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Capacitive compensation for power–factor control Different types of power capacitors shunt and series capacitors Effect of shunt capacitors (Fixed and switched) Power factor correction Capacitor allocation Economic justification Procedure to determine the best capacitor location.

Dr. Rohit BabuSuivre

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THE POWER FACTOR amitmaholiya1

- LENDI INSTITUTE OF ENGINEERING AND TECHNOLOGY Jonnada, Andhra Pradesh- 535005 UNIT–V: Compensation for Power Factor Improvement Department of Electrical and Electronics Engineering
- Syllabus Department of Electrical and Electronics Engineering Capacitive compensation for power–factor control Different types of power capacitors shunt and series capacitors Effect of shunt capacitors (Fixed and switched) Power factor correction Capacitor allocation Economic justification Procedure to determine the best capacitor location.
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering Power Factor Correction (Power Factor Compensation) uses parallel connected capacitors to oppose the effects of inductive elements and reduce the phase shift between the voltage and current Power Factor Correction is a technique which uses capacitors to reduce the reactive power component of an AC circuit in order to improve its efficiency and reduce current. When dealing with direct current (DC) circuits, the power dissipated by the connected load is simply calculated as the product of the DC voltage times the DC current, that is V*I, given in watts (W). For a fixed resistive load, current is proportional to the applied voltage so the electrical power dissipated by the resistive load will be linear. For an AC circuit, the power dissipated in watts at any instant in time is equal to the product of the volts and amperes at that exact same instant.
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering In a DC circuit the average power is simply V*I, but the average power of an AC circuit is not the same value as many AC loads have inductive elements, such as coils, windings, transformers, etc. where the current is out of phase with the voltage by some degrees resulting in the actual power dissipated in watts being less than the product of the voltage and current. For an AC circuit containing an inductor, coil, or solenoid or some other form of inductive load, its inductive reactance (XL) creates a phase angle with the current lagging behind the voltage by 90o.
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering RL Series Circuit R and L Vector Diagrams
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering Phasor Diagram for the Series RL Circuit
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering Power Factor Correction Example No1 An RL series circuit consists of a resistance of 15Ω and an inductor which has an inductive reactance of 26Ω. If a current of 5 amperes flows around the circuit, calculate: 1) the supply voltage. 2) the phase angle between the supply voltage and circuit current. 3) Draw the resulting phasor diagram. Solution: 1). The supply voltage VS
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering We can double check this answer of 150Vrms using the impedances of the circuit as follows: 2). The phase angle (phi) using Trigonometry functions is:
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering 3). The resulting phasor diagram showing VS
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering Power in a RL Series Circuit The power consumed by the resistor in watts is called the “real power” so is given the symbol “P” (or W). Watts can also be calculated as I2R, where R is the total resistance of the circuit. Real Power, P = V*I cos(Θ) Inductive Power Triangle
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering Capacitive Power Triangle
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering Power Triangle Equations
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering Power Factor Correction Example No. 2 A coil has a resistance of 10Ω and an inductance of 46mH. If it draws a current of 5 Amperes when connected to a 100Vrms, 60Hz supply, calculate: 1) the voltages across the components. 2) the phase angle of the circuit. 3) the different powers consumed by the series RL circuit.
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering First find the impedances 1). The voltages across the resistor, VR and inductor, VL
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering 2). The phase angle of the circuit 3). The circuit power
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering Power Factor Correction • Power Factor Correction (Power Factor Compensation) improves the phase angle between the supply voltage and current while the real power consumption in watts remains the same. • Adding an impedance in the form of capacitive reactance in parallel with the coil above will decrease Θ and thus increases the power factor which in turn reduces the circuits rms current drawn from the supply. • The power factor of an AC circuit can vary from between 0 and 1 depending on the strength of the inductive load but in reality it can never be less than about 0.2 for the heaviest of inductive loads. • Adding a capacitor in parallel with the coil will not only reduce this unwanted reactive power, but will also reduce the total amount of current taken from the source supply.
