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4.8.1 TRANSFORMERS Harmonics can affect transformers primarily in two ways. Voltage harmonics produce additional losses in the transformer core as the higher frequency harmonic voltages set up hysteresis loops, which superimpose on the fundamental loop. Each loop represents higher magnetization power requirements and higher core losses. A second and a more serious effect of harmonics is due to harmonic frequency currents in the transformer windings. The harmonic currents increase the net RMS current flowing in the transformer windings which results in additional I2R losses. Winding eddy current losses are also increased. Winding eddy currents are circulating currents induced in the conductors by the leakage magnetic flux. Eddy current concentrations are higher at the ends of the windings due to the crowding effect of the leakage magnetic field at the coil extremities. The winding eddy current losses increase as the square of the harmonic current and the square of the frequency of the current. Thus, the eddy loss (EC) is proportional to I h2 × h2, where Ih is the RMS value of the harmonic current of order h, and h is the harmonic frequency order or number. Eddy currents due to harmonics can significantly increase the transformer winding temperature. Transformers that are required to supply large nonlinear loads must be derated to handle the harmonics. This derating factor is based on the percentage of the harmonic currents in the load and the rated winding eddy current losses. One method by which transformers may be rated for suitability to handle harmonic loads is by k factor ratings. The k factor is equal to the sum of the square of the harmonic frequency currents (expressed as a ratio of the total RMS current) multiplied by the square of the harmonic frequency numbers: 2 2 2 2 2 2 2 2 2 k = I 1(1) + I 2(2) + I 3(3) + I 4(4) + … + I n(n) 2 (4.25) where I1 is the ratio between the fundamental current and the total RMS current. I2 is the ratio between the second harmonic current and the total RMS current. I3 is the ratio between the third harmonic current and the total RMS current. Equation (4.25) can be rewritten as: 2 2 k = Σ I n h ( h = 1, 2, 3, …, n ) (4.26) Example: Determine the k rating of a transformer required to carry a load consisting of 500 A of fundamental, 200 A of third harmonics, 120 A of fifth harmonics, and 90 A of seventh harmonics: Total RMS current (I) = I1 = 500/559 = 0.894 © 2002 by CRC Press LLC 2 2 2 2 ( 500 + 200 + 120 + 90 ) = 559 A I3 = 200/559 = 0.358 I5 = 120/559 = 0.215 I7 = 90/559 = 0.161 k = (0.894)212 + (0.358)232 + (0.215)252 + (0.161)272 = 4.378 The transformer specified should be capable of handling 559 A of total RMS current with a k factor of not less than 4.378. Typically, transformers are marked with k ratings of 4, 9, 13, 20, 30, 40, and 50, so a transformer with a k rating of 9 should be chosen. Such a transformer would have the capability to carry the full RMS load current and handle winding eddy current losses equal to k times the normal rated eddy current losses. The k factor concept is derived from the ANSI/IEEE C57.110 standard, Recommended Practices for Establishing Transformer Capability When Supplying NonSinusoidal Load Currents, which provides the following expression for derating a transformer when supplying harmonic loads: I max.(pu) = [PLL–R(pu)/1 + (Σf h2 h2/Σf h2 )PEC–R(pu)]1/2 (4.27) where I max.