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current-carrying

By Michelle Moore,2014-12-19 04:03
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current-carrying

Passive Filter Design for Adjustable Speed Drive to

    ththEliminate the 5 and 7 Harmonics

Introduction

     Now more and more induction motors are used in the utility system. That is because the induction motor is rugged, reliable, and single-fed machine, it can directly absorb the reactive power from the utility, and needn’t additional magnetic field provider. In order to control the speed of induction motor, many methods have been developed. The most popular way is to use the Adjustable Speed Drive (ASD), usually it is an AC inverter. With this device, we can get two advantages: one is that we can get a low start current; the other is that we can change the motor speed conveniently by controlling the output frequency of the ASD.

    But at the same time, AC inverter can also causes harmonics in utility and becomes the harmonic source in utility system. These harmonics will reduce the power factor and, if seriously enough, it will destroy the motor in this system. In our project, we need to build ththa passive filter to eliminate the 5 and 7 harmonic.

    In our project, we use a system including AC source, Adjustable Speed Driver, Induction motor, DC motor and a resistor to observe the harmonics, then design harmonic filters to ththeliminate the 5 and 7 harmonic.

Induction Motor

    Induction motors are the simplest and most rugged machine of all electric motors. There are only two main components: the stator and the rotor. The rotor is built from a number of conducting bars running parallel to the axis of the motor and two conducting rings at the ends. Figure 1 below shows an example of the stator bars and the two rings of a squirrel cage induction motor. The stator contains a pattern of copper coils arranged in windings. As alternating current is passing through the windings, a moving magnetic field is form near the stator which induces a current in the rotor, and creating its own magnetic field.

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    Figure 1: Stator Bars of a Squirrel Cage Induction Motor

An induction motor constitute of the following

    1. It is a single-fed motor which means that it does not require a commutator, slip-

    rings, or brushes.

    2. Since there is no brushes in an induction motor, it therefore operates at high

    efficiency

    3. It has the same characteristic as a transformer where the secondary winding is

    energized when the machine rotates.

    of an induction motor is defined as: The synchronous speed NS

    120f (1.1) SNP

    Is the frequency of the current in the stator winding andPis the number of f

    poles. And the Motor Speed = Synchronous Speed Motor Slip.

    The revolving field induces electromotive force (EMF) in the rotor winding. Since the rotor winding forms a closed loop, the induced EMF in each coil gives rise to an induced current in that coil. When a current-carrying coil is immersed in a magnetic field, it experiences a force (or torque) that tends to rotate it. The torque thus developed is called the starting torque. If the load torque is less than the starting torque, the rotor starts rotating. The force developed and the thereby the rotation of the rotor are in the same direction as the revolving field.

    This is in accordance with Faraday’s law of induction. Under no load, the rotor soon achieves a speed nearly equal to the synchronous speed. However, the rotor can never rotate at the synchronous speed because the rotor coils would appear stationary with respect to the revolving field and there would be no induced EMF in them. In the absence of an induced EMF in the rotor coils, there would be no current in the rotor conductors and consequently no force would be experienced by them. In the absence of a force, the rotor would tend to slow down. As soon as the rotor slows down, the induction process takes over again. In summary, the rotor receives its power by induction only

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    when there is a relative motion between the rotor speed and the revolving field. Since the rotor rotates at a speed lower than the synchronous speed of the revolving field, an induction motor is also called an asynchronous motor.

Now let’s use an example to show how to get the equivalent circuit parameters. The rated

    values of the induction motor are shown as follows,

    CLASS S CODE J

    HP 3 R.P.M. 1750

    VOLTS 208 PHASE 3

    CYCLE 60 AMPS 8.9

    Table 1: Nameplate Data for Induction Motor

Determine Equivalent Circuit Parameters

    Equivalent circuit parameters are useful for determining the response to changes in the load. The parameters are determined by the stator resistance, the blocked-rotor, the no load, and load test.

