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The Story of Electrical Energy, Pt.14 This month, we look into the methods that make electrical energy use more efficient. For optimum voltage regulation and lowest running costs, the current and voltage need to he in phase or close to it. If they are not in phase, the energy losses are higher than they should he. By BRYAN MAHER Inductive electric machines and appliances draw lagging out-of-phase currents . Naturally, we would like the power station alternators to operate with minimum electrical losses and best power efficiency. And the same may be said for the high voltage transmission lines and transformers. To this end, the loads supplied should ideally draw a current in phase with the voltage. Why should that follow? Just read on. In most industries, AC squirrel cage induction motors form a major percentage of the load. Induction motors produce mechanical rotation by the interaction of the stator's rotating magnetic field with the conductors of the rotor. Being highly inductive , such motors draw currents which rise later than the voltage in each cycle. This is illustrated in Fig.1. In this diagram, the current is lagging the voltage by some phase angle <j>; ie, the current peaks come after the voltage peaks and so on. Resistive loads such as heaters, ovens, filament lamps and electrolytic baths (used for plating, etching, etc) draw a current which is exactly in phase with the voltage as shown in Fig.2(a). In other words, the current and voltage rise and fall together throughout each cycle. (The word "phase" originally meant "a division of time".) Inductive loads lag. Because the back voltage (or back-EMF) generated by an inductance opposes the applied voltage, a pure inductor takes a cur- SUPPLY VOLTAGE - . ._ ✓ MOTOR CURRENT '' ' TIME AND ANGLE i LAG ANGLE ONE CYCLE =360- =20 MILLISECONDS (50Hz) Fig.I: this diagram illustrates the fundamental nature of an induction motor load; its current waveform lags the voltage waveform. This causes problems because the distribution system must supply additional current. SILICON CI-IIP 0 Power & power factor The power as_sociated with a current is equal to the product of voltage, current and the cosine of the phase angle. This is expressed by the following formula: P = V.I.cos<j> The cos<j> term is also called the power factor (PF). Readers who remember their trigonometry may recall that the cosine of zero degrees is unity and that cos goo = o. Since there is zero phase angle between the voltage and current through resistive loads, their power factor is unity (because cos O = 1). This means that the power dissipated in a resistive load is simply the product of the voltage multiplied by the current. By contrast, all purely reactive (inductive or capacitive) loads do not use any power at all! This is because the cosine of their go phase angle (lag or lead) is zero and therefore they have zero power factor. 0 <I> 86 rent which is delayed by go with respect to the voltage, as shown in Fig.2(b). By contrast, capacitive loads lead. Because capacitive current is proportional to the rate-of-change of voltage, a pure capacitor draws a current which is exactly goo ahead of the voltage. Fig.2(c) shows that capacitor current is greatest where the voltage is changing at maximum rate; ie, at the zerocrossing crossing points on the voltage curve. Real machines In the real world, all inductive components have some resistance in addition to their inductance. Thus, iron cored coils have a power factor greater than zero but less than unity. Motor windings show both simple ohmic resistance and also inductance. But there's more here than meets the eye. A motor's job in life is to convert electrical current into mechanical ro- - / CURRENT IN PHASE .,,. - '- / " CURRENT LAGGING '\ / / I '\ I 90' LAG ' ' j",_ 90' LEAD CURRENT LEADING Fig.2: this diagram shows the phase difference between the voltage and current for three types ofload: (a} for resistive loads, the voltage and current are in phase; (b} for pure inductive loads, the current lags the voltage by exactly 90°; and (c} for pure capacitive loads, the current leads the voltage by exactly 90°. tation of the shaft. The total induction motor current can be visualised as having five components: (1). An inductive component to provide the magnetic field. This component has 90° lag angle;, ie PF = 0. (2). The simple ohmic resistance of the copper windings. (3). The eddy current iron losses. (4). · The mechanical friction and windage losses . (5). The mechanical power output at the shaft. Note that the last four components all have a power factor of 1. The first four of the above components are reasonably constant. However, that fifth component of motor current clearly varies with any changes in mechanical load. Factory example Suppose a factory receives a 