PERFORMANCE OF TRANSMISSION LINES:

By
D.Rajesh Asst. Prof.
EEE Department
20 January 2021 D.Rajesh Asst.Prof. EEE Department 1

PERFORMANCE OF TRANSMISSION LINES
Generalized network constants and equivalent
circuits of short, medium, long transmission line.
Line performance: regulation and efficiency,
Ferranti effect.

 l

 By performance of lines is meant the determination of efficiency
and regulation of lines.
 The efficiency of lines is defined as
 % efficiency =
Power delivered at the receiving end
Power sent from the sending end
× 100
 % efficiency =
Power delivered at the receiving end
Powerdelivered at thereceiving end+ losses
× 100
 The end of the line where load is connected is called the receiving
end and where source of supply is connected is called the sending
end.

 The regulation of a line is defined as the change in the
receiving end voltage, expressed in percent of full load
voltage, from no load to full load, keeping the sending
end voltage and frequency constant. Expressed
mathematically,
% regulation =
𝑉′
𝑟−𝑉𝑟
𝑉𝑟
× 100
where 𝑉′
𝑟 is the receiving end voltage under no load
condition and 𝑉
𝑟 the receiving end voltage under full load
condition. It is to be noted here that 𝑉′
𝑟 and 𝑉
𝑟 are the
magnitudes of voltages.

 The performance of a T.L mainly based on 4
parameters
1.series resistance
2.Inductance
3.shunt capacitance
4.conductance.

 The shunt conductance (G) is mainly due to flow
of relative current over the surface of the insulators
especially during bad weather.
 Effect of “R” IR drop, 𝐼2
𝑅 loss in T.L
 Effect of “L” I𝑋𝐿 drop in T.L in 900
with the
current flowing through the conductor (𝑉𝐿=2 fL*I)
 Effect of “C” charging current 𝐼𝐶=2 fc𝑉
𝑠 which is
900
with the voltage

The equivalent circuit of a single phase short transmission line
is shown in Fig. Here, the total line resistance and inductance are
shown as concentrated or lumped instead of being distributed.
The circuit is a simple a.c. series circuit.
Let I = load current
R = loop resistance i.e., resistance of both conductors
𝑋𝐿 = loop reactance
𝑉𝑅 = receiving end voltage
cos φ𝑅 = receiving end power factor (lagging)
𝑉𝑆 = sending end voltage
cos φ𝑆 = sending end power factor

 The phasor diagram of the line for lagging load power
factor is shown in Fig.
 Current I is taken as the reference phasor. OA
represents the receiving end voltage 𝑉𝑅 leading I by φ𝑅 .
 AB represents the drop IR in phase with I. BC
represents the inductive drop I𝑋𝐿 and leads I by 900
.
OC represents the sending end voltage 𝑉𝑆 and leads I
by φ𝑆.

 (OC)2
= (OD)2
+ (DC)2
or
 (𝑉𝑆)2
= (OE + ED)2
+ (DB + BC)2
 = (𝑉𝑅 cos φ𝑅 + 𝐼R)2
+ (𝑉𝑅 sinφ𝑅+ 𝐼𝑋𝐿)2
 ∴ 𝑉𝑆 = (𝑉𝑅 cos φ𝑅 + 𝐼R)2+ (𝑉𝑅 sinφ𝑅+ 𝐼𝐶)2
 (i)percentage Voltage regulation =
𝑉𝑆−𝑉𝑅
𝑉𝑅
× 100
 (ii) Sending end p.f., cos φ𝑆 =
OD
OC
=
𝑉𝑅 cos φ𝑅 + 𝐼R
𝑉𝑆
 (iii) Power delivered = 𝑉𝑅𝐼𝑅cos φ𝑅
 Line losses = I2
𝑅
 Power sent out = 𝑉𝑅𝐼𝑅cos φ𝑅 + I2
𝑅
 %age Transmission efficiency =
Power delivered
Power sent out
× 100
=
𝑉𝑅𝐼𝑅cos φ𝑅
𝑉𝑅𝐼𝑅cos φ𝑅 + I2𝑅
× 100

An approximate expression for the sending end voltage 𝑉𝑆
can be obtained as follows. Draw perpendicular from B
and C on OA produced as shown in Fig. Then OC is nearly
equal to OF
OC = OF = OA + AF = OA + AG + GF
= OA + AG + BH
= 𝑉𝑅 +𝐼Rcos φ𝑅 + 𝐼𝑋𝐿 sinφ𝑅

 Solution in complex notation. It is often convenient and profitable to make
the line calculations in complex notation.
 Taking 𝑉𝑅as the reference phasor, draw the phasor diagram as shown in Fig.
It is clear that 𝑉𝑆 is the phasor sum of 𝑉𝑅 and Ԧ
𝐼 Ԧ
𝑍
 𝑉𝑅= 𝑉𝑅+j0
 Ԧ
𝐼= Ԧ
𝐼 ∠ − φ𝑅=I(cos φ𝑅 − j sinφ𝑅)
 Ԧ
𝑍=R+ j 𝑋𝐿
 𝑉𝑆= 𝑉𝑅+ Ԧ
𝐼 Ԧ
𝑍
=( 𝑉𝑅+j0)+I(cos φ𝑅 − j sinφ𝑅)(R+ j 𝑋𝐿)
= (𝑉𝑅+ 𝐼Rcos φ𝑅 + 𝐼𝑋𝐿 sinφ𝑅)+j(𝐼𝑋𝐿 cos φ𝑅 − 𝐼Rsin φ𝑅)
∴ 𝑉𝑆= (𝑉𝑅+ 𝐼Rcos φ𝑅 + 𝐼𝑋𝐿 sinφ𝑅)2+(𝐼𝑋𝐿 cos φ𝑅 − 𝐼Rsin φ𝑅) 2
The second term under the root is quite small and can be neglected with
reasonable accuracy.
Therefore, approximate expression for 𝑉𝑆 becomes :
𝑉𝑆 = 𝑉𝑅 +𝐼Rcos φ𝑅 + 𝐼𝑋𝐿 sinφ𝑅

 This system may be regarded as consisting of
three single phase units, each wire transmitting
one-third of the total power.

 1. Effect on regulation. The expression for voltage
regulation of a short transmission line is given by :
 percentage Voltage regulation =
𝑉𝑆−𝑉𝑅
𝑉𝑅
× 100
=
𝐼Rcos φ𝑅+𝐼𝑋𝐿 sinφ𝑅
𝑉𝑅
(for lagging p.f.)
 percentage Voltage regulation =
𝑉𝑆−𝑉𝑅
𝑉𝑅
× 100
=
𝐼Rcos φ𝑅−𝐼𝑋𝐿 sinφ𝑅
𝑉𝑅
(for leading p.f.)

 The following conclusions can be drawn from the above
expressions :
 (i) When the load p.f. is lagging or unity or such leading
that 𝐼Rcos φ𝑅 > 𝐼𝑋𝐿 sinφ𝑅, then voltage regulation is
positive i.e., receiving end voltage 𝑉𝑅 will be less than the
sending end voltage 𝑉𝑆.
 (ii) For a given 𝑉𝑅 and I, the voltage regulation of the line
increases with the decrease in p.f. for lagging loads.
 (iii) When the load p.f. is leading to this extent that
𝐼𝑋𝐿 sinφ𝑅 > 𝐼Rcos φ𝑅, then voltage regulation is negative
i.e. the receiving end voltage 𝑉𝑅 is more than the sending
end voltage 𝑉𝑆.
 (iv) For a given 𝑉𝑅 and I, the voltage regulation of the line
decreases with the decrease in p.f. for leading loads.

 2. Effect on transmission efficiency. The power delivered to
the load depends upon the power factor.
 P=𝑉𝑅𝐼cos φ𝑅 (For 1-phase line)
 ∴I=
𝑃
𝑉𝑅cos φ𝑅
 P=3 𝑉𝑅𝐼cos φ𝑅 (For 3-phase line)
 ∴I=
𝑃
3𝑉𝑅cos φ𝑅
 It is clear that in each case, for a given amount of power to be
transmitted (P) and receiving end voltage (𝑉𝑅), the load
current I is inversely proportional to the load p.f. cos φ𝑅.
 Consequently, with the decrease in load p.f., the load current
and hence the line losses are increased. This leads to the
conclusion that transmission efficiency of a line decreases
with the decrease in load p.f. and vice-versa,

1. A single phase overhead transmission line delivers 1100 kW at
33 kV at 0·8 p.f. lagging. The total resistance and inductive
reactance of the line are 10 Ω and 15 Ω respectively. Determine :
(i) sending end voltage (ii) sending end power factor and (iii)
transmission efficiency.
Load power factor, cos φ𝑅 = 0·8 lagging
Total line impedance, Ԧ
𝑍=R+ j 𝑋𝐿 = 10 + j 15
Receiving end voltage, 𝑉𝑅 = 33 kV = 33,000 V
P=1100Kw cos φ𝑅=0.8 then sin φ𝑅 = 0.6
Line current, I=?
P=𝑉𝑅𝐼cos φ𝑅
∴I=
𝑃
𝑉𝑅cos φ𝑅
=
1100K
33,000∗0·8
=41.67A

 The equivalent circuit and phasor diagram of the line are shown in
Figs.
 (i) Sending end voltage, 𝑉𝑆= 𝑉𝑅+ Ԧ
𝐼 Ԧ
𝑍
𝑉𝑅= 𝑉𝑅+j0
Ԧ
𝐼= Ԧ
Ԧ
𝑍=R+ j 𝑋𝐿
𝑉𝑆= 𝑉𝑅+ Ԧ
𝐼 Ԧ
𝑍
𝑉𝑆=33000+ 41·67 (0·8 − j 0·6)(10 + j 15)
=33,708·3 + j 250
∴ Magnitude of 𝑉𝑆= (33708.3)2+ (250)2=33709V

 (ii) Sending end p.f., cos φ𝑆
φ𝑆= φ𝑅 + α
Angle between 𝑉𝑆 and 𝑉𝑅 is α=tan−1 250
33708.3
=0.420
φ𝑅=cos−1 0.8= 36.870
φ𝑆= 37.290
Sending end p.f., cos φ𝑆=cos(37.290
)=0.7956 lagging
(iii) Transmission efficiency
=
Power delivered
Power sent
× 100
Power sent= Power delivered + losses
=1100Kw+𝐼2
𝑅
= 1100Kw+ (41.67)2
∗ 10
=1117.364Kw
Transmission efficiency =
1100
1117.364
× 100
=98.44%

