### Power Factor Correction Capacitors Sizing Calculations – Part Three

Today, we will continue explaining the technical part for Power Factor Correction Capacitors Sizing Calculations. We will explain the following topics:

• What causes low power factor?
• Bad impacts of low power factor,
• Benefits of Power Factor correction.

 1- What causes Low Power Factor?

 There are many causes for getting a low power factor for some equipments as follows: Equipment Design, Operating Conditions.

 1.A Equipment Design A low power factor can be a result of the design of the equipment for example: Welder’s construction which needs very high reactive power to work. Traditional motors designs have a lower power factor than the energy efficient electric motors that are available today. Traditional electromagnetic ballasts have a lower power factor than the electronic ballasts

 1.B Operating Conditions  The following operating conditions will affect the value of PF: Loading, Line voltage.

The power factor of an electrical motor reaches its maximum value under full load. The power factor decreases rapidly when the load decreases. Fig.1 below illustrates the effect of the loading on a motor’s power factor and Table-1 gives values of PF for different motor loading factors.

 1.B.2 Line voltage Increasing the line voltage on motors and transformers above the rated voltage will increase the consumption of reactive energy. The result will be reduction of power factor. For example, an increase of 10% on the rated voltage can result in 20% reduction of the power factor.

 2- Bad Impacts of Low Power Factor

 There are many bad impacts of low power factor as follows: Loss in efficiency of the equipment, Loss in distribution capacity, Larger Investment for future Expansion, Increase the running costs of the installation.

 2.1 Loss in efficiency of the equipment When an installation operates with a low power factor, the amount of useful power available inside the installation at the distribution transformers is considerably reduced due to the amount of reactive energy that the transformers have to carry. In this case, we say that there is a loss in the efficiency of the transformer. Fig.2 below indicates the available actual power of distribution equipment designed to supply power to 1000 KW installation. Fig.2 So, if you have distribution equipment designed to supply a 1000 KW installation at PF=1, the same equipment will supply only 80% of the power (800 KW) for the same installation at PF=0.8. The loss in the efficiency of the distribution equipment in this case will be 20%.

 2.2 Loss in distribution capacity In general terms, as the power factor of a three phase system decreases the current rises. The heat dissipation in the system rises proportionately by a factor equivalent to the square of the current rise. Fig.3 below graphically displays the variation of the I2R losses in feeders and branches. Losses are expressed in percent as a function of power factor. Fig.3

 2.3 Larger Investment for future Expansion In case of expansion, a larger investment is required in the equipment needed to increase distribution capability of the installation, such as oversized transformers, cables and switchgears. 2.3.A Transformers For an installation which requires 800KW, the transformer should be approximately: 800KVA if power factor = 1 1000 KVA if power factor = 0.8 1600 KVA if power factor = 0.5 Low power factor means you’re not fully utilizing the electrical power you’re paying for. Another way to look at it is that at 0.5 power factor, it takes 100% more KVA to do the same work when power factor = 1. This means that with a high power factor, the effective capacity of your local electricity network potentially deferring future investment in electrical infrastructure and allowing you to connect more machinery to the same utility connection. 2.3.B Large size conductors Fig.4 below shows a variation of a cross section of a conductor as a function of the power factor for a given useful power. This illustrates that when the power factor of an installation is low, the surcharge on the electricity bill is only part of the problem. Fig.4 So, for an installation with a power factor = 0.7, the cross-section of the conductor must be twice as large as it would be if the power factor =1.

 2.4 Increase the running costs of the installation The running costs of an installation with low power factor will be increased due to the following reasons: A ‘poor power factor penalty’ from the supplier, sometimes called a ‘reactive power charge’ will be applied, Draws more power from the network to compensate for the reactive power and achieve the same tasks will increase your electricity bill, It can reduce the life expectancy of electrical equipment in extreme cases.

 3- Benefits of Power Factor correction

 There are many benefits of power factor correction like: Increase the efficiency of distribution equipment, Release more system capacity to connect additional loads, Reduced power losses, Voltage improvement Due to Reduced Line Voltage Drop, Reduction in Size of Transformers, Cables and Switchgear in New Installations and Less Investment.

3.1 Increase the efficiency of distribution equipment

If the efficiency of a distribution equipment = Active power used by equipment /equipment’s total working power ---- (equation-1)

And

Equipment’s total working power = Active power used by equipment + Reactive power produced in equipment ----- (equation-2)

So, in equation-2, if we decrease/ compensate for the reactive power by increasing the PF, the equipment’s total working power will be decreased when Active power used by equipment is fixed. Hence, from equation-1, the efficiency of the equipment will be increased.

Decreasing the total power will make good savings in the utility bill.

Example#1:

Table-2 shows 600 KW working power vs KVA required for different power factors

 Power factor 0.6 0.7 0.8 0.9 1 Active power 600 KW 600 KW 600 KW 600 KW 600 KW Reactive power 800 KVAR 612 KVAR 450 KVAR 291 KVAR 0 KVAR Total power 1000 KVA 875 KVA 750 KVA 667 KVA 600 KVA
Table-2

When power factor capacitors are located at the terminals of an inductive load, they will deliver all or most of the reactive power required by the load. This means a reduction in system current will occur, permitting additional load to be connected to the system without increasing the size of the transformers, switchboards and other distribution equipment.

Table-3 shows the increased available power at the transformer output by improvement of Power Factor from 0.7 to 1.

 Power factor Increased available power 0.7 0% 0.8 +14% 0.85 +21% 0.9 +28% 0.95 +36% 1 +43%
Table-3

Often, this release in the system capacity is reason enough to warrant an improvement in power factor. Especially when conductors or panels are overheating or where overcurrent devices frequently opened.