3-phase supply at 11kV. This company would provide its own 11kV/415V transformer for all its machines. Perhaps this installation contains 100 induction motors. Let's further suppose that the average motor current is 134.54A. If we made careful measurements, we might find the power or in-phase component to be 100A, while the magnetising or lagging out-of-phase current component is 90A. Assuming this motor to be typical of all 100 machines, the total factory current supplied by the transformer is 13,454 amps, which amounts to quite a sizeable installation. The vector diagram of Fig.3 shows how the current components relate. In this example, the 42° phase angle is typical for the whole factory, where some motors are on full load and some are lightly loaded. If a motor is on no load, the inductive component of current predominates, so the motor current would have a large lag angle (about 80°} and a low power factor (around 0.2). But on full load, the power component (number 5 in the list above) exceeds all others. Thus, the total motor current would have a small lag angle (around 30°). The full load power factor would be about 0.85. You might wonder what lag angle and power factor has got to do with the price of energy. In fact, the factory electricity bill is charged only on the in-phase or power component. The meter measures kilowatt-hours. The kilowatt is truly a measure of power and the kWh reading is the energy used at a certain power over some number of hours. None of these indicate anything of the reactive outof-phase current component, because its power factor equals zero. So why should the consumer worry? Why should the factory manager care? He should and does. For consumers like our industrial example, the outof-phase component of current is a very big part of the total amperes flowing into and through the factory's 11kV/415 transformer. We call the product of voltage and out-of-phase component of current by the name Volt Amps Reactive, or VAR. Naturally a thousand of these we call kVAR, (kiloVARs},oramillionMVAR, (megaVARs). As we have already seen, the main factory transformer must supply 415 V, 3-phase, at 13,454A. That's quite a big unit and its size is expressed by the product of voltage and line current. This product is (✓ 3 x 415V x 13,454A) = 9.6707 million volt amps. This would normally be expressed as 9.67MVA. Looking back, we see that MW is simply the product of MVA and PF. However, if built to provide almost 10MVA, that factory transformer is likely to be a lot more expensive than "--""'go""'oo""A""M""Ps'"'1""'Lo""u=-TO:::F""P""HA""SE:,---,-8 This is a vector diagram showing an induction motor load in which the phase angle bet\\!;een the load current and the applied voltage is 42 °. If a suitable bank of capacitors was connected in parallel, the phase angle between current and voltage could be reduced to zero and the resulting current drawn from the distribution system also reduced. OCT0BER1991 87 would have to flow in the transformer's 1 lkV primary, in the state grid system and all the way back to the alternator stator windings in the power station. Without power factor correction somewhere in the system, all power line conductors, circuit breakers, transformers and metering equipment would have to be of higher current capacity, just to supply that outof-phase current component for which the customer would otherwise pay no extra. Therefore, most supply authorities apply penalty charges to the electricity bill to cover the cost of the additional plant they must provide to cover low power factor loads. Usually a "maximum demand" recording ammeter is installed and the price per kWh of power used is increased as a penalty for currents above a predetermined value. Control of capacitors You might wonder what happens if half the motors are switched off for some of the time? In this circumstance, This is a large outdoor bank of capacitors for power factor correction. There are three vertical banks, one for each phase of the mains supply. it needs to be. Observe that only the 10,000A in-phase component of the current produces any power. The other component, the 9000A magnetising current (although vitally necessary to form the magnetic fields) produces no power. Neither does it consume any power from the supply. Wouldn't it be nice if the transformer only had to supply the 10,000A in-phase component of current? This can be done by adding capacitors to the circuit. Capacitors Recall that capacitors take a leading current; ie, a perfect capacitor takes a current which leads the voltage by 90°. Fig.2(c) illustrates this. Ifwe add enough capacitors to the secondary windings of the transformer, we can 88 SILICON CHIP cancel out the lagging current of the motors. Because it is a 3-phase system, we will require three identical capacitor banks. The photos show some typical