CHAPTER 3 TRANSMISSION LINES

DET30053
POWER SYSTEM

CHAPTER 3
TRANSMISSION LINES

LEARNING OUTCOME :

At the end of lessons, students should be able to ;

3.1 Understand the nature of transmission lines
3.2 Apply the understanding of transmission lines
3.3 understand the voltage regulation concept
3.4 apply the understanding of voltage regulation
concept.
3.5 understand the losses in transmission lines
3.6 understand the function of transmission’s
insulators.
3.7 Apply the function of transmission’s insulators

FUNCTION OF TRANSMISSION LINES

A transmission line is used for the
transmission of electrical power from
generating substation to the various
distribution units. It transmits the wave of
voltage and current from one end to
another. The transmission line is made
up of a conductor having a uniform
cross-section along the line.

FUNCTION OF TRANSMISSION LINES

➢To transmit the electrical energy from one
area to another area.

Transmission lines tower

Type of Transmission Line

1. Overhead Transmission line
2. Under ground Transmission line
3. Submerine Transmission line

Types of transmission lines

• Short transmission line : Shorter then 80 Km
• Medium transmission line : 80 – 240 Km
• Long transmission line : Longer then 240 Km

Types of transmission line

Short line model

Length : 0 – 80KM
Voltage : < 20kV

Short transmission line

The per-phase equivalent circuit of a short line

R Is X

Ir
Line
Vs VR Load

Neutral

Where: Single phase line

R - loop resistance of line

X - loop reactance of the line

VS - voltage at the sending end
VR - voltage at the receiving end
Is = Sending end current

Ir = Receiving end current

Assumption of no line admittance leads to

IS = IR

We can relate voltages through the Kirchhoff’s voltage
law

VS = VR + ZI = VR + RI + jXLI

Z = R + jXL

VR = VS − RI − jX LI

Phasor Diagram

C

Vs Ir cos r IX
G
H

AA
Vr IR B IX sin m

s Vr cos r

r

O Vr cos r DF I

Figure : Short line phasor diagram Where;

OA - Voltage at the receiving end
OI - Load current
AB - The voltage drop in the resistance of line
BC - The voltage drop in the reactance of line
OC - Voltage at the sending end

Or

Vs = VR + IR cos R + IX sin R

Example 1

A single phase load of 200 kVA is delivered at
2500V over transmission line having R = 1.4
ohm, X = 0.8 ohm.
Calculate the current, voltage and power factor
at the sending end when the power factor of
the load is 0.8 lagging.

Answer

Kuasa ketara, S = kVA
S = 200kVA
VR = 2500V
R=1.4
XL=0.8
Diberi power factor, cos R = 0.8
Current ???
S= IR x VR
IR = S / VR

= 200000 / 2500 = 80 A

Voltage sending end???

Vs = (VR cos R + IR)2 + (VR sin R + IXL)2
Find sin R
cos R = 0.8

R = cos-1 0.8
= 36.87 

sin 36.87 = 0.6
Vs =  (2500 (0.8) + 80(1.4))2 + (2500 (0.6) + (80)(0.8))2

= 2628 V

Answer

Power factor at sending end

Cos s = 2500(0.8) + (80)(1.4) / 2628
= 0.8

Self exercise 1

A single phase overhead transmission line delivers
1100kW at 33kV at 0.8 p.f lagging. The total
resistance and inductive reactance of the line are
10Ω and 15Ω respectively. Determine:
i. Sending end voltage
ii. Sending end power factor
iii. Transmission efficiency

Pout / Pin x 100

Ps = VsIs cos s adalah Pin
PR = VRIR cos R adalah Pout

Solution

Given
P = 1100kW, VR = 33kV
Cos R = 0.8, R = 10Ω, X = 15Ω

i. Sending end voltage

Vs = VR + IR cos R + IX sin R
First find I ??
PR = VRIR cos R
1100k = (33k) IR (0.8)
1100k = 26.4k IR
IR = 1100k / 26.4k = 41.67A

Cos R = 0.8
R = cos -1 0.8
= 36.87

Sin 36.87 = 0.6

Vs = VR + IR cos R + IX sin R
Vs = 33k + (41.67)(10)(0.8) + (41.67)(15) (0.6)

= 33k +333.36 +375.03
Vs = 33.71kV

Cos s = 0.79

iii.  = Pout / Pin x 100%
= VRIR cos R / VsIs cos s x 100%
= (33k)(41.67)(0.8)/ (33.71k)(41.67)(0.79) x100
= 99%

Short transmission line: phasor
diagram

Effect Of Power Factor
i. Load with lagging power factor.

ii. Load with unity power factor.

iii. Load with leading power factor.