The DC Stator-Resistance Test

    This test determines each phase resistance. If R is the resistance between two terminals, then the per-phase resistance is:

    RRforYConnection1,?0.5 RRforConnection1,?105

    Figure 2: DC Resistance Test Circuit

     In this test, we observe the DC voltage and the DC current, the values are

    VV10.6dc I,!8.9dc

     We know this induction motor is Y- Connected, then

    V10.6dc (1.2) R,,,!0.5961228.9I?dc

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The Blocked-Rotor Test

     This test is used to measure the total series impedance. It is similar to the short circuit test of a transformer. In this case, the rotor is held stationary by applying external torque to the shaft. The stator field winding is connected to a variable three-phase power supply. The voltage is carefully increased from zero to a level at which the motor draws the rated current. At this time, the readings of the line current, the applied line voltage, and the power input are taken by using the two-wattmeter method, as illustrated in Figure 3.

    Figure 3: Blocked-Rotor Test Circuit

     We assume the excitation current is quite small and can be neglected, the equivalent circuit of the motor is given in Figure 4.

    Figure 4: Equivalent of Blocked-Rotor Test

     The total series impedance is

     (1.3) ZRjX,?eee

The values we measured are:

    PW268.4br

    VV22.878br

    IA8.6br

    Now calculate the series resistance and the reactance,

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    V22.87brZ,,,!2.66brI8.6br

    P268.4/3brph,PF,,,0.455brS22.878.6?br

     then

    ~RZ,?,?,!cos2.660.4551.21brbr

    2,?,??,!XZsin2.6610.4552.37~brbr

From the DC Test, R=0.596, 1

    '(,?,!R1.210.5960.6142

     (1.4) X2.37'brXX,,,,!1.182122

The No-Load Test

     This test is used to measure the magnetization impedance. In this case the rated voltage is impressed upon the stator windings and the motor operates freely without any load. Therefore this test is similar to the open-circuit test on the transformer except that friction and windage loss is associated with an induction motor. Since the slip is nearly zero, the impedance of the rotor circuit is almost infinite. The no-load test circuit is show in figure 5.

    Figure 5: No-Load Test Circuit

The values we measured are:

    PW135nl

    VV119.93nl (1.5) IA3.94nl

    QVar1410nl

    Now let’s calculate the magnetization impedance.

    First let’s assume the mechanical losses are 1% of the rated power. From the nameplate of the motor, P=3HP=3×746=2238W.

    Then the mechanical loss is

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    PW,?,22381%22.38mec (1.6)

    The effective power in the motor is

    PP13522.38'nlmec (1.7) PW,,,37.54nl33

    Now we can get the power loss in R1 22 (1.8) PIRW,?,?,3.940.5969.25Rnl11

    Figure 6: Phase Current and Voltage at No-load

The per-phase power in the core Rm,

    ' (1.9) PPPW,?,?,37.549.2528.29mnlR1

    Power factor of no-load circuit,

    'P37.54nl,,,,~cos0.079PFnl0' (1.10) ?119.933.94Snl

    ((,85.4~0

    The voltage across the magnetization circuit is

    ''(((EVIRjXj,??,,?,??,,119.9303.9485.40.5961.18115.150.98 (1.11) ;;;;nlnl022

    P28.29'm~,,,cos0.062360?EI115.153.94 nl0

    ('(,86.425~0

    The per-phase reactive power in the core

    '( (1.12) QPVar,?,?,tan28.29tan86.425452.8~mm0

    From the equation above, we can finally find the value of the magnetization impedance.

    22E115.150R,,,!468.7mP28.29m (1.13) 22E115.150X,,,!29.28mQ452.8m

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     Now let’s check quickly if our results are correct. The reactive power we measured is Qnl

    =1410Var.

    The total per-phase reactive power we calculated is

    2'2 (1.14) QQIXVar?,??,??,452.8452.83.941.18471.1mR12

    The total reactive power is

    Q = 3×471.1= 1413.3Var

    This value is close to the value we measured, that means our calculation is correct.

Adjustable Speed Drive

    The Adjustable Speed Drive (ASD) we used is an AC Inverter named E-TRAC[4]. The principle of operation of this device is to provide both an adjustable voltage and an adjustable frequency to the AC induction motor. The AC inverter automatically maintains the required volts/hertz ratio, allowing the Ac motor to run at its optimum efficiency and providing rated torque capability throughout the motor’s speed range. The basic formula that relates the output frequency to motor speed is:

    120?f (2.1) NsP

    Ns: Synchronous Speed (RPM)

     f: Frequency (Hertz)

     P: Number of Poles in Motor. In our project, the number is 4.

