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CHAPTER

CURRENT AND VOLTAGE

RELATIONS ON A

TRANSMISSION LINE

have examined the parameters of a transmission line and are ready to s an element of a power system. Figure 6.1 shows a 500-kV

aving bundled conductors. In overhead lines the conductors are suspended the tower and insulated from it and from each other by insulators, the

determined by the voltage of the line. Each insulator string Fig. 6.1 has 22 insulators. The two shorter arms above the phase conductors

ly made of steel. These wires, much smaller in diameter than rs, are not visible in the picture, but they are electrically

er and are therefore at ground potential. These wires are or ground wires and shield the phase conductors from

A very important problem in the design and operation of a power system e of the voltage within specified limits at various points in the

m. In this chapter we develop formulas by which we can calculate the ge, current, and power at any point on a transmission line, provided we

these values at one point, usually at one end of the line. The purpose of this chapter, however, is not merely to develop the

ns, but also to provide an opportunity to understand the ters of the line on bus voltages and the flow of power. In

193

CHAPTER 6 CURRENT AND VOI-TAGE RELATIONS ON A 1'RANSMISSION LINE

FIGURE 6.1 A 500-kV transmission line. Conductors are 76/19 ACSR with aluminum cross section of 2,515, cmil. Spacing between phases is 30 ft 3 in and the two conductors per bundle are 18 in apa (Courtesy Carolina Power and Light Company.)

6.1 REPRESENTATION OF LINES 195

this way we can see the importance of the design of the line and better understand the developments to come in later chapters. This chapter also provides an introduction to the study of transients on lossless lines in order to indicate how problems arise due to surges caused by lightning and switching.

In the modern power system data from all over the system are being fed continuously into on-line computers for control and information purposes. Power-flow studies performed by a computer readily supply answers to ques- tions concerning the effect of switching lines into and out of the system or of changes in line parameters. Equations derived in this chapter remain important, however, in developing an overall understanding of what is occurring on a system and in calculating efficiency of transmission, losses, and limits of power flow over a line for both steady-state and transient conditions.

6.1 REPRESENTATION OF LINES

The general equations relating voltage and current on a transmission line recognize the fact that all four of the parameters of a transmission line discussed in the two preceding chapters are uniformly distributed along the line. We derive these general equations later, but first we use lumped parameters which give good accuracy for short lines and for lines of medium length. If an overhead line is classified as short, shunt capacitance is so small that it can be omitted entirely with little loss of accuracy, and we need to consider only the series resistance R and the series inductance L for the total length of the line.

A medium-length line can be represented sufficiently well by R and L as lumped parameters, as shown in Fig. 6.2, with half the capacitance to neutral of the line lumped at each end of the equivalent circuit. Shunt conductance G , as mentioned previously, is usually neglected in overhead power transmission lines when calculating voltage and current. The same circuit represents the short line if capacitors are omitted.

Insofar as the handling of capacitance is concerned, open-wire 60-Hz lines less than about 80 km (50 mi) long are short lines. Medium-length lines are roughly between 80 km (50 mi) and 240 km (150 mi) long. Lines longer than 240 km (150 mi) require calculations in terms of distributed constants if a high degree of accuracy is required, although for some purposes a lumped-parameter representation can be used for lines up to 320 km (200 mi) long.

kc- 2 C ' FIGURE 6.2 2 Single-phase equivalent of a medium- length line. The capacitors are omitted - 1 V - for a short line. - -

196 CHAPTER 6 CURRENT A N D VOLTAGE RELATIONS ON A TRANSMISSION LINE

Normally, transmission lines are operated with balanced three-phase loads. Although the lines are not spaced equilaterally and not transposed, the resulting dissymmetry is slight and the phases are considered to be balanced.

In order to distinguish between the total series impedance of a line and the series impedance per unit length, the following nomenclature is adopted:

z = series impedance per unit length per phase

y = shunt admittance per unit length per phase to neutral

1 = length of line

Z = zl = total series impedance per phase

Y = yl = total shunt admittance per phase to neutral

6.2 THE SHORT TRANSMISSION LINE

The equivalent circuit of a short transmission line is shown in Fig. 6.3, where I, and IR are the sending- and receiving-end currents, respectively, and Vs and V, are the sending- and receiving-end line-to-neutral voltages.

The circuit is solved as a simple series ac circuit. So,

where Z is zl, the total series impedance of the line. The effect of the variation of the power factor of the load on the voltage

regulation of a line is most easily understood for the short line and therefore will be considered at this time. Voltage regulation of a transmission line is the rise in voltage at the receiving end, expressed in percent of full-load voltage, when full load at a specified power factor is removed while the sending-end

Gen.

FIGURE 6.3 Equivalent circuit of a short transmission line where the resistance R and inductance L are values for the entire length of the line.

6.2 THE SHORT TRANSMISSION LINE 197

( b ) Load p.f.=100% ( c ) Load p.f.=70% lead

r diagrams of a short transmission line. A11 diagrams are drawn for the same magnitudes of VR

ltage is held constant. Corresponding to Eq. (2.33) we can write

~ V R , NLI - ~ V R , F L ~ loo Percent regulation = JVR, FLJ

ere IVR, ,I is the magnitude of receiving-end voltage at no load and IVR, ,I is age at full load with IV,I constant. After the

a short transmission line, represented by the circuit of Fig. 6.3, is , the voltage at the receiving end is equal to the voltage at the sending

cted, the receiving-end voltage is designated g-end voltage is V,, and IVs) = IVR,,I. The

for the same magnitudes of the receiving- t a larger value of the sending-end voltage iving-end voltage when the receiving-end hen the same current and voltage are in ltage is required to maintain the given ing-end current leads the voltage. The

ltage drop is the same in the series impedance of the line in all cases; because the different power factors, however, the voltage drop is added to the

e in each case. The regulation is greatest ven negative, for leading power factors. n line is larger than the resistance, and

rinciple of regulation illustrated in Fig. 6.4 is true for any load supplied by dominantly inductive circuit. The magnitudes of the voltage drops IRR and for a short line have been exaggerated with respect to VR in drawing the

diagrams in order to illustrate the point more clearly. The relation longer lines is similar to that for short

Example 6.1. A 300-MVA 20-kV three-phase generator has a subtransient reactance of 20%. The generator supplies a number of synchronous motors over a 64-km transmission line having transformers at both ends, as shown on the one-line diagram of Fig. 6.5. The motors, all rated 13.2 kV, are represented by just

198 CHAPTER 6 CURRENT AND VOLTAGE RELATIONS ON A TRANSMISSION LINE

FIGURE 6.5 One-line diagram for Example 6.1.

two equivalent motors. The neutral of one motor M, is grounded througl reactance. The ncutral of thc second motor M2 is not connected to ground (ar unusual condition). Rated inputs to the motors are 200 MVA and 100 kVA for M and M2, respectively. For both motors X: = 20%. The three-phase transformel T, is rated 350 MVA, 230/20 kV with leakage reactance of 10%. Transformer T; is composed of three single-phase transformers, each rated 127/13.2 kV, 1(H MVA with leakage reactance of 10%. Series reactance of the transmission line is 0.5 fk/km. Draw the reactance diagram with all reactances marked in per unit Select the generator rating as base in the generator circuit.

Solution. The three-phase rating of transformer T, is

and its line-to-line voltage ratio is

A base of 300 MVA, 20 kV in the generator circuit requires a 300-MVA base in all parts of the system and the following voltage bases:

In the transmission line: 230 kV (since T , is rated 230/20 kV)

13.2 In the motor circuit: 230- = 13.8 kV

220

These bases are shown in parentheses on the one-line diagram of Fig. 6.5. The reactances of the transformers converted to the proper base are

Transformer T, :

Transformer T, :

300 X = 0.1 X - = 0.0857 per unit

350

X = 0.1 - = 0.0915 per unit ( :::: l2

6.2 M E SHORT TRANSMISSION LINE 199

FIGURE 6.6 Reactance diagram for Example 6.1. Reactances are in per unit on the specified base.

The base impedance of the transmission line is

and the reactance of the line is

0.5 X 64 = 0.1815 per unit

176.3

Reactance Xs of motor M, = 0.2 ( - ) (KI2 = 0.2745 per unit

Reactance Xi of motor M, = 0.2 ( - ::: ) (g)' = 0.5490 per unit

Figure 6.6 is the required reactance diagram when transformer phase shifts are omitted.

Example 6.2. If the motors M, and M , of Example 6.1 have inputs of 120 and 60 MW, respectively, at 13.2 kV, and both operate at unity power factor, find the voltage at the terminals of the generator and the voltage regulation of the line.

