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  1. 1. CHAPTER 2TRANSMISSION LINE MODELSAs we have discussed earlier in Chapter 1 that the transmission line parametersinclude series resistance and inductance and shunt capacitance. In this chapter we shalldiscuss the various models of the line. The line models are classified by their length. Theseclassifications areShort line approximation for lines that are less than 80 km long.Medium line approximation for lines whose lengths are between 80 km to 250 km.Long line model for lines that are longer than 250 km.These models will be discussed in this chapter. However before that let us introduce theABCD parameters that are used for relating the sending end voltage and current to thereceiving end voltage and currents.2.1 ABCD PARAMETSRSConsider the power system shown in Fig. 2.1. In this the sending and receiving endvoltages are denoted by VS and VR respectively. Also the currents IS and IR are entering andleaving the network respectively. The sending end voltage and current are then defined interms of the ABCD parameters asRRS BIAVV (2.1)RRS DICVI (2.2)From (2.1) we see that0RIRSVVA (2.3)This implies that A is the ratio of sending end voltage to the open circuit receiving endvoltage. This quantity is dimension less. Similarly,0RVRSIVB (2.4)i.e., B, given in Ohm, is the ratio of sending end voltage and short circuit receiving endcurrent. In a similar way we can also define0RIRSVIC mho (2.5)
  2. 2. 1.310RVRSIID (2.6)The parameter D is dimension less.Fig. 2.1 Two port representation of a transmission network.2.2 SHORT LINE APPROXIMATIONThe shunt capacitance for a short line is almost negligible. The series impedance isassumed to be lumped as shown in Fig. 2.2. If the impedance per km for an l km long line isz0 = r + jx, then the total impedance of the line is Z = R + jX = lr + jlx. The sending endvoltage and current for this approximation are given byRRS ZIVV (2.7)RS II (2.8)Therefore the ABCD parameters are given by0and,1 CZBDA (2.9)Fig. 2.2 Short transmission line representation.2.2 MEDIUM LINE APPROXIMATIONMedium transmission lines are modeled with lumped shunt admittance. There are twodifferent representations nominal- and nominal-T depending on the nature of the network.These two are discussed below.2.2.1 Nominal- RepresentationIn this representation the lumped series impedance is placed in the middle while theshunt admittance is divided into two equal parts and placed at the two ends. The nominal-representation is shown in Fig. 2.3. This representation is used for load flow studies, as weshall see later. Also a long transmission line can be modeled as an equivalent -network forload flow studies.
  3. 3. 1.32Fig. 2.3 Nominal- representation.Let us define three currents I1, I2 and I3 as indicated in Fig. 2.3. Applying KCL atnodes M and N we getRRsRsIVYVYIIIIII223121(2.10)AgainRRRRRRsZIVYZVIYVZVZIV1222(2.11)Substituting (2.11) in (2.10) we getRRRRRRsIYZVYZYIVYZIVYZYI12142122(2.12)Therefore from (2.11) and (2.12) we get the following ABCD parameters of thenominal- representation12YZDA (2.13)ZB (2.14)mho14YZYC (2.15)2.2.1 Nominal-T RepresentationIn this representation the shunt admittance is placed in the middle and the seriesimpedance is divided into two equal parts and these parts are placed on either side of theshunt admittance. The nominal-T representation is shown in Fig. 2.4. Let us denote themidpoint voltage as VM. Then the application of KCL at the midpoint results in
  4. 4. 1.33Fig. 2.4 Nominal-T representation.22 ZVVYVZVV RMMMSRearranging the above equation can be written asRSM VVYZV42(2.16)Now the receiving end current is given by2ZVVI RMR (2.17)Substituting the value of VM from (2.16) in (2.17) and rearranging we getRRs IYZZVYZV 1412(2.18)Furthermore the sending end current isRMS IYVI (2.19)Then substituting the value of VM from (2.16) in (2.19) and solvingRRR IYZYVI 12(2.20)Then the ABCD parameters of the T-network are12YZDA (2.21)14YZZB (2.22)mhoYC (2.23)2.3 LONG LINE MODEL