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering A power factor of 0.95 is equal to a phase angle of: cos(0.95) = 18.2o thus the amount of VAR required is: If the original uncorrected VAR value was 433VAR and the new calculated value is 82.2VAR, we need a reduction of 433 – 82.2 = 350.8 VAR(capacitive). Therefore:
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering Therefore the capacitance of the capacitor is calculated as: So to improve the power factor of the coil in example no2 from 0.5 to 0.95 requires a parallel connected capacitor of 93uF. New Volt-Amperes Value
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering Power Triangle If the circuits apparent power has been reduced from 500VA to just 263VA, we can calculate the rms current supplied as: S = V*I, therefore: I = S/V = 263/100 = 2.63 Amperes
- Capacitive compensation for power–factor control Department of Electrical and Electronics Engineering Final Power Factor Correction Circuit This has the effect of reducing the circuits power factor, that is the ratio of active power to the apparent power, as well as improving the power quality of the circuit and reduces the amount of source current required. This technique is called “Power Factor Correction”.
- ADVANTAGES AND BENEFITS OF POWER FACTOR IMPROVEMENT Department of Electrical and Electronics Engineering (i) The cost and bill for the energy is reduced since there is penalty for poor power factor (<0.85) (ii) Voltage improvement and better regulation (iii) Overall system losses becomes reduced and hence is important from conservation of energy Table 1: Typical power factor values of industrial loads and plants
- Different types of power capacitors Department of Electrical and Electronics Engineering • A power capacitor is a highly technical and complex device in that very thin dielectric materials and high electric stresses are involved, coupled with highly sophisticated processing techniques. • In the past, most power capacitors were constructed with two sheets of pure aluminum foil separated by three or more layers of chemically impregnated kraft paper. • Capacitor sizes have increased from the 15–25 kvar range to the 200–300 kvar range (capacitor banks are usually supplied in sizes ranging from 300 to 1800 kvar). • Capacitors are getting more attention today than ever before, partly due to a new dimension added in the analysis: changeout economics. • Under certain circumstances, even replacement of older capacitors can be justified on the basis of lower-loss evaluations of the modern capacitor design.
- Different types of power capacitors Department of Electrical and Electronics Engineering Power capacitors are passive electronic components that provide a static source of reactive power in electrical distribution systems. Capacitance, a measure of energy storage ability, is typically expressed as C = K A/D, where A is the area of the electrodes, D is their separation, and K is a function of the dielectric between the electrodes. Power capacitors are used in: •the aerospace and automotive industries •power factor correction and lighting circuits •power supplies •motor starters
- Different types of power capacitors Department of Electrical and Electronics Engineering 1. Aluminum Electrolytic Capacitors: Aluminum electrolytic capacitors use an electrolytic process to form the dielectric. Wet electrolytic capacitors have a moist electrolyte. Dry or solid electrolytic capacitors do not. 2. Ceramic Capacitors: Ceramic capacitors have a dielectric made of ceramic materials. 3. Chip Capacitors: Chip capacitors or surface mount capacitors do not have leads. 4. Film Capacitors: Film capacitors are insulated with polyester, polycarbonate, polypropylene, polystyrene, or other dielectric materials. 5. High Voltage Capacitors: High voltage capacitors are used for storing charge and energy in high voltage applications. 6. Supercapacitors: Ultracapacitors store charges (energy) by physically separating positive and negative charges (unlike batteries which do so chemically). Very high power densities can be achieved by this method.
- Different types of power capacitors Department of Electrical and Electronics Engineering 7. Tantalum Capacitors: Tantalum capacitors are used in smaller electronic devices including portable telephones, pagers, personal computers, and automotive electronics.
- Shunt and Series capacitors Department of Electrical and Electronics Engineering 1. Series Capacitors Series capacitors, that is, capacitors connected in series with lines, have been used to a very limited extent on distribution circuits due to being a more specialized type of apparatus with a limited range of application. Fig. 1: Voltage phasor diagrams for a feeder circuit of lagging power factor: (a) and (c) without and (b) and (d) with series capacitors. • A series capacitor is a negative (capacitive) reactance in series with the circuit’s positive (inductive) reactance with the effect of compensating for part or all of it.