For a given source voltage VS and magnitude of
the line current, the received voltage is lower
for lagging loads and higher for leading loads.

Types of transmission line

Medium line model
Length : 80KM – 240KM
Voltage : 20kV – 100kV

Types of transmission line

Long line model
Length : 240km and above
Voltage : 100kV and above

Transmission line characteristics

The real power input to a 3-phase transmission line can be computed as

2. Power flow in a transmission line

Pin = 3VS IS cosS = 3VLL,S IS cosS (9.43.1)

where VS is the magnitude of the source (input) line-to-neutral voltage and VLL,S is
the magnitude of the source (input) line-to-line voltage. Note that Y-connection is

assumed! Similarly, the real output power from the transmission line is

Pout = 3VR IR cosR = 3VLL,R IR cosR (9.43.2)

The reactive power input to a 3-phase transmission line can be computed as

Qin = 3VS IS sinS = 3VLL,S IS sin S (9.43.3)

Transmission line characteristics

And the reactive output power is

(9.44.1)

The apparent power input to a 3-phase transmission line can be computed as

Sin = 3VS IS = 3VLL,S IS (9.44.2)

And the apparent output power is

Sout = 3VR IR = 3VLL,R IR (9.44.3)

Transmission line characteristics

3Transmission line efficiency

The efficiency of the transmission line is (9.47.1)

 = Pout 100%

Pin



Example 1

A three phase transmission line system carrying a power
of 2200 kW to consumer with voltage of 11kV in lagging power
factor 0.85. This line has a resistance of 2.5 and inductance coil of
4 . Calculate the sending end voltage of the system.

Vs = VR + IR cos R + IX sin R (1 phase)
Solution :
P = 2200 kW
VR = 11kV
Cos R = 0.85
R= 2.5
XL = 4 
VL = 3 Vp
Vp = VL / 3

P = 3 VRIR cos R

2200 kW = 3 (11k) IR (0.85)
IR = 135.85A

VR(1phase) = VL / 3 = 11k/ 3 = 6.35kV

Cos R = 0.85
R = 31.79, sin 31.79 = 0.53

Vs = VR + IR cos R + IX sin R
Vs = 6.35k + (135.85)(2.5)(0.85) +
(135.85)(4)(0.53)
Vs = 6.35k + 288.68 + 288
Vs = 6926.68 V (single phase)
Tukar kpd three phase

VL = 3 x Vp
VL = 3 x 6926.68
VL = 11.99kV
Vs = 11.99kV (three phase)

Example 2

15000 kVA is received at 33kV, o.85 power factor
(lagging) over an 8 km three phase overhead line. Each
line has R = 0.29 Ohm per km and

X =0.65 Ohm per km.
Calculated
i. the voltage at the sending end
ii. The power loss in line
iii. The percentage regulation
iv. The power input to the sending end
v. The efficiency of transmission
vi. The power factor at the sending end line

Solution

Given
Apparent power, S = 15000kVA
VR = 33kV (voltage line)
Cos R = 0.85
R = 0.29 x 8km = 2.32
X = 0.65 x 8km = 5.2
Find VR in single phase
VL = 3Vp
Vp = VL / 3

= 33k / 3 = 19.05kV (single phase)

VR = 19.05kV
Cos R = 0.85

R = Cos-1 0.85
= 31.79

Sin 31.79 = 0.53
Find current, IR = ??

S = 3IRVR(1) atau S = 3 IRVR (3)
15000k = 3IR (19.05k)

IR = 15000k / 57.15k = 262.5A

(i) Vs = ???
Vs = VR + IR cos R + IX sin R

Vs = 19.05k + (262.5)(2.32)(0.85) + (262.5)(5.2) (0.53)

= 20.10kV (single phase)
Vs = 3 x Vp
Vs = 3 x 20.10k
Vs = 34.81kV

(ii) Power loss in line
Pin= Pout + Plosses
Plosses = Pin - Pout
Pin = 3VsIs coss
Pin = 3(34.81k)(262.5)(???)