(pu) = ratio of the maximum nonlinear current of a specified harmonic makeup that the transformer can handle to the transformer rated current. PLL–R(pu) = load loss density under rated conditions (per unit of rated load I2R loss density. PEC–R(pu) = winding eddy current loss under rated conditions (per unit of rated I2R loss). fh = harmonic current distribution factor for harmonic h (equal to harmonic h current divided by the fundamental frequency current for any given load level). h = harmonic number or order. As difficult as this formula might seem, the underlying principle is to account for the increased winding eddy current losses due to the harmonics. The following example might help clarify the IEEE expression for derating a transformer. Example: A transformer with a full load current rating of 1000 A is subjected to a load with the following nonlinear characteristics. The transformer has a rated winding eddy current loss density of 10.0% (0.10 pu). Find the transformer derating factor. Harmonic number (h) 1 3 5 7 © 2002 by CRC Press LLC fh (pu) 1 0.35 0.17 0.09 Maximum load loss density, PLL–R(pu) = 1 + 0.1 = 1.1 Maximum rated eddy current loss density, PEC–R(pu) = 0.1 Σf h2 h2 = 12 + (0.35)232 + (0.17)252 + (0.09)272 = 3.22 Σf h2 = 12 + 0.352 + 0.172 + 0.092 = 1.16 I max.(pu) = [1.1/1 + (3.22 × 0.1/1.16)]1/2 = 0.928 The transformer derating factor is 0.928; that is, the maximum nonlinear current of the specified harmonic makeup that the transformer can handle is 928 A. The ANSI/IEEE derating method is very useful when it is necessary to calculate the allowable maximum currents when the harmonic makeup of the load is known. For example, the load harmonic conditions might change on an existing transformer depending on the characteristics of new or replacement equipment. In such cases, the transformer may require derating. Also, transformers that supply large third harmonic generating loads should have the neutrals oversized. This is because, as we saw earlier, the third harmonic currents of the three phases are in phase and therefore tend to add in the neutral circuit. In theory, the neutral current can be as high as 173% of the phase currents. Transformers for such applications should have a neutral bus that is twice as large as the phase bus. 4.8.2 AC MOTORS Application of distorted voltage to a motor results in additional losses in the magnetic core of the motor. Hysteresis and eddy current losses in the core increase as higher frequency harmonic voltages are impressed on the motor windings. Hysteresis losses increase with frequency and eddy current losses increase as the square of the frequency. Also, harmonic currents produce additional I2R losses in the motor windings which must be accounted for. Another effect, and perhaps a more serious one, is torsional oscillations due to harmonics. Table 4.1 classified harmonics into one of three categories. Two of the more prominent harmonics found in a typical power system are the fifth and seventh harmonics. The fifth harmonic is a negative sequence harmonic, and the resulting magnetic field revolves in a direction opposite to that of the fundamental field at a speed five times the fundamental. The seventh harmonic is a positive sequence harmonic with a resulting magnetic field revolving in the same direction as the fundamental field at a speed seven times the fundamental. The net effect is a magnetic field that revolves at a relative speed of six times the speed of the rotor. This induces currents in the rotor bars at a frequency of six times the fundamental frequency. The resulting interaction between the magnetic fields and the rotor-induced currents produces torsional oscillations of the motor shaft. If the frequency of the oscillation coincides with the natural frequency of the motor rotating members, severe damage to the motor can result. Excessive vibration and noise in a motor operating in a harmonic environment should be investigated to prevent failures. © 2002 by CRC Press LLC Motors intended for operation in a severe harmonic environment must be specially designed for the application. Motor manufacturers provide motors for operation with ASD units. If the harmonic levels become excessive, filters may be applied at the motor terminals to keep the harmonic currents from the motor windings. Large motors supplied from ASDs are usually provided with harmonic filters to prevent motor damage due to harmonics. 