Solution. Together the motors take 180 MW, or

180 - = 0.6 per unit 300

Therefore, with V and I at the motors in per unit,


With phase-a voltage at the motor terminals as reference, we have

13.2 v = - = 13.8

0.9565& per unit

0.6 1 = - -

0.9565 - 0.6273/00 per unit

Phase-a per-unit voltages at other points of Fig. 6.6 arc

At m : V = 0.9565 + 0.6273(j0.0915)

0.9565 + j0.0574 = 0.958213.4340 per unit

V = 0.9565 + 0.6273(j0.0915 + j0.1815)

0.9565 + j0.1713 = 0.97171 10.1540 per unit

V = 0.9565 + 0.6273(j0.0915 + j0.1815 + j0.0857)

0.9565 + j0.2250 = 0.98261 13.237' per unit

The voltage regulation of the line is

0.9826 - 0.9582 Percent regulation = x 100 = 2.55%

0.9582

and the magnitude of the voltage at thc gcncrator terminals is

If it is desired to show the phase shifts due to the Y - A transformers, the angles of the phase-a voltages at rn and 1 should be increased by 30'. Then the angle of the phase-a current in the line should also be increased by 30" from 0'.

6.3 THE MEDIUM-LENGTH LINE

The shunt admittance, usually pure capacitance, is included in the calculations for a line of medium length. If the total shunt admittance of the line is divided into two equal parts placed at the sending and receiving ends of the line, the circuit is called a nominal T. We refer to Fig. 6.7 to derive equations. To obtain an expression for V,, we note that the current in the capacitance at the

6.3 THE MEDIUM-LENGTH LINE 201

receiving end is VRY/2 and the current in the series arm is IR + VRY/2. Then,

's z 5-

t - vs - 2 - Y - I; t +

To derive I,, we note that the current in the shunt capacitance at the sending end is VsY/2, which added to the current in the series arm gives

I -

Substituting Vs, as given by Eq. ( 6 . 9 , in Eq. (6.6) yields

FIGURE 6.7 Nominal* circuit of a medium-length

Equations (6.5) and (6.7) may be expressed in the general form

- - transmission line.

where

These ABCD constants are sometimes called the generalized circuit constants of the transmission line. In general, they are complex numbers. A and D are dimensionless and equal each other if the line is the same when viewed from either end. The dimensions of B and C are ohms and mhos o r siemens,


respectively. The constants apply to any linear, passive, and bilateral four-terminal network having two pairs of terminals. Such a network is called a two-port net work.

A physical meaning is easily assigned to the constants. By letting IR be zero in Eq. (6.8), we see that A is the ratio Vs/VR at no load. Similarly, B is the ratio Vs/IR when the receiving end is short-circuited. The constant A is useful in computing regulation. If V,, ., is the receiving-end voltage at full load for a sending-end voltage of V,, Eq. (6.3) becomes

Ivsl/lAl - IVR.FLI loo Percent regulation = 1 % . FLI

Table A.6 in the Appendix lists ABCD constants for various networks and combinations of networks.

6.4 THE LONG TRANSMISSION LINE: SOLUTION OF THE DIFFERENTIAL EQUATIONS

The exact solution of any transmission line and the one required for a high degree of accuracy in calculating 60-Hz lines more than approximately 150 mi long must consider the fact that the parameters of the lines are not lumped but, rather, are distributed uniformly throughout the length of the line.

Figure 6.8 shows one phase and the neutral connection of a three-phase line. Lumped parameters are not shown because we are ready to consider the solution of the line with the impedance and admittance uniformly distributed. In Fig. 6.8 we consider a differential element of length dx in the line at a distance x from the receiving end of the line. Then z d3c and y d u are, respectively, the series impedance and shunt admittance of the elemental section. V and I are phasors which vary with x.

Gen.

FIGURE 6.8 Schematic diagram of a transmission line showing one phase and the neutral return. Nomenclature for the line and the elemental length are indicated.

Average line current in the element is ( I + I + d I ) / 2 , and the increase of V in the distance dx is quite accurately expressed as

when products of the differential quantities are neglected. Similarly,

V + V + d V d l =

2 y d x = vydx

Then, from Eqs. (6.12) and (6.13) we have

and

Let us differentiate Eqs. (6.14) and (6.15) with respect to x , and we obtain

and

If we substitute the values of d l / & and dV/dx from Eqs. (6.15) and (6.14) in Eqs. (6.16) and (6.17), respectively, we obtain

and

Now we have Eq. (6.18) in which the only variables are V and x and Eq. (6.19) in which the only variables are I and x. The solutions of those equations for V and I , respectively, must be expressions which when differentiated twice with respect to x yield the original expression times the constant yz. For instance, the solution for V when differentiated twice with respect to x must yield yzV.

204 CHAPTER 6 CURRENT A N D VOLTAGE RE1.ATIONS ON A TRANSMISSION LINE

This suggests an exponential form of solution. Assume that the solution of Eq. (6.18) is

Taking the second derivative of V with respect to x in Eq. (6.20) yields

which is yz times the assumed solution for V . Therefore, Eq. (6.20) is the solution of Eq. (6.18). When we substitute the value given by Eq. (6.20) for V in Eq. (6.14), we obtain

The constants A, and A , can be evaluated by using the conditions at the receiving end of the line; namely, when x = 0, V = VR and I = I,. Substitution of these values in Eqs. (6.20) and (6.22) yields

1 V R = A , + A 2 and I R = - ( A , - A 2 ) m

Substituting Zc = and solving for A , give

------------

A , = vR + IRZc 7 - 1 R L ,

2 and A, =

2

Then, substituting the values found for A , and A, in Eqs. (6.20) and (6.22) and letting y = 6, we obtain

where Zc = and is called the characteristic impedance of the line, and y = \/zy and is called the propagation constant.

Equations (6.23) and (6.24) give the rms values of V and I and their phase angles at any specified point along the line in terms of the distance x from the receiving end to the specified point, provided I.',, I,, and the parameters of the line are known.

i 6.5 THE LONG TRANSMISSION LINE: ? INTERPRETATION OF THE EQUATIONS

. Both y and Zc are complex quantities. The real part of the propagation L constant y is called the attenuation constant a and is measured in nepers per (.

$ unit length. The quadrature part of y is called the phase constant P and is f measured in radians per unit length. Thus, 2

I y = a + j p

and Eqs. (6.23) and (6.24) become

I v~ + I R Z c E a x E j p x + VR - I~zc -ax -]pX v = 2

E E 2

v ~ / z c + IR E a ~ E I P ~ - v ~ / z C - IR -axE-jpx

/ and

I = 2 2

E

f i e properties of eaX and &IPx help to explain the variation of the phasor , values of voltage and current as a function of distance along the line. The term 4 changes in magnitude as x changes, but &Ipx (identical to cos p x + j sin p x ) I t, always has a magnitude of 1 and causes a shift in phase of P radians per unit , length of line.

The first term in Eq. (6.26), [(VR + IRZc)/2]~"x&'Px, increases in magni- ! tude and advances in phase as distance x from the receiving end increases. ! Conversely, as progress along the line from the sending end toward the 1. receiving end is considered, the term diminishes in magnitude and is retarded in 5 phase. This is the characteristic of a traveling wave and is similar to the ! behavior of a wave in water, which varies in magnitude with time at any point,

whereas its phase is retarded and its maximum value diminishes with distance

I from the origin. The variation in instantaneous value is not expressed in the term but is understood since VR and IR are phasors. The first term in Eq. (6.26) is called the incident voltage.

I The second term in Eq. (6.261, [(VR - I , Z ~ ) / ~ ] E - " ~ E - ~ ~ ~ , diminishes in magnitude and is retarded in phase from the receiving end toward the sending

t, end. It is called the reflected voltage. At any point along the line the voltage is [ the sum of the component incident and reflected voltages at that point.

Since the equation for current is similar to the equation for voltage, the current may be considered to be composed of incident and reflected currents.

If a line is terminated in its characteristic impedance Zc, receiving-end voltage VR is equal to I,Zc and there is no reflected wave of either voltage or current, as may be seen by substituting IRZc for VR in Eqs. (6.26) and (6.27). A line terminated in its characteristic impedance is called a flat line or an infinite line. The latter term arises from the fact that a line of infinite length cannot have a reflected wave. Usually, power lines are not terminated in their characteristic impedance, but communication lines are frequently so terminated in


order to eliminate the reflected wave. A typical value of Z, is 400 R for a single-circuit overhead line and 200 R for two circuits in parallel. The phase angle of 2, is usually between 0 and - 15". Bundled-conductor lines have lower values of Z, since such lines have lower L and higher C than lines with a single conductor per phase.