- Shunt and Series capacitors Department of Electrical and Electronics Engineering • A series capacitor can even be considered as a voltage regulator that provides for a voltage boost that is proportional to the magnitude and power factor of the through current. • A series capacitor produces more net voltage rise than a shunt capacitor at lower power factors, which creates more voltage drop.
- Shunt and Series capacitors Department of Electrical and Electronics Engineering Fig. 1: Voltage phasor diagrams for a feeder circuit of lagging power factor: (a) and (c) without and (b) and (d) with series capacitors. Consider the feeder circuit and its voltage phasor diagram as shown in Figure 1 (a) and (c). The voltage drop through the feeder can be expressed approximately as cos sin L VD IR IX (1) where R is the resistance of the feeder circuit XL is the inductive reactance of the feeder circuit cos θ is the receiving-end power factor sin θ is the sine of the receiving-end power factor angle
- Shunt and Series capacitors Department of Electrical and Electronics Engineering However, when a series capacitor is applied, as shown in Figure 1 (b) and (d), the resultant lower voltage drop can be calculated as cos ( )sin L C VD IR I X X (2) where Xc is the capacitive reactance of the series capacitor. 1.1 Overcompensation The series-capacitor size is selected for a distribution feeder application in such a way that the resultant capacitive reactance is smaller than the inductive reactance of the feeder circuit. However, in certain applications (where the resistance of the feeder circuit is larger than its inductive reactance), the reverse might be preferred so that the resultant voltage drop is cos ( )sin C L VD IR I X X (3) The resultant condition is known as overcompensation.
- Shunt and Series capacitors Department of Electrical and Electronics Engineering Fig. 2 Overcompensation of the receiving-end voltage: (a) at normal load and (b) at the start of a large motor.
- Shunt and Series capacitors Department of Electrical and Electronics Engineering 1.2 Leading Power Factor To decrease the voltage drop considerably between the sending and receiving ends by the application of a series capacitor, the load current must have a lagging power factor. Fig. 3 Voltage phasor diagram with leading power factor: (a) without series capacitors and (b) with series capacitors. As can be seen from the figure, the receiving end voltage is reduced as a result of having series capacitors. When cos θ = 1.0, sin θ ≅ 0, and therefore, ( )sin 0 L C I X X hence, Equation (2) becomes VD IR (4) Thus, in such applications, series capacitors practically have no value.
- Shunt and Series capacitors Department of Electrical and Electronics Engineering 2. Shunt Capacitors • Shunt capacitors, that is, capacitors connected in parallel with lines, are used extensively in distribution systems. • Shunt capacitors supply the type of reactive power or current to counteract the out of- phase component of current required by an inductive load. Fig. 4 Voltage phasor diagrams for a feeder circuit of lagging power factor: (a) and (c) without and (b) and (d) with shunt capacitors.