  5. 5. 1.34For accurate modeling of the transmission line we must not assume that theparameters are lumped but are distributed throughout line. The single-line diagram of a longtransmission line is shown in Fig. 2.5. The length of the line is l. Let us consider a small stripx that is at a distance x from the receiving end. The voltage and current at the end of thestrip are V and I respectively and the beginning of the strip are V + V and I + Irespectively. The voltage drop across the strip is then V. Since the length of the strip is x,the series impedance and shunt admittance are z x and y x. It is to be noted here that thetotal impedance and admittance of the line arelyYlzZ and (2.24)Fig. 2.5 Long transmission line representation.From the circuit of Fig. 2.5 we see thatIzxVxIzV (2.25)Again as x 0, from (2.25) we getIzdxdV(2.26)Now for the current through the strip, applying KCL we getxVyxVyxyVVI (2.27)The second term of the above equation is the product of two small quantities and thereforecan be neglected. For x 0 we then haveVydxdI(2.28)Taking derivative with respect to x of both sides of (2.26) we getdxdIzdxdVdxdSubstitution of (2.28) in the above equation results
  6. 6. 1.35022yzVdxVd(2.29)The roots of the above equation are located at (yz). Hence the solution of (2.29) is of theformyzxyzxeAeAV 21 (2.30)Taking derivative of (2.30) with respect to x we getyzxyzxeyzAeyzAdxdV21 (2.31)Combining (2.26) with (2.31) we haveyzxyzxeyzAeyzAdxdVzI 211(2.32)Let us define the following two quantitiesimpedancesticcharacterithecallediswhichyzZC (2.33)constantnpropagatiothecallediswhichyz (2.34)Then (2.30) and (2.32) can be written in terms of the characteristic impedance andpropagation constant asxxeAeAV 21 (2.35)xCxCeZAeZAI 21(2.36)Let us assume that x = 0. Then V = VR and I = IR. From (2.35) and (2.36) we then get21 AAVR (2.37)CCRZAZAI 21(2.38)Solving (2.37) and (2.38) we get the following values for A1 and A2.2and221RCRRCR IZVAIZVAAlso note that for l = x we have V = VS and I = IS. Therefore replacing x by l and substitutingthe values of A1 and A2 in (2.35) and (2.36) we get
  7. 7. 1.36lRCRlRCRS eIZVeIZVV22(2.39)lRCRlRCRS eIZVeIZVI22(2.40)Noting thatleelee llllcosh2andsinh2We can rewrite (2.39) and (2.40) aslIZlVV RCRS sinhcosh (2.41)lIZlVI RCRS coshsinh(2.42)The ABCD parameters of the long transmission line can then be written aslDA cosh (2.43)lZB C sinh (2.44)CZlCsinhmho (2.45)Example 2.1: Consider a 500 km long line for which the per kilometer line impedanceand admittance are given respectively by z = 0.1 + j0.5145 and y = j3.1734 10 6mho.Therefore5.54024.40629079101734.35241.090101734.3795241.0101734.35145.01.0666jjyzZCand6419.00618.05.846448.029079500101734.35241.0 6jlyzlWe shall now use the following two formulas for evaluating the hyperbolic formssincoshcossinhsinhsinsinhcoscoshcoshjjjjApplication of the above two equations results in the following values5998.00495.0sinhand037.08025.0cosh jljl
  8. 8. 1.37Therefore from (2.43) to (2.45) the ABCD parameters of the system can be written as0015.01001.272.2404.43037.08025.05jCjBjDA2.3.1 Equivalent- Representation of a Long LineThe -equivalent of a long transmission line is shown Fig. 2.6. In this the seriesimpedance is denoted by Z while the shunt admittance is denoted by Y . From (2.21) to(2.23) the ABCD parameters are defined as12ZYDA (2.46)ZB (2.47)mho14ZYYC (2.48)Fig. 2.6 Equivalent representation of a long transmission line.Comparing (2.44) with (2.47) we can writellZyzllzllyzlZZ Csinhsinhsinhsinh (2.49)where Z = zl is the total impedance of the line. Again comparing (2.43) with (2.46) we get1sinh212cosh lZYZYl C (2.50)Rearranging (2.50) we get22tanh222tanh22tanh2tanh1sinh1cosh12llYyzllyllzylZllZYCC(2.51)where Y = yl is the total admittance of the line. Note that for small values of l, sinh l = l andtanh ( l/2) = l/2. Therefore from (2.49) we get Z = Z and from (2.51) we get Y = Y . This
  9. 9. 1.38implies that when the length of the line is small, the nominal- representation with lumpedparameters is fairly accurate. However the lumped parameter representation becomeserroneous as the length of the line increases. The following example illustrates this.Example 2.2: Consider the transmission line given in Example 2.1. The equivalentsystem parameters for both lumped and distributed parameter representation are given inTable 2.1 for three different line lengths. It can be seen that the error between the parametersincreases as the line length increases.Table 2.1 Variation in equivalent parameters as the line length changes.Length ofthe line(km)Lumped parameters Distributed parametersZ ( ) Y (mho) Z Y (mho)100 52.41 79 3.17 10 490 52.27 79 3.17 10 489.98250 131.032 79 7.93 10 490 128.81 79.2 8.0 10 489.9500 262.064 79 1.58 10 390 244.61 79.8 1.64 10 389.62.4 CHARACTERIZATION OF A LONG LOSSLESS LINEFor a lossless line, the line resistance is assumed to be zero. The characteristicimpedance then becomes a pure real number and it is often referred to