Chapter 6. Characteristics and Performance of Transmission Lines

Characteristics and Performance of Transmission Lines 125



This is a simple series circuit. The relationship

between sending-end, receiving-end voltages and

currents can be written as:

1

LMV OP = LM0 Z OP LMV OP

I Q N

1Q N I Q

N

S



R



S



R



...(6.4)



The phasor diagram for the short-line is shown in

Fig. 6.2 for lagging load current.

From Fig. 6.2, we can write



Fig. 6.1: Short line model.



|VS| cos (dS – dR) = |I|R cos dR + |I|

X sin dR + |VR|

(dS – dR) is very small,

\



...(6.5)



\ cos (dS – dR) » 1.0



|VS| = |VR| + |I| (R cos dR

+ X sin dR)

...(6.6)



Equation (6.6) is quite accurate for the normal range

of load.

Fig. 6.2: Phasor diagram.



6.3 VOLTAGE REGULATION



Voltage regulation of the transmission line may be defined as the percentage change in voltage

at the receiving end of the line (expressed as percentage of full-load voltage) in going from noload to full-load.

Percent voltage regulation =

where



NL

FL

|VR |-|VR |

FL

|VR |



´ 100



...(6.7)



NL

|VR | = magnitude of no-load receiving end voltage

FL

|VR | = magnitude of full-load receiving end voltage



NL

At no load, IR = 0, VR = VR and from eqn. (6.3),

NL

VR =



VS

A



...(6.8)



Using eqns. (6.7) and (6.8), we get,

Percentage Voltage regulation =



|VS| -|A |VR |

| FL

|A |VR |

| FL



´ 100



...(6.9)



FL

For a short line, |A| = 1.0, |VR | = |VR|



\ Percent voltage regulation =



|VS|-|VR|

´ 100

|VR|



Using eqns. (6.10) and (6.6), we get,



...(6.10)



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Electrical Power Systems



Percent voltage regulation =



|I |( R cos d R + X sin d R )

´ 100

|V R|



...(6.11)



In the above derivation, dR has been considered positive for a lagging load. dR will be

negative, for leading load. Therefore, for leading power factor load,

Percent voltage regulation =



|I|( R cos d R - X sin d R )

|VR|



...(6.12)



From the above equations, it is clear that the voltage regulation is a measure of line voltage

drop and depends on the load power factor.



6.4 MEDIUM TRANSMISSION LINE

For the lines more than 80 km long and below 250 km in length are treated as medium length

lines, and the line charging current becomes appreciable and the shunt capacitance must be

considered. For medium length lines, half of the shunt capacitance may be considered to be

lumped at each end of the line. This is referred to as the nominal p model as shown in Fig. 6.3.

The sending end voltage and current for the nominal p model are obtained as follows:



Fig. 6.3: Medium length line, nominal p representation.



From KCL, the current in the series impedance designated by IL is

Y

VR

2

From KVL, the sending end voltage is



IL = IR +



...(6.13)



VS = VR + ZIL



...(6.14)



From eqns. (6.14) and (6.13), we get,



FG

H



VS = 1 +



IJ

K



ZY

V R + ZIR

2



...(6.15)



The sending end current is,



Y

VS

2

From eqns. (6.16), (6.15) and (6.13), we get,

IS = IL +



FG

H



IS = Y 1 +



IJ FG

K H



...(6.16)



IJ

K



ZY

ZY

VR + 1 +

IR

4

2



...(6.17)



Characteristics and Performance of Transmission Lines 127



Eqns (6.15) and (6.17) can be written in matrix form.



LMFG1 + ZY IJ Z OP LV O

LMV OP = MMH 2ZYK ZY PP MM PP

N I Q MY FGH1 + IJK FGH1 + IJK P MNM I PQP

2 P

MN 4

Q

R



S



S



...(6.18)



R



Therefore, the ABCD constants for the nominal p model are given by



FG

H



A= 1+



FG

H



IJ

K



ZY

, B = Z,

2



C=Y 1+



IJ

K



FG

H



ZY

ZY

, D= 1+

2

4



IJ

K



6.5 LONG TRANSMISSION LINE

For short and medium length lines, accurate models were obtained by assuming the line

parameters to be lumped. In case the lines are more than 250 km long, for accurate solutions

the parameters must be taken as distributed uniformly along the length as a result of which the

voltages and currents will vary from point to point on the line. In this section, expressions for

voltage and current at any point on the line are derived. Then, based on these equations, an

equivalent p model is obtained for long transmission line. Figure 6.4 shows one phase of a

distributed line of length l km.



Fig. 6.4: Schematic diagram of a long transmission line with

distributed parameters.