The percent released capacity resulting from an improvement in PF is:

% released system capacity = 100 x (1 – PFO / PFf)

Where:
PFO = original power factor before improvement,
PFf = final power factor after improvement.

Example#2:

Determine the system capacity released by improving power factor from 0.5 to 0.8.

Solution:

% released system capacity = 100 x (1 – PFO / PFf)
% released system capacity = 100 x (1 – 0.5 / 0.8) = 37.5 %

This means that the KVA load or line current is reduced by 37.5% of what it was before PF correction which means that we add 37.5% KVA capacity to the system without exceeding the total power used by the system before PF correction.

Also, we can get % released system capacity by using Laws as follows:

KVA Released = kW (1/ PFO - 1/ PFf)
% KVA Released = KVA Released / KVA Original

Example#3:

Calculate the system capacity released for above example using Laws noting that the load is 200 KW.

Solution:

From laws:
KVA Released = kW (1/ PFO - 1/ PFf) = 200 (1/0.5 – 1/0.8) = 200 x 0.75 = 150 KVA

Initial KVA = 200 KW/0.5 = 400 KVA

% released system capacity = 150/400 = 37.5% which is the same answer in above example.

Also, from the following fig.5 you can find the KVA Released factor for any power factor improvement.

Fig.5

Example#4:

From example#3, determine the %KVA Released

Solution:

KVA Released factor = 0.75
KVA Released = 200 KW x 0.75 = 150 KVA
Initial KVA = 200 KW/0.5 = 400 KVA
% released system capacity = 150/400 = 37.5% which is the same answer in above example.

Another curve (Fig.6) can be used to find the % reduction in current for any power factor improvement.

Fig.6

Example#5:

Determine the system capacity released by improving power factor from 0.7 to 0.95.

Solution:

% released system capacity = 100 x (1 – PFO / PFf)
% released system capacity = 100 x (1 – 0.7 / 0.95) = 27 %

From the curve you can get the same answer.

 3.3 Reduced power losses For the older power system networks, the kilowatt losses (I2R) can account for 2 to 5 % of the total load. Since the power losses are proportional to I2 and the current is inversely proportional to PF, so an improvement in power factor will cause a reduction in system current, a reduction in system losses and reduced power bills. This reduction can be approximated as: % loss reduction in power = 100 x (1 – PFO2 / PFf2) Where: PFO = original power factor before improvement, PFf = final power factor after improvement. From the following curve (Fig.7) you can find the % reduction in losses for any power factor improvement. Fig.7 Example#6: Improve power factor from 70 percent to 95 percent, calculate the reduction in power losses. Solution: % loss reduction in power = 100 x (1 – PFO2 / PFf2) % loss reduction in power = 100 x (1 –0.72 / 0.952) = 46% From the above curve, you can get the same solution. Also, you can calculate the reduction in power losses at transformers from the following curve (Fig.8): Fig.8 Where: 3= transformer with normal losses 2= transformer with reduced losses 1= transformer with low losses In any transformer, there are two types of losses: Iron (or core losses) power dissipated in the transformer under no load conditions, Copper (or winding losses). Example#7: 500 kVA Transformer with Actual load 300 kW, if we correct PF from 0.7 to 0.95, calculate the % reduction in power losses. Solution: Situation before power factor correction: Iron losses: 1150 kW (independent of PF) from above curve Copper losses = 6000* (actual load/rated load)² = 6000* (300/0.7 / 500)² = 4410 W Situation after power factor correction to 0.9: Iron losses = 1150 kW (independent of PF) from above curve Copper losses = 6000* (300/0.95 / 500)² = 2393.35 W Reduction in Copper losses = 4410 – 2393.35 = 2016.65 W %Reduction in power losses = 2016.65 / 4410 =45.7% The same result we got it from above example#6.

 3.4 Voltage improvement Due to Reduced Line Voltage Drop When capacitors are added to the power system, the voltage level will increase. The percent voltage rise associated with an improvement in PF can be approximated as: % voltage rise = (capacitor KVAR x Transformer % IZ) / Transformer KVA Where: Transformer % IZ = Transformer impedance % from nameplate. From the following curve (fig.9) you can find the % voltage drop for any power factor improvement. Fig.9 Example#8: Improve power factor from 60 percent to 90 percent, calculate the reduction in voltage drop. Solution: From the above curve, at PF = 0.6 the, Voltage drop = 5.1% And at PF = 0.9, the voltage drop = 3.6% So, the reduction in voltage drop = 5.1-3.6 = 1.5% Notes: This benefit will result in more efficient performance of motors and other electrical equipment. Under normal operating conditions, the percent voltage rise will be only a few percent. Therefore, voltage improvement shouldn’t be regarded as a primary consideration for PF correction.

 3.5 Reduction in Size of Transformers, Cables and Switchgear in New Installations and Less Investment Installing power factor correction equipment allows the conductors cross-section to be reduced, since less current is absorbed by the compensated installation for the same active power. Fig-10 below represents the increasing size of conductors required to carry the same 100 KW at various power factors. Fig-10 So, for an installation with a 100 KW load, at a power factor = 1, the cross-section of the conductor is half size as it would be if the power factor =0.7. This will lead to less Investment for new installations.

 Finally, Fig-11 below summarizes the benefits of power factor correction. Fig-10

In the next article, we will continue explaining the technical part for Power Factor Correction Capacitors Sizing Calculations. Please, keep following.

The previous and related articles are listed in below table:

 Subject Of Previous Article Article Glossary of Power Factor Correction Capacitors Types of Loads, The Power Triangle, What is a power factor? Types of power factor Why utilities charge a power factor penalty? Billing Structure.

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