8.4 CAUSES OF LOW POWER FACTOR

Circuits containing purely resistive loads such as filament lamps, strip heaters, cooking stoves, etc., operate with a power factor of unity. When the power factor is 1, all the energy supplied by the source is consumed by the load. Circuits containing inductive or capacitive elements usually have a power factor below 1. When the power factor is equal to 0, the energy flow is entirely reactive, and stored energy in the load returns to the source at the end of each cycle. Good power factor is considered to be greater than 0.85 or 85%.

1. Most A.C motors (any type of wound coil) are of the induction type i.e., single-phase or three-phase induction motors generate reactive power with the current waveform lagging the voltage, and consequently operating at a poor power factor.
2. Lighting loads such as arc lamps and electric discharge lamps operate at a low lagging power factor.
3. Industrial heating furnaces such as arc furnaces usually operate on the principle of striking an arc, and operate at a low lagging power factor.
4. Normal power factor (NPF) ballasts typically have a value of (0.4) – (0.6). Ballasts with a power factor greater than (0.9) are considered to be high power factor ballasts (HPF).

8.4.1 Disadvantages of Low Power Factor

Power factor plays a vital role in A.C circuits since the power consumed depends upon the operating power factor.

It is evident that Real power P = V ICosϕ

From the above expression, for a fixed power and voltage the load current is inversely proportional to the power factor. Lower the power factor, higher will be the load current and vice versa.

Low power factor results in the following disadvantages:

1.    Increase in copper losses

When a load has a power factor lower than 1, more current is required to deliver the same amount of useful energy. Copper losses (I²R losses) increase with increase in current. This results in poor efficiency with increase in losses.

2.    Large conductor size

To transmit a given power at a constant voltage, the conductor will have to carry relatively more current at a lower power factor. This obliges greater conductor size i.e., the cross-sectional area of transmission lines, cables and motor conductors has to be designed based on the increased current and entails more cost. This is illustrated in Example 8.1 for better understanding.

3.    Large KVA rating

Electric machinery like alternators, transformers, motors and switchgear are always rated in KVA. And KVA = KW / Cos ϕ. From this relation it is clear that KVA rating of the equipment is inversely proportional to the power factor. The lower the power factor, the larger will be the KVA rating. Therefore at low power factor, the KVA rating of the equipment should be made more, making the equipment size larger and expensive.

4.    Poor voltage regulation

A large current with a low lagging power factor results in greater voltage drops (IZ) in power system components like alternators, transformers, transmission and distribution lines. This increased voltage drop results in poor voltage regulation and decreased voltage at the receiving end, and impairs the performance of consumer loads. In order to maintain receiving end or consumer end voltages within permissible limits, additional equipment like voltage regulators, booster transformers and on-load tap changing transformers are required, which are expensive in nature.

5.    Reduced handling capacity

Since an increase in the reactive component of the current prevents full utilization of installed capacity, the lagging power factor reduces the handling capacity of all the elements of the system.

6.    Increase generation and transmission costs

The significance of the power factor lies in the fact that utility companies supply customers with Volt-Amperes, but bill them for Watts. Power factors below 1.0 require a utility to generate more than the minimum Volt-Amperes necessary to supply the working power i.e., the real power (Watts). This increases generation and transmission costs. Utilities may charge additional costs or penalize the customers (not domestic loads but large consumers like industries) who have a power factor below some limit. This case is illustrated in Example 8.2.

Example 8.1

Consider a 10 KW, single phase A.C motor having an efficiency of 90 percent, operating at a terminal voltage of 230 V. Calculate the current drawn by the motor when (i) it is operating at a power factor 1 (unity p.f) and (ii) it is operating at a 0.7 lagging power factor.

Solution:

Given that

Output of the motor = 10000 W

Efficiency of the motor = 90 percent = 0.9

Inputofthemotor =

KVA input of the motor at unity power factor = 11111.11/1 = 11111.11 KVA.

Input full load current drawn by the motor operating at unity power factor is

KVA input of the motor at 0.7 power factor = 11111.11 / 0.7 = 15873.01 KVA.

Input full load current drawn by the motor operating at 0.7 power factor is

Therefore, the motor working at unity power factor is drawing a current of 48.31 A while that which works with a lower power factor of 0.7 is drawing a current of 69.01 A, which was approximately 20 A more.

Example 8.2

An alternator is supplying a load of 500 KW at unity power factor. If the same machine is operated at a power factor 0.7 lagging, calculate the KW capacity of the supplying load.

Solution:

Given that the alternator is supplying 500 KW at a power factor = 1

Therefore its KVA rating at unity power factor is,

Now, the same machine is operated at a p.f of 0.7 lag,

Therefore we can conclude that,

A 500 KVA alternator working at unity power factor can supply a load of 500 KW and the same alternator working at a power factor 0.7 can supply a load of 350 KW only. Decrease in KW power supplied by the alternator = 500 – 350 = 150 kW.

Example 8.3

Assume a 4 KV single-phase circuit, which feeds a load of 450 kW and operates at a lagging power factor of 0.75. If it is desired to improve the power factor, determine the following:

• The reactive power consumption.
• The new corrected power factor after installing a shunt capacitor bank with a rating of 400 KVAR

Solution:

Before installing the shunt capacitors for power factor correction,

The current drawn by the load can be calculated by

1. Reactive power consumption can be calculated by

       Q₁ = S₁* Sin ϕ₁    [Cos ϕ₁ = 0.70, therefore ϕ₁ = 45.57⁰]

           = 642.85 × Sin(45.57°)

           ≅ 459 KVAR

   After power factor correction by installing a 450 KVAR capacitor bank

    

   Q₂ = Q₁ − Q_c

    

           = 459 − 400

    

         = 59 KVAR

    

2. Therefore the new corrected power factor can be calculated by

    

           Cos ϕ₂ = P /[P² + (Q₁ − Q_C)²]^1/2

    

               = 450/[450² + 59²]^1/2

    

   = 450/453.85

    

                       = 0.9915 or 99.15 percent.