E2-E3: ELECTRICAL CHAPTER-12 ENERGY MANAGEMENT SYSTEM REACTIVE POWER MANAGEMENT BSNL, India For Internal Circulation Only Page: 1
Reactive Power Management 1.0 Introduction : In today s power scenario, we are facing a major power crunch. Day by day, the gap between demand and supply of electric energy is widening. In the Indian Energy Scenario, even though the generation capacity increased from 1,400 MW in 1947 to 1,65,000 MW by Dec 2010 and the gross generation increased 100-fold from 5 billion units per year to 535 billion units per year in the same period, still the per capita electricity consumption is one of the lowest in the world @ 612 kwh per annum. And we have a very low optimum utilization of existing generation capacity our average Plant Load Factor is only 72%. Also, the T & D Losses are very high at app. 30% as against International standards of 8 to 10%. Add to this the inefficient utilization of electric power by end consumer; the Planning Commission estimates that over 25, 000MW equivalent capacity creation could be achieved through efficient utilization of electric energy by the end user. In this scenario, reactive power management or power factor improvement measures can result in considerable energy saving. This Chapter aims to discuss the causes, effects and remedies of the reactive power problems in industry and the considerable energy saving potential of Reactive Power Management. 2.0 Objective: At the end of this chapter, participants will be able to understand: Problems of poor Power factor Benefits of power factor Improvement Methods to improve power factor and compensation types 3.0 Reactive Power: There are three types of loads in electrical networks (viz.) Resistive, Inductive & Capacitive. In resistive loads, the current is in phase with the voltage. In (pure) inductive loads, the current lags behind the voltage by 90 o. In (pure) capacitive loads, the current leads the voltage by 90 o. But, in practice, the loads are a mix up of two or more of the above three components. Hence, the degree of the current lag or lead with respect to the voltage also varies, depending upon the proportion of the above components in the connected load. Electric power is utilized to do some useful work like pumping of a fluid or conveying of some material or compression or expansion of a gas, etc. The power required to do this useful work is called the Active Power or Real Power. The unit of active power is kilowatt (kw). But, to do this useful work, the electrical machine, as it needs to convert the electrical energy into motive energy, as in the case of motors, draws a certain other power called Reactive Power. This is the inherent property of the induction machine and thus cannot be avoided. The unit of reactive power is kilovolt ampere (reactive) i.e. kvar. Strictly speaking as this power is not doing any useful work this is unwanted power. BSNL, India For Internal Circulation Only Page: 2
We also know that electric power is a product of voltage and current. So, for any useful work to be done, the power that is consumed from the mains by the machine can be calculated by multiplying the voltage (as read from a voltmeter) and the current (as read from an ammeter). This power is termed the Apparent Power, as this is the power, which is apparent or visible as readings in the meters. The unit of apparent power is kilovolt ampere (kva). Ideality requires that there shall be no difference between the apparent power and the active power, implying that the entire power that is drawn from the mains is used to do only useful work. But, in practice it is not so. Strangely, in the case of induction machines, it can be observed that there is always a difference between active power and apparent power. This means that not all the power drawn from the mains is used for doing actual work and some power is used in the induction machine itself. If this cannot be avoided altogether, one should at least know the ratio of actual power to apparent power (i.e.) to know what percentage of the power drawn from the mains is converted to useful work. This ratio is denoted by a factor called Power Factor. This is the ratio of Actual Power to the Apparent Power. Obviously, the higher the power factor, the better is the