- Shunt and Series capacitors Department of Electrical and Electronics Engineering Voltage drop in feeders, or in short transmission lines, with lagging power factor can be approximated as R X L VD I R I X (5) where R is the total resistance of the feeder circuit, Ω XL is the total inductive reactance of the feeder circuit, Ω IR is the real power (or in-phase) component of the current, A IX is the reactive (or out-of-phase) component of the current lagging the voltage by 90°, A
- Shunt and Series capacitors Department of Electrical and Electronics Engineering Voltage drop in feeders, or in short transmission lines, with lagging power factor can be approximated as R X L VD I R I X (5) where R is the total resistance of the feeder circuit, Ω XL is the total inductive reactance of the feeder circuit, Ω IR is the real power (or in-phase) component of the current, A IX is the reactive (or out-of-phase) component of the current lagging the voltage by 90°, A
- Shunt and Series capacitors Department of Electrical and Electronics Engineering Voltage drop in feeders, or in short transmission lines, with lagging power factor can be approximated as R X L VD I R I X (5) where R is the total resistance of the feeder circuit, Ω XL is the total inductive reactance of the feeder circuit, Ω IR is the real power (or in-phase) component of the current, A IX is the reactive (or out-of-phase) component of the current lagging the voltage by 90°, A
- Shunt and Series capacitors Department of Electrical and Electronics Engineering Example 1 Consider the right-angle triangle shown in Figure 5 (b). Determine the power factor of the load on a 460 V three-phase system, if the ammeter reads 100 A and the wattmeter reads 70 kW. Fig. 5 (a) Phasor diagram and (b) power triangle for a typical distribution load. Solution: 3( )( ) 1000 V I S 3(460 )(100 ) 1000 V A 79.67 kVA
- Shunt and Series capacitors Department of Electrical and Electronics Engineering Thus, cos P pf S 70 79.67 kW kVAR 0.88 88 % or
- Effect of shunt capacitors (Fixed and switched) Department of Electrical and Electronics Engineering Fixed Capacitor Bank • There are certain loads mainly certain industrial loads which need fixed reactive power to meet power factor. • In this type feeder fixed capacitor bank is used. • These banks do not have separate control system to switch ON or OFF. • These banks run with feeders. • The banks are connected to the feeders as long as the feeders are live.
- Effect of shunt capacitors (Fixed and switched) Department of Electrical and Electronics Engineering Switched Capacitor Banks • In high voltage power system, compensation of reactive power is mainly required during peak load condition of system. There may be reverse effect if the bank is connected to the system at mean load condition. • At low load condition, the capacitive effect of bank may increase the reactive power of the system instead of decreasing it. • In this situation capacitors bank must be switched ON during peak load poor power factor condition and must also be switched OFF during low load and high power factor condition. Here switched capacitor banks are used.
- Power Factor Correction Department of Electrical and Electronics Engineering The cosine of the angle between current and sending voltage is known as the power factor of the circuit. Assume that a load is supplied with a real power P, lagging reactive power Q1, and apparent power S1 at a lagging power factor of (1) or When a shunt capacitor of Qc kVA is installed at the load, the power factor can be improved from cos θ1 to cos θ2, where
- Power Factor Correction Department of Electrical and Electronics Engineering or (2) 2.1 Concept of Leading and Lagging Power Factors In general, for a given load, the power factor is lagging if the load withdraws reactive power; on the other hand, it is leading if the load supplies reactive power. Fig. 6 Examples of some of the sources of leading and lagging reactive power at the load.
- Power Factor Correction Department of Electrical and Electronics Engineering 2.2 Economic Power Factor • As can be observed from Figure 7 (b), the apparent power and the reactive power are decreased from S1 to S2 kVA and from Q1 to Q2 kVAR (by providing a reactive power of Q), respectively. • The reduction of reactive current results in a reduced total current, which in turn causes less power losses. Fig. 7 Illustration of power factor correction. • The economic power factor is the point at which the economic benefits of adding shunt capacitors just equal the cost of the capacitors. • In the past, this economic power factor was around 95%. • Today’s high plant and fuel costs have pushed the economic power factor toward unity.
- Power Factor Correction Department of Electrical and Electronics Engineering Example 2 Assume that a substation supplies three different kinds of loads, mainly, incandescent lights, induction motors, and synchronous motors, as shown in Figure 8. The substation power factor is found from the total reactive and real powers of the various loads that are connected. Based on the given data in Figure 8, determine the following: a. The apparent, real, and kVARs of each load b. The total apparent, real, and reactive powers of the power that should be supplied by the substation c. The total power factor of the substation Fig. 8: For Example 8.2: (a) connection diagram, (b) phasor diagrams of individual loads, and (c) phasor diagram of combined loads.
