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Typology: Exercises
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Power Factor in Electrical
Energy Management
PDH Online | PDH Center
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Power factor is the percentage of electricity that is being used to do useful work. It is defined as the ratio of ‘active or actual power’ used in the circuit measured in watts or kilowatts (W or KW), to the ‘apparent power’ expressed in volt-amperes or kilo volt-amperes (VA or KVA).
The apparent power also referred to as total power delivered by utility company has two components.
Reactive Power produces no productive work. An inductive motor with power applied and no load on its shaft should draw almost nil productive power, since no output work is being accomplished until a load is applied. The current associated with no-load motor readings is almost entirely "Reactive" Power. As a load is applied to the shaft of the motor, the "Reactive" Power requirement will change only a small amount. The ‘Productive Power’ is the power that is transferred from electrical energy to some other form of energy (i.e. such as heat energy or mechanical energy). The apparent power is always in always in excess of the productive power for inductive loads and is dependent on the type of machine in use. The working power (KW) and reactive power (KVAR) together make up apparent power, which is measured in kilovolt-amperes (KVA). Graphically it can be represented as:
Auto Parts 75- Brewery 75- Cement 80- Chemical 65- Coal Mine 65- Clothing 35- Electroplating 65- Foundry 75- Forging 70- Hospital 75- Machine Manufacturing 60- Metalworking 65- Office Building 80- Oil field Pumping 40- Paint Manufacturing 65- Plastic 75- Stamping 60- Steel Works 65- Tool, dies, jigs industry 65-
Typical uncorrected industrial power factor is 0.8. This means that a 1MVA transformer can only supply 800KW or that a consumer can only draw 80 useful Amps from a 100Amp supply. To put it the other way, a 3-phase 100KW load would draw 172A per phase instead of the 139A expected. For inherently low power factor equipment, the utility company has to generate much more current than is theoretically required. This excess current flows through generators, cables, and transformers in the same manner as the useful current. If steps are
not taken to improve the power factor of the load, all the equipment from the power station to the installation sub-circuit wiring, has to be larger than necessary. This results in increased capital expenditure and higher transmission and distribution losses throughout the whole network.
To discourage these inefficiencies the electricity companies charge for this wasted power. These charges appear on electricity bills as "reactive power charges", "KVA maximum demand" or "KVA availability charges". For instance known information taken from billing about electrical system:
KVA = 1000, KW = 800, KVAR = 600, PF =.
Typical Utility Billing Structure Examples:
I) 90% Billing Structure - Where demand billed is based on 90% of the KVA or 100% of the KW - Whichever is greater. Because the facility has a power factor of 0.80 they will pay demand rates on 90% of the KVA 1000 x .90 = 900 KVA because it is the larger number (900 KVA > 800 KW). Thus the facility is paying a penalty on 100 KVA of unproductive power. Correcting the facility’s Power Factor to 90% + will eliminate this penalty cost.
II) 100% KVA + 100% KW Billing Structure - Where one rate is applied to 100% of the KVA and another rate is applied to 100% of the KW. Both are then added together to determine the total demand charged on the bill. If we correct the power factor to unity (KVA = KW or 800 KVA = 800 KW) we can recover costs paid on 200 KVA at *KVA rates. Assuming an equal rate is being paid for KVA and KW
Rather than pay demand costs on 1000 KVA + 800 KW = 1800 if the Power Factor = Unity we will pay demand costs on 800 KVA + 800 KW = 1600. Savings = 1800 -1600 = 200. (More examples are provided later in this paper).
*Note: Generally the cost per KVA is greater than the cost for KW. Thus the savings would be greater by correcting the power factor to unity.
The reactive power charges levied as penalties in the billing should always be regulated. The excess reactive currents and associated charges can be removed by a well-established technology called "Power factor correction". Simply put, this technology offsets the inductive reactive currents by introducing equal and opposite capacitive reactive currents. Typically this can reduce electricity bills by 5-8%, with a payback period of 12 to 18 months. In addition, the consumer shall gain from improved supply availability, improve voltage and reduced power losses.
1000 800 600 400 200 0 1 0.8 0.6 0.4 0.
POWER FACTOR
KW
Loss in distribution capacity
The figure below graphically displays the variation of the I 2 R losses in feeders and branches. Losses are expressed in percent as a function of power factor.
20
10 5
1 1 0.6 0. POWER FACTOR
LOSS(%)
Larger Investment
In case of expansion, a larger investment is required in the equipment needed to increase distribution capability of the installation, such as oversized transformers and switchgears.
Transformers
For an installation which requires 800KW, the transformer should be approximately:
800KVA for power factor = 100%
1000 KVA for power factor = 80%
1600 KVA for power factor = 50%
Large size conductors
The figure 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.