utilization of the power drawn. But, all induction machines have a very poor power factor. Typical Example : When the need arises to correct for poor power factor in an AC power system, you probably won't have the luxury of knowing the load's exact inductance in henrys to use for your calculations. You may be fortunate enough to have an instrument called a power factor meter to tell you what the power factor is (a number between 0 and 1), and the apparent power (which can be figured by taking a voltmeter reading in volts and multiplying by an ammeter reading in amps). In less favorable circumstances you may have to use an oscilloscope to compare voltage and current waveforms, measuring phase shift in degrees and calculating power factor by the cosine of that phase shift. Most likely, you will have access to a wattmeter for measuring true power, whose reading you can compare against a calculation of apparent power (from multiplying total voltage and total current measurements). From the values of true and apparent power, you can determine reactive power and power factor. Let's do an example problem to see how this works: (Figure below) BSNL, India For Internal Circulation Only Page: 3
Wattmeter reads true power; product of voltmeter and ammeter readings yields apparent power. First, we need to calculate the apparent power in kva. We can do this by multiplying load voltage by load current: As we can see, 2.308 kva is a much larger figure than 1.5 kw, which tells us that the power factor in this circuit is rather poor (substantially less than 1). Now, we figure the power factor of this load by dividing the true power by the apparent power: Using this value for power factor, we can draw a power triangle, and from that determine the reactive power of this load: (Figure 1) BSNL, India For Internal Circulation Only Page: 4
Figure 1 Reactive power may be calculated from true power and appearant power. To determine the unknown (reactive power) triangle quantity, we use the Pythagorean Theorem backwards, given the length of the hypotenuse (apparent power) and the length of the adjacent side (true power): If this load is an electric motor, or most any other industrial AC load, it will have a lagging (inductive) power factor, which means that we'll have to correct for it with a capacitor of appropriate size, wired in parallel. Now that we know the amount of reactive power (1.754 kvar), we can calculate the size of capacitor needed to counteract its effects: BSNL, India For Internal Circulation Only Page: 5
Rounding this answer off to 80 µf, we can place that size of capacitor in the circuit and calculate the results: (Figure 2) Figure -2 BSNL, India For Internal Circulation Only Page: 6
4.0 Problems of low power factor: i) If power factor will be low, Current drawn will be higher: As we know, for 3-phase induction motors, KW = 3 * V * I * Cosφ I = [(kw) /( 3 * V * Cosφ)] For same kilowatt to be delivered and for the same applied voltage, the current drawn by the motor will be inversely proportional to the power factor. (i.e.) poorer the power factor, higher will be the current drawn and vice versa. ii) As the current drawn is higher at low power factor, the power loss (I 2 R loss) in cables will be higher too. iii) So also will be the power loss in the power transformer. iv) Because of increased current flow at low power factor, the impedance voltage (IZ) drop in the transformer will increase and will lead to poor voltage regulation in the transformer. v) As power factor is the ratio of kw to kva, at low power factors, for the same connected load (kw), the kva demand of the system increases (kva =kw/pf). That is, for the same connected load, the requirement of contracted maximum demand (CMD) would increase at low power factor. Increased CMD means increased deposit with the electricity utility company as also increased demand charges every month under the two-part tariff structure. On the other hand, if a certain power factor is assumed and a certain maximum demand is contracted from the utility company, and if the power factor falls below the assumed value, now, the recorded demand will exceed the CMD and will lead to heavy penalties imposed by the utility company, which might even lead to ultimate disconnection of power supply, in certain cases. BSNL, India For Internal Circulation Only Page: 7