- Power Factor Correction Department of Electrical and Electronics Engineering Solution a. 1. For a 100 kVA lighting load Since incandescent lights are basically a unity power factor load, it is assumed that all the current is kilowatt current. Hence, 1 1 100 100 S P kVA kW 2. For 400 hp of connected induction motor loads Assume that for the motor loads,
- Power Factor Correction Department of Electrical and Electronics Engineering 3. 200 hp motor with a 0.8 leading power factor At full load, assume kVA = motor-hp rating = 200 kVA:
- Power Factor Correction Department of Electrical and Electronics Engineering b. At the substation, the total real power is
- Power Factor Correction Department of Electrical and Electronics Engineering c. The kVA of the substation is The power factor of the substation is
- Power Factor Correction Department of Electrical and Electronics Engineering Example 3 Assume that a load withdraws 80 kW and 60 kvar at a 0.8 power factor. It is required that its power factor is to be improved from 80% to 90% by using capacitors. Determine the amount of the reactive power to be provided by using capacitors. Without capacitors at PF = 0.8 Solutions kW = 80 kvar = 60 Thus, the kVA requirement of the load is kVA = (802 + 602)1/2 = 100 kVA
- Power Factor Correction Department of Electrical and Electronics Engineering With capacitors at PF = 0.9 kW = 80 kVA ≅ 88.9 Hence, the line supplies 38.7 kvar and the load needs 60 kvar, and the capacitor supplies the difference, or
- Power Factor Correction Department of Electrical and Electronics Engineering Example 4 Assume that a certain load withdraws a kilowatt current of 2 A and kilovar current of 2 A. Determine the amount of total current that it withdraws. Solution: (kvar current)^2+ (kW current)^2= (total current)^2 (2 A)^2+ (2 A)^2= (total current)^2 or 4 + 4 = (total current)^2 Hence, Total current = 8^1/2= 2.83 A
- Power Factor Correction Department of Electrical and Electronics Engineering Example 5 Assume that a 460 V cable circuit is rated at 240 A but is carrying a load of 320 A at 0.65 power factor. Determine the kvar of capacitor that is needed to reduce the current to 240 A. Solution The kVA corresponding to 240 A is
- Power Factor Correction Department of Electrical and Electronics Engineering Thus, the operating power factor corresponding to the new load is The capacitor kvar required is
- Power Factor Correction Department of Electrical and Electronics Engineering Example 6 Assume that a 1000 kW turbine unit (turbogenerator set) has a turbine capability of 1250 kW. It is operating at a rated load of 1250 kVA at 0.85 power factor. An additional load of 150 kW at 0.85 power factor is to be added. Determine the value of capacitors needed in order not to overload the turbine nor the generator. Solution: Original load
- Power Factor Correction Department of Electrical and Electronics Engineering Additional load Total load
- Power Factor Correction Department of Electrical and Electronics Engineering The minimum operating power factor for a load of 1150 kW and not exceeding the kVA rating of the generator is The maximum load kvar for this situation is
- Power Factor Correction Department of Electrical and Electronics Engineering where 0.426 is the tangent corresponding to the maximum power factor of 0.935. Thus, the capacitors must provide the difference between the total load kvar and the permissible generator kvar, or
- OPTIMUM POWER FACTOR FOR DISTRIBUTION SYSTEMS Department of Electrical and Electronics Engineering For a given power Pk the peak power supplied, Qc required capacitor bank kVAR, is obtained by Where tan is the original p.f and tan the optimum p.f. tan is obtained with the following procedure. 1 (i) The best desired p.f (a value chosen between 0.90 to 0.98) is set. (ii) Additional savings in ‘kW’ losses at the set p.f value, when all capacitors are connected to the substation or distribution bus is determined. (iii) Reduction in losses due to capacitors applied to substation buses and additional losses due to capacitors are computed. (iv) Reduction in KVA demand and hence the additional capacitors needed at buses and feeders is computed.