4.8.3 CAPACITOR BANKS Capacitor banks are commonly found in commercial and industrial power systems to correct for low power factor conditions. Capacitor banks are designed to operate at a maximum voltage of 110% of their rated voltages and at 135% of their rated kVARS. When large levels of voltage and current harmonics are present, the ratings are quite often exceeded, resulting in failures. Because the reactance of a capacitor bank is inversely proportional to frequency, harmonic currents can find their way into a capacitor bank. The capacitor bank acts as a sink, absorbing stray harmonic currents and causing overloads and subsequent failure of the bank. A more serious condition with potential for substantial damage occurs due to a phenomenon called harmonic resonance. Resonance conditions are created when the inductive and capacitive reactances become equal at one of the harmonic frequencies. The two types of resonances are series and parallel. In general, series resonance produces voltage amplification and parallel resonance results in current multiplication. Resonance will not be analyzed in this book, but many textbooks on electrical circuit theory are available that can be consulted for further explanation. In a harmonic-rich environment, both series and parallel resonance may be present. If a high level of harmonic voltage or current corresponding to the resonance frequency exists in a power system, considerable damage to the capacitor bank as well as other power system devices can result. The following example might help to illustrate power system resonance due to capacitor banks. Example: Figure 4.17 shows a 2000-kVA, 13.8-kV to 480/277-V transformer with a leakage reactance of 6.0% feeding a bus containing two 500-hp adjustable speed drives. A 750-kVAR Y-connected capacitor bank is installed on the 480-V bus for power factor correction. Perform an analysis to determine the conditions for resonance (consult Figure 4.18 for the transformer and capacitor connections and their respective voltages and currents): Transformer secondary current (I) = 2000 × 103/ 3 × 480 = 2406 A Transformer secondary volts = (V) = 277 Transformer reactance = I × XL × 100/V = 6.0 Transformer leakage reactance (XL) = 0.06 × 277/2406 = 0.0069 Ω XL = 2πfL, where L = 0.0069/377 = 0.183 × 10–4 H © 2002 by CRC Press LLC 13.8 KV SOURCE TRANSFORMER 2000 KVA, 13.8 KV-480/277 6% REACTANCE 750 KVAR CAPACITOR BANK C I H 500 HP, ASD 500 HP, ASD FIGURE 4.17 Schematic representation of an adjustable speed drive and a capacitor bank supplied from a 2000-kVA power transformer. 6F C= 27 7V 00 18 L= 0.0 0 480 V FIGURE 4.18 Transformer and capacitor bank configuration. 0.0 08 3H TS OL 27 7V 480 VOLTS TRANSFORMER © 2002 by CRC Press LLC OL TS 902 A 2406 A CAPACITOR BANK For the capacitor bank, 3 × 480 × IC = 750 × 103, where IC = 902 A Capacitive reactance (XC) = V/IC = 277/902 = 0.307 Ω XC = 1/2πfC, where C = 1/(377 × 0.307) = 86 × 10–4 F For resonance, XL = XC; therefore, 2πfRL = 1/2πfRC where fR is the resonance frequency fR = 1/2π LC ≅ 401 Hz The resonance frequency is 401 Hz or the 6.7th (401/60) harmonic frequency. The resonance frequency is close to the seventh harmonic frequency, which is one of the more common harmonic frequency components found in power systems. This condition can have very serious effects. The following expression presents a different way to find the harmonic resonance frequency: Resonance frequency order = Rn = ( MV A SC ÷ MVAR C ) (4.28) where MVASC is the available symmetrical fault MVA at the point of connection of the capacitor in the power system, and MVARC is the rating of the capacitor bank in MVAR. In the above example, neglecting the source impedance, the available fault current = 2406 ÷ 0.06 ≅ 40,100 A. Available fault MVA = 3 × 480 × 40, 100 × 10 –6 = 33.34 Capacitor MVAR = 0.75 Therefore, the resonance frequency number = 33.34 ÷ 0.75 = 6.67, and the harmonic frequency = 6.67 × 60 = 400.2. This proves that similar results are