as the surgeimpedance. The propagation constant becomes a pure imaginary number. Defining thepropagation constant as = j and replacing l by x we can rewrite (2.41) and (2.42) asxIjZxVV RCR sincos (2.52)xIZxjVI RCR cossin(2.53)The term surge impedance loading or SIL is often used to indicate the nominalcapacity of the line. The surge impedance is the ratio of voltage and current at any pointalong an infinitely long line. The term SIL or natural power is a measure of power deliveredby a transmission line when terminated by surge impedance and is given byCnZVPSIL20(2.54)where V0 is the rated voltage of the line.At SIL ZC = VR/IR and hence from equations (2.52) and (2.53) we getxjRxR eVeVV (2.55)xjRxR eIeII (2.56)
  10. 10. 1.39This implies that as the distance x changes, the magnitudes of the voltage and current in theabove equations do not change. The voltage then has a flat profile all along the line. Also asZC is real, V and I are in phase with each other all through out the line. The phase angledifference between the sending end voltage and the receiving end voltage is then = l. Thisis shown in Fig. 2.7.Fig. 2.7 Voltage-current relationship in naturally loaded line.2.4.1 Voltage and Current Characteristics of an SMIB SystemFor the analysis presented below we assume that the magnitudes of the voltages at thetwo ends are the same. The sending and receiving voltages are given bySS VV and 0RR VV (2.57)where is angle between the sources and is usually called the load angle. As the total lengthof the line is l, we replace x by l to obtain the sending end voltage from (2.39) assincos22RCRjRCRjRCRSS IjZVeIZVeIZVVV (2.58)Solving the above equation we getsincosCRSRjZVVI (2.59)Substituting (2.59) in (2.52), the voltage equation at a point in the transmission line that is ata distance x from the receiving end is obtained assinsinsinsinsincoscosxVxVxjZjZVVxVV RSCCRSR (2.60)In a similar way the current at that point is given bysincoscos xVxVZjI RSC(2.61)
  11. 11. 1.40Example 2.3: Consider a 500 km long line given in Example 2.1. Neglect the lineresistance such that the line impedance is z = j0.5145 per kilometer. The line admittanceremains the same as that given in Example 2.1. Then6524.402101734.35145.06jjyzZCandrad/km0013.0101734.35145.0 6jjjyzTherefore = l = 0.6380 rad. It is assumed that the magnitude of the sending and receivingend voltages are equal to 1.0 per unit with the line being unloaded, i.e., VS = VR = 1 0 perunit. The voltage and current profiles of the line for this condition is shown in Fig. 2.8. Themaximum voltage is 1.0533 per unit, while the current varies between –0.3308 per unit to0.3308 per unit. Note that 1 per unit current is equal to 1/ZC.Fig. 2.8 Voltage and current profile over a transmission line.When the system is unloaded, the receiving end current is zero (IR = 0). Therefore wecan rewrite (2.58) ascosRSS VVV (2.62)Substituting the above equation in (2.52) and (2.53) we get the voltage and current for theunloaded system asxVV Scoscos(2.63)
  12. 12. 1.41xVZjI SCsincos(2.64)Example 2.4: Consider the system given in Example 2.3. It is assumed that the systemis unloaded with VS = VR = 1 0 per unit. The voltage and current profiles for the unloadedsystem is shown in Fig. 2.9. The maximum voltage of 1.2457 per unit occurs at the receivingend while the maximum current of 0.7428 per unit is at the sending end. The current fallsmonotonically from the sending end and voltage rises monotonically to the receiving end.This rise in voltage under unloaded or lightly loaded condition is called Ferranti effect.Fig. 2.9 Voltage and current profile over an unloaded transmission line.2.4.2 Mid Point Voltage and Current of Loaded LinesThe mid point voltage of a transmission line is of significance for the reactivecompensation of transmission lines. To obtain an expression of the mid point voltage, let usassume that the line is loaded (i.e., the load angle is not equal to zero). At the mid point ofthe line we have x = l/2 such that x = /2. Let us denote the midpoint voltage by VM. Let usalso assume that the line is symmetric, i.e., VS = VR = V. We can then rewrite equation(2.60) to obtainsin2sinVVVMAgain noting that22cos2cos1sintancos221sincos1VVjVVV