From KVL, we can write,

V(x + Dx) = z.Dx.I(x) + V(x)

\

As



V ( x + Dx) - V ( x)

= z.I(x)

Dx



...(6.19)



Dx ® 0



dV ( x)

= z.I(x)

dx



...(6.20)



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Electrical Power Systems



From KCL, we can write,

I(x + Dx) = I(x) + y. Dx .V(x + Dx)

\

As



I ( x + Dx) - I ( x)

= y.V(x + Dx)

Dx



...(6.21)



Dx ® 0



dI ( x)

= y.V(x)

dx



...(6.22)



Differentiating eqn. (6.20) and substituting from eqn. (6.22), we get,

d 2 V ( x)

dx



\



2



d 2 V ( x)

dx 2



dI ( x)

= z. yV(x)

dx



– zy V(x) = 0



g2 = zy



Let

Therefore,



= z×



d 2 V ( x)

dx 2



...(6.23)

...(6.24)



– g2V(x) = 0



...(6.25)



The solution of the above equation is

V(x) = C1 eg x + C2 e–g x



...(6.26)



where, g, known as the propagation constant and is given by,

g = a + jb =



zy



...(6.27)



The real part a is known as the attenuation constant, and the imaginary part b is known as

the phase constant. b is measured in radian per unit length.

From eqn. (6.20), the current is,

I(x) =



1 dV ( x)

×

Z

dx



\



I(x) =



g

(C1 egx – C2 e–gx)

z



\



I(x) =



\



I(x) =



y

(C1 egx – C2 e–gx)

z

1

(C egx – C2 e–gx)

ZC 1



...(6.28)



where, ZC is known as the characteristic impedance, given by

ZC =



z

y



...(6.29)



Characteristics and Performance of Transmission Lines 129



Now note that, when x = 0, V(x) = VR and from eqn. (6.26), we get

VR = C1 + C2



...(6.30)



also when x = 0, I(x) = IR and from eqn. (6.28), we get,

IR =



1

(C1 – C2)

ZC



...(6.31)



Solving eqns. (6.31) and (6.32), we obtain,

C1 =



VR + ZC IR

2



...(6.32)



C2 =



(VR - ZC IR )

2



...(6.33)



Substituting the values of C1 and C2 from eqns. (6.32) and (6.33) into eqns. (6.26) and (6.28),

we get

V(x) =



(VR + Z C I R ) gx (V R - Z C I R ) - gx

e +

e

2

2



...(6.34)



I(x) =



(VR + Z C I R ) g x (V R - ZC I R ) - g x

e e

2 ZC

2 ZC



...(6.35)



The equations for voltage and currents can be rearranged as follows:



( e gx + e - gx )

(e gx - e - gx )

VR + Z C

IR

V(x) =

2

2

I(x) =

or



( e gx - e - gx )

(e gx + e - g x )

VR +

IR

2 ZC

2



...(6.36)

...(6.37)



V(x) = cosh(gx)VR + ZC sinh(gx) IR



...(6.38)



1

sinh(gx)VR + cosh(gx)IR

ZC



...(6.39)



I(x) =



Our interest is in the relation between the sending end and the receiving end of the line.

Therefore, when x = l, V(l) = VS and I(l) = IS. The result is

VS = cosh (g l)VR + ZC sinh(g l )IR

IS =

Therefore, ABCD constants are:



1

sinh (gl ) VR + cosh(g l )IR

ZC



A = cosh(g l ) ; B = ZC sinh(g l)



...(6.40)

...(6.41)

...(6.42)



1

sinh(g l ) ; D = cosh(g l )

...(6.43)

ZC

It is now possible to find an accurate equivalent p model for long transmission line as shown

in Fig. 6.5.

C=



130



Electrical Power Systems



Similar to the expressions of eqns. (6.15) and (6.17) obtained for the nominal p, for equivalent

p model we have,



FG

H



VS = 1 +



FG

H



IJ

K



Z¢Y ¢

VR + Z ¢ I R

2



IS = Y ¢ 1 +



IJ

K



FG

H



...(6.44)



IJ

K



Z¢Y ¢

Z ¢Y ¢

VR + 1 +

IR

4

2



...(6.45)



Now comparing eqns. (6.44) and (6.45) with eqns. (6.40) and (6.41), respectively and making

use of the identity

tanh



FG gl IJ = cosh(g l) - 1

H 2 K sinh (gl)



...(6.46)



the parameters of equivalent p model are obtained as:

Z¢ = ZC sinh(g l) =



FG IJ

H K



Z × sinh(g l)

gl



1

Y tanh (g l 2)

gl

Y¢

tanh

=

=

ZC

2

2

gl 2

2



...(6.47)

...(6.48)



Fig. 6.5: Equivalent p model for long transmission line.