In power system work characteristic impedance is sometimes called surge impedance. The term "surge impedance," however, is usually reserved for the special case of a lossless line. If a line is lossless, its series resistance and shunt conductance are zero and the characteristic impedance reduces to the real number d m , which has the dimensions of ohms when L is the series inductance of the line in henrys and C is the shunt capacitance in farads. Also, the propagation constant y = fi for the line of length 1 reduces to the imaginary number jp = j w m / l since the attenuation constant a resulting from line losses is zero. When dealing with high frequencies or with surges due to lightning, losses are often neglected and the surge impedance becomes important. Surge-impedance loading (SIL) of a line is the power delivered by a line to a purely resistive load equal to its surge impedance. When so loaded, the line supplies a current of

where IV,I is the line-to-line voltage at the load. Since the load is pure resistance,

IVLl SIL = J?;Iv,I 6 x d m W

or with IVLI in kilovolts,

IVL12 SIL = - dL/C MW

Power system engineers sometimes find it convenient to express the power transmitted by a line in terms of per unit of SIL, that is, as the ratio of the power transmitted to the surge-impedance loading. For instance, the permissi- ble loading of a transmission line may be expressed as a fraction of its SIL, and SIL provides a comparison of load-carrying capabilities of lines.'

A wavelength A is the distance along a line between two points of a wave which differ in phase by 360°, or 27r rad. If p is the phase shift in radians per

' see R. D. Dunlop, R. Gutman, and P. P. Marchenko, "Analytical Development of Loadability Characteristics for EHV and UHV Transmission Lines," lEEE Transactions on Power Apparatus and Systems, vol. PAS-98, no. 2, 1979, pp. 606-617.

6.6 THE LONG TRANSMISSION LINE: HYPERBOLIC FORM OF THE EQUATIONS 207

mile, the wavelength in miles is

The velocity of propagation of a wave in miles per second is the product of the wavelength in miles and the frequency in hertz, or

2 ~ f Velocity = A f = -

P

For the lossless line of length I meters P = 2 7 ~ f m / l and Eqs. (6.29) and (6.30) become

A = - fm velocity = - m When values of L and C for low-loss overhead lines are substituted in these equations, it is found that the wavelength is approximately 3000 mi at a frequency of 60 Hz and the velocity of propagation is very nearly the speed of light in air (approximately 186,000 mi/s or 3 x lo8 m/s).

If there is no load on a line, I, is equal to zero, and as determined by Eqs. (6.26) and (6.271, the incident and reflected voltages are equal in magnitude and in phase at the receiving end. In this case the incident and reflected currents are equal in magnitude but are 180" out of phase at the receiving end. Thus, the incident and reflected currents cancel each other at the receiving end of an open line but not at any other point unless the line is entirely lossless so that the attenuation a is zero.

6.6 THE LONG TRANSMISSION LINE: HYPERBOLIC FORM OF THE EQUATIONS

The incident and reflected waves of voltage are seldom found when calculating the voltage of a power line. The reason for discussing the voltage and the current of a line in terms of the incident and reflected components is that such an analysis is helpful in obtaining a better understanding of some of the phenomena of transmission lines. A more convenient form of the equations for computing current and voltage of a power line is found by introducing hyperbolic functions. Hyperbolic functions are defined in exponential form

E 6 - &-6

sinh 8 = 2

E~ + & - e cosh 8 =

2


By rearranging Eqs. (6.23) and (6.24) and substituting hyperbolic functions for the exponential terms, we find a new set of equations. The new equations, giving voltage and current anywhere along the line, are

V = VR cosh yx + IRZ, sinh yx

VR I = I, cosh yx + - sinh yx

zc Letting x = 1 to obtain the voltage and the current at the sending end, we have

Vs = VR cosh yl + IRZc sinh yl

From examination of these equations we see that the generalized circuit constants for a long line are

A = cosh y l sinh y l c=-

zc B = Zcsinh y l D = cosh yl

Solving Eqs. (6.35) and (6.36) for VR and I, in terms of Vs and Is, we obtain

VR = Vs cosh yl - IsZc sinh y l

vs IR = Is cosh yl - - sinh yl

zc For balanced three-phase lines the currents in the above equations are

line currents and the voltages are line-to-neutral voltages, that is, line voltages divided by a. In order to solve the equations, the hyperbolic functions must be evaluated. Since yl is usually complex, the hyperbolic functions are also complex and can be evaluated with the assistance of a calculator or computer.

For solving an occasional problem without resorting to a computer there are several choices. The following equations give the expansions of hyperbolic sines and cosines of complex arguments in terms of circular and hyperbolic functions of real arguments:

cosh(a1 + jpl ) = cosh a1 cos p l + j sinh a1 sin p l

sinh(a1 + jpl ) = sinh a1 cos p l + j cosh a1 sin p l

Equations (6.40) and (6.41) make possible the computation of hyperbolic func-


: tions of complex arguments. The correct mathematical unit for /3l is the radian,

! and the radian is the unit found for pl by computing the quadrature component - of 71. Equations (6.40) and (6.41) can be verified by substituting in them the t exponential forms of the hyperbolic functions and the similar exponential forms ; of the circular functions. I Another method of evaluating complex hyperbolic functions is suggested

by Eqs. (6.31) and (6.32). Substituting a + j/3 for 6, we obtain

Example 6.3. A single-circuit 60-Hz transmission line is 370 km (230 mi) long. The conductors are Rook with flat horizontal spacing and 7.25 m (23.8 ft) between conductors. The load on the line is 125 MW at 215 kV with 100% power factor. Find the voltage, current, and power at the sending end and the voltage regulation of the line. Also, determine the wavelength and velocity of propagation of the line.

Solution. Feet and miles rather than meters and kilometers are chosen for the calculations in order to use Tables A.3 through A.5 in the Appendix:

and from the tables for Rook

215,000 v, = ----- =

6 1 2 4 , 1 3 0 b V to neutral


From Eqs. (6.42) and (6.43) and noting that 0.4750 rad = 27.22'

sinh yl = 0.4654 + j0.2394 - 0.4248 + j0.2185

Then, from Eq. (6.35)

and from Eq. (6.36)

At the sending end

Line voltage = fi X 137.86 = 238.8 kV

Line current = 332.3 A

Power factor = cos(27.77' - 26.33") = 0.9997 z 1.0

Power = 6 x 238.8 x 332.3 x 1.0 = 137,443 kW

From Eq. (6.35) we see that at no load ( I , = 0)


So, the voltage regulation is

The wavelength and velocity of propagation are computed as follows:

Velocity = fA = 60 x 3043 = 182,580 mi/s

We note particularly in this example that in the equations for Vs and I, the value of voltage must be expressed in volts and must be the line-to-neutral voltage.

Example 6.4. Solve for the sending-end voltage and the current found in Example 6.3 using per-unit calculations.

Solution. We choose a base of 125 MVA, 215 kV to achieve the simplest per-unit values and to compute base impedance and base current as follows:

215' Base impedance = - =

125 370 R

125,000 Base current = = 335.7 A 6 X 215

2, = 406.4/ - 5 .4g0

370 = 1.0981 - 5.48O per unit

215 2 1 5 / 6 v --=-- - 215 2 1 5 / 6

- 1.0 per unit

For use in Eq. (6.35) we chose V, as the reference voltage. So,

V, = 1 .o/ per unit (as a line-to-neutral voltage)

and since the load is at unity power factor,

= 1.1102/27.75~ per unit

and by Eq. (6.36)

= 0.990/ 26.350 per unit

At the sending end

Line voltage = 1.1102 X 215 = 238.7 kV

Line current = 0.990 x 335.7 = 332.3 A

Note that we multiply line-to-line voltage base by the pei-unit magnitude of the voltage to find the line-to-line voltage magnitude. We could have multiplied the line-to-neutral voltage base by the per-unit voltage to find the line-to-neutral voltage magnitude. The factor 6 does not enter the calculations after we have expressed all quantities in per unit.