For instance, in an installation where no correction has been made and which has a power factor of 70%, the cross-section of the conductor must be twice as large as it would be if the power factor were 100%.
0.5 0.6 0.7 0.8^1
0.3 0.4 0.
11.09 (^) 6.25 (^4) 2.79 (^) 2.04 (^) 1. 1.21 1
Practically speaking, when an installation uses its rated power capacity, the distribution cables within the installation are rapidly loaded to their full carrying capacity if the power factor decreases. Most often, as the need for energy in an installation expands, the first reaction is to double the distribution system although it would be less expensive to perform a correction of power factor on each load or group of loads.
Benefit 1 - Reduce Utility Power Bills In areas where a KVA demand clause or some other form of low power factor penalty is incorporated in the electric utility's power rate structure, removing system KVAR improves the power factor, reduce power bills by reducing the KVA. Most utility bills are influenced by KVAR usage.
Benefit 2 – Increase System Capacity
The power factor improvement releases system capacity and permits additional loads (motors, lighting, etc.) to be added without overloading the system. In a typical system with a 0.80 PF, only 800 KW of productive power is available out of 1000 KVA installed. By correcting the system to unity (1.0 PF), the KW = KVA. Now the corrected system will support 1000 KW, versus the 800 KW at the .80 PF uncorrected condition; an increase of 200 KW of productive power. This is achieved by adding capacitors which furnish the necessary magnetizing current for induction motors and transformers. Capacitors reduce the current drawn from the power supply; less current means lesser load on transformers and feeder
It is useful to have an idea of the value of the power factor of commonly used electrical equipment. This will give an idea as to the amount of reactive energy that the network will have to carry. Find below is the summary of power factor of commonly used electrical equipment.
Lighting
Incandescent Lamps: The power factor is equal to unity.
Fluorescent Lamps: Usually have a low power factor, for example, 50% power factor would not be unusual. They are sometimes supplied with a compensation device to correct low power factor.
Mercury Vapor Lamps: The power factor of the lamp is low; it can vary between 40% to 60%, but the lamps are often supplied with correction devices.
Distribution Transformer
The power factor varies considerably as a function of the load and the design of the transformer. A completely unloaded transformer would be very inductive and have a very low power factor.
Electrical Motors
Induction Motors: The power factor varies in accordance with the load. Unloaded or lightly loaded motors exhibit a low power factor. The variation can be 30% to 90%.
Synchronous Motors: Very good power factor when the excitation is properly adjusted. Synchronous motors can be over excited to exhibit a leading power factor and can be used to compensate a low power system.
Industrial Heating
With resistance, as in ovens or dryers, the power factor is often closed to 100%.
Welding
Electric arc welders generally have a low power factor, around 60%.
Other types of machinery or equipment those are likely to have a low power factor include:
Air Compressor & Pumps (external Motors) (^) 75-
Hermetic Motors (compressors) 50- Arc Welding 35- Resistance Welding 40- Machining 40- Arc Furnaces 75- Induction Furnaces (60Hz) 100 Standard Stamping 60- High Speed Stamping 45- Spraying 60-
From the above list, we can see that a low power factor can be a result of the design of the equipment, as in the case of welders, or it can be result from the operating conditions under which the equipment is used, as in lightly loaded induction motors which are probably the worst offenders.
Equipment Design
In an old installation, one is limited by the inefficiency of the existing system. However, given the opportunity to expand and purchase new equipment, one should consider some of the energy efficient electric motors that are available today.
Operating Conditions
Load: The power factor of an electrical motor reaches its maximum value under full load. The power factor decreases rapidly when the load decreases. The figure below symbolically illustrates the effect of the load on the power factor of a motor.
is used to measure the load current. The voltage is also measured. Now using the clamp on, power factor meter with appropriate CT, the power factor reading is noted.
The necessary data for desired power factor correction is current, line voltage and existing power factor. Now that the survey has been completed and it has been determined that power factor is a problem, the final step is to improve it. There are several approaches:
Power factor correction can be made in two ways:
When installing equipment, the following points are normally considered:
Generally the cost of rotating machinery, both synchronous and phase advancing, makes its use uneconomical, except where one is using rotating plant for a dual function – drive and
power factor correction. In addition the wear and tear inherent in all rotating machines involves additional expense for upkeep and maintenance.
Capacitors have none of these disadvantages. Compared with other forms of correction, the initial cost is very low, upkeep costs are minimal and they can be used with the same high efficiency on all sizes of installation. They are compact, reliable, highly efficient & convenient to install and lend themselves to individual, group or automatic method of correction. These facts indicate that generally speaking, power factor correction by means of capacitors is the most satisfactory and economical methods.
The static capacitor owing to its low losses, simplicity and high efficiency is now used almost universally for power factor correction.