installations. So by adding suitable capacitors to the installation, we have reduced the transformer secondary current from 13,454 amps down to 10,000 amps. This enables a transformer rated at around 7MVA to be used instead of one rated at lOMVA. That's a big reduction in size and cost! Running costs are also reduced. Smaller transformer currents mean smaller losses and therefore less continual costs. Penalty charges There are other benefits too. Without capacitors installed, extra current Banks of power factor correction capacitors can be very large or relatively modest as with these units made at Asea Brown Boveri's Capacitor Division at Lilydale in Victoria. You can now afford a sate II ite TV system For many years you have probably looked at satellite TV systems and thought "one day". You can now purchase the following K band system for only $995. 00 This is about 1/3 the price of corn parable systems Here's what you get: _. A 1.8 metre pressed steel prime focus dish antenna, complete with all the mounting hardware - as well as a self supporting ground stand. The induction motor is the cause of most power factor problems in the AC power distribution network, along with fluorescent lamps. Because it is an inductive load, the motor current lags the AC voltage waveform and that means extra current has to be provided. the remaining motors only take about 4500A of out-of-phase lagging current but the capacitor bank would still be taking its full 9000A of leading current! This would amount to having over-compensation and so the total factory current would be higher tha.n necessary, with a leading power factor! There are three ways to prevent such a situation: (1) Rather than have one large bank of capacitors for the whole factory, we could use separate capacitors for each motor. This way, when the motor is switched off, so too is the capacitor. This method can have a severe disadvantage, though. Each time the motor circuit breaker is opened, if the capacitor and motor inductance find resonance at the wavefront frequency, a large overvoltage spike may be generated. In extreme cases, this has been known to puncture the motor winding insulation. When next the motor is switched on, explosive breakdown can occur. (2) Alternatively, the large bank of capacitors may be divided into sections, each brought on line as required according to the number of motors in use. Computers are now used to monitor the power factor and control the switching of capacitor banks. (3) Finally, another method of power factor correction involves using a unloaded synchronous motor. However, that m ethod is beyond the scope of this article. Fluorescent lights It is not only induction motors that produce lagging power factor problems. Fluorescent lamps also have a lagging power factor, because of the current limiting inductor (known as a ballast). This typically results in a power factor of between 0. 7 and 0.8. When you consider the thousands of fluorescent lamp fittings used in all commercial buildings, factories and public buildings such as schools and hospitals, the leading power factor becomes a major problem for the supply authorities. For this reason, in commercial and public buildings, the supply authority requires that each fluorescent lamp fitting contain a power factor correction capacitor. This is connected in parallel with the 240VAC input. Power correction capacitors are not required for fluorescent lamps used in homes, however. Acknowledgements Grateful thanks to the NSW Electricity Commission and to ASEABrown Boveri for data and photos and permission to publish. SC _. One super low noise LNB (low noise block converter) 1.4dB or better. _. One KU band feedhorn and all the mounting hardware as well as a magnetic signal polariser. _. 30 metres oflow loss coaxial cable with a single pair control line. _. A 99 channel infrared control satellite receiver with adjustable IF and audio bandwidth, polarity, and dual digital readout. The IR control unit has a range of approx. 10 metres. Before you receive your system the unit is pre-programmed to the popular AUSSAT transponders via the internal EEPROM memory. This unit is also suitable for C band applications. CALL, FAX or WRITE to AV-COMM PTY LTD. PO BOX 386, NORTHBRIDGE NSW 20.63 PHONE (02) 949 7417 FAX (02) 949 7095 All items are available seperately. Ask about our low noise 'C' band LNB, and other interesting products. All systems are provided with dish pointing details. ----------- 1 Yes Garry, Please send me more information I on your K band satellite systems. II Name ....................................... I Address ............... .................... I ................................................... II ...........................P/Code .. ....... . I Telephone .. ... .. .. .. ... ... . .. .. .. .. ..... I I II I I II I .~N=~!:.------0=· OCT0BER1991 89