6.7 THE E Q U I V U N T CIRCUIT OF A LONG LINE

The nomina1-r circuit does not represent a transmission line exactly because it does not account for the parameters of the line being uniformly distributed. The discrepancy between the nominal T and the actual line becomes larger as the length of line increases. It is possible, however, to find the equivalent circuit of a long transmission line and to represent the line accurately, insofar as measure- ments at the ends of the line are concerned, by a network of lumped parameters. Let us assume that a T circuit similar to that of Fig. 6.7 is the equivalent circuit of a long line, but let us call the series arm of our equivalent* circuit 2' and the shunt arms Y'/2 to distinguish them from the arms of the nominal-n '

circuit. Equation (6.5) gives the sending-end voltage of a symmetrical-rr circuit , in terms of its series and shunt arms and the voltage and current at the receiving end. By substituting 2' and Y'/2 for Z and Y/2 in Eq. (6.51, we obtain the i

sending-end voltage of our equivalent circuit in terms of its series and shunt I

6.7 THE EQUIVALENT CIRCUIT OF A LONG LINE 213

arms and the voltage and current at the receiving end:

For our circuit to be equivalent to the long transmission line the coefficients of V, and IR in Eq. (6.44) must be identical, respectively, to the coefficients of VR and I, in Eq. (6.35). Equating the coefficients of I, in the two equations yields

Z' = Z, sinh yl

sinh yl Z' = fi sinh yl = il-

f i l

sinh yl Z' = 2---

yl

, where Z is equal to 21, the total series impedance of the line. The term (sinh y l ) / y l is the factor by which the series impedance of the nominal rr must

, be multiplied to convert the nominal rr to the equivalent T. For small values of , yl, both sinh yl and yl are almost identical, and this fact shows that the nominal T represents the medium-length transmission line quite accurately, insofar as the series arm is concerned.

To investigate the shunt arms of the equivalent-T circuit, we equate the coefficients of VR in Eqs. (6.35) and (6.44) and obtain

Z'Y'

2 - + 1 = coshyl

I Substituting 2, sinh yl for 2' gives

Y'Z, sinh yl

2 + 1 = cosh yl

Y' 1 cosh yl - 1 - - - - 2 Z, sinh yl

Another form of the expression for the shunt admittance of the equivalent circuit can be found by substituting in Eq. (6.49) the identity

yl cosh yl - 1 tanh - =

2 sinh yl

t The identity can be verified by substituting the exponential forms of Eqs. (6.31)

2 Y tanh 71/2 6'9

= 2 r1/2 Equivalent-a circuit of a transmission line.

and (6.32) for the hyperbolic functions and by recalling that tanh 8 =

sinh 8/cosh 8. Now

Y' 1 - - 71

- - tanh - 2 2, 2

where Y is equal to yl, the total shunt admittance of the line. Equation (6.52) shows the correction factor used to convert the admittance of the shunt arms of the nominal T to that of the equivalent T. Since tanh(yl/2) and y1/2 are very nearly equal for small values of yl, the nominal rr represents the medium-length transmission line quite accurately, for we have seen previously that the correction factor for the series arm is negligible for medium-length lines. The equivalent-rr circuit is shown in Fig. 6.9. An equivalent-T circuit can also be found for a transmission line.

Example 6.5. Find the equivalent-r circuit for the line described in Example 6.3 and compare it with the nominal-.rr circuit.

Solution. Since sinh yl and cosh yl are already known from Example 6.3, Eqs. (6.45) and (6.49) are now used.

2' = 406.41 - 5.48' x 0.45971 84.930 = 186.821 79.45. R in series arm

= 0.000599/ 89.820 S in each shunt arm

Using the values of z and y from Example 6.3, we find for the nominal-.rr circuit a series impedance of

6.8 POWER FLOW T H R O U G H ATRANSMISSION LINE 215

and equal shunt arms of

For this line the impedance of the series arm of the nominal IT exceeds that of the equivalent 71- by 3.8%. The conductance of the shunt arms of the nominal 71- is 2.0% less than that of the equivalent IT.

We conclude from the preceding example that the nominal .rr may represent long lines sufficiently well if a high degree of accuracy is not required.

6.8 POWER FLOW THROUGH A .TRANSMISSION LINE

:Although power flow at any point along a transmission line can always be found if the voltage, current, and power factor are known or can be calculated, very interesting equations for power can be derived in terms of ABCD constants. The equations apply to any network of two ports or two terminal pairs. Repeating Eq. (6.8) and solving for the receiving-end current IR yields

ye obtain

ken, the complex power VR I,* at the receiving end is

216 CHAPTER 6 CURRENT AND VOLTAGE RELATIONS ON A TKANSMISSION LINE

FIGURE 6.10 Phasors of Eq. (6.56) plotted in the complex plane, with

I magnitudes and angles as indicated.

and real and reactive power at the receiving end are

Noting that the expression for complex power PR + jQR is shown by Eq. (6.56) to be the resultant of combining two phasors expressed in polar form, we can plot these phasors in the complex plane whose horizontal and vertical coordinates are in power units (watts and vars). Figure 6.10 shows the two complex quantities and their difference as expressed by Eq. (6.56). Figure 6.11 shows the same phasors with the origin of the coordinate axes shifted. This figure is a power diagram with the resultant whose magnitude is IPR + jQRl, or IVRI IIR1, at an angle OR with the horizontal axis. As expected, the real and imaginary components of 1 PR + jQRl are

where 6, is the phase angle by which VR leads I,, as discussed in Chap. 1. The sign of Q is consistent with the convention which assigns positive values to Q when current is lagging the voltage.

Now let us determine some points on the power diagram of Fig. 6.11 for various loads with fixed values of IVsl and IVRI. First, we notice that the position of point n is not dependent on the current IR and will not change so long as

6.8 POWER FLOW THROUGH A TRANSMISSION LINE 217

var + I

FIGURE 6.11 Power diagram obtained by shifting the origin of the coordinate axes of Fig. 6.10.

IV,I is constant. We note further that the distance from point n to point k is constant for fked values of IVsI and IVRI. Therefore, as the distance 0 to k changes with changing load, the point k, since it must remain a t a constant distance from the fixed point n, is constrained to move in a circle whose center is at n. Any change in PR will require a change in QR to keep k on the circle. If a different value of IVsI is held constant for the same value of IV,I, the location of point n is unchanged but a new circle of radius nk is found.

Examination of Fig. 6.11 shows that there is a limit to the power that can be transmitted to the receiving end of the line for specified magnitudes of sending- and receiving-end voltages. An increase in power delivered means that the point k will move along the circle until the angle P - 6 is zero; that is, more power will be delivered until S = P. Further increases in S result in less power received. The maximum power is

- I I - IAI lvRI2 'R, max - IBI IBI

C O S ( ~ - a)

The load must draw a large leading current to achieve the condition of maximum power received. Usually, operation is limited by keeping S less than about 35" and IVsI/IVRI equal to or greater than 0.95. For short lines thermal ratings limit the loading.

In Eqs. (6.53) through (6.61) IVs( and IVRI are line-to-neutral voltages and coordinates in Fig. 6.11 are watts and vars per phase. However, if lVsl and IVR( are line-to-line voltages, each distance in Fig. 6.11 is increased by a factor of 3 and the coordinates on the diagram are total three-phase watts and vars. If the voltages are kilovolts, the coardinates are megawatts and megavars.


6.9 REACTIVE COMPENSATION OF TRANSMISSION LINES

The performance of transmission lines, especially those of medium length and longer, can be improved by reactive compensation of a series or parallel type. Series compensation consists of a capacitor bank placed in series with each phase conductor of the line. Shunt compensation refers to the placement of inductors from each line to neutral to reduce partially or completely the shunt susceptance of a high-voltage line, which is particularly important at light loads when the voltage at the receiving end may otherwise become very high.

Series compensation reduces the series impedance of the line, which is the principal cause of voltage drop and the most important factor in determining the maximum power which the line can transmit. In order to understand the effect of series impedance Z on maximum power transmission, we examine Eq. (6.61) and see that maximum power transmitted is dependent on the reciprocal of the generalized circuit constant B, which for the nominal-.rr equals Z and for the equivalent-rr equals Z (sinh y l ) / y l . Because the A, C, and D constants are functions of Z, they will also change in value, but these changes will be small in comparison t o the change in 5.

The desired reactance of the capacitor bank can be determined by compensating for a specific amount of the total inductive reactance of the line. This leads to the term "compensation factor," which is defined by Xc/XL, where Xc is the capacitive reactance of the series capacitor bank per phase and X, is the total inductive reactance of the line per phase.