For the induction motor we used:

    Motor RPM= Synchronous Speed- Motor Slip (RPM) = 1800-75.

    From the above equation, we can easily find that the rotating speed of the induction motor is proportional to the frequency. Now let’s explain briefly how the AC Inverter can

    output a variable frequency.

    First we need to decide how many pulse rectifiers this ASD have. We know that the harmonic order number can be express as following:

    KNi,)1 (2.2)

    K: harmonic order

    N: pulse number

    : Integer

    From the harmonic diagram we got in our experiment (Figure 13, 14 and 15), the harmonic order is 5, 7, 11, 13 If we set = 1, we can find N=6. So the ASD is a 6-pulse rectifier.

    We simplify the basic circuit as follow:

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    Figure 7: AC-DC-AC inverter

    The ASD is a complicated electronic device. The practical circuit is more complicated than this one. We only use this to explain how to get variable frequency.

    Figure 8: Three Phase Input Voltage

    The input of rectifier is a 3-Φ sine wave figure above. As long as the amplitude of phase A is greater than phase B and C, the diode 1 and 4 conduct; when phase B is greater than A and C, diode 2 and 5 conduct; diode 3 and 6 conduct when the amplitude of phase C is greater than phase A and B. then at the output of the rectifier, we observe a waveform with a triple. This triple is determined by the amplitude of the input voltage and the trigger time of the diode. This tripled voltage is smoothed by the DC Link, if the value of the capacitor is large enough, then the output of the DC Ling is an ideal DC voltage. The DC voltage goes into the PWM Inverter, by controlling the switch time of these six thyristors, we can get 3- AC output signal.

The current of the 6-pulse AC Inverter is I,

    50:?2311A??ii1???? (2.3) ???Itmtntcos1cos1cos,??????;;??;;,;?????,???mn????1????

In this equation, the fundamental current is Is,

    23A ?Itcos;;s

The total harmonic current is I, h

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    50:?11ii1???? (2.4) Imtnt??,????1cos1cos????,;,;?h,?????mn????i1??

mi,?61

    ni,?61

From the equation (1.4), we can also get the harmonic order, which is 5, 7, 11, 13…

     ththIn our project, our purpose is to eliminate the 5 and the 7 harmonics.

Harmonic Limits

    Harmonics has become a major factor in power systems and standards were developed to set limits on harmonic produce by the customer and limit harmonics the utility provide. Nonlinear loads and devices are the major causes of harmonics in power systems. The IEEE 519-1992 and IEEE P519A/D7 [6] harmonic limits are applied at the point of common coupling (PCC). Figure 9 shows the PCC where multiple customers can be served. The PCC is the point between the customer and the utility and is the point where multiple customers can be served. PCC can be at either the primary or the secondary of a supply transformer depending on the amount of customers being served by the transformer.

    These standards set limits on the harmonic currents generated by individual customers. With these limits in place, the over all power system harmonic voltage remains acceptable. However, to accomplish this, the customer and the utility are equally responsible on the quality of power being provided.

    Figure 9: PCC where multiple customers is being served

    The customers use the IEEE 519-1992 guidelines to limit the level of harmonic current that gets injected at PCC. The utility responsibility is to provide every customer with voltage that’s a pure sine wave. From the limits provided in 519-1992, this limit is

    5% Total Harmonic Distortion (THD) that is allowed. Two main limits concerns to be addressed when talking about harmonics is Voltage and Current Distortion Limits. Table 2 shows the values used to design for worst case operation of the power system at PCC.

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    Table 2: Voltage Distortion Limits

    ?2Vh?2h (3.1) ,?100%VnTHDnV

    Where:

    Vh = magnitude of individual harmonic components

    h = harmonic order

    Vn = nominal system rms voltage

Current Distortion Limits are shown in Table 3 the line for various voltage levels. This

    harmonic current is evaluated at the PCC.

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