Example 6.1: A single phase 60 Hz generator supplies an inductive load of 4500 kW at a power

factor of 0.80 lagging by means of an 20 km long overhead transmission line. The line resistance

and inductance are 0.0195 W and 0.60 mH per km. The voltage at the receiving end is required

to be kept constant at 10.2 kV.

Find (a) the sending end voltage and voltage regulation of the line; (b) the value of the

capacitors to be placed in parallel with the load such that the regulation is reduced to 60% of that

obtained in part (a); and (c) compare the transmission efficiencies in parts (a) and (b).

Solution: The line constants are:

R = 0.0195 × 20 = 0.39 W

X = 0.60 × 10–3 × 2p × 60 × 20 = 4.52 W

(a) This is a short line with I = IR = IS given by

|I| =



4500

Amp = 551.47 Amp.

10.2 ´ 0.80



Characteristics and Performance of Transmission Lines 131



From eqn. (6.6),

|VS| » |VR| + |I|(R cos dR + X sin dR)

Here



|VR| = 10.2 kV = 10200 Volt

cosdR = 0.8, sindR = 0.6



\



|VS| = 10200 + 551.47 (0.39 × 0.8 + 4.52 × 0.6)



\



|VS| = 11.867 kV



\



Voltage regulation =



(11.867 - 10.2)

´ 100 = 16.34%

10.2



(b) Voltage regulation desired = 0.60 × 16.34 = 9.804%

Therefore, under this condition we can write

|VS| - 10.2

= 0.09804

10.2



\



|VS| = 11.2 kV



Figure 6.6 shows the equivalent circuit of the

line with a capacitor in parallel with the load.

Assuming combined power factor of the load and

capacitor = cos dR¢



Fig. 6.6



By using eqn. (6.6), we can write,

(11.2 – 10.2) × 10–3 = |IR| (R cos dR¢ + X sin dR¢)



...(i)



Since the capacitance does not draw any real power, we have,

|IR| =



4500

10.2 cosd ¢R



From eqns. (i) and (ii), we get

4.52 tan dR¢ = 1.876

\



tan dR¢ = 0.415



\



dR¢ = 22.5°



\

\

Now



cos dR¢ = 0.9238

|IR| = 477.56 Amp.

IC = IR – I,

IR = 477.56 -22.5° = 441.2 – j182.75

I = 551.47 -36.87° = 441.2 – j330.88



\



IC = 441.2 – j182.75 – 441.2 + j330.88



\



IC = j148.13 Amp.



Now



XC =



V

1

10.2 ´ 1000

= R =

2p ´ 60 ´ C

148.13

IC



...(ii)



132



Electrical Power Systems



\



C = 38.5 mF Ans.



(c) Efficiency of transmission

Output

4500

=

Output + losses 4500 + (55147 )2 ´ 0.39 ´ 10 -3

.

= 97.43%



Case (a)



h=



Case (b)



h=



4500

= 98.06%

4500 + (477.56) 2 ´ 0.39 ´ 10 -3



It is to be noted that by placing a capacitor in parallel with the load, the receiving end power

factor improves from 0.80 to 0.9238.

Example 6.2: A 220 kV, three phase transmission line is 60 km long. The resistance is 0.15 W/

Km and the inductance 1.4 mH/Km. Use the short line model to find the voltage and power at

the sending end and the voltage regulation and efficiency when the line is supplying a three

phase load of

(a) 300 MVA at 0.8 pf lagging at 220 kV

(b) 300 MVA at 0.8 pf leading at 220 kV

Solution:

R = 0.15 × 60 = 9 W

Assuming

\



f = 50 Hz.

X = 2p × 50 × 1.4 × 10–3 × 60 = 26.39 W.



(a) Receiving end voltage per phase is

VR =



220 0°

3



= 127 0° kV



The three phase apparent power is 300 MVA at 0.8 pf lagging

\



f = 36.87°



\



S = 300 36.87° = (240 + j180) MVA



The current per phase is given by

IR =

\



S*

*

3 VR



=



300 -36.87°

3 ´ 127 0°



´ 10 3 Amp



IR = 787.4 -36.87° Amp



From eqn. (6.6), the sending end voltage magnitude is

|VS| = |VR| + |I| (R cos dR + X sin dR)

|VR| = 127 kV, |I| = 787.4 Amp = 0.7874 kA,

R = 9 W, X = 26.39 W

cosdR = 0.8, sindR = 0.6

\



|VS| = 127 + 0.7874 (9 × 0.8 + 26.39 × 0.6)



Characteristics and Performance of Transmission Lines 133



\



|VS| = 145.13 kV.



\ Sending end line-to-line voltage

|VS|L–L =

Voltage regulation =



3 × 145.13 = 251.37 kV

25137 - 220

.

= 14.26%

220



Per phase real power loss in the line

PLoss = |I|2 R = (787.4)2 × 9 × 10–6 MW = 5.58 MW.