Simply put, a capacitor is an electric device that can store electric charge for later release. Generally, capacitors are used in one of the three ways: to store and release energy, to discriminate between DC (direct current) and AC (alternating current) frequencies, and to discriminate between higher and lower AC frequencies.
A simple capacitor consists of two metal plates that are held parallel to each other with a small place between them. An insulating material called dielectric occupies the space. This insulating material can be made of many materials including oil, paper, glass, ceramics, and mica, plastic, or even air. Capacitance is a measure of the energy that a capacitor is capable of storing. The capacitance of a device is directly proportional to the surface areas of the plates and inversely proportional to the plates' separation.
Induction motors, transformers and many other electrical loads require magnetizing current (KVAR) as well as actual power (KW). By representing these components of apparent power (KVA) as the sides of a right triangle, we can determine the apparent power from the right triangle rule: KVA^2 = KW 2 + KVAR^2. To reduce the KVA required for any given load, you must shorten the line that represents the KVAR. This is precisely what capacitors do.
KW
KVA Vk A
Supply this kilovars w ith
Elim inate this KVA from KVA demandcharge
Static or fixed Power Factor correction
Compensation on the load side of the AC motor starter (motor switched or "at the load"). Fixed capacitors provide a constant amount of reactive power to an electrical system. Primarily, fixed capacitors are applied to individual motor loads, but they can also be applied to the main power bus with proper treatment. Fixed capacitors are suitable for indoor or outdoor use. Fixed capacitors are available in low voltages (832 volt and below), from. KVAR up to 400 KVAR (If more than 400 KVAR is required, smaller units are paralleled together).
Central or Bulk Power Factor correction
Central power factor compensation is applied for electrical systems with fluctuating loads. The central power factor correction is usually installed at the main power distribution. The capacitors are controlled by a microprocessor-based relay, which continuously monitors the power factor of the total current supplied to the distribution board. The relay then connects or disconnects capacitors to supply capacitance as needed in a fashion to maintain a power factor better than a preset limit (typically 0.95). Ideally, the power factor should be as close to unity as possible.
When harmonic distortion is a concern, systems are built based on the principles explained under ‘Harmonic Distortion and Power Factor Correction’ later in this paper.
The total KVAR rating of capacitors required to improve the power factor to any desired value can be calculated by using the tables published by leading power factor capacitor manufacturers.
To properly select the amount of KVAR required to correct the lagging power factor of a 3- phase motor you must follow the steps as stated. Step #1: Determine KW and Existing Power Factor. Step # 2: Existing Power Factor on Table, move across table to Desired Power Factor. The number represented is your multiplier number.
Example-
An energy audit for a facility indicates following measurements at the load side of the transformer; 480V, 1200A and 800 KW operating load.
i. What is the Power Factor? ii. How much Reactive Power (KVAR) is in the system?
Solution
i) To calculate the Power Factor, we must first calculate the KVA in the system.
Substitute the KVA into the Power Factor Formula
ii) To calculate the Reactive Power (KVAR) in the system requires re-arranging the formula
and solving for KVAR.
Example - 4
The measurement at the main distribution board of a manufacturing industry indicates 1000 KVA and 800 KW. Determine the system KVAR and PF of the facility. Determine also the KVAR required for achieving power factor of 0.95 while providing the same productive power of 800 KW?
Solution
Measured KVA = 1000
Measured KW = 800
i) System KVAR and PF of the facility
ii) System KVAR after power factor correction to.
System KVA after correction
System KVAR after correction
iii) Power capacitor KVAR rating
Power Capacitor KVAR = KVAR (uncorrected) – KVAR (corrected)
= 600 – 265 = 335 KVAR
We can use the multiplier table for capacitor selection (refer above) straight away when the KW load, uncorrected power factor and the desired power factor are known as shown in examples above.
Example - 5:
Billing based on KW Demand Charges An industrial plant has a demand of 1000 KW and operates at 80% power factor. The utility company supplying power to this unit requires minimum power factor of 85% and levies a KW demand charge of $8.00 in the electricity bill. Determine the savings possible by improving the power factor to a minimum required target of 0.85 along with the payback period of putting any investment on power factor correction.
Solution
i) The monthly KW billing is determined by the ratio of target power factor to the existing power factor times KW demand.
KW billing on power factor of 0.
The amount of monthly KW billing: 1000KW x 0.85 target PF / 0.80 existing PF = 1062 KW
Total demand charge @ $ 8.00 = 1062KW x $ 8.00 = $
ii) KVAR required to increase power factor from 0.8 to 0.
The multiplying factor = 0.13 (from the capacitor estimation table above)
Therefore KVAR required = 0.13 x 1000 = 130 KVAR
iii) Capacitor Investment