 Alternatively
 𝑉𝑆 = 𝑉𝑅 +𝐼Rcos φ𝑅 + 𝐼𝑋𝐿 sinφ𝑅
=33708.39V
 Sending end p.f., cos φ𝑆 =
𝑉𝑆
=0.7958 lagging

2. An overhead 3-phase transmission line delivers 5000 kW at 22 kV at 0·8 p.f. lagging. The
resistance and reactance of each conductor is 4 Ω and 6 Ω respectively. Determine : (i)
sending end voltage (ii) percentage regulation (iii) transmission efficiency.
Load power factor, cos φ𝑅 = 0·8 lagging
Then Sin φ𝑅 =0.6
Receiving end voltage/phase, 𝑉𝑅 = 22,000 / 3 = 12,700 V
Impedance/phase, Ԧ
𝑍=R+ j 𝑋𝐿 = 4 + j 6
Line current, I=
𝑃
3𝑉𝑅cos φ𝑅
=
5000𝐾𝑤
3∗12700∗0.8
=164A
(i) Sending end voltage, 𝑉𝑆= 𝑉𝑅+ Ԧ
𝐼 Ԧ
𝑍
Ԧ
𝐼= Ԧ
𝑉𝑆=12700+ {164 (0·8 − j 0·6)(4 + j 6)}
=13815.2+j393.6
∴ Magnitude of 𝑉𝑆= (13815.2)2+ (393.6)2=13820.8V
Line value of 𝑉𝑆= 3*13820.8= 23938 V = 23·938 kV

(ii) percentage Voltage regulation =
𝑉𝑆−𝑉𝑅
𝑉𝑅
× 100
=
13820.8−12700
12700
× 100=8.825%
=
Power delivered
Power sent
× 100
=5000Kw+3𝐼2
𝑅
= 5000Kw+ 3(164)2
∗ 4
= 5000Kw + 322.752Kw
=5322.752Kw
5000
5322.752
× 100
=93.94%

3. A 3-phase line delivers 3600 kW at a p.f. 0·8 lagging to a load. If the sending end voltage
is 33 kV, determine (i) the receiving end voltage (ii) line current (iii) transmission efficiency.
The resistance and reactance of each conductor are 5·31 Ω and 5·54 Ω respectively.
Resistance of each conductor, R = 5·31 Ω
Reactance of each conductor, X𝐿 = 5·54 Ω
Load power factor, cos φ𝑅 = 0·8 (lagging)
Sending end voltage/phase, V𝑆 = 33,000 / 3 = 19,052 V
Let V𝑅 be the phase voltage at the receiving end.
Line current, I=
𝑃
3𝑉𝑅cos φ𝑅
=
3600 𝐾𝑤
3𝑉𝑅∗0.8
=
15∗ 105
𝑉𝑅
(i) Using approximate expression for 𝑉𝑆, we get,
19052= 𝑉𝑅 +
15∗ 105
𝑉𝑅
*5.31*0.8+
15∗ 105
𝑉𝑅
*5.54*0.6
𝑉𝑅
2
-19052 𝑉𝑅+1,13,58,000=0
Solving this equation, we get, 𝑉𝑅 = 18,435 V
∴ Line voltage at the receiving end = 3 × 18,435 = 31,930 V = 31·93 kV

(ii) Line current, I=
𝑃
3𝑉𝑅cos φ𝑅
=
3600 𝐾𝑤
3𝑉𝑅∗0.8
=
15∗ 105
𝑉𝑅
=
15∗ 105
18,435
= 81.36A
=
Power delivered
Power sent
× 100
=3600Kw+3𝐼2
𝑅
= 3600Kw+ 3(81.36)2
∗ 5.31
= 3600Kw + 105.447Kw
=3705.447Kw
3600
3705.447
× 100
=97.15%

4. A short 3-φ transmission line with an impedance of (6 + j 8) Ω
per phase has sending and receiving end voltages of 120 kV and
110 kV respectively for some receiving end load at a p.f. of 0·9
lagging. Determine (i) power output and (ii) sending end power
factor.
I=650.2A
(i) power output
= 3𝑉𝑅𝐼cos φ𝑅
=1,11,490Kw
(ii) sending end power factor
cos φ𝑆 =
𝑉𝑆
=0.88 lag

5. An 11 kV, 3-phase transmission line has a resistance of
1·5 Ω and reactance of 4 Ω per phase. Calculate the
percentage regulation and efficiency of the line when a total
load of 5000 kVA at 0.8 lagging power factor is supplied at
11 kV at the distant end.
6. A 3-phase, 50 Hz, 16 km long overhead line supplies 1000
kW at 11kV, 0·8 p.f.lagging. The line resistance is 0·03 Ω per
phase per km and line inductance is 0·7 mH per phase per
km. Calculate the sending end voltage, voltage regulation
and efficiency of transmission.

7. A three phase T.L 50 Km consists of 3 copper conductors in 1.2m delta. Load
conditions at R.E are 10,000KVA at 0.8 pf lag, 33000V. 50 Hz. Line is designed so
that T.L losses are approximately 10% of the power delivered. Find (i) sending
end voltage and pf (ii) percentage regulation (iii) transmission efficiency.
Power supplied (S)= 10000KVA
𝑉𝑅=33000V cos φ𝑅=0.8
T.L losses are approximately 10% of the power delivered.
Length = 50Km
Line current (I)=
𝑆
3𝑉𝐿
=
10000𝐾
3∗33000
= 174.95𝐴 𝑆= 3𝑉𝐿𝐼𝐿=3𝑉𝑝ℎ𝐼𝑝ℎ
Losses=3𝐼2
𝑅=3 ∗ (174.95)2
𝑅
Losses= 10% of the power delivered
3 ∗ (174.95)2
𝑅= 10%(S* cos φ𝑅)
=
10
100
∗ 10000𝐾 ∗ 0.8
R=8.712 Ω

Inductance/conductor/m =2 ∗ 10−7log𝑒
𝐷𝑚
𝐷𝑠
where 𝐷𝑠 = GMR or self-GMD = 0·7788 r
𝐷𝑚 = Spacing between conductors = d
 Resistance R =
ρ𝑙
𝑎
a= X-area of conductor
R =
ρ𝑙
Π𝑟2 r= radius of conductor
r =
ρ𝑙
Π𝑅
ρ = 1.73 ∗ 10−6
Ω- cm 𝑙 =50Km=50*103
∗ 102
cm R= 8.712 Ω
r =
1.73∗10−6∗50∗103∗ 102
Π∗8.712
r=0.562cm
 Inductance/conductor/m =2 ∗ 10−7log𝑒
𝐷𝑚
𝐷𝑠
𝐷𝑚=1.2m=120cm 𝐷𝑠 = 0·7788 r
= 2 ∗ 10−7log𝑒
120
0.7788∗0.562
=11.227 * 10−7
H/m
 Inductance/conductor/Km=1.1227mH/Km
 Total Inductance of line (L)=1.1227*50=56.134mH
 Inductive reactance (𝑋𝐿 )=2ΠfL=17.635Ω

(i) sending end phase voltage and power factor
𝑉𝑅=
33000
3
I=174.95 R=8.712Ω 𝑋𝐿=17.635Ω
𝑉𝑆 =22123V
Sending end p.f., cos φ𝑆 =
𝑉𝑆
=0.757 lag
(ii) Transmission efficiency
=
Power delivered
Power sent
× 100
=10000k*0.8+{
10
100
∗ 10000𝑘 ∗ 0.8 }
=8800k
10000𝑘∗0.8
8800k
× 100
=90.91%
(iii) percentage Voltage regulation =
𝑉𝑆−𝑉𝑅
𝑉𝑅
× 100
=
22123−19052
19052
× 100=16.12%

8. A 3-phase load of 2000 kVA, 0·8 p.f. is supplied at 6·6 kV, 50 Hz by means of a
33 kV transmission line 20 km long and 33/6·6 kV step-down transfomer. The
resistance and reactance of each conductor are 0·4 Ω and 0·5 Ω per km
respectively. The resistance and reactance of transformer primary are 7.5 Ω and
13.2 Ω, while those of secondary are 0.35 Ω and 0.65 Ω respectively. Find the
voltage necessary at the sending end of transmission line when 6.6 kV is
maintained at the receiving end. Determine also the sending end power factor
and transmission efficiency.

Medium transmission lines have sufficient length (50-
150 km) and usually operate at voltages greater than
20 kV, the effects of capacitance cannot be neglected.
The most commonly used methods for the solution of
medium transmissions lines are :
(i) End condenser method
(ii) Nominal T method
(iii) Nominal π method.

(i) End condenser method
In this method, the capacitance of the line is lumped or
concentrated at the receiving or load end
Let
𝐼𝑅 = load current per phase
R = resistance per phase
𝑋𝐿 = inductive reactance per phase
C = capacitance per phase
𝑉𝑆 = sending end voltage per phase
𝑉𝑅 = receiving end voltage per phase
𝐼𝑆 = sending end current per phase
𝐼𝐶 = capacitive current per phase

φ𝑅
 The phasor diagram for the circuit
 Taking the receiving end voltage 𝑉𝑅 as the reference phasor,
𝐼𝑅= 𝐼𝑅 ∠ − φ𝑅= 𝐼𝑅 (cos φ𝑅 − j sinφ𝑅)
 Capacitive current, 𝐼𝐶 = j 𝑉𝑅 ω C = j 2 π f C𝑉𝑅
 The sending end current 𝐼𝑠 is the phasor sum of load current 𝐼𝑅 and
capacitive current 𝐼𝐶 i.e.,
Sending end current,𝐼𝑠=𝐼𝑅+𝐼𝐶
𝑉𝑹
𝐼𝑺
𝐼𝑹
𝐼𝐶

 Sending end current,𝐼𝑠=𝐼𝑅+𝐼𝐶
=𝐼𝑅 (cos φ𝑅 − j sinφ𝑅)+ j 2 π f C𝑉𝑅
=𝐼𝑅cos φ𝑅+j ( −𝐼𝑅 sinφ𝑅+2 π f C𝑉𝑅)
Sending end voltage, 𝑉𝑆= 𝑉𝑅+𝐼𝑠
Ԧ
𝑍
= 𝑉𝑅+𝐼𝑠(R+ j 𝑋𝐿)
Thus, the magnitude of sending end voltage 𝑉𝑆 can be calculated.
percentage Voltage regulation =
𝑉𝑆−𝑉𝑅
𝑉𝑅
× 100
Transmission efficiency
=
Power delivered / 𝑝ℎ𝑎𝑠𝑒
Power delivered / 𝑝ℎ𝑎𝑠𝑒+𝑙𝑜𝑠𝑠𝑒𝑠/𝑝ℎ𝑎𝑠𝑒
× 100
=
𝑉𝑅𝐼𝑅cos φ𝑅
𝑉𝑅𝐼𝑅cos φ𝑅+𝐼𝑠
2
𝑅
× 100

 Limitations.
Although end condenser method for the solution of
medium lines is simple to work out calculations, yet
it has the following drawbacks :
(i) There is a considerable error (about 10%) in
calculations because the distributed capacitance
has been assumed to be lumped or concentrated.
(ii) This method overestimates the effects of line
capacitance.