- OPTIMUM POWER FACTOR FOR DISTRIBUTION SYSTEMS Department of Electrical and Electronics Engineering v. Total annual savings in demand reduction due to additional capacitors applied, is computed (Rs/year). vi. Total annual saving due to additional released capacity per year and hence the annual savings due to energy losses (Rs/year). vii. Total annual losses due to cost of capacitors, energy losses in capacitors and other operating expenses are computed. viii.Net savings (annual) is arrived at (Rs/year) ix. Is the power factor set economical? For this, the procedure is repeated by setting a new value. x. Best p.f is one which gives maximum value for step viii.
- OPTIMUM POWER FACTOR FOR DISTRIBUTION SYSTEMS Department of Electrical and Electronics Engineering 1. Best Capacitor Location and Optimum Capacitor Size The capacitor size and location is usually determined by minimizing (i) power loss (ii) voltage regulation, and (iii) additional cost and other expenses for illustration. The procedure usually followed is (i) to determine the desired p.f for the given load or circuit from the data, viz., kW or KVA and p.f (ii) the feeder or line voltage and improvement needed (iii) to determine correction factor needed to correct either load or feeder circuit p.f from the existing one (iv) to check the existing capacitors or other p.f correcting devices, their capacity, location, etc., and improvement given
- OPTIMUM POWER FACTOR FOR DISTRIBUTION SYSTEMS Department of Electrical and Electronics Engineering Optimal Location of Capacitor • To obtain the best location for the capacitor banks, the feeder line is decided into a number of sections with a combination of both uniformly distributed loads and concentrated loads. • This component is counterbalanced by a capacitor bank in each section or segment such that power loss due to inductive current is reduced by the capacitor allocated in that section. • The ratio of the reactive (capacitive) ‘kVAR’ of the capacitor to the total reactive load is determined along with the ratio reactive current at the end of the line segment to the beginning of the line segment (a). • The distance of capacitor bank from the feeding point of the line segment is obtained by considering (a) the ratio, (b) original losses due to reactive current, (c) capacitance, and (d) compensation ratio. • The optimum number of capacitor banks is obtained by making the cost function of the above variables a minimum.
- OPTIMUM POWER FACTOR FOR DISTRIBUTION SYSTEMS Department of Electrical and Electronics Engineering 2 Benefits with Capacitor Installation in Distribution Systems In general, the benefits from installation of capacitors are (i) The generation plant capacity, together with transmission and distribution substations and network capacity is released and hence increased to that extent (ii) Reduced energy losses (power losses) (iii) Reduced voltage drop and improved voltage regulation (iv) Increase in revenue due to voltage improvement, reduced losses and hence supply of more energy Quantitative estimate of the above benefit and hence the revenue increase is a more involved procedure and complex calculations are needed. The total benefits due to capacitors installation can be put as total financial (benefit) obtained due to (i) Demand reduction + (ii) Energy savings (reduction) + Revenue increase due to release of additional capacity.
- OPTIMUM POWER FACTOR FOR DISTRIBUTION SYSTEMS Department of Electrical and Electronics Engineering Review Questions 1. Explain the need for p.f improvement in distribution systems. 2. How in the capacitor bank ratings obtained when the load p.f is to be improved from cos theta1 to cos theta 2? 3. Explain how reduction in line current and hence power losses are obtained with p.f improvement. 4. What are the different locations for p.f improvement capacitors? Discuss their relative advantages and disadvantages. 5. How does p.f improvement help in reduction in % VD and hence voltage regulation of distribution transformers? With transformer of 6% reactance and 2% magnetizing current, what will be the probable ratings of capacitor bank to compensate for magnetizing current and improve regulation by 2%? 6. What is series capacitor compensation in feeder lines ? How does it improve the regulation of the lines? Discuss with suitable examples.
- OPTIMUM POWER FACTOR FOR DISTRIBUTION SYSTEMS Department of Electrical and Electronics Engineering 8. Discuss how voltage profi le of a long feeder can be improved by connecting shunt capacitor banks at the end of the feeders. 9. How is economical power factor arrived at for a given distribution system with different loads ? 10. What is the justification for p.f improve and what are the benefits ? 11. Compare and explain role of shunt and series capacitors in p.f correction.
- Thank You

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