obtained by using Eq. (4.28). The expression in Eq. (4.28) is derived as follows: The available three-phase fault current at the common bus is given by ISC = V ÷ X, where V is the phase voltage in kilovolts and X is the total reactance of the power system at the bus. ISC is in units of kiloamperes. ISC = V ÷ 2πf1L, where f1 is the fundamental frequency Short circuit MVA = MVASC = 3 × V × ISC = 3V2 ÷ 2πf1L © 2002 by CRC Press LLC From this, L = 3V 2 ÷ 2πf1(MVASC) At resonance, XLR = 2πfRL = 3V 2fR ÷ f1(MVASC) Because fR ÷ f1 = resonance frequency order, Rn, then XLR = 3V 2Rn ÷ (MVASC) For the capacitor bank, IC = V ÷ XC, and capacitor reactive power MVARC = 3 × V × IC = 3V 2(2πf1C). We can derive an expression for the capacitive reactance at resonance XCR = 3V 2 ÷ Rn(MVARC). Equating XLR and XCR, the harmonic order at resonance is the expression given by Eq. (4.28). The capacitor bank and the transformer form a parallel resonant circuit with the seventh harmonic current from the ASDs acting as the harmonic source. This condition is represented in Figure 4.19. Two adjustable speed drives typically draw a current of 550 A each, for a total load of 1100 A. If the seventh harmonic current is 5.0% of the fundamental (which is typical in drive applications), the seventh harmonic current seen by the parallel resonant circuit is 55 A = I7. If the resistance of the transformer and the associated cable, bus, etc. is 1.0%, then R ≅ 0.0012 Ω. The quality factor, Q, of an electrical system is a measure of the energy stored in the inductance and the capacitance of the system. The current amplification factor (CAF) of a parallel resonance circuit is approximately equal to the Q of the circuit: Q = 2π(maximum energy stored)/ energy dissipated per cycle Q = (2π)(1/2)LIm2 ÷ (I 2R)/f, where Im = 2I Q = X/R I H=HARMONIC CURRENT FROM ASD(S) IH L C (CAF)I H FIGURE 4.19 Parallel resonance circuit formed by transformer inductance and capacitor bank capacitance at harmonic frequency fH. © 2002 by CRC Press LLC For the seventh harmonic frequency, CAF = X7/R = 7 × 0.0069/0.0012 = 40.25. Therefore, current IR = 40.25 × 55 = 2214 A. The net current through the capacitor 2 2 bank = ( I C + I R ) = 2390 A. It is easy to see that the capacitor bank is severely overloaded. If the capacitor protective device does not operate to isolate the bank, the capacitor bank will be damaged. In the above example, by changing the capacitor bank to a 500-kVAR unit, the resonance frequency is increased to 490 Hz, or the 8.2 harmonic. This frequency is potentially less troublesome. (The reader is encouraged to work out the calculations.) In addition, the transformer and the capacitor bank may also form a series resonance circuit as viewed from the power source. This condition can cause a large voltage rise on the 480-V bus with unwanted results. Prior to installing a capacitor bank, it is important to perform a harmonic analysis to ensure that resonance frequencies do not coincide with any of the characteristic harmonic frequencies of the power system. 4.8.4 CABLES Current flowing in a cable produces I2R losses. When the load current contains harmonic content, additional losses are introduced. To compound the problem, the effective resistance of the cable increases with frequency because of the phenomenon known as skin effect. Skin effect is due to unequal flux linkage across the cross section of the conductor which causes AC currents to flow only on the outer periphery of the conductor. This has the effect of increasing the resistance of the conductor for AC currents. The higher the frequency of the current, the greater the tendency of the current to crowd at the outer periphery of the conductor and the greater the effective resistance for that frequency. The capacity of a cable to carry nonlinear loads may be determined as follows. The skin effect factor is calculated first. The skin effect factor depends on the skin depth, which is an indicator of