  13. 13. 1.42we obtain the following expression of the mid point voltage22cos2cosVVM (2.65)The mid point current is similarly given by22sin2sinsin2cosCCMZVVVZjI (2.66)The phase angle of the mid point voltage is half the load angle always. Also the mid pointvoltage and current are in phase, i.e., the power factor at this point is unity. The variation inthe magnitude of voltage with changes in load angle is maximum at the mid point. Thevoltage at this point decreases with the increase in . Also as the power through a lossless lineis constant through out its length and the mid point power factor is unity, the mid pointcurrent increases with an increase in .Example 2.5: Consider the transmission line discussed in Example 2.4. Assuming themagnitudes of both sending and receiving end voltages to be 1.0 per unit, we can compute themagnitude of the mid point voltage as the load angle ( ) changes. This is given in Table 2.2.The variation in voltage with is shown in Fig. 2.10.Table 2.2 Changes in the mid point voltage magnitude with load anglein degree VM in per unit20 1.037325 1.028330 1.0174It is of some interest to find the Thevenin equivalent of the transmission line lookingfrom the mid point. It is needless to say that the Thevenin voltage will be the same as the midpoint voltage. To determine the Thevenin impedance we first find the short circuit current atthe mid point terminals. This is computed through superposition principle, as the short circuitcurrent will flow from both the sources connected at the two ends. From (2.52) we computethe short circuit current due to source VS (= V ) as2sin1CSCjZVI (2.67)Similarly the short circuit current due to the source VR (= V) is2sin2CSCjZVI (2.68)Thus we have
  14. 14. 1.43Fig. 2.10 Variation in voltage profile for a loaded line22sin2cos22sin21CCSCSCSCjZVjZVVIII (2.69)The Thevenin impedance is then given by2tan2CSCMTHTHZjIVjXZ (2.70)2.4.3 Power in a Lossless LineThe power flow through a lossless line can be given by the mid point voltage andcurrent equations given in (2.66) and (2.67). Since the power factor at this point is unity, realpower over the line is given bysinsin2CMMeZVIVP (2.71)If V = V0, the rated voltage we can rewrite the above expression in terms of the natural powerassinsinnePP (2.72)For a short transmission line we haveXllcclZZ CC sin (2.73)
  15. 15. 1.44where X is the total reactance of the line. Equation (2.71) then can be modified to obtain thewell known power transfer relation for the short line approximation assin2XVPe (2.74)In general it is not necessary for the magnitudes of the sending and receiving endvoltages to be same. The power transfer relation given in (2.72) will not be valid in that case.To derive a general expression for power transfer, we assumeSS VV and 0RR VVIf the receiving end real and reactive powers are denoted by PR and QR respectively, we canwrite from (2.52)sincossincosRRRCRSSVjQPjZVjVVEquating real and imaginary parts of the above equation we getsincoscosRRCRSVQZVV (2.75)andsinsinRRCSVPZV (2.76)Rearranging (2.76) we get the power flow equation for a losslees line assinsinCRSRSeZVVPPP (2.77)To derive expressions for the reactive powers, we rearrange (2.75) to obtain thereactive power delivered to the receiving end assincoscos2CRRSRZVVVQ (2.78)Again from equation (2.61) we can writesincos RSCSVVZjI (2.79)The sending end apparent power is then given by
  16. 16. 1.45sinsincossincos2CRSCSRSCSSSSSZVVjZVjVVZjVIVjQPEquating the imaginary parts of the above equation we get the following expression for thereactive generated by the sourcesincoscos2CRSSSZVVVQ (2.80)The reactive power absorbed by the line is thensincos2cos22CRSRSRSLZVVVVQQQ (2.81)It is important to note that if the magnitude of the voltage at the two ends is equal, i.e.,VS = VR = V, the reactive powers at the two ends become negative of each other, i.e.,QS = QR. The net reactive power absorbed by the line then becomes twice the sending endreactive power, i.e., QL = 2QS. Furthermore, since cos 1for small values of , the reactivepowers at the two ends for a short transmission line are given byRCS QXVZVVQ cos1sincoscos 222(2.82)The reactive power absorbed by the line under this condition is given bycos12 2XVQL (2.83)Example 2.5: Consider a short, lossless transmission line with a line reactance of 0.5per unit. We assume that the magnitudes of both sending and receiving end voltages to be 1.0per unit. The real power transfer over the line and reactive power consumed by the line areshown in Fig. 2.11. The maximum real power is 2.0 per unit and it occurs for = 90 . Alsothe maximum reactive power consumed by the line occurs at = 180 and it has a value of 8per unit.
  17. 17. 1.46Fig. 2.11 Real power flow and reactive power consumed by a transmission line.