Per phase receiving end power

PR =



300

´ 0.8 = 80 MW

3



\ Per phase sending end power

PS = (80 + 5.58) = 85.58 MW.

Transmission line efficiency is

h=

(b)



PR

80

=

= 93.47%.

PS 85.58



3 |VR||IR| = 300



\

\

as:



|VR| = 220 kV

|IR| = 787.4 Amp



Load is at 0.8 power factor leading. For leading power factor load, eqn. (6.6) can be written

|VS| = |VR|+ |I| (R cos dR – X sin dR)

\

\



|VS| = 127 + 787.4 (9 × 0.8 – 26.39 × 0.6) = 120.2 kV

|VS|L–L =

Voltage regulation =



3 × 120.2 = 208.2 kV

208.2 - 220

= –5.36%

220



Per phase real power loss = (787.4)2 × 9 × 10–6 = 5.58 MW

Per phase receiving end power,

PR =



300

´ 0.8 = 80 MW

3



Per phase sending end power,

PS = (80 + 5.58) = 85.58 MW

Transmission line efficiency

h=



PR

80

=

= 93.47%

PS 85.58



134



Electrical Power Systems



Example 6.3: Determine the efficiency and regulation of a 3-phase, 150 km long, 50 Hz

transmission line delivering 20 MW at a power factor of 0.8 lagging and 66 kV to a balanced

load. Resistance of the line is 0.075 W/km, 1.5 cm outside dia, spaced equilaterally 2 meters

between centres. Use nominal p method.

Solution:

R = 0.075 × 150 = 11.25 W

diameter of the conductor = 1.5 cm

1.5

= 0.75 cm

\

radius r =

2

d = 2 mt = 200 cm



FGH 200 IJK Henry

0.75



\



L = 2 × 10–7 × (150 × 1000) ln



\

\



L = 0.1675 Henry

X = 2 × p × 50 × 0.1675 = 52.62 W.



The capacitance per phase =

\

\

\



2 ´ p ´ 8.854 ´ 10 -12

´ (150 ´ 1000) = 1.49 mF.

200

ln

0.75



FGH IJK



Y = jwC = j 2p × 50 × 1.49 × 10–6 mho

Y = j 468.1 × 10–6 mho



Y

= j 234 × 10–6 mho

2

Z = (11.25 + j52.62) = 53.809 77.9° W.



Now



3 × |IR|× 66 × 0.8 = 20 × 1000

\



|IR| = 218.7 Amp. at 0.8 pf lagging



Receiving end phase voltage

|VR| =

From eqn. (6.15), we have



66

3



FGH



= 38.104 kV



IJK



VS = 1 + ZY VR + Z IR

2

IR = 218.7 -36.87°

VR = 38.104 0°



ZY

= 53.809 × 234 × 10–6 77.9° + 90°

2

= 0.01259 167.9° = (–0.0123 + j0.00264)



Characteristics and Performance of Transmission Lines 135



VS = (1 – 0.0123 + j0.00264) × 38.104 0° +



\



(53.809 77.9° ´ 218.7 -36.87°)

1000



\



.

VS = (0.9877 + j0.00264) × 38.104 0° + 11.76 4103°



\



VS = 37.63 + j0.1 + 8.87 + j7.72



\



VS = (46.5 + j7.82) = 47.15 9.54° kV



\



3 × 47.15 9.54° kV = 8166 9.54° kV

.



VS(L – L) =



FGH



IJK



81.66

|VS|

- 66

-|VR|

0.9877

|

Voltage regulation VR = |A

= 25.26%

=

66

|VR|

Power loss per phase = |I|2 R = (218.7)2 × 11.25 × 10–6 MW = 0.538 MW

Per phase receiving end power

PR =



20

MW

3



Per phase sending end power

PS =



20

+ 0.538 = 7.204 MW

3



Transmission efficiency

h=



20 3

= 92.54%.

7.204



Example 6.4: Determine the voltage, current and power factor at the sending end of a 3 phase,

50 Hz, overhead transmission line 160 km long delivering a load of 100 MVA at 0.8 pf lagging

and 132 kV to a balanced load. Resistance per km is 0.16 W, inductance per km is 1.2 mH and

capacitance per km per conductor is 0.0082 mF. Use nominal p method.

Solution:



R = 0.16 × 160 = 25.6 W

X = 1.2 × 10–3 × 2p × 50 × 160 = 60.3 W.

Y = j2p × 50 × 0.0082 × 10–6 × 160 = j4.12 × 10–4 mho

Z = R + jX = 25.6 + j60.3 = 65.51 67° W.



From eqn. (6.15),



FGH



VS = 1 +



IJK



ZY

VR + ZIR

2



Phase voltage at the receiving end,

VR =



132

3



0° kV = 76.21 0° kV