# PS-2 PPT

```1
Subject :- Electrical Power System-2
Topic :- Power Flow Throuh Transmission Lines
Topics Covered
3

Introduction.

Power flow through transmission line.

Single - line diagram of three phase transmission.

Derivation.

Circle diagram

Analytical method.

Graphical method.

Summary.
Introduction
4

The electric power generated in the generating station is transmitted using
transmission lines.

Transmission lines are conductors designed to carry electricity over a long
distances with minimum losses and distortion.

The parameters associated with these transmission lines are inductance,
capacitance, resistance and conductance.
5
Power Flow Through Transmission
Lines
SS =PS +
jQS
G
Generating
station
VS ∠ δ
VR ∠0
Transmission SR =PR + jQR
line
ABCD
Bus-1
LOA
D
Bus-2
Fig:- Single line diagram of three phase
transmission
Assuming,
VR = Receiving End voltage
= |VR| ∠0&deg;(VR is reference phasor)
VS = |VS| ∠δ&deg;= Sending End voltage(δ is the phase angle between sending and
receiving end voltage)
6
Power Flow Through Transmission
Lines

Generalised line constants are :
A = |A|∠ α ; B = |B| ∠β; C = |C| ∠γ; D = |D| ∠Δ;

Complex power at receiving end
SR =PR + jQR = VR IR*

…eq (1)
Here,
IR* is conjugate of Receiving end current IR
We know that
VS = AVR + BIR
7
Power Flow Through Transmission
Lines

From the above equation
⸫I R =
VS −AVR
B
=
|VS| ∠δ −|A|∠ α|VR| ∠0&deg;
|B| ∠ β
=
|VS|
|A||VR|
∠(δ – β ) ∠(α – β)
|B|
|B|
|V |
|A||V |
i.e. = IR* = |B|S ∠(β − δ ) - |B| R ∠(β − α)
…eq (2)
Power Flow Through Transmission
Lines
8

Now we put the value IR* in equation of …eq (1), we get
SR = VR IR*
2
= |Vs||VR| ∠(β − δ ) - |A||VR | ∠(β − α)
|B|
|B|

…eq (3)
Now, we separate real and imaginary parts, then we get the values of PR and
QR So, Receiving end True power,
2
PR = |Vs||VR| cos (β − δ ) - |A||VR | cos(β −
|B|
|B|
α)
Receiving end Reactive power,
2
QR = |Vs||VR| sin (β − δ ) - |A||VR | sin(β −
|B|
|B|
α)
…eq (4)
…eq (5)
Power Flow Through Transmission
Lines
9

For fixed values of Vs and VR, Power Received will be maximum when
cos(β − δ) =1 or when δ= β, So
|Vs||V R| |A||V R2|
PR(max) =
- |B| cos(β −
|B|
α)
and

QR(max) = -
|A||V R2|
sin(β −
|B|
α)
In transmission line
A=D=1∠0&deg;
B=Z ∠θ
…eq (7)
…eq (8)
Power Flow Through Transmission
Lines
10

Now we substitute above values in eq (4)&amp;eq (5), We get
⸫P = Vs VR cos(θ − δ) - VR 2 cos θ
R
Z
Z
Q = Vs VR sin(θ − δ) - VR 2 sin θ
β =
0
Resistance of transmission line is usually very small as compared to
R

⸫α = 0
Z
Z
reactance. Hence Z = X and θ = 90 &deg;
VsItVis
R sin δ
⸫ Pangle.
=
R
Z
QR =
small
δ=1)
V V
s
X
R
2
- VR
X
(⸫ δ is the power angle. It is usually very small ⸫ cos
Methods Of Finding The Performance
Of
Transmission
Line.
Basically two methods
11


Analytical method.

Graphical method.

Analytical methods are found to be laborious, while graphical method is
convenient.

Graphical method or circle diagram are helpful for determination of active
power P, Reactive power Q, power angle δ and power factor for given load
condition.

Relations between the sending end and receiving end voltage and currents
are given below.
VS = AVR + BIR A, B, C, D are generalised constants of transmission. IS
= CVR + DIR VS = sending endvoltage,
Methods Of Finding The Performance
Of Transmission Line.
12
VR = Receiving end voltage
IS = Sending end current,
IR =Receiving end current.

By taking either VS, VR, IS or IR as a reference these characteristics can be
plotted.

These characteristics are nothing but representing circles, hence such
diagrams are called circle diagrams.

Circle diagram is drawn by taking active power P on X- axis and reactive
power on Y- axis.
Receiving End Power Circle
Diagram :
13

Receiving end true power – Horizontal coordinates

Reactive power component – Vertical coordinates

From the equation,

VS = AVR +BIR

⸫IR =
=
VS −AVR
B
VS ∠δ A ∠δ
V ∠0
B ∠ β B ∠β R
= Vs ∠(δ − β) - AVR ∠(α−β)
B
B
Receiving End Power Circle
Diagram :
14
|A||VR|
∠(β − α)
|B|

|Vs|
IR* = |B| ∠(β − δ ) -

Volt- ampere at the receiving end will be
SR = PR + jQR = VR IR*
2
|Vs||V
|
|A||V
|
R
R
=
∠(β − δ ) ∠(β − α)
|B|
|B|
R| [cos (β − δ ) + j sin (β − δ )] = |Vs||V
2
|A||V|B|
R|
|B|
[ cos(β − α)+j sin (β − δ
)]
2
|Vs||V
|
|A||V
|
R
R
SR =
cos (β − δ ) cos(β − α) +
|B|
|B|
|Vs||VR| j sin (β − δ ) - |A||VR|2 j sin (β − δ )
|B|
|B|
Receiving End Power Circle
Diagram :
15

By separating real and imaginary parts, we have

|Vs||VR|
|A||VR|2
PR =
cos (β − δ )
cos(β −
|B|
|B|
α)

|Vs||VR|
|A||VR|2
QR =
sin (β − δ )
sin(β −
|B|
|