Cos s= VR cos R + IR / Vs
= 19.05k (0.85) + (262.5)(2.32) / 20.10k
= 0.84

Pin = 3(34.81k)(262.5)(???)
= 3(34.81k)(262.5)(0.84)
= 13.29 MW

Pout = 3VRIR cos R
= 3(33k)(262.5)(0.85)
= 12.75MW

Ploss = Pin-Pout
= 13.29M – 12.75M
= 0.54MW

iii. % regulation
% regulation = Vs – VR / VR x 100%

= 34.81k – 33k/ 33k X 100%
= 5.48%

iv. Power input sending end

Pin = 3VsIs coss
= 3(34.81k)(262.5)(0.84)
= 13.29 MW

v. Efficiency = Pout / Pin x 100%
= 12.75M/13.29M x 100%
= 95.9%

vi. Power factor at sending end
Cos s= VR cos R + IR / Vs
= 19.05k (0.85) + (262.5)(2.32) / 20.10k
= 0.84

Self assessment 1

A 30 km, three phase 50Hz transmission line is
transferring power to a 500 kW load with power
factor of 0.9 lagging. The load voltage is 10 kV. Given
resistance and inductance of the line per phase of
0.1Ω per km, respectively, calculate:
i. Sending end current
ii. Sending end voltage
iii. Percent of voltage regulation
iv. Efficiency
[Is = 32.08<-25.84, Vs = 5.899Kv < 0.43, %VR = 2.24%,

%η = 98.22%]

Medium-length transmission line

Medium line model
Length : 80Km – 240Km
Voltage : 20kV – 100kV

 model

two capacitors of equal values
(each corresponding to a half of
total admittance) placed at the
sending and receiving ends.

The current through the receiving end capacitor can be found as

IC 2 = VR Y
2

And the current through the series impedance elements is

I ser = VR Y + IR
2

Medium-length transmission line

Long-length transmission line

Long line model
Length : 240 Km and above
Voltage : 100kV and above

Corona

The Phenomenon Of corona

Corona has been derived from glow surrounding the
Conductors when operation voltage is sufficiently high
For an overhead transmission system the atmospheric air ,
which is the dielectric medium, behaves practically insulator
when the potential difference between the conductor is small





Methods to reduce corona effects.

The following factor can be used to control corona;

1. Increase diameter of the conductor by using hollow
conductors of steel core aluminium conductor (ACSR)

2. Use Bundle conductors – consists of two or more
parallel subconductors at a spacing of several diameter.
(produced less resistances and reduce losses)

3. Increase the spacing between the conductors

Overhead line Insulator

Function : Overhead line insulator are use to separate
line conductor from each other and from the
supporting structure electrically

The consideration are kept in view in the design of an insulator;

1. The insulator should be able to withstand the over voltages due
to lighning, switching or other causes under severe wheather
conditions in addition to the normal working voltage.

2. It should posess high mechanical strength to bear the conductor
load under worst loading conditions.

3. It needs to have a high resistance to temperature change to
reduce damages from power flasover.

4. The leakage of current to earth should be minimum to keep the
corona loss and radio interference.

Types of Insulator

There are three main types of insulator use for
overhead line

1.Pin type insulator
2.Suspension type insulator
3.Tension type insulator

Overhead line insulator s are produced from
toughened glass or high quality we process
porcelain.

Pin type insulator

Pin type insulator

The pin insulator is supported on a forged steel
pin or bolt which is secured to the cross arm of
supporting structure.

The conductor is tied to the insulator on the top
groove on straight line positions and side groove
in angle position by annealed binding wire

A single piece pin insulator are used for lower
voltages, 33 kV but for higher voltage two or
more pieces are cemented together

Pin type insulator

Advantages of Pin Insulator

O Low cost
O Easy to install
O Easy to do maintenance

Disadvantages of Pin Insulator

O If any fault occur at 1 disc need replace all
the disc because all the disc is in one part.

O High cost for maintenance

Suspension type insulator

A suspension insulator consists of a number of separate insulator
unit connected with each other by metal links to form a flexible chain
or string.

The insulator string is suspended from the cross arm of the support

Suspension type insulator

• In suspension insulator numbers of
insulators are connected in series to
form a string and the line conductor is
carried by the bottom most insulator.

• Each insulator of a suspension string is
called disc insulator because of their
disc like shape.