vi) Heavy penalty by Utility Companies : Almost all electricity supply companies, today, impose a heavy penalty for low power factor. A benchmark power factor is fixed (say 0.9) and for every one percent (i.e.) 0.01 drop in the power factor, a percentage (say about 1.0%) of the total electrical bill amount is charged as power factor penalty. This expenditure is totally unnecessary. 5.0 Benefits of power factor improvement: i) Reduction in kvar Demand ii) iii) iv) Reduction in kva Demand Reduction in Transformer Rating Reduction in Line Current v) Reduction in Cable Size vi) vii) Reduction in Switchgear rating Reduction in Power Loss 5.1 How to improve power factor? As we have seen above, the problem of low power factor is due to the large reactive power drawn by the induction machines. So, if one wants to improve the power factor, one has to bring down the reactive power drawn by the induction machines. But, the reactive power that is drawn by an induction machine cannot be removed or reduced from the machine, as it is an inherent property of the machine. So, one has to think of any other method to improve the power factor. The fact that comes to our aid is that the induction machines draw lagging reactive power. Fortunately, we have a circuit element capacitor which draws leading reactive power, thus behaving in an opposite fashion to the induction machine. Thus, in a network if both inductive and capacitive components are present, then the net reactive power drawn from the mains will be the difference of the reactive powers of the inductive and capacitive components. Hence, to overcome the lagging reactive power problem in networks with induction machines, one can connect capacitive loads, which would draw reactive power in the opposite direction and thus would neutralize the lagging reactive power of induction machines. There are many such capacitive devices such as Synchronous Condensers, Static Capacitors and the like. Among them, Static Capacitors are most popular because of their cost advantage, ease of installation and maintenance. BSNL, India For Internal Circulation Only Page: 8
5.2 How to calculate the kvar required? Capacitor kvar required = kw (tanφ 1 tanφ 2 ), where kw is the connected load φ 1 = cos -1 pf 1 φ 2 = cos -1 pf 2 pf 1 = Existing power factor pf 2 = Target power factor Normally, the term (tanφ 1 tanφ 2 ) is given as a multiplication factor for various existing power factors and for various target power factors, in a tabular form. The same is enclosed herewith as Annexure 1. 6.0 Types of Compensation: 6.1 Fixed Compensation: Here the switching ON and switching OFF of the capacitor units is done manually depending upon the requirement. This is suitable only for steady loads and for the no-load compensation of transformers & motors. 6.2 Variable Compensation: This is most suitable for varying loads, as is the case with most of the process plants. Here the switching ON and the switching OFF of the Capacitor units is done automatically by an Automatic Power Factor Correction (APFC) Relay or a Reactive Power Management (RPM) Relay, in which the target power factor is set by the operator. The micro controller in the relay automatically calculates the existing power factor from the voltage and the current inputs given to it and switches ON the required number of capacitor steps to achieve the desired target power factor. 7.0 Modes of compensation: 7.1 Central Compensation: This is the installation and connection of the entire required kvar of capacitors as a single bank near the transformer. Advantages: i) Improves PF at the metering point; thus eliminates low PF penalty ii) iii) iv) Optimizes kva demand of the installation. Reduces load on the Transforme. Reduces transformer losses v) Reduces ACB Rating BSNL, India For Internal Circulation Only Page: 9