When the nominal-T circuit is used to represent the line and capacitor bank, the physical location of the capacitor bank along the line is not taken into account. If only the sending- and receiving-end conditions of the line are of interest, this will not create any significant error. However, when the operating conditions along the line are of interest, the physical location of the capacitor ----- --------

bank must be taken into account, This c a n b e accomp7isEiimosfxasily%y determining ABCD constants of the portions of line on each side of the capacitor bank and by representing the capacitor bank by its ABCD constants. The equivalent constants of the combination (actually referred to as a cascaded connection) of line-capacitor-line can then be determined by applying the equations found in Table A.6 in the Appendix.

In the southwestern part of the United States series compensation is especially important because large generating plants are located hundreds of miles from load centers and large amounts of power must be transmitted over long distances. The lower voltage drop in the line with series compensation is an additional advantage. Series capacitors are also useful in balancing the voltage drop of two parallel lines.

Example 6.6. In order to show the relative changes in the B constant with respect to the change of the A, C, and D constants of a line as series compensation is applied, find the constants for the line of Example 6.3 when uncompensated and for a series compensation of 70%.

Solution. The equivalent< circuit and found in Examples 6.3 and 6.5 can be used with Eqs. (6.37) to find, for the uncompensated line

sinh yl 0.45961 84.940 c=-= 406.4/ - 5.48"

The series compensation alters only the series arm of the equivalentq circuit. The new series arm impedance is also the generalized constant B. So,

and by Eqs. (6.10)

The example shows that compensation has reduced the constant B to about one-third of its value for the uncompensated line without affecting the A and C constants appreciably. Thus, maximum power which can be transmitted is increased by about 300%.

When a transmission line, with or without series compensation, has the desired load transmission capability, attention is turned to operation under light loads or at no load. Charging current is an important factor to be considered and should not be allowed to exceed the rated full-load current of the line.

Equation (5.25) shows us that the charging current is usually defined as B,IVI if Bc is the total capacitive susceptance of the line and IV) is the rated voltage to neutral. As noted following Eq. (5.251, this calculation is not an exact determination of charging current because of the variation of I V ( along the line. If we connect inductors from line to neutral at various points along the line so


that the total inductive susceptance is BL, the charging current becomes

We recognize that the charging current is reduced by the term in parentheses. The shunt compensation factor is BL/Bc.

The other benefit of shunt compensation is the reduction of the receiving- end voltage of the line which on long high-voltage lines tends to become too high at no load. In the discussion preceding Eq. (6.11) we noted that IVsl/lAI equals IV,,,,I. We also have seen that A equals 1.0 when shunt capacitance is neglected. In the medium-length and longer lines, however, the presence of capacitance reduces A. Thus, the reduction of the shunt susceptance to the value of (Bc - BL) can limit the rise of the no-load voltage at the receiving end of the line if shunt inductors are introduced as load is removed.

By applying both series and shunt compensation to long transmission lines, we can transmit large amounts of power efficiently and within the desired voltage constraints. Ideally, the series and shunt elements should be placed at intervals along the line. Series capacitors can be bypassed and shunt inductors can be switched off when desirable. As with series compensation, ABCD constants provide a straightforward method of analysis of shunt compensation.

Example 6.7. Find the voltage regulation of the line of Example 6.3 when a shunt inductor is connected at the receiving end of the line during no-load conditions if the reactor compensates for 70% of the total shunt admittance of the line.

Solution. From Example 6.3 the shunt admittance of the line is

and for the entire line

B, = 5.105 x x 230 = 0.001174 S

For 70% compensation

We know the ABCD constants of the line from Example 6.6. Table A.6 of the Appendix tells us that the inductor alone is represented by the generalized constants

A = D = 1 B = 0 C = -jB, = -jO.O00822S

The equation in Table A.6 for combining two networks in series tells us that for

1 i / the line and inductor

6.10 TRANSMISSION-LINE TRANSIENTS 221

b: 6 The voltage regulation with the shunt reactor connected at no load becomes

137.86/1.0411 - 124.13 = 6.67%

124.13

which is a considerable reduction from the value of 24.7% for the regulation of the uncompensated line.

k10 TRANSMISSION-LINE TRANSIENTS

e transient overvoltages which occur on a power system are either of external in (for example, a lightning discharge) or generated internally by switching ations. In general, the transients on transmission systems are caused by any en change in the operating condition or configuration of the systems.

ightning is always a potential hazard to power system equipment, but switching erations can also cause equipment damage. At voltages up to about 230 kV,

the insulation level of the lines and equipment is dictated by the need to protect against lightning. On systems where voltages are above 230 kV but less than 700 kV switching operations as well as lightning are potentially damaging to insulation. At voltages above 700 kV switching surges are the main determinant of the level of insulation. ~ - - - O f c e u ~ ~ e , ~ r g r ~ ~ c a b k ~ e i m r n u n e to direct lightning strokes and fcan be protected against transients originating on overhead lines. K o w e t ~ ; f m $economic and technical reasons overhead lines at transmission voltage levels

i prevail except under unusual circumstances and for short distances such as (under a river.

t Overhead lines can be protected from direct strokes of lightning in most ;cases by one or more wires at ground potential strung above the power-line :conductors as mentioned in the description of Fig. 6.1. These protecting wires, ;called ground wires, or shield wires, are connected to ground through the transmission towers supporting the line. The zone of protection is usually considered to be 3 on each side of vertical beneath a ground wire; that is, the power lines must come within this 6 sector. The ground wires, rather than the power line, receive the lightning strokes in most cases.

Lightning strokes hitting either ground wires or power conductors cause an injection of current, which divides with half the current flowing in one direction and half in the other. The crest value of current along the struck conductor varies widely because of the wide variation in the intensity of the strokes, Values of 10,000 A and upward are typical. In the case where a power

222 CHAPTER 6 CURRENT AND VOLII.AC3E RELA-TIONS ON A ,I'RANSMISSION LINE

line receives a direct stroke the damage to equipment at line terminals is caused by the voltages between the line and the ground resulting from the injected charges which travel along the line as current. These voltages are typically above a million volts. Strokes to the ground wires can also cause high-voltage surges on the power lines by electromagnetic induction.

6.11 TRANSIENT ANALYSIS: TRAVELING WAVES

The study of transmission-line surges, regardless of their origin, is very complex and we can consider here only the case of a lossless lines2

A lossless line is a good representation for lines of high frequency where w L and oC become very large compared to R and G. For lightning surges on a power transmission line the study of a lossless line is a simplification that enables us to understand some of the phenomena without becoming too involved in complicated theory.

Our approach to the problem is similar to that used earlier for deriving the steady-state voltage and current relations for the long line with distributed constants. We now measure the distance x along the line from the sending end (rather than from the receiving end) to the differential element of length Ax shown in Fig. 6.12. The voltage v and the current i are functions of both x and t so that we need to use partial derivatives. The series voltage drop along the elemental length of line is

and we can write

The negative sign is necessary because v + (dv/dx) A x must be less than v for positive values of i and ai/at. Similarly,

or further study, see A. Greenwood, Electrical Transients irr Power Systems, 2d ed., Wiley-Intersci- ence, New York, 1991.

ai i i + = 4 x - I ! -

+ I 1 I I

i v I av FIGURE 6.12 I j + - ax Ax Schematic diagram of an elemental section of a I I transmission line showing one phase and neu-

I - I I - tral return. Voltage v and current i are func- I I tions of both x and t . The distance x is mea- - + A x + x sured from the sending end of the line.

, We can divide through both Eqs. (6.63) and (6.64) by Ax, and since we are onsidering only a lossless line, R and G will equal zero to give

iow we can eliminate i by taking the partial derivative of both terms in Eq. 5.65) with respect to x and the partial derivative of both terms in Eq. (6.66) iith respect to t . This procedure yields d2i/dx at in both resulting equations, nd eliminating this second partial derivative of i between the two equations ields

, Equation (6.67) is the so-called traueling-waue equation of a lossless ransrnission line. A solution of the equation is a function of (x - vt), and the oltage is expressed by

v = f(x - vt) I

l e function is undefined but must be single valued. The constant v must have he dimensions of meters per second if x is in meters and t is in seconds. We an verify this solution by substituting this expression for v into Eq. (6.67) to letermine v. First, we make the change in variable

ad write


FIGURE 6.13 A voltage wave which is a function of ( x - vt) is shown for values of t equal to t , and t2 .