the penetration of the current in a conductor. Skin depth (δ) is inversely proportional to the square root of the frequency: δ=S÷ f where S is a proportionality constant based on the physical characteristics of the cable and its magnetic permeability and f is the frequency of the current. If Rdc is the DC resistance of the cable, then the AC resistance at frequency f, (Rf) = K × Rdc. The value of K is determined from Table 4.9 according to the value of X, which is calculated as: X = 0.0636 f µ ÷ R dc (4.29) where 0.0636 is a constant for copper conductors, f is the frequency, µ is the magnetic permeability of the conductor material, and Rdc is the DC resistance per mile of the conductor. The magnetic permeability of a nonmagnetic material such as copper is approximately equal to 1.0. Tables or graphs containing values of X and K are available from cable manufacturers. © 2002 by CRC Press LLC TABLE 4.9 Cable Skin Effect Factor X K X K X K 0 0.1 0.2 0.3 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1 1 1 1.00004 1.00032 1.00067 1.00124 1.00212 1.0034 1.00519 1.00758 1.01071 1.0147 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.1 2.3 2.4 2.5 2.6 1.01969 1.02558 1.03323 1.04205 1.0524 1.0644 1.07816 1.09375 1.11126 1.13069 1.15207 1.17538 1.20056 2.7 2.8 2.9 3.0 3.1 3.1 3.3 3.4 3.5 3.6 3.7 3.8 3.9 1.22753 1.2662 1.28644 1.31809 1.35102 1.38504 1.41999 1.4577 1.49202 1.52879 1.56587 1.60312 1.64051 Example: Find the 60-Hz and 420-Hz resistance of a 4/0 copper cable with a DC resistance of 0.276 Ω per mile. Using Eq. (4.29), X60 = 0.0636 ( 60 × 1 ÷ 0.276 ) = 0.938 From Table 4.2, K ≅ 1.004, and R60 = 1.004 × 0.276 = 0.277 Ω per mile. Also, X420 = 0.0636 ( 420 × 1÷ 0.276 ) = 2.48 From Table 4.2, K ≅ 1.154, and R420 = 1.154 × 0.276 = 0.319 Ω per mile. The ratio of the resistance of the cable at a given frequency to its resistance at 60 Hz is defined as the skin effect ratio, E. According to this definition, E2 = resistance at second harmonic frequency ÷ resistance at the fundamental frequency = R120 ÷ R60 E3 = resistance at third harmonic frequency ÷ resistance at the fundamental frequency = R180 ÷ R60 Also, remember that the general form expression for the individual harmonic distortions states that In is equal to the RMS value of the nth harmonic current divided by the RMS value of the fundamental current, thus an expression for the current rating factor for cables can be formulated. The current rating factor (q) is the equivalent fundamental frequency current at which the cable should be rated for carrying nonlinear loads containing harmonic frequency components: 2 2 2 2 q = I 1E1 + I 2E2 + I 3E3 + … + I nEn © 2002 by CRC Press LLC (4.30) where I1, I2, I3 are the ratios of the harmonic frequency currents to the fundamental current, and E1, E2, E3 are the skin effect ratios. Example: Determine the current rating factor for a 300-kcmil copper conductor required to carry a nonlinear load with the following harmonic frequency content: Fundamental = 250 A 3rd harmonic = 25 A 5th harmonic = 60 A 7th harmonic = 45 A 11th harmonic = 20 A The DC resistance of 300-kcmil cable = 0.17 Ω per mile. Using Eq. (4.29), X60 = 0.0636 ( 60 × 1 ÷ 0.17 ) = 1.195, K ≅ 1.0106 X180 = 0.0636 ( 180 × 1 ÷ 0.17 ) = 2.069, K ≅ 1.089 X300 = 0.0636 ( 300 × 1 ÷ 0.17 ) = 2.672, K ≅ 1.220 X420 = 0.0636 ( 420 × 1 ÷ 0.17 ) = 3.161, K ≅ 1.372 X660 = 0,0636 ( 660 × 1 ÷ 0.17 ) = 3.963, K ≅ 1.664 R60 = 1.0106 × 0.17 = 0.1718 Ω/mile R180 = 1.089 × 0.17 = 0.1851 Ω/mile R300 = 1.220 × 0.17 = 0.2074 Ω/mile R420 = 1.372 × 0.17 = 0.2332 Ω/mile R660 = 1.664 × 0.17 = 0.2829 Ω/mile Skin effect ratios are: E1 = 1, E3 = 1.077, E5 = 1.207, E7 = 1.357, E11 = 1.647 The individual harmonic distortion factors are: I1 = 1.0, I3 = 25/250 = 0.1, I5 = 60/250 = 0.24, I7 = 0.18, I11 = 20/250 = 0.08 The current rating factor from Eq. (4.30) is given by: q = 1 + (0.1)2(1.077) + (0.24)2(1.207) + (0.18)2(1.357) + (0.08)2(1.647) = 1.135 © 2002 by CRC Press LLC