here is the formula anyway:
P.F = KW/KVA
From the power factor triangle, we see that:
KVA2 = KW2 + KVAR2
Using this formula you can easily calculate the KVA and KVAR component of an electrical system given the power factor (P.F) and the KW component.
Types of Electrical Loads and The Power Type They Consume
The various types of electrical loads and the type of power they consume are summarized in the table below:
Types of Power | Common Names | Typical Load |
KW - kilowatt (produces active current) | Active power Kilowatt/Watt power Real power Resistive power | Resistors Incandescent lights Toasters |
KVAR - Kilovolt Amperes Reactive (produces reactive current) | Reactive power Imaginary power KVAR/VAR power Inductive/Capacitive power | Reactors/Inductors Capacitors |
KVA - Kilovolt Amperes (KVA =vector sum of Active and Reactive power) | Apparent power Complex power Total power KVA/VA power | All industrial loads: Motors Variable Speed Drive Lighting loads Welders |
The reactive component (KVAR) of any electrical distribution system can easily be reduced in order to improve power factor by using capacitors. Capacitors are basically reactive loads. They tend to generate reactive power hence they find good use in power factor correction application. So instead of having the utility company supply the reactive power that you will end up paying for, get a capacitor bank and have them supply the reactive energy component as shown below:
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| Power Factor Improvement with Capacitors |
As seen from the diagram above, at a power factor of 0.7, the KVA requirement for the loads is at 142KVA while the reactive power required is 100KVAR. With power factor improvement capacitors installed and the power factor improved to 0.95, the KVA requirement drops to 105KVA while the reactive required is now at 33KVAR, the balance of 67KVAR is now being supplied by the capacitor with significant impact on utility bills.
Benefits of Improving Power Factor with Capacitors
When capacitors are used to improve power factor , the following benefits will accrue:
1. Reduced electrical power bills
2. Reduces I2R losses in electrical conductors
3. Reduces loading on transformers by releasing system capacity
4. Improves voltage on the electrical distribution system thereby allowing motors to run more efficiently and cooler. This helps to prolong the operation and life to the motor.
Capacitors Reduce Utility Bills
As detailed below, improving power factor with capacitors will have significant impact on utility bills over time as shown by a breakdown of the utility billing system in the table below:
Billing Type | Billing Concept | How Capacitors Reduce Cost |
KVA | Utility companies bill for every amp of current both active and reactive. Bill is typically based on peak current | Capacitors reduce reactive current and therefore peak current |
KW demand with power factor adjustment | Utility companies bill for KW demand plus a surcharge for low power factor. For example, you could pay for any power factor below 0.85 | Capacitors increase power factor to the minimum required eliminating surcharge. Sometimes you could get credit for high power factor. |
KW demand with reactive demand charge | Utility companies bill for KW demand plus a surcharge for excessive reactive demand | Capacitors reduce reactive demand thereby eliminating surcharge |
Power Factor Sample Calculations
Sample Problem:
A Factory is operating several induction motors plus other loads. Its power factor is 0.65 and it consumes an average of 195KW for a given billing period. Calculate:
(a) The KVA consumption
(b) The reactive power, KVAR
(c) If it is required to improve power factor to 0.95, what size of capacitor in KVAR is required?
(d) Suppose the utility company supplying this factory electricity has the following billing regime:
1. Energy Rate = $3.6 per KWH
2. Demand Charge = $1.9 per KW
3. Power Factor penalty = $0.18 per KVARH
Calculate the total energy bill for the month and the savings when power factor improvement was made. Assume a 30 day billing period.
Sample Solution:
(a) P.F = KW/KVA;
KVA = KW/P.F; now KW = 195, P.F = 0.65
Therefore KVA = 195/0.65 = 300KVA
(b) Now KVA2 = KW2 + KVAR2 ;
KVAR = SQRT[(300*300) � (195*195)] = 227.98
(c) At P.F = 0.95, KVA = 195/0.95 = 205.26
KVAR = SQRT[(205.26*205.26) � (195*195)] = 64.09
KVAR of capacitors required for power factor correction is
= 227.98 � 64.09 = 163.89
(d) Total hours per month = 30 x 24 = 720
With power factor of 0.65:
Energy charge per month = 195 x 720 x $3.6 = $505,440
Demand charge per month = 195 x $1.9 = $370.5
Power factor penalty charge per month = 227.98 x 720 x $0.18 = $29,546.2
Total bill for the month = $505,440 + $370.5 + $29,546.2 = $535,356.7
Now with power factor now improved to 0.95:
Energy charge per month is the same = $505,440
Demand charge is the same = $370.5
Power factor penalty charge per month = 64.09 x 720 x $0.18 = $8,306.06
Total bill for the month = $505,440 + $370.5 + $8,306.06 = $514,116.56
Savings in utility bills = $535,356.7 - $514,116.56 = $21,240.14
Mr. Muhammad Saqib, M.Sc. (PAK), Research Fellow (PAK), a member of IEEE, is a Fulbright Alumnus and earned his Master’s Degree in Electrical and Power Engineering from University Of Engineering And Technology Taxila, PAKISTAN.
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