1. A (medium) single phase transmission line 100 km long has the following
constants :
Resistance/km = 0·25 Ω ; Reactance/km = 0·8 Ω
Susceptance/km = 14 × 10−6
siemen ; Receiving end line voltage = 66,000 V
Assuming that the total capacitance of the line is localized at the receiving end
alone, determine (i) the sending end current (ii) the sending end voltage (iii)
regulation and (iv) supply power factor. The line is delivering 15,000 kW at 0.8
power factor lagging. Draw the phasor diagram to illustrate your calculations.
Total resistance, R = 0·25 × 100 = 25 Ω
Total reactance, 𝑋𝐿 = 0·8 × 100 = 80 Ω
Total susceptance, Y = 14 × 10−6
× 100 = 14 × 10−4
S
Receiving end voltage, 𝑉𝑅 = 66,000 V
cos φ𝑅=0.8 sin φ𝑅 =0.6

φ𝑅
 The phasor diagram for the circuit
 Taking the receiving end voltage 𝑉𝑅 as the reference phasor,
𝑉𝑅= 𝑉𝑅+j0=66000v
Load current, 𝐼𝑅 =
𝑃
𝑉𝑅cos φ𝑅
=
15000𝐾
66000∗0.8
=284A
=284 (0·8 − j 0·6) = 227 − j 170
𝑉𝑹
𝐼𝑺
𝐼𝑹
𝐼𝐶

 Capacitive current= 𝐼𝐶=jY* 𝑉𝑅
= 14 × 10−4*66000=j92
(i) Sending end current, 𝐼𝑠=𝐼𝑅+𝐼𝐶
= 227 − j 170+j92
= 227 − j 78 =240.02∠ −18.960
Magnitude of 𝐼𝑠= (227)2+(78)2
=240A
(ii) Sending end voltage, 𝑉𝑆= 𝑉𝑅+𝐼𝑠
Ԧ
𝑍
= 𝑉𝑅+𝐼𝑠(R+ j 𝑋𝐿)
=66000+(227 − j 78 )(25+j80)
=77915+j16210= 79583∠11.750
Magnitude of 𝑉𝑆 = (77915)2+(16210)2
=79583V
(iii) percentage Voltage regulation =
𝑉𝑆−𝑉𝑅
𝑉𝑅
× 100
=
79583−66000
66000
× 100=20.58%
(iv) supply power factor.
Phase angle between sending end voltage and sending end current
φ𝑆= 11.750
-(−18.960
)= 30.720
supply power factor cos φ𝑆=0.859 lag

(ii)Nominal T Method:
In this method, the whole line capacitance is assumed to be concentrated at the middle
point of the line and half the line resistance and reactance are lumped on its either side.
Therefore, in this arrangement, full charging current flows over half the line.
Let
𝐼𝑅 = load current per phase R = resistance per phase
𝑋𝐿 = inductive reactance per phase C = capacitance per phase
𝑉𝑆 = sending end voltage per phase 𝑉𝑅 = receiving end voltage per phase
𝑉1 = voltage across capacitor C 𝐼𝑆 = sending end current per phase
𝐼𝐶 = capacitive current per phase

φ𝑅
 The phasor diagram for the circuit Taking the receiving end voltage 𝑉𝑅 as the reference phasor,
𝑉𝑅= 𝑉𝑅+j0 𝐼𝑅= 𝐼𝑅 ∠ − φ𝑅= 𝐼𝑅 (cos φ𝑅 − j sinφ𝑅)
𝑉𝑹
𝐼𝑺
𝐼𝑹
𝐼𝐶
voltage across capacitor C, 𝑉1 = 𝑉𝑅+ 𝐼𝑅
Ԧ
𝑍
2
= 𝑉𝑅+𝐼𝑅 (cos φ𝑅 − j sinφ𝑅) (
R
2
+ j
𝑋𝐿
2
)
 Capacitive current, 𝐼𝐶 = j 𝑉1 ω C = j 2 π f C𝑉1
Sending end voltage, 𝑉𝑆= 𝑉1+ 𝐼𝑆
Ԧ
𝑍
2
= 𝑉1+ 𝐼𝑆(
R
2
+ j
𝑋𝐿
2
)

A 3-phase, 50-Hz overhead transmission line 100 km long has the following constants :
Resistance/km/phase = 0.1 Ω
Inductive reactance/km/phase = 0·2 Ω
Capacitive susceptance/km/phase = 0·04 ×10−4 siemen
Determine (i) the sending end current (ii) sending end voltage (iii) sending end power
factor and (iv) transmission efficiency when supplying a balanced load of 10,000 kW at 66
kV, p.f. 0·8 lagging. Use nominal T method.
Total resistance/phase, R = 0·1 × 100 = 10 Ω
Total reactance/phase. 𝑋𝐿 = 0·2 × 100 = 20 Ω
Capacitive susceptance, Y = 0·04 × 10−4 × 100 = 4 × 10−4 S
Receiving end voltage/phase, 𝑉𝑅 = 66,000/ 3 = 38105 V
𝑃
3𝑉𝑅cos φ𝑅
=
10000𝐾
3∗66𝑘∗0.8
=109A

φ𝑅
 (i) the sending end current The phasor diagram for the circuit Taking the receiving end
voltage 𝑉𝑅 as the reference phasor,
𝑉𝑅= 𝑉𝑅+j0= 38105 V 𝐼𝑅= 𝐼𝑅 ∠ − φ𝑅= 𝐼𝑅 (cos φ𝑅 − j sinφ𝑅)= 87·2 − j 65·4
𝑉𝑹
𝐼𝑺
𝐼𝑹
𝐼𝐶
voltage across capacitor C, 𝑉1 = 𝑉𝑅+ 𝐼𝑅
Ԧ
𝑍
2
= 𝑉𝑅+𝐼𝑅 (cos φ𝑅 − j sinφ𝑅) (
R
2
+ j
𝑋𝐿
2
)= 39,195 + j 545
 Capacitive current, 𝐼𝐶 = j 𝑉1 ω C = j 2 π f C𝑉1= − 0 ⋅218 + j15⋅6
Sending end current,𝐼𝑠=𝐼𝑅+𝐼𝐶= 87·0 − j 49·8 = 100 ∠ − 29.780
A
(ii)Sending end voltage, 𝑉𝑆= 𝑉1+ 𝐼𝑆
Ԧ
𝑍
2
= 𝑉1+ 𝐼𝑆(
R
2
+ j
𝑋𝐿
2
)= 40128 + j 1170=40145 ∠1.670
V

 (iii) Referring to phasor diagram,
 θ1 = angle between 𝑉𝑅 and 𝑉𝑆 = 1.670
 θ2= angle between 𝑉𝑅 and 𝐼𝑆 = 29.780
∴ φ𝑆 = angle between
𝑉𝑆 and 𝐼𝑆
= θ1 + θ2 = 1.670 + 29.780
= 31.450
sending end power factor cos φ𝑆=0.853 lag
(iv) transmission efficiency
Sending end power = 3 𝑉𝑆 𝐼𝑆 cos φ𝑆 = 3 × 40,145 × 100 × 0·853
= 10273105 W = 10273·105 kW
Power delivered = 10,000 kW
∴ Transmission efficiency =
10 000
10273·105
×100
= 97·34%

A 3-phase, 50 Hz transmission line 100 km long delivers 20 MW
at 0·9 p.f. lagging and at 110 kV. The resistance and reactance of
the line per phase per km are 0·2 Ω and 0·4 Ω respectively, while
capacitance admittance is 2·5 ×10−6
siemen/km/phase.
Calculate : (i) the current and voltage at the sending end (ii)
efficiency of transmission. Use nominal T method.

(iii) Nominal Method
In this method, capacitance of each conductor is divided into two halves; one
half being lumped at the sending end and the other half at the receiving end. It is
obvious that capacitance at the sending end has no effect on the line drop.
However, its charging current must be added to line current in order to obtain
the total sending end current.
Let
𝐼𝑅 = load current per phase
R = resistance per phase
𝑋𝐿 = inductive reactance per phase
C = capacitance per phase
𝑉𝑆 = sending end voltage per phase
𝐼𝑆 = sending end current per phase
𝐼𝐶1 = charging current at load end per phase
𝐼𝐶2 = charging current at sending end per phase

φ𝑅
𝑉𝑅= 𝑉𝑅+j0 𝐼𝑅= 𝐼𝑅 ∠ − φ𝑅= 𝐼𝑅 (cos φ𝑅 − j sinφ𝑅)
𝑉𝑹
𝐼𝑳
𝐼𝑹
𝐼𝐶1
 Charging current at load end is
𝐼𝐶1 = j 𝑉𝑅 ω( C/2) = jπ f C𝑉𝑅
Line current= 𝐼𝐿=𝐼𝑅+𝐼𝐶1
Sending end voltage, 𝑉𝑆= 𝑉𝑅+ 𝐼𝐿
Ԧ
𝑍= 𝑉𝑅+ 𝐼𝐿(𝑅+ j𝑋𝐿)
 Charging current at sending end is
𝐼𝐶2 = j 𝑉𝑆 ω( C/2) = jπ f C𝑉𝑆
Sending end current,𝐼𝑠=𝐼𝐿+𝐼𝐶2
𝐼𝐶2
𝐼𝑺

A 3-phase, 50Hz, 150 km line has a resistance, inductive reactance
and capacitive shunt admittance of 0·1 Ω, 0·5 Ω and 3 ×10−6
S per
km per phase. If the line delivers 50 MW at 110 kV and 0·8 p.f.
lagging, determine the sending end voltage and current. Assume
a nominal π circuit for the line.
Total reactance/phase. 𝑋𝐿 = 0·5 × 150 = 75 Ω
Capacitive susceptance, Y = 3 × 10−6
× 150 = 45 × 10−5
S
Receiving end voltage/phase, 𝑉𝑅 = 110,000/ 3 = 63,508 V
𝑃
3𝑉𝑅cos φ𝑅
=
50∗106
3∗110,000∗0.8
=328A

φ𝑅
𝑉𝑅= 𝑉𝑅+j0=63,508V 𝐼𝑅= 𝐼𝑅 ∠ − φ𝑅= 𝐼𝑅 (cos φ𝑅 − j sinφ𝑅)=262.4-j196.8
𝑉𝑹
𝐼𝑳
𝐼𝑹
𝐼𝐶1
 Charging current at load end is
𝐼𝐶1 = 𝑉𝑅 j( Y/2) =j14.3
Line current= 𝐼𝐿=𝐼𝑅+𝐼𝐶1=262.4-j182.5
Ԧ
𝑍= 𝑉𝑅+ 𝐼𝐿(𝑅+ j𝑋𝐿)=81,131+j16,942.5=82,881 ∠11.790
 Charging current at sending end is
𝐼𝐶2 = 𝑉𝑆 j( Y/2) =-3.81+j8.25
Sending end current,𝐼𝑠=𝐼𝐿+𝐼𝐶2=258.6-j164.25=306.4 ∠32.420
A
𝐼𝐶2
𝐼𝑺

A 100-km long, 3-phase, 50-Hz transmission line has following line constants:
Resistance/phase/km = 0·1 Ω Reactance/phase/km = 0·5 Ω
Susceptance/phase/km = 10 × 10−6 S
If the line supplies load of 20 MW at 0·9 p.f. lagging at 66 kV at the receiving
end, calculate by nominal π method :
(i) sending end power factor (ii) regulation (iii) transmission efficiency

The line constants are considered as uniformly distributed throughout
the length of the line.
The equivalent circuit of a 3-phase long transmission line on a phase-
neutral basis. The whole line length is divided into n sections, each
section having line constants
1
𝑛
th of those for the whole line.

 The following points may by noted :
(i) The line constants are uniformly distributed over the entire length of line as is
actually the case.
(ii) The resistance and inductive reactance are the series elements.
(iii) The leakage susceptance (B) and leakage conductance (G) are shunt
elements. The leakage susceptance is due to the fact that capacitance exists
between line and neutral. The leakage conductance takes into account the energy
losses occurring through leakage over the insulators or due to corona effect
between conductors. Admittance = 𝐺2 + 𝐵2.
(iv) The leakage current through shunt admittance is maximum at the sending
end of the line and decreases continuously as the receiving end of the circuit is
approached at which point its value is zero.

Analysis of Long Transmission Line (Rigorous method)
Single phase and neutral connection of a 3-phase line with impedance and
shunt admittance of the line uniformly distributed.
Consider a small element in the line of length dx situated at a distance x from
the receiving end.
Let
z = series impedance of the line per unit length
y = shunt admittance of the line per unit length
V = voltage at the end of element towards receiving end
V + dV = voltage at the end of element towards sending end
I + dI = current entering the element dx
I = current leaving the element dx
Then for the small element dx,
z dx = series impedance
y dx = shunt admittance

 Obviously the rise in voltage over the element length in the direction of increasing x
dV = I z dx Voltage drop in the element dx
𝑑𝑉
𝑑𝑥
= 𝑧𝐼 (1)
Now, the current entering the element is I + dI whereas the current leaving the element is I. The
difference in the currents flows through shunt admittance of the element i.e.,
dI = Current through shunt admittance of element = V y dx
Or
𝑑𝐼
𝑑𝑥
= 𝑉𝑦 (2)
Differentiating eq. (1) w.r.t. x, we get,
𝑑2𝑉
𝑑𝑥2 = 𝑧
𝑑𝐼
𝑑𝑥
𝑑2𝑉
𝑑𝑥2 = =z 𝑉𝑦 (3)
The solution of this differential equation is
V=𝐴1𝑒 𝑦𝑧 𝑥 + 𝐴2𝑒− 𝑦𝑧 𝑥 (4)
Where 𝐴1 𝑎𝑛𝑑 𝐴2 are unknown constants. Differentiating eq. (4) w.r.t. x, we get,
𝑑𝑉
𝑑𝑥
= 𝑦𝑧(𝐴1𝑒 𝑦𝑧 𝑥
− 𝐴2𝑒− 𝑦𝑧 𝑥
) (5)
From eq 1 we get I=
1
𝑧
𝑑𝑉
𝑑𝑥
=
1
𝑧
( 𝑦𝑧(𝐴1𝑒 𝑦𝑧 𝑥
)=
𝑦
𝑧
(𝐴1𝑒 𝑦𝑧 𝑥
) (6)

Equations (4) and (5) give the expressions for V and I in the form of unknown constants 𝐴1 and 𝐴2.
The values of 𝐴1 and 𝐴2 can be found by applying end conditions as under :
At receiving end x = 0, V = 𝑉𝑅 and I = 𝐼𝑅
Putting these values in eq. (4) & (6), we have,
𝑉𝑅= 𝐴1 + 𝐴2 (7)
𝐼𝑅 =
𝑦
𝑧
( 𝐴1 - 𝐴2 ) (8)
By solving above equations we get
𝐴1=
1
2
𝑉𝑅+ 𝐼𝑅
𝑧
𝑦
𝐴2 =
1
2
𝑉𝑅 − 𝐼𝑅
𝑧
𝑦
Putting these values in eq. (4) & (6), we have,
V=
1
2
𝑉𝑅+ 𝐼𝑅
𝑧
𝑦
(𝑒 𝑦𝑧 𝑥)+
1
2
𝑧
𝑦
𝑒− 𝑦𝑧 𝑥)
I=
𝑦
𝑧
1
2
𝑉𝑅+ 𝐼𝑅
𝑧
𝑦
𝑒 𝑦𝑧 𝑥
−
1
2
𝑧
𝑦
𝑒− 𝑦𝑧 𝑥
For a transmission line
𝑧
𝑦
is a constant, called the characteristic constant, 𝑍𝐶 and 𝑦𝑧 is another
constant, called the propagation constant, γ, both complex quantities.

V=(
𝑉𝑅+ 𝐼𝑅𝑍𝐶
2
)(𝑒γ𝑥
)+(
𝑉𝑅− 𝐼𝑅𝑍𝐶
2
)(𝑒−γ𝑥
) (9)
I=
1
𝑍𝐶
{ (
1
2
𝑉𝑅+ 𝐼𝑅𝑍𝐶)𝑒γ𝑥
−
1
2
𝑉𝑅 − 𝐼𝑅𝑍𝐶 𝑒−γ𝑥
}
I=
1
2
𝑉𝑅
𝑍𝐶
+ 𝐼𝑅 𝑒γ𝑥
-
1
2
𝑉𝑅
𝑍𝐶
− 𝐼𝑅 𝑒−γ𝑥
(10)
Expanding and re arranging above expressions, we get
V= 𝑉𝑅
𝑒
γ𝑥
+𝑒
−γ𝑥
2
+ 𝐼𝑅𝑍𝐶
𝑒
γ𝑥
−𝑒
−γ𝑥
2
V = 𝑉𝑅 cosh γ𝑥+ 𝐼𝑅𝑍𝐶 sinh γ𝑥 (11)
I=
1
𝑍𝐶
𝑉𝑅
𝑒γ𝑥
−𝑒−γ𝑥
2
+ 𝐼𝑅𝑍𝐶
𝑒γ𝑥
+𝑒−γ𝑥
2
=
1
𝑍𝐶
𝑉𝑅 sinh γ𝑥 + 𝐼𝑅𝑍𝐶 cosh γ𝑥
I =
𝑉𝑅
𝑍𝐶
sinh γ𝑥 + 𝐼𝑅 cosh γ𝑥 (12)
the sending end voltage 𝑉𝑆 and sending end current 𝐼𝑆 can b obtained by substituting x=l
𝑉𝑆 = 𝑉𝑅 cosh γ𝑙+ 𝐼𝑅𝑍𝐶 sinh γ𝑙
𝐼𝑆 =
𝑉𝑅
𝑍𝐶
sinh γ𝑙 + 𝐼𝑅 cosh γ𝑙
γl= 𝑦𝑧 l = 𝑦𝑙 ∗ 𝑧𝑙= 𝑌𝑍
Where Z is the total impedance of the line and Y is the total admittance of the line.
𝑉𝑆 = 𝑉𝑅 cosh 𝑌𝑍+ 𝐼𝑅𝑍𝐶 sinh 𝑌𝑍
𝐼𝑆 = 𝐼𝑅 cosh 𝑌𝑍+
𝑉𝑅
𝑍𝐶
sinh 𝑌𝑍

Surge Impedance:
The square root of the line impedance Z and shunt admittance Y is called the ‘surge
impedance’ 𝑍0 of the line
𝑍0=
𝑍
𝑌
Z=R+iX =R+jωL and Y=G+jB=G+j ωC
For a loss less line R=0, G=0
𝑍0=
𝐿
𝐶
which is pure resistance
Its value varies between 400 Ω and 600 Ω in case of overhead transmission lines and
between 40 Ω and 60 Ω in case of underground cables.
Surge Impedance Loading:
This is defined as the load (of unity power factor) that can be delivered by the line of
negligible resistance.
Power transmitted 𝑃𝑅 =
𝑉𝑅
2
𝑍0
Where 𝑉𝑅 is receiving end voltage in Kv and 𝑍0 is surge impedance in ohms
𝑃𝑅 is called Surge Impedance Loading or natural power of the line

In order to increase the power transmitted through a long transmission line either value
of receiving end voltage is to be increased, surge impedance Zo is to be decreased.
(i) Increase in voltage 𝑉𝑅 :
Nowadays the trend is for higher and higher voltages so that this is the most widely
adopted method to increase the power limit for heavily loaded long transmission lines.
But there are some Practical difficulties in this method and it is expensive .
(ii) Decrease of surge impedance Zo:
since the spacing between the conductors cannot be decreased much, it being dependent
in the line voltages and corona etc., the value of Zo cannot be varied as such. But some
are artificial means are employed to decrease Zo.
For a lose less transmission Zo=
𝐿
𝐶
and γ= jω 𝐿𝐶=jβ.
β is the phase Shift. The latter determines the torque angle, between Vs and Vr and
hence system stability. To decrease Zo either L is decreased using series capacitor or C is
increased using shunt capacitance
Power transmitted
𝑃𝑅 =
𝑉𝑅
2
𝑍0

A 3-φ transmission line 200 km long has the following constants :
Resistance/phase/km = 0·16 Ω Reactance/phase/km = 0·25 Ω
Shunt admittance/phase/km = 1·5 × 10−6 S Calculate by rigorous method the sending
end voltage and current when the line is delivering a load of 20 MW at 0·8 p.f. lagging.
The receiving end voltage is kept constant at 110 kV.
Total reactance/phase, X = 0·25 × 200 = 50 Ω
Total shunt admittance/phase, Y = j 1·5 × 10−6 × 200 = 0·0003 ∠ 90º
Series Impedance/phase, Z = R + j X = 32 + j 50 = 59·4 ∠ 58º
The sending end voltage 𝑉𝑆 per phase is given by :
𝑉𝑆 = 𝑉𝑅 cosh 𝑌𝑍+ 𝐼𝑅𝑍𝐶 sinh 𝑌𝑍
𝑉𝑅
𝑍𝐶
sinh 𝑌𝑍
Cosh x=1+
𝑥2
2!
+
𝑥4
4!
+
𝑥6
6!
+………..
sinh x=x+
𝑥3
3!
+
𝑥5
5!
+
𝑥7
7!
+………..
=x(1+
𝑥2
3!
+
𝑥4
5!
+
𝑥6
7!
+………..)

Cosh 𝑌𝑍 =1+
𝑌𝑍
2!
+
𝑌2𝑍2
24
approx….
sinh 𝑌𝑍 = 𝑌𝑍 (1+
𝑌𝑍
6
approx….)
𝑌𝑍 = 59⋅ 4 ∠ 58º × 0 ⋅ 0003 ∠ 90º = 0 ⋅133 ∠ 74º
Z Y = 0·0178 ∠ 148º
𝑌2𝑍2 = 0·00032 ∠ 296º
the characteristic constant, 𝑍𝐶 =
𝑧
𝑦
=
59⋅ 4 ∠ 58º
0 ⋅ 0003 ∠ 90º
= 445 ∠ 16º
Cosh 𝑌𝑍= 0·992 + j 0·00469 = 0·992 ∠ 0·26º
sinh 𝑌𝑍= 0·0362 + j 0·1275 = 0·1325 ∠ 74.14º
𝑃
3𝑉𝑅cos φ𝑅
=
20∗106
3∗110,000∗0.8
=131A
𝑉𝑆 = 𝑉𝑅 cosh 𝑌𝑍+ 𝐼𝑅𝑍𝐶 sinh 𝑌𝑍=67018 + j 6840 = 67366 ∠ 5.82ºV
𝑉𝑅
𝑍𝐶
sinh 𝑌𝑍=129·83 + j 19·42 = 131·1 ∠ 8.5º A

In any four terminal network, the input voltage and input current can be
expressed in terms of output voltage and output current.
𝑉𝑆= Ԧ
𝐴 𝑉𝑅+ 𝐵 𝐼𝑅
𝐼𝑆= Ԧ
𝐶 𝑉𝑅+ 𝐷 𝐼𝑅
Where 𝑉𝑆 = sending end voltage per phase
𝐼𝑆 = sending end current
𝐼𝑅 = receiving end current
and Ԧ
𝐴 , 𝐵, Ԧ
𝐶 and 𝐷 (generally complex numbers) are the
constants known as generalised circuit constants of the transmission
line.

The following points may be kept in mind :
(i) The constants Ԧ
𝐴 , 𝐵, Ԧ
𝐶 and 𝐷 are generally complex numbers.
(ii) The constants Ԧ
𝐴 and 𝐷 are dimensionless whereas the
dimensions of 𝐵 and Ԧ
𝐶 are ohms and siemen respectively.
(iii) For a given transmission line, Ԧ
𝐴 = 𝐷
(iv) For a given transmission line, Ԧ
𝐴 𝐷 − 𝐵 Ԧ
𝐶 = 1

(i) Short lines.
In short transmission lines, the effect of line capacitance is neglected. Therefore, the line is
considered to have series impedance.
𝐼𝑆= 𝐼𝑅 (1)
𝑉𝑆= 𝑉𝑅+𝐼𝑅
Ԧ
𝑍 (2)
Comparing these with eqs.
𝑉𝑆= Ԧ
𝐴 𝑉𝑅+ 𝐵 𝐼𝑅 (3)
𝐼𝑆= Ԧ
𝐶 𝑉𝑅+ 𝐷 𝐼𝑅 (4)
Ԧ
𝐴 = 1 ; 𝐵 = Ԧ
𝑍, Ԧ
𝐶 = 0 and 𝐷 = 1
Incidentally ; Ԧ
𝐴 = 𝐷 and Ԧ
𝐶 = 1 × 1 − Ԧ
𝑍 × 0 = 1

(ii) Medium lines- End Condenser method:
In this method the capacitance of the line is assumed to be lumped at the load end
𝐼𝑠=𝐼𝑅+𝐼𝐶
𝐼𝑠 = 𝐼𝑅+ 𝑌 𝑉𝑅 (5)
𝑉𝑆= 𝑉𝑅+𝐼𝑆
Ԧ
𝑍
= 𝑉𝑅+(𝐼𝑅+ 𝑌 𝑉𝑅) Ԧ
𝑍
𝑉𝑆 = 𝑉𝑅(1+ Ԧ
𝑍𝑌)+ Ԧ
𝑍 𝐼𝑅 (6)
Comparing eqs (5) &(6) with eqs.
𝑉𝑆= Ԧ
𝐼𝑆= Ԧ
We get
Ԧ
𝐴 = 1+ Ԧ
𝑍𝑌 ; 𝐵 = Ԧ
𝑍, Ԧ
𝐶 = 𝑌 and 𝐷 = 1
Incidentally ; Ԧ
𝐶 = 1

(iii) Medium lines – Nominal T method.
In this method, the whole line to neutral capacitance is assumed to be concentrated at the middle point of the line
and half the line resistance and reactance are lumped on either side
𝑉𝑆= 𝑉1+ 𝐼𝑆
Ԧ
𝑍
2
(7)
𝑉1 = 𝑉𝑅+ 𝐼𝑅
Ԧ
𝑍
2
(8)
𝐼𝑠=𝐼𝑅+𝐼𝐶 (9)
𝐼𝐶 = 𝑌 𝑉1 (10)
Put eq(10), (8) in (9)
𝐼𝑠=𝐼𝑅+𝐼𝐶 = 𝐼𝑅+ 𝑌 (𝑉𝑅+ 𝐼𝑅
Ԧ
𝑍
2
)
𝐼𝑠 = 𝑌 𝑉𝑅+ 𝐼𝑅(1+
𝑌 Ԧ
𝑍
2
) (11)
Put eq (11) , (8) in (7)
𝑉𝑆= 𝑉1+ 𝐼𝑆
Ԧ
𝑍
2
= 𝑉𝑅+ 𝐼𝑅
Ԧ
𝑍
2
+[ 𝑌 𝑉𝑅+ 𝐼𝑅(1+
𝑌 Ԧ
𝑍
2
) ]
Ԧ
𝑍
2
𝑉𝑆 = (1+
𝑌 Ԧ
𝑍
2
) 𝑉𝑅+( Ԧ
𝑍+
𝑌 Ԧ
𝑍2
4
) 𝐼𝑅 (12)
Comparing eqs (11) &( 12) with eqs.
𝑉𝑆= Ԧ
𝐼𝑆= Ԧ
Ԧ
𝐴 = 𝐷= 1+
𝑌 Ԧ
𝑍
2
; 𝐵 = Ԧ
𝑍 (1+
𝑌 Ԧ
𝑍
4
), and Ԧ
𝐶 = 𝑌
Incidentally ; Ԧ
𝐶 = 1+
𝑌 Ԧ
𝑍
2
2
− 𝑌 Ԧ
𝑍(1+
𝑌 Ԧ
𝑍
4
)= 1

(iv) Medium lines – Nominal method.
In this method, line-to-neutral capacitance is divided into two halves ; one half
being concentrated at the load end and the other half at the sending end
𝐼𝐿=𝐼𝑅+𝐼𝐶1 (13)
Ԧ
𝑍 (14)
Sending end current 𝐼𝑠=𝐼𝐿+𝐼𝐶2 (15)
𝐼𝐶2= 𝑉𝑆
𝑌
2
(16) 𝐼𝐶1= 𝑉𝑅
𝑌
2
(17)
Put eqs (13) (17) in eq (13)
𝑉𝑆= 𝑉𝑅+(𝐼𝑅+𝑉𝑅
𝑌
2
) Ԧ
𝑍
𝑉𝑆 = (1+
𝑌 Ԧ
𝑍
2
) 𝑉𝑅+ 𝐼𝑅
Ԧ
𝑍 (18)
Put eqs (16), (17),(13),(18) in eq (15)
𝐼𝑠=𝐼𝑅+𝑉𝑅
𝑌
2
+ (1+
𝑌 Ԧ
𝑍
2
) 𝑉𝑅+ 𝐼𝑅
Ԧ
𝑍
𝑌
2
𝐼𝑠 = (1+
𝑌 Ԧ
𝑍
4
) 𝑌 𝑉𝑅+ (1+
𝑌 Ԧ
𝑍
2
) 𝐼𝑅 (19)

𝑉𝑆= Ԧ
𝐼𝑆= Ԧ
Ԧ
𝐴 = 𝐷= 1+
𝑌 Ԧ
𝑍
2
; 𝐵 = Ԧ
𝑍, and Ԧ
𝐶 = (1+
𝑌 Ԧ
𝑍
4
) 𝑌
Incidentally ; Ԧ
𝐶 = 1+
𝑌 Ԧ
𝑍
2
2
− Ԧ
𝑍𝑌(1+
𝑌 Ԧ
𝑍
4
)=1

(v) Long lines—Rigorous method.
By rigorous method, the sending end voltage and current of a long
transmission line are given by
𝑉𝑆 = 𝑉𝑅 cosh 𝑌𝑍+ 𝐼𝑅𝑍𝐶 sinh 𝑌𝑍 (20)
𝐼𝑆 =
𝑉𝑅
𝑍𝐶
sinh 𝑌𝑍+ 𝐼𝑅 cosh 𝑌𝑍 (21)
𝑉𝑆= Ԧ
𝐼𝑆= Ԧ
Ԧ
𝐴= 𝐷 = cosh 𝑌𝑍 𝐵= 𝑍𝐶 sinh 𝑌𝑍 Ԧ
𝐶 =
1
𝑍𝐶
sinh 𝑌𝑍
Incidentally ;
Ԧ
𝐶 = cosh 𝑌𝑍 *cosh 𝑌𝑍- 𝑍𝐶 sinh 𝑌𝑍*
1
𝑍𝐶
sinh 𝑌𝑍
=𝑐𝑜𝑠ℎ2
𝑌𝑍- 𝑠𝑖𝑛ℎ2
𝑌𝑍=1

A 3 phase overhead transmission line has a total series impedance per phase of 200∠800
ohms and total shunt admittance of 0.0013 ∠900
mho per phase. The line delivers a load of
80MW at 0.8 p.f. lagging and 220kV between the lines. Calculate
(a) The ABCD constants of the line
(b) The sending end voltage, current and p.f. of the line
(c) the efficiency of transmission
Total series impedance per phase Z= 200∠800 ohms
Total shunt admittance Y= 0.0013 ∠900 mho per phase.
Receiving end voltage/phase, 𝑉𝑅 = 220,000/ 3 = 1,27,000 V
𝑃
3𝑉𝑅cos φ𝑅
=
80∗106
3∗220,000∗0.8
=262.43A
Taking the receiving end voltage 𝑉𝑅 as the reference phasor,
𝑉𝑅= 𝑉𝑅+j0= 1,27,000 ∠ 00 V
𝐼𝑅= 𝐼𝑅 ∠ − φ𝑅= 𝐼𝑅 (cos φ𝑅 − j sinφ𝑅)=262.43(0.8-j0.6)=262.43 ∠ −36.870

 Using nominal T method
Ԧ
𝐴 = 𝐷= 1+
𝑌 Ԧ
𝑍
2
;
=1+
1
2
(0.0013 ∠900 × 200∠800)=1+0.13 ∠1700
=0.872+j0.02257=0.872 ∠1.50
𝐵 = Ԧ
𝑍 (1+
𝑌 Ԧ
𝑍
4
),
= 200∠800 ×(1+
1
4
(0.0013 ∠900 × 200∠800)
=187.2 ∠80.690
and Ԧ
𝐶 = 𝑌
= 0.0013 ∠900
sending end voltage 𝑉𝑆= Ԧ
=0.872 ∠1.50 × 1,27,000 ∠ 00+ 187.2 ∠80.690 × 262.43 ∠ −36.870
=1,46,146+j36.915=1,50,736 ∠ 14.180
Sending end line voltage= 3*150736=2,61,000V
sending end current 𝐼𝑆= Ԧ
=186.6+j32.64=189.43 ∠ 9.920

Sending end power factor=cos φ𝑆
=cos (14.180- 9.920)
=0.9972 lag
Transmission efficiency=
𝑝𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡
𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡
× 100
Power input= 3 V𝑆𝐿I𝑆 cos φ𝑆
= 3 ×2,61,000 ×189.43×0.9972
=85.4MW
Transmission efficiency=
𝑝𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡
𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡
× 100
=
80
85.4
× 100
=93.68%

Find the following for a single circuit transmission line delivering a load of 50MVA at 110kV and 0.8 p.f lagging
(i)The sending end voltage,
(ii) The sending end current
(iii) The sending end p.f. of the line
(iv) the efficiency of transmission
Given Ԧ
𝐴 = 𝐷= 0.98 ∠30
𝐵= 110 ∠750
ohm 𝐶= 0.0005 ∠800
mho
A balanced 3-phase load of 30 MW is supplied at 132 kV, 50 Hz and 0·85 p.f. lagging by means of a transmission
line. The series impedance of a single conductor is (20 + j52) ohms and the total phase-neutral admittance is 315 ×
10−6
siemen. Using nominal T method, determine:
(i) the A, B, C and D constants of the line (ii) sending end voltage (iii) regulation of the line.
To determine the no load voltage at receiving end when load is suddenly thrown off
𝑉𝑆= Ԧ
𝑉𝑆= Ԧ
𝐴 𝑉′
𝑅+ 𝐵 0 𝑉′
𝑅=
𝑉𝑆
Ԧ
𝐴
Regulation=
𝑉𝑆−𝑉𝑅
𝑉𝑅
× 100
Under no load 𝑉𝑆=𝑉′
𝑅=
𝑉𝑆
𝐴
Regulation=
𝑉𝑆
𝐴
−𝑉𝑅
𝑉𝑅
× 100

A 132 kV, 50 Hz, 3-phase transmission line delivers a
load of 50 MW at 0·8 p.f. lagging at the receiving end.
The generalised constants of the transmission line are :
Ԧ
𝐴 = 𝐷= 0·95 ∠1·4º ; 𝐵 = 96 ∠78º ; 𝐶 = 0·0015 ∠90º
Find the regulation of the line and charging current. Use
Nominal-T method.

A three phase 300km long T.L has following parameters
r=0.15 ohm/km; x=0.5ohm/km; y=3× 10−6
mho/km if the line is represented by nominal ,
determine
(a) The ABCD constants of the line
(b) The power at UPF that can be received if the voltage at each end is maintained at 220Kv
(c) The sending end voltage required if a load of 200MW at 0.85 lagging p.f. is to be delivered with
the receiving end voltage remains at 220Kv
Total reactance/phase, X = 0·5 × 300 = 150 Ω
Total shunt admittance/phase, Y = j 3× 10−6
× 300 = 9 × 10−4
∠ 90º
Series Impedance/phase, Z = R + j X = 45 + j 150 = 156.6 ∠ 73.3º
nominal ,
Ԧ
𝐴 = 𝐷= 1+
𝑌 Ԧ
𝑍
2
; 𝐵 = Ԧ
𝑍, and Ԧ
𝐶 = (1+
𝑌 Ԧ
𝑍
4
) 𝑌
Ԧ
𝐴 = 𝐷= 1+
𝑌 Ԧ
𝑍
2
= 1+
9 ×10−4 ∠ 90º×156.6 ∠ 73.3º
2
=0.932 ∠ 1.244º
𝐵 = 156.6 ∠ 73.3º
Ԧ
𝐶 = (1+
𝑌 Ԧ
𝑍
4
) 𝑌= 1+
9 ×10−4 ∠ 90º×156.6 ∠ 73.3º
4
× 9 × 10−4 ∠ 90º
= 8.69 × 10−4
∠ 90.6º S

(b) The power at UPF that can be received if the voltage at each end is maintained at 220Kv
Receiving end voltage/phase, 𝑉𝑅 = 220,000/ 3 = 1,27,000 V
𝑉𝑅= 𝑉𝑅 ∠0=127kv= 𝑉𝑆
Load current 𝐼𝑅= 𝐼𝑅 ∠ − φ𝑅
= 𝐼𝑅 ∠0= 𝐼𝑅
𝑉𝑆= Ԧ
𝑉𝑆 = (0.932 ∠ 1.244º× 127k)+( 156.6 ∠ 73.3º × 𝐼𝑅)
𝑉𝑆 =118364 ∠ 1.244º+( 156.6 ∠ 73.3º × 𝐼𝑅)
𝑉𝑆 =118425+j2563+[(45+j150)] × 𝐼𝑅
𝑉𝑆 =(118425+45𝐼𝑅)+j(2563+150𝐼𝑅)
𝑉𝑆= 118425+45𝐼𝑅
2 + 2563+150𝐼𝑅
2
𝑉𝑆 =
220𝑘
3
Squaring on both sides & on solving
We get
16133× 106
=14024.5 × 106
+6.57 × 106
+10.658 × 106
𝐼𝑅+0.7689 × 106
𝐼𝑅 +0.024525 × 106
𝐼𝑅
2
0.024525 × 𝐼𝑅
2
+11.427 𝐼𝑅 -2102=0
𝐼𝑅=141.175A
Power delivered=3 𝑉𝑅 𝐼𝑅cos φ𝑅
=3*
220𝑘
3
*141.175*1=53.8MW

(c) 𝑉𝑆= Ԧ
Receiving end voltage/phase, 𝑉𝑅 = 220,000/ 3 = 1,27,000
cos φ𝑅=0.85 sinφ𝑅=0.5267
𝑃
3𝑉𝑅cos φ𝑅
=
200∗106
3∗220,000∗0.8
=617.57A
=617.5 ∠ -31.75º
𝑉𝑆=(0.932 ∠ 1.244º× 127k)+( 156.6 ∠ 73.3º ×617.24 ∠ -31.75º)
=202158 ∠ 19.27º
Line value = 3* 𝑉𝑆=350.15KV

 Evaluation of ABCD constants of long transmission lines
(i) Convergent series complex Angle method
(ii) Convergent series Real Angle method
Ԧ
𝐴= 𝐷 = cosh 𝑌𝑍 𝐵= 𝑍𝐶 sinh 𝑌𝑍 Ԧ
𝐶 =
1
𝑍𝐶
sinh 𝑌𝑍
Cosh x=1+
𝑥2
2!
+
𝑥4
4!
+
𝑥6
6!
+……….. sinh x=x+
𝑥3
3!
+
𝑥5
5!
+
𝑥7
7!
+………..=x(1+
𝑥2
3!
+
𝑥4
5!
+
𝑥6
7!
+………..)
Ԧ
𝐴= 𝐷 = Cosh 𝑌𝑍 =1+
𝑌𝑍
2
+
𝑌2𝑍2
24
+
𝑌3𝑍3
720
+….….
𝐵= 𝑍𝐶 sinh 𝑌𝑍 =
𝑍
𝑌
𝑌𝑍 (1+
𝑌𝑍
6
𝑌2𝑍2
120
+
𝑌3𝑍3
5040
+….….)
=Z (1+
𝑌𝑍
6
𝑌2𝑍2
120
+
𝑌3𝑍3
5040
+….….)
Ԧ
𝐶 =
1
𝑍𝐶
sinh 𝑌𝑍=
𝑌
𝑍
𝑌𝑍 (1+
𝑌𝑍
6
𝑌2𝑍2
120
+
𝑌3𝑍3
5040
+….….)
=Y (1+
𝑌𝑍
6
𝑌2𝑍2
120
+
𝑌3𝑍3
5040
+….….)

In this method the hyperbolic sines and cosines are
expanded by the trigonometrical formulae
𝑌𝑍=α+jβ α, β are real in rad
Then Cosh(α+jβ)= Cosh α Cosh jβ +sinh α sinh jβ
=Cosh α Cos β +jsinh α sin β
sinh(α+jβ)= sinh α Cosh jβ +sinh α sinh jβ
= sinh α Cos β +j Cosh α sin β

A three phase 50Hz,1000km long T.L has following line constants per phase per km uniformly
distributed r=0.22ohm; x=0.45ohm; g=4× 10−9
s; b=2.53 × 10−6
s Determine the auxiliary constants
by using
Given r=0.22ohm; x=0.45ohm; g=4× 10−9s; b=2.53 × 10−6s
z=r+jx=0.22+j0.45 y=g+jb=(0.004+j2.53) × 10−6s
Total series impedance per phase Z=z ×l= 500.9∠63.950
ohms
Total shunt admittance Y= y ×l =2.53 × 10−3
∠89.910
mho per phase.
Ԧ
𝐴= 𝐷 = Cosh 𝑌𝑍 =1+
𝑌𝑍
2
+
𝑌2𝑍2
24
+
𝑌3𝑍3
720
+….….
=0.524 ∠25.850
𝐵= 𝑍𝐶 sinh 𝑌𝑍 =Z (1+
𝑌𝑍
6
𝑌2𝑍2
120
+
𝑌3𝑍3
5040
+….….)
= 500.9∠63.950
(1+0.211 ∠153.860
+0.0133 ∠ −52.280
+4.0355 × 10−4
∠101.580
= 500.9∠63.950 ×0.822 ∠5.770
=411.7 ∠69.720
Ԧ
𝐶 =
1
𝑍𝐶
sinh 𝑌𝑍= Y (1+
𝑌𝑍
6
𝑌2𝑍2
120
+
𝑌3𝑍3
5040
+….….)
= 2.53 × 10−3
∠89.910
×0.822 ∠5.770
= 2.08 × 10−3
∠95.680

𝑌𝑍=α+jβ α, β are real in rad
Total series impedance per phase Z=z ×l= 500.9∠63.950
ohms =500.9 ∠1.118 rad
Total shunt admittance Y= y ×l =2.53 × 10−3
∠89.910
mho per phase.
Z=500.9 ∠1.118 rad Y= 2.53 × 10−3 ∠1.945 rad
𝑌𝑍=α+jβ= 500.9 ∠1.118 × 2.53 × 10−3 ∠1.945
= 1.1256 ∠3.063 rad =0.2545+j1.096
α =0.2545 rad β =1.096 rad
Then Cosh(α+jβ)=Cosh α Cos β +jsinh α sin β
=Cosh 0.2545 Cos 1.096 +jsinh 0.2545 sin 1.096
=0.472+j0.2288=0.5232 ∠0.446 rad
=0.5232 ∠25.550
sinh(α+jβ)=sinh α Cos β +j Cosh α sin β
=0.926 ∠1.443 rad
= 0.926 ∠82.670
Ԧ
𝐴= 𝐷 = Cosh 𝑌𝑍= 0.5232 ∠25.550
𝐵= 𝑍𝐶 sinh 𝑌𝑍=
𝑍
𝑌
× 0.926 ∠82.670
= 197984.18∠ −25.960× 0.926 ∠82.670=444.95 ∠ −12.980 × 0.926 ∠82.670
=412 ∠69.70 ohm
Ԧ
𝐶 =
1
𝑍𝐶
sinh 𝑌𝑍=2.247 × 10−3
∠12.980
× 0.926 ∠82.670
=2.68 × 10−3
∠95.650
mho

A single circuit 50 Hz, 3-phase transmission line has the following parameters per km: R =
0.2 ohm, L = 1.3 mH and C = 0.01 μF The voltage at the receiving end is 132 kV. If the line
is open at the receiving end, find the rms value and phase angle of the following:
(i) The incident voltage to neutral at the receiving end (reference).
(ii) The reflected voltage to neutral at the receiving end.
(iii) The incident and reflected voltages to neutral at 120 km from the receiving end.
The series impedance per unit length of the line
z = r + jx = (0.2 + j1.3 × 314 × 10–3) = (0.2 + j0.408)
= 0.454 ∠ 63.88°
The shunt admittance = jωC = j314 × 0.01 × 10−6
= 3.14 × 10−6 ∠+90°
V=(
2
)(𝑒γ𝑥
)+(
2
)(𝑒−γ𝑥
)
The first term in the above expression is called the incident voltage wave as its value
increases as x is increased. Since we are taking receiving end as the reference and as x
increases the value of voltage increases that means a voltage wave decreases in magnitude
as it travels from the sending end towards the receiving end, that is why this part of the
voltage in the above expression is called incident voltage. For similar reason the second
part is called the reflected voltage. At any point along the line, voltage is the sum of these
two components i.e., sums of incident and reflected voltages.

The characteristic impedance Zc =
𝑍
𝑌
=
0.454 ∠ 63.88°
3.14 × 10−6 ∠+90°
=380 ∠–13.06°
γ = 𝑌𝑍 = 0.314 ×10−6× 0.454 ∠(90 + 63.88)°/2
= (0.2714 + j1.169) × 10−3
= (α + jβ)
Receiving end voltage/phase, 𝑉𝑅 = 132,000/ 3 = 76200 V
and receiving end current under open circuited condition 𝐼𝑅 = 0
(i) The incident voltage to neutral at the receiving end (x = 0)
=
2
Since it is no load condition, 𝐼𝑅 = 0.
∴ Incident voltage =
𝑉𝑅
2
=
76200
2
=38100 V
(ii) Similarly, the reflected voltage to neutral at the receiving end
=
2
=
𝑉𝑅
2
=38100V

(iii) The incident voltage at a distance of 120 km from the receiving end
=(
2
)(𝑒γ𝑥
)
=(
2
)(𝑒(α + jβ) 𝑥
)
=(
76200
2
)(𝑒(0.2714 + j1.169)× 10−3×120)
=38100*(1.0229+j0.1444)
=38100*1.033∠8.0350
=39.358 ∠8.0350 KV
The reflected voltage at a distance of 120 km from the receiving end
=(
2
)(𝑒−γ𝑥
)
=(
2
)(𝑒−(α + jβ) 𝑥
)
=(
76200
2
)(𝑒−(0.2714 + j1.169)× 10−3×120)
= 36.88 ∠ −8.0350
KV

A single circuit 50 Hz, 3-phase transmission line has the
following parameters per km: R = 0.2 ohm, L = 1.3 mH and C =
0.01 μF .Determine the efficiency of the line if the line is 120 km
long and delivers 40 MW at 132 kV and 0.8 p.f. lagging.
The series impedance per unit length of the line
z = r + jx = (0.2 + j1.3 × 314 × 10–3) = (0.2 + j0.408)
= 0.454 ∠ 63.88°
The shunt admittance = jωC = j314 × 0.01 × 10−6
= 3.14 × 10−6
∠+90°

Two networks are said to be connected in tandem when the output of one
network is connected to the input of the other network. Let the constants of these
networks be 𝐴1, 𝐵1, 𝐶1, 𝐷1 and 𝐴2, 𝐵2, 𝐶2, 𝐷2 which are connected in tandem as
shown in Fig.
The net constants of the system relating the terminal conditions can be found as
follows:
V = 𝐷1 𝑉
𝑠 – 𝐵1 𝐼𝑠 1
I = – 𝐶1 𝑉
𝑠 + 𝐴1 𝐼𝑠 2
and V = 𝐴2 𝑉
𝑟 + 𝐵2 𝐼𝑟 3
I = 𝐶2 𝑉
𝑟 + 𝐷2 𝐼𝑟 4
From equations, we get
𝐷1𝑉
𝑠 – 𝐵1𝐼𝑠= 𝐴2𝑉
𝑟 + 𝐵2𝐼𝑟 5
– 𝐶1𝑉
𝑠 + 𝐴1𝐼𝑠= 𝐶2𝑉
𝑟 + 𝐷2𝐼𝑟 6

Multiplying equation (5) by 𝐴1 and (6) by 𝐵1 adding the resulting equations
𝐴1𝐷1𝑉
𝑠 – 𝐴1𝐵1𝐼𝑠= 𝐴1𝐴2𝑉
𝑟 + 𝐴1𝐵2𝐼𝑟 7
– 𝐵1𝐶1𝑉
𝑠 + 𝐴1𝐵1𝐼𝑠= 𝐵1𝐶2𝑉
𝑟 + 𝐷2𝐵1𝐼𝑟 8
adding the resulting equations
𝐴1𝐷1 − 𝐵1𝐶1 𝑉
𝑠 = 𝐴1𝐴2 + 𝐵1𝐶2 𝑉
𝑟 + 𝐴1𝐵2 + 𝐵1𝐷2 𝐼𝑟 9
Multiplying equation (5) by 𝐶1 and (6) by 𝐷1 adding the resulting equations
𝐶1𝐷1𝑉
𝑠 – 𝐵1𝐶1𝐼𝑠= 𝐴2𝐶1𝑉
𝑟 + 𝐵2𝐶1𝐼𝑟 10
– 𝐶1𝐷1𝑉
𝑠 + 𝐴1𝐷1𝐼𝑠= 𝐶2𝐷1𝑉
𝑟 + 𝐷1𝐷2𝐼𝑟 11
adding the resulting equations
𝐴1𝐷1 − 𝐵1𝐶1 𝐼𝑠= 𝐴2𝐶1 + 𝐶2𝐷1 𝑉
𝑟 + 𝐵2𝐶1 + 𝐷1𝐷2 𝐼𝑟
Since 𝐴1𝐷1 − 𝐵1𝐶1 = 1, the constants for the two networks in tandem are
A= 𝐴1𝐴2 + 𝐵1𝐶2
B= 𝐴1𝐵2 + 𝐵1𝐷2
C= 𝐴2𝐶1 + 𝐶2𝐷1
D= 𝐵2𝐶1 + 𝐷1𝐷2
The relation is given in matrix form as follows:
𝐴 𝐵
𝐶 𝐷
=
𝐴1 𝐵1
𝐶1 𝐷1
𝐴2 𝐵2
𝐶2 𝐷2
If network 2 is at the sending end and 1 is at the receiving end the overall constants for
the two networks in tandem can be obtained by interchanging the subscripts in equations.

In case two networks are connected in parallel as shown in Fig. and
currents at the sending end and receiving end be 𝐼𝑠1 and 𝐼𝑟1 for network
1 and 𝐼𝑠2 and 𝐼𝑟2 for network 2
𝐼𝑠= 𝐼𝑠1 + 𝐼𝑠2 1
𝐼𝑟= 𝐼𝑟1+ 𝐼𝑟2 2
For network 1
𝑉
𝑠 = 𝐴1 𝑉
𝑟 + 𝐵1 𝐼𝑟1 3
𝐼𝑠1 = 𝐶1 𝑉
𝑟 + 𝐷1 𝐼𝑟1 4
For network 2
𝑉
𝑠 = 𝐴2 𝑉
𝑟 + 𝐵2 𝐼𝑟2 5
𝐼𝑠2 = 𝐶2 𝑉
𝑟 + 𝐷2 𝐼𝑟2 6

Comparing equations 3 & 5, we get
𝐴1 𝑉
𝑟 + 𝐵1 𝐼𝑟1= 𝐴2 𝑉
𝑟 + 𝐵2 𝐼𝑟2
𝐴1 − 𝐴2 𝑉
𝑟= 𝐵2 (𝐼𝑟 − 𝐼𝑟1) − 𝐵1 𝐼𝑟1 ( 𝐼𝑟= 𝐼𝑟1+ 𝐼𝑟2)
𝐴1 − 𝐴2 𝑉
𝑟= 𝐵2 𝐼𝑟 − 𝐼𝑟1(𝐵1 + 𝐵2 )
𝐼𝑟1=
𝐵2 𝐼𝑟− 𝐴1−𝐴2 𝑉𝑟
𝐵1 + 𝐵2
Substituting the value of 𝐼𝑟1in equation 3, we get
𝑉
𝑠 = 𝐴1 𝑉
𝑟 + 𝐵1
𝐵1 + 𝐵2
= 𝐴1 −
𝐵1 𝐴1−𝐴2
𝐵1 + 𝐵2
𝑉
𝑟+
𝐵1𝐵2
𝐵1 + 𝐵2
𝐼𝑟
=
𝐴1𝐵2+𝐵1𝐴2
𝐵1 + 𝐵2
𝑉
𝑟+
𝐵1𝐵2
𝐵1 + 𝐵2
𝐼𝑟
Substituting the value of𝐼𝑠1, 𝐼𝑠2 in equation 1, we get
𝐼𝑠= 𝐼𝑠1 + 𝐼𝑠2
= 𝐶1 𝑉
𝑟 + 𝐷1 𝐼𝑟1+ 𝐶2 𝑉
𝑟 + 𝐷2 𝐼𝑟2
=(𝐶1 + 𝐶2 ) 𝑉
𝑟 + 𝐷1𝐼𝑟1+ 𝐷2 (𝐼𝑟 − 𝐼𝑟1)
=(𝐶1 + 𝐶2 ) 𝑉
𝑟+(𝐷1 - 𝐷2 ) 𝐼𝑟1+ 𝐷2 𝐼𝑟

Substituting the value of 𝐼𝑟1
𝐼𝑠 =(𝐶1 + 𝐶2 ) 𝑉
𝑟+(𝐷1 - 𝐷2 )
𝐵1 + 𝐵2
+ 𝐷2 𝐼𝑟
= 𝐶1 + 𝐶2 −
𝐴1−𝐴2 (𝐷1 − 𝐷2 )
𝐵1 + 𝐵2
𝑉
𝑟+
𝐵1𝐷2+𝐷1𝐵2
𝐵1 + 𝐵2
𝐼𝑟
Therefore, for the two networks connected in parallel, the
generalised constants of the equivalent network are
A=
𝐵1 + 𝐵2
B=
𝐵1𝐵2
𝐵1 + 𝐵2
C= 𝐶1 + 𝐶2 −
𝐴1−𝐴2 (𝐷1 − 𝐷2 )
𝐵1 + 𝐵2
D=
𝐵1 + 𝐵2

Two identical three phase transmission lines are connected in parallel to supply a load of
100MW at 132 kV and 0.8 p.f lagging at the receiving end. The ABCD constants of each
transmission line are as follows:
A=D=0.98∠10, B=100 ∠750 Ω per phase; C=0.0005 ∠900 siemens per phase. Determine (a)
the ABCD constants of the combined network (b) the sending end power factor
A=
𝐵1 + 𝐵2
=
0.98∠10×100 ∠750+0.98∠10×100 ∠750
100 ∠750+100 ∠750
= 0.98∠10
B=
𝐵1𝐵2
𝐵1 + 𝐵2
= 50∠750
C= 𝐶1 + 𝐶2 −
𝐴1−𝐴2 +(𝐷1 − 𝐷2 )
𝐵1 + 𝐵2
= 𝐶1 + 𝐶2
=0.001 ∠900
D=
𝐵1 + 𝐵2
= 0.98∠10

𝑃
3𝑉𝑅cos φ𝑅
=
100∗106
3∗132,000∗0.8
=546.7A
Taking the receiving end voltage 𝑉𝑅 as the reference phasor,
𝑉𝑅= 𝑉𝑅+j0= 76,210 ∠ 00
V
𝐼𝑅= 𝐼𝑅 ∠ − φ𝑅= 𝐼𝑅 (cos φ𝑅 − j sinφ𝑅)=546.7 (0.8-j0.6)=546.7∠ −36.870
sending end voltage 𝑉𝑆= Ԧ
= 0.98∠10 ×76,210 ∠ 00+ 50∠750 ×546.7∠ −36.870
=97,879.8 ∠ 10.70
sending end current 𝐼𝑆= Ԧ
=495 ∠ −28.70
Sending end power factor=cos φ𝑆
=cos (10.70-(−28.70))
=0.7727 lagging

Two three phase transmission lines have the generalized constants:
𝐴1 = 𝐷1 =0.98∠20
; 𝐵1 = 28∠690
Ω; 𝐶1= 0.0002∠880
𝑆
𝐴2 = 𝐷2 =0.95∠30
; 𝐵2 = 40∠850
Ω; 𝐶1= 0.0004∠900
𝑆
They are connected in series and delivers a load current of 200A at 0.95
pf (lagging) at 110kV. Determine the sending end voltage and sending
end current.

When a long line is operating under no load or
light load condition, the receiving end voltage is
greater than the sending end voltage. This is
known as Ferranti-effect.

(i) Assume no load condition.
V=(
2
)(𝑒γ𝑥
)+(
2
)(𝑒−γ𝑥
)
put γ=α+jβ
V=(
2
)(𝑒α𝑥. 𝑒jβ𝑥
)+(
2
)(𝑒−α𝑥. 𝑒−jβ𝑥
)
When 𝑥 = 𝑙 𝑎𝑛𝑑 𝐼𝑅=0
𝑉
𝑠=(
𝑉𝑅
2
)(𝑒α𝑙
. 𝑒jβ𝑙
)+(
𝑉𝑅
2
)(𝑒−α𝑙
. 𝑒−jβ𝑙
)
At 𝑙=0
𝑉𝑅=
𝑉𝑅
2
+
𝑉𝑅
2
As 𝑙 increases, the incident component of sending end voltage increases
exponentially and turns the vector anti-clockwise through an angle β𝑙, whereas
the reflected part of sending end voltage decreases by the same amount and is
rotated clockwise through the same angle β𝑙.
The sum of these two components of sending end voltage gives a voltage which
is smaller than 𝑉𝑅.

(ii) A simple explanation of Ferranti-effect can be given by
approximating the distributed parameters of the line by lumped
impedance as shown in Fig.
Since usually the capacitive reactance of the line is quite large as
compared to the inductive reactance, under no load or lightly loaded
condition the line current is of leading p.f. The phasor diagram is given
below for this operating condition.
𝐼𝑆= 𝐼𝑅 + 𝐼𝐶
under no load 𝐼𝑅 =0
𝐼𝑆= 𝐼𝐶
𝑉𝑆= 𝑉𝑅 + 𝐼𝐶𝑅 + 𝐼𝐶𝑋
The charging current produces drop in the reactance of the line which
is in phase opposition to the receiving end voltage and hence the
sending end voltage becomes smaller than the receiving end voltage.

❖ Ferranti-effect is based on the net reactive power flow on the line.
❖ If the reactive power generated at a point is more than the reactive power
absorbed, the voltage at that point becomes higher than the normal value
and vice versa.
❖ The inductive reactance of the line is a sink for the reactive power whereas
the shunt capacitances generate reactive power.
❖ In fact, if the line loading corresponds to the surge impedance loading, the
voltage is same everywhere as the reactive power absorbed then equals the
reactive power generated by the line.
❖ If the loading is less than SIL, the reactive power generated is more than
absorbed, therefore, the receiving end voltage is greater than the sending
end voltage. This explains, therefore, the phenomenon due to Ferranti-
effect.

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