vi) Extremely easy to maintain vii) Less capital expenditure Disadvantages: i) All the problems associated with low PF still prevails downstream of the capacitor installation ii) Full benefits of capacitor provision not utilised 7.2 Group Compensation: This is the installation and connection of the required kvar of capacitors at each PCC and/or MCC Panel bus bars. Advantages: i) All the advantages listed above ii) Cable losses minimized further Disadvantages : i) Still the problems of low PF persist between the PCC/MCC Bus bars and the ultimate load. ii) As the number of units is more, cost per kvar of capacitors will increase relatively. iii) More maintenance required as more number of units are installed. 7.3 Individual Compensation: This is the installation and connection of the required kvar of capacitors directly at the load terminals. Advantages: (i) Maximum benefit of capacitor installation attained. Disadvantages: i) Many number of discreet capacitor units required. ii) iii) iv) Cost per kvar of capacitor units will be maximum. More maintenance required. Suitable only for constant loads v) May lead to other more serious problems, if opted for varying loads. 8.0 Types of Capacitors : Depending upon the type of dielectric used in a capacitor, they are classified as: i) Metalized Polypropylene (MPP) BSNL, India For Internal Circulation Only Page: 10
ii) Mixed Dielectric (MD) iii) Film-Foil or All Polypropylene (FF or APP) iv) Mixed Dielectric Low loss (MD-XL) The relative merits and de-merits of each type are given in Annexure - 2. BSNL, India For Internal Circulation Only Page: 11
Annexure 1 Multiplying Factors for calculating the capacitor kvar required for various present and final power factors Final PF Present PF 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1.00 0.4 1.807 1.836 1.865 1.896 1.928 1.963 2.000 2.041 2.088 2.149 2.291 0.45 1.500 1.529 1.559 1.589 1.622 1.656 1.693 1.734 1.781 1.842 1.985 0.5 1.248 1.276 1.306 1.337 1.369 1.403 1.440 1.481 1.529 1.590 1.732 0.55 1.034 1.063 1.092 1.123 1.156 1.190 1.227 1.268 1.315 1.376 1.518 0.6 0.849 0.878 0.907 0.938 0.970 1.005 1.042 1.083 1.130 1.191 1.333 0.65 0.685 0.714 0.743 0.774 0.806 0.840 0.877 0.919 0.966 1.027 1.169 0.7 0.536 0.565 0.594 0.625 0.657 0.692 0.729 0.770 0.817 0.878 1.020 0.75 0.398 0.426 0.456 0.487 0.519 0.553 0.590 0.631 0.679 0.739 0.882 0.8 0.266 0.294 0.324 0.355 0.387 0.421 0.458 0.499 0.547 0.608 0.750 0.85 0.135 0.164 0.194 0.225 0.257 0.291 0.328 0.369 0.417 0.477 0.620 0.9 0.029 0.058 0.089 0.121 0.156 0.193 0.234 0.281 0.342 0.484 0.91 0.030 0.060 0.093 0.127 0.164 0.205 0.253 0.313 0.456 0.92 0.031 0.063 0.097 0.134 0.175 0.223 0.284 0.426 0.93 0.032 0.067 0.104 0.145 0.192 0.253 0.395 0.94 0.034 0.071 0.112 0.160 0.220 0.363 0.95 0.037 0.078 0.126 0.186 0.329 0.96 0.041 0.089 0.149 0.292 0.97 0.048 0.108 0.251 0.98 0.061 0.203 0.99 0.142 Procedure to find out the capacitor kvar required: Capacitor kvar required = Connected Load x Multiplying Factor Example: Present power factor: 0.80 Final power factor: 0.98 Connected load: 100kW Capacitor kvar required = Connected Load x Multiplying Factor = 100 x 0.547 ( as shown in grey area) BSNL, India For Internal Circulation Only Page: 12
= 54.7 kvar, say, about 55kVAR Annexture -2 Comparison of various capacitor technologies MPP-S Rati ng MPP-H Rating MD Rating MD- XL Rating FF Rating Life Non linear load capabilit y Optimu m Up to 10% Initial Cost Lowest 10 Operatin g Cost Total 3 1 Long life Up to 15% Mediu m Lowest 10 Lowest 10 5 6 4 Long life Up to 25% Highe st Highe st 10 Long 10 10 Up to 25% 10 1 High 1 Lon g life Up to 25% High est 1 Lowest 10 High 4 24 25 22 31 25 10 10 1 BSNL, India For Internal Circulation Only Page: 13
Questions 1. The load of a building is 1500 KW. The power factor of load is to be improved from 0.75 to 1.0. Calculate impact on:- (i) Transformer rating & Maximum demand (ii) Switchgear capacity. 2. List out the four problems of Low Power Factor. 3. List out the various Benefits of Good Power Factor. 4. Write down the three ways to improve Power Facor. 5. What are the types of power factor Compensation. 6. What are the modes of Compensation. What are the advantages and disadvantages of Central compensation 7. List out the various Types of Capacitors. Discuss their merits/ demerits. 8. What are T&D losses. How much are the T& D losses in India. 9. What is Plant Load Factor? How to increase the Plant Load Factor. 10. If PF is to improve from 0.6 ( lag)to 1.00 ( lag), how much KVAR will be required? Load in KW is 250 KW. BSNL, India For Internal Circulation Only Page: 14