Then,

and

Similarly, we obtain

Substituting these second partial derivatives of v in Eq. (6.67) yields

and we see that Eq. (6.68) is a solution of Eq. (6.67) if

The voltage as expressed by Eq. (6.68) is a wave traveling,in the positiv direction. Figure 6.13 shows a function of ( x - vt), which is similar to the sb of a wave of voltage traveling along a line which has been struck by lightn: The function is shown for two values of time t , and t,, where t , > t , .

observer traveling with the wave and staying at the same point on the wave s no change in voltage at that point. To the observer

x - vt = a constant

6.11 TRANSIENT ANALYSIS: TRAVELING WAVES 225

Ifor L and C in henrys per meter and farads per meter, respectively. Thus, the jvoltage wave travels in the positive x direction with the velocity v. [, A function of ( x + vt) can also be shown to be a solution of Eq. (6.67) !and, by similar reasoning, can be properly interpreted as a wave traveling in the ;negative x direction. The general solution of Eq. (6.67) is c

v = f l ( x - vt) + f 2 ( x + vt)

which is a solution for simultaneous occurrence of forward and backward ;components on the line. Initial conditions and boundary (terminal) conditions :determine the particular values for each component. 1 If we express a forward traveling wave, also called an incident waue, as

v + = f l ( x - vt)

a wave of current will result from the moving charges and will be expressed by

which can be verified by substitution of these values of voltage and current in Eq. (6.65) and by the fact that v is equal to 1/ m.

Similarly, for a backward moving wave of voltage where

the corresponding current is

From Eqs. (6.78) and (6.79) we note that

226 CHAPTER 6 C U R R E N T A N D VOLTAGE RELATIONS O N A I'RANSMISSION LINE

and from Eqs. (6.80) and (6.81) that

If we had decided to assume the positive direction of current for i- to be in the direction of travel of the backward traveling wave the minus signs would change to plus signs in Eqs. (6.81) and (6.83). We choose, however, to keep the positive x direction as the direction for positive current for both forward and backward traveling waves.

The ratio of v + to i+ is called the characteristic impedance Z, of the line. We have encountered characteristic impedance previously in the steady-state solution for the long line where 2, was defined as \/z/y, which equals \ILIC when R and G are zero.

6.12 TRANSIENT ANALYSIS: REFLECTIONS

We now consider what happens when a voltage is first applied to the sending end of a transmission line which is terminated in an impedance 2,. For our very simple treatment we consider ZR to be a pure resistance. If the termination is other than a pure resistance, we would resort to Laplace transforms. The transforms of voltage, current, and impedance would be functions of the Laplace transform variable s.

When a switch is closed applying a voltage to a line, a wave of voltage v t accompanied by a wave of current i+ starts to travel along the line. The ratio of the voltage v, at the end of the line at any time to the current iR at the end of the line must equal the terminating resistance 2,. Therefore, the arrival of vt and i+ at the receiving end where their values are v: and i; must result in backward traveling or reflected waves v - and i- having values v, and i, at the receiving end such that

where v, and i, are the reflected waves v- and i- measured at the receiving end.

If we let 2, = we find from Eqs. (6.82) and (6.83) that

6.12 TRANSIENT ANALYSIS: REFLECTIONS 227

en, substituting these values of i,+ and i, in Eq. (6.84) yields 6 The voltage v, at the receiving end is evidently the same function of t as v i (but with diminished magnitude unless 2, is zero or infinity). The reflection coefficient p, for voltage at the receiving end of the line is defined as v,/vi, so, for voltage

I We note from Eqs. (6.85) and (6.86) that

and therefore the reflection coefficient for current is always the negative of the reflection coefficient for voltage.

If the line is terminated in its characteristic impedance Z,, we see that the reflection coefficient for both voltage and current will be zero. There will be no reflected waves, and the line will behave as though it is infinitely long. Only when a reflected wave returns to the sending end does the source sense that the line is neither infinitely long nor terminated in 2,.

I l l e m i n a t i o n in a short circuit results in a p, for voltage of - 1. If the -----7------

termination is an open circuit, Z, is infinite and p, is fouiZ-6yTiiGfig7he numerator and denominator in Eq. (6.88) by 2, and by allowing Z, to approach infinity to yield p, = 1 in the limit for voltage.

We should note at this point that waves traveling back toward the sending end will cause new reflections as determined by the reflection coefficient at the sending end p,. For impedance at the sending end equal to 2, Eq. (6.88) becomes

With sending-end impedance of Zs, the value of the initial voltage impressed across the line will be the source voltage multiplied by Zc/(Zs + 2,). Equation (6.82) shows that the incident wave of voltage experiences a line impedance of Z,, and at the instant when the source is connected to the line 2, and Z, in series act as a voltage divider.


FIGURE 6.14 Circuit diagram, lattice diagram, and plot of voltage versus time for Example 6.8, where the receiving-end resistance is 90 a.

Example 6.8. A dc source of 120 V with negligible resistance is connected through a switch S to a lossless transmission line having 2, = 30 R. The line is terminated in a resistance of 90 a. If the switch closes at t = 0, plot v, versus time until t = ST, where T is the time for a voltage wave to travel the length of the line. The circuit is shown in Fig. 6.14(a).

Sdution. When switch S is closed, the incident wave of voltage starts to travel along the line and is expressed as

where U(vt - x ) is the unit step function, which equals zero when (vt - X ) ir negative and equals unity when (vt - x ) is positive. There can be no reflected wave until the incident wave reaches the end of the line. With impedance to the incident wave of 2, = 30 R, resistance of the source zero and v+= 120 V, the reflection coefficient becomes

t

When v + reaches the end of the line, a reflected wave originates of value

and so

When t = 2T, the reflected wave arrives at the sending end where the sending-end reflection coefficient p, is calculated by Eq. (6.90). The line termination for the reflected wave is Z,, the impedance in series with the source, or zero in this case. so,

and a reflected wave of -60 V starts toward the receiving end to keep the sending-end voltage equal to 120 V. This new wave reaches the receiving end at t = 3T and reflects toward the sending end a wave of

and the receiving-end voltage becomes

vR = 180 - 60 - 30 = 90V

I An excellent method of keeping track of the various reflections as they occur is the lattice diagram shown in Fig. 6.14(b). Here time is measured along the vertical axis in intervals of T. On the slant lines there are recorded the values of

I 1 the incident and reflected waves. In the space between the slant lines there are r shown the sum of all the waves above and the current or voltage for a point in that - area of the chart. For instance, at x equal to three-fourths of the line length and

t = 4.25T the intersection of the dashed lines through these points is within the area which indicates the voltage is 90 V.

, Figure 6.14(c) shows the receiving-end voltage plotted against time. The 'I

voltage is approaching its steady-state value of 120 V.

Lattice diagrams for current may also be drawn. We must remember, however, that the reflection coefficient for current is always the negative of the reflection coefficient for voltage.

If the resistance at the end of the line of Example 6.8 is reduced to 10 cR as shown in the circuit of Fig. 6.15(a), the lattice diagram and plot of voltage are as shown in Figs. 6.15(b) and 6.15(c). The resistance of 10 IR gives a negative value for the reflection coefficient for voltage, which always occurs for resistance 2, less than Z,. As we see by comparing Figs. 6.14 and 6.15, the negative p, causes the receiving-end voltage to build up gradually to 120 V, while a positive pR causes an initial jump in voltage to a value greater than that of the voltage originally applied at the sending end.


FIGURE 6.15 Circuit diagram, lattice diagram, and plot of voltae versus time when the receiving-end resistance for E ample 6.8 is changed to 10 a.

Reflections do not necessarily arise only at the end of a line. If one line i joined to a second line of different characteristic impedance, as in the case of a overhead line connected to an underground cable, a wave incident to th junction will behave as though the first line is terminated in the 2, of th second line. However, the part of the incident wave which is not reflected wi travel (as a refracted wave) along the second line at whose termination reflected wave will occur. Bifurcations of a line will also cause reflected an refracted waves.

I t should now be obvious that a thorough study of transmission-lin transients in general is a complicated problem. We realize, however, that voltage surge such as that shown in Fig. 6.13 encountering an impedance at th end of a Iossless line (for instance, at a transformer bus) will cause a voltag wave of the same shape to travel back toward the source of the surge. Th reflected wave wiIl be reduced in magnitude if the terminal impedance is othe than a short or open circuit, but if 2, is greater than Z,, our study has show that the peak terminal voltage will be higher than, often close to double tb peak voltage of the surge.

Terminal equipment is protected by surge arresters, which are also called [lightning arresters and surge diverters. An ideal arrester connected from the line I to a grounded neutral would (1) become conducting at a design voltage above the arrester rating, (2) limit the voltage across its terminals to this design value, and (3) become nonconducting again when the line-to-neutral voltage drops below the design value.

Originally, an arrester was simply an air gap. In this application when the ,surge voltage reaches a value for which the gap is designed, an arc occurs that ;causes an ionized path to ground, essentially a short circuit. However, when the (surge ends, the 60-Hz current from the generators still flows through the arc to 'ground. The arc has to be extinguished by the opening of circuit breakers.

I Arresters capable of extinguishing a 60-Hz current after conducting surge .current to ground were developed later. These arresters are made using 'anonlinear resistors in series with air gaps to which an arc-quenching capability has been added. The nonlinear resistance decreases rapidly as the voltage across it rises. Typical resistors made of silicon carbide conduct current propor- .tional to approximately the fourth power of the voltage across the resistor. When the gaps arc over as a result of a voltage surge, a low-resistance current path to ground is provided through the nonlinear resistors. After the surge ends

and the voltage across the arrester returns to the normal line-to-neutral level the resistance is sufficient to limit the arc current to a value which can be quenched by the series gaps. Quenching is usually accomplished by cooling and deionizing the arc by elongating it magnetically between insulating plates.

The most recent development in surge arresters is the use of zinc oxide in place of silicon carbide. The voltage across the zinc oxide resistor is extremely constant over a very high range of current, which means that its resistance at normal line voltage is so high that a series air gap is not necessary to limit the drain of a 60-Hz current at normal ~ o l t a g e . ~

$ 6.13 DIRECT-CURRENT TRANSMISSION

The transmission of energy by direct current becomes economical when compared to ac transmission only when the extra cost of the terminal equipment required for dc lines is offset by the lower cost of building the lines. Converters at the two ends of the dc lines operate as both rectifiers to change the generated alternating to direct current and inverters for converting direct to alternating 1 current so that power can flow in either direction.

b The year 1954 is generally recognized as the starting date for modern

high-voltage dc transmission when a dc line began service at 100 kV from I Vastervik on the mainland of Sweden to Visby on the island of Gotland, a

5 ;. : 'see E. C. Sakshaug, J. S. Kresge, and S. A. Miske, Jr., "A New Concept in Station Arrester

Design," IEEE Trunsactions on Power Apparatus arid Systems, vol. PAS-96, no. 2, March/ApriI ' 1977, pp. 647-656.

232 CHAPTER 6 CLIIXRENT AND V01.7.AOE RI:I.A.I'IONS ON A .I'IXANSMISSION I.INE

distance of 100 km (62.5 mi) across the Baltic Sea. Static conversion equipment was in operation much earlier to transfer energy between systems of 25 and 60 Hz, essentially a dc transmission line of zero length. In the United States a dc line operating at 800 kV transfers power generated in the Pacific northwest to the southern part of California. As the cost of conversion equipment decreases with respect to the cost of line construction, the economical minimum length of dc lines also decreases and at this time is about 600 km (375 mi).

Operation of a dc line began in 1977 to transmit power from a mine-mouth generating plant burning lignite at Center, North Dakota, to near Duluth, Minnesota, a distance of 740 km (460 mi). Preliminary studies showed that the dc line including terminal facilities would cost about 30% less than the cornpa- rable ac line and auxiliary equipment. This line operates at +250 kV (500 kV line to line) and transmits 500 MW.

Direct-current lines usually have one conductor which is at a positive potential with respect to ground and a second conductor operating at an equal negative potential. Such a line is said to be bipolar. The line could be operated with one energized conductor with the return path through the earth, which has a much lower resistance to direct current than to alternating current. In this case, or with a grounded return conductor, the line is said to be monopolar.

In addition to the lower cost of dc transmission over long distances, there are other advantages. Voltage regulation is less of a problem since at zero frequency the series reactance w L is no longer a factor, whereas it is the chief contributor to voltage drop in an ac line. Another advantage of direct current is the possibility of monopolar operation in an emergency when one side of a bipolar line becomes grounded.

Due to the fact that underground ac transmission is limited to about 5 km because of excessive charging current at longer distances, direct current was chosen to transfer power under the English Channel between Great Britain and France. The use of direct current for this installation also avoided the difficulty of synchronizing the ac systems of the two countries.

No network of dc lines is possible at this time because no circuit breaker is available for direct current that is comparable to the highly developed ac breakers. The ac breaker can extinguish the arc which is formed when the breaker opens because zero current occurs twice in each cycle. The direction and amount of power in the dc line is controlled by the converters in which grid-controlled mercury-arc devices are being displaced by the semiconductor rectifier (SCR). A rectifier unit will contain perhaps 200 SCRs.

Still another advantage of direct current is the smaller amount of right of way required. The distance between the two conductors of the North Dakota- Duluth 500-kV line is 25 ft. The 500-kV ac line shown in Fig. 6.1 has 60.5 ft between the outside conductors. Another consideration is the peak voltage of the ac line, which is x 500 = 707 kV. So, the line requires more insulation between the tower and conductors as well as greater clearance above the earth.

We conclude that dc transmission has many advantages over alternating current, but dc transmission remains very limited in usage except for long lines

nce there is no dc device which can provide the excellent switching operations nd protection of the ac circuit breaker. There is also no simple device to hange the voltage level, which the transformer accomplishes for ac systems.

.14 SUMMARY

long-line equations given by Eqs. (6.35) and (6.36) are, of course, valid for a of any length. The approximations for the short- and medium-length lines e analysis easier in the absence of a computer. Circle diagrams were introduced because of their instructional value in

howing the maximum power which can be transmitted by a line and also in howing the effect of the power factor of the load or the addition of capacitors.

ABCD constants provide a straightforward means of writing equations in a ore concise form and are very convenient in problems involving network

eduction. Their usefulness is apparent in the discussion of series and shunt eactive compensation.

a The simple discussion of transients, although confined to lossless lines and dc sources, should give some idea of the complexity of the study of transients which arise from lightning and switching in power systems.

6.1. An 18-krn, 60-Hz, single-circuit, three-phase line is composed of Partridge conductors equilaterally spaced with 1.6 m between centers. The line delivers 2500 k W at 11 kV to a balanced load. Assume a wire temperature of 5C. (a) Determine the per-phase series impedance of the line. ( b ) What must be the sending-end voltage when the power factor is

(i) 80% lagging,

cent regulation of the line at the above power factors.

i ( d ) Draw phasor diagrams depicting the operation of the line in each case. 6.2. A 100-mi, single-circuit, three-phase transmission line delivers 55 MVA at 0.8

E power-factor lagging to the load at 132 kV (line to line). The line is composed of Drake conductors with flat horizontal spacing of 11.9 ft between adjacent conduc-

i tors. Assume a wire temperature of 5C. Determine I (a) The series impedance and the shunt admittance of the line.

( b ) The ABCD constants of the line. ( c ) The sending-end voltage, current, real and reactive powers, and the power

factor. ( d ) The percent regulation of the line.

6.3. Find the ABCD constants of a rr circuit having a 600-0 resistor for the shunt branch at the sending end, a 1-kR resistor for the shunt branch at the receiving end, and an 80-0 resistor for the series branch.


6.4. The ABCD constants of a three-phase transmission line are

The load at the receiving end is 50 MW at 220 kV with a power factor of 0.9 lagging. Find the magnitude of the sending-end voltage and the voltage regulation. Assume that the magnitude of the sending-end voltage remains constant.

6.5. A 70-mi, single-circuit, three-phase line composed of Ostrich conductors is ar- ranged in flat horizontal spacing with 15 ft between adjacent conductors. The line delivers a load of 60 MW at 230 kV with 0.8 power-factor lagging. ( a ) Using a base of 230 kV, 100 MVA, determine the series impedance and the

shunt admittance of the line in per unit. Assume a wire temperature of 50°C. Note that the base admittance must be the reciprocal of base impedance.

(6) Find the voltage, current, real and reactive power, and the power factor at the sending end in both per unit and absolute units.

I c ) What is the percent regulation of the line? 6.6. .A single-circuit, three-phase transmission line is composed of Parakeet conductors

with flat horizontal spacing of 19.85 ft between adjacent conductors. Determine the characteristic impedance and the propagation constant of the line at 60 Hz and 50' C temperature.

6.7. Using Eqs. (6.23) and (6.241, show that if the receiving end of a line is terminated by its characteristic impedance Z,, then the impedance seen at the sending end of the line is also Z , regardless of line length.

6.8. A 200-mi transmission line has the following parameters at 60 Hz:

Resistance r = 0.21 O/mi per phase

Series reactance x = 0.78 Cl/mi per phase

Shunt susceptance b = 5.42 X S/mi per phase

(a) Determine the attenuation constant a, wavelength A , and the velocity of propagation of the line at 60 Hz.

( b ) If the line is open-circuited at the receiving end and the receiving-end voltage is maintained at 100 kV line to line, use Eqs. (6.26) and (6.27) to determine the incident and reflected components of the sending-end voltage and current.

( c ) Hence, determine the sending-end voltage and current of the line. 6.9. Evaluate cosh 6 and sinh 6 for 6 = 0.5&.

6.10. Using Eqs. (6.1), (6.21, (6.101, and (6.371, show that the generalized circuit constants of all three transmission-line models satisfy the condition that

6.11. The sending-end voltage, current, and power factor of the line described in ! Ejrample 6.3 are found to be 260 kV (line to line), 300 A, and 0.9 lagging,

respectively. Find the corresponding receiving-end voltage, current, and power factor.

6.12. A 60-Hz three-phase transmission line is 175 mi long. It has a total series impedance of 35 + j140 fl and a shunt admittance of 930 X 1 0 - ~ b S. It delivers 40 MW at 220 kV, with 90% power-factor lagging. Find the voltage at the sending end by ( a ) the short-line approximation, ( b ) the nominal- approximation, and ( c ) the long-line equation.

6.13. Determine the voltage regulation for the line described in Prob. 6.12. Assume that ' the sending-end voltage remains constant.

6.14. A three-phase, 60-Hz transmission line is 250 mi long. The voltage at the sending

Y'i ~.3~skYmiTk~~%H8~~~~&di~r~&fit%n'efl?n&/e%* d ~&PoQ4~kl*1f@

' 6.15. If the load on the line described in Prob. 6.14 is 80 MW at 220 kV, with unity , power factor, calculate the current, voltage, and power at the sending end. Assume

that the sending-end voltage is held constant and calculate the voltage regulation of the line for the load specified above.

6.16. A three-phase transmission line is 300 mi long and serves a load of 400 MVA, with 0.8 lagging power factor at 345 kV. The ABCD constants of the line are

( a ) Determine the sending-end line-to-neutral voltage, the sending-end current, and the percent voltage drop at full load.

( b ) Determine the receiving-end line-to-neutral voltage at no load, the sending-end current at no load, and the voltage regulation.

6.17. Justify Eq. (6.50) by substituting for the hyperbolic functions the equivalent exponential expressions.

6.18. Determine the equivalent* circuit for the line of Prob. 6.12. 6.19. Use Eqs. (6.1) and (6.2) to simplify Eqs. (6.57) and (6.58) for the short transmission

line with ( a ) series reactance X and resistance R and ( b ) series reactance X and negligible resistance.

6.20. Rights of way for transmission circuits are difficult to obtain in urban areas, and existing lines are often upgraded by reconductoring the line with larger conductors or by reinsulating the line for operation at higher voltage. Thermal considerations and maximum power which the line can transmit are the important considerations. A 138-kV line is 50 km long and is composed of Partridge conductors with flat horizontal spacing of 5 m between adjacent conductors. Neglect resistance and find

236 CHAPTER 6 CURRENT AND VOLTAGE RBI.AT1ONS ON A TRANSMISSION LINE

the percent increase in power which can bc transmitted for constant IV,I and IV,I while 6 is limitcd to 45" (a) If the Partridge conductor is replaced by Osprey, which has more than twice

the area of aluminum in square millimeters, (b) If a second Partridge conductor is placed in a two-conductor bundle 40 cm

from the original conductor and a ccnter-to-center distance between bundles of 5 m, and

(c) If the voltage of the original line is raised to 230 kV with increased conductor spacing of 8 m.

6.21. Construct a receiving-end power-circle diagram similar to Fig. 6.11 for the line of Prob. 6.12. Locate the point corresponding to the load of Prob. 6.12, and locate the center of circles for various values of lVsl if IV,I = 220 kV. Draw the circle passing through the load point. From the measured radius of the latter circle determine IV,I and compare this value with the values calculated for Prob. 6.12.

6.22. A synchronous condenser is connected in parallel with the load described in Prob. 6.12 to improve the overall power factor at the receiving end. The sending-end voltage is always adjusted so as to maintain the receiving-end voltage fixed at 220 kV. Using the power-circle diagram constructed for Prob. 6.21, determine the sending-end voltage and the reactive power supplied by the synchronous condenser when the overall power factor at the receiving end is (a) unity ( b ) 0.9 leading.

6.23. A series capacitor bank having a reactance of 146.6 R is to be installed at the midpoint of the 300-mi line of Prob. 6.16. The ABCD constants for each 150-mi portion of line are

( a ) Determine the equivalent ABCD constants for the cascade combination of the line-capacitor-line. (See Table A.6 in the Appendix.)

( b ) Solve Prob. 6.16 using these equivalent ABCD constants. 6.24. The shunt admittance of a 300-mi transmission line is

Determine the ABCD constants of a shunt reactor that will compensate for 60% of the total shunt admittance.

6.25. A 250-Mvar, 345-kV shunt reactor whose admittance is 0.0021/-90' S is connected to the receiving end of the 300-mi line of Prob. 6.16 at no load. (a) Determine the equivalent ABCD constants of the line in series with the shunt

reactor. (See Table A.6 in the Appendix.) (b) Rework part ( b ) of Prob. 6.16 using these equivalent ABCD constants and the

sending-end voltage found in Prob. 6.16.

S h u h $ o de uma linha de 500kV, corn uma carga igual A sua impechcia caracterfstica "Z", variando o cornprimento da mesma de acordo corn o comprimento de onda h = 4874,47krn. 0 s m t r o s de linha adotados form obtidos a partir da sub-rotinha "LINE CONSTANTS" do ATP e s?lo apresentados a seguir:

SeqGncia positiva: RI = 0,0272Rkm; XI = 0,4070R/km, C1= 0,0108pFh

Sequencia zero: R,-, = 0,2093Qflun; Xo = 0,9 1 59Rflun; CO = 0,008OpFlkm.

Comprirnentos de linha adotados: L = W8 = 609,3 1 krq L = W4 = 1218,62 krq L = 3W4 = 1827,93 km; L = W2 = 2437,24 km; L = 5hl2 = 3046,54 km; L = 3hJ4 = 3655,85 km; L = 7W8 = 4265,16km9 L = h = 4874,47 km.

Ten& da fonte: V1 = 500KV " = 408,248kV 47

Circuito equivalente para a linha construido no ATPDRAW para posterior simula@io no ATP:

P A ~ M E T R O S DE LINHA DE TRANSMISSAO

Simukqh de urna linha de 500kV em aberto(sem carga), variando o cornprimento da mesma de acordo corn o cornprimento de onda h = 4874,47km. 0 s padmetros de linha adotados foram obtidos a partir da sub-rotinha "LINE CONSTANTS" do ATP e d o apresentados a seguir:

Sequencia positiva: R1 = 0,02721R/km; XI= O,407OC2/km; C1 = O,O108pF/km.

Sequencia zero: Ro = 0,2093- & = 0,9159Qflun; co = o,oo8op/km.

Comprimentos de linha adotados: L = W8 = 609,3 1 km; L = hl4 = 1218,62 km, L = 3hJ4 = 1827,93 km,

, L = h/2 = 2437,24 km, ------- -

L = 5W2 = 3046,54 km; L = 3hl4 = 3655,85 krn; L = 7W8 = 4265,16km; L = h = 4874,47 krn.

Te& da fonte: V1 = 500KY IIT = 408,248kV &

Circuito equivalente para a linha construido no ATPDRAW para posterior simulqiio no ATP:

S h u h $ o de uma linha de 500kV, corn uma carga igual A sua impechcia caracterfstica "Z", variando o cornprimento da mesma de acordo corn o comprimento de onda h = 4874,47krn. 0 s m t r o s de linha adotados form obtidos a partir da sub-rotinha "LINE CONSTANTS" do ATP e s?lo apresentados a seguir:

SeqGncia positiva: RI = 0,0272Rkm; XI = 0,4070R/km, C1= 0,0108pFh

Sequencia zero: R,-, = 0,2093Qflun; Xo = 0,9 1 59Rflun; CO = 0,008OpFlkm.

Comprirnentos de linha adotados: L = W8 = 609,3 1 krq L = W4 = 1218,62 krq L = 3W4 = 1827,93 km; L = W2 = 2437,24 km; L = 5hl2 = 3046,54 km; L = 3hJ4 = 3655,85 km; L = 7W8 = 4265,16km9 L = h = 4874,47 km.

Ten& da fonte: V1 = 500KV " = 408,248kV 47

Circuito equivalente para a linha construido no ATPDRAW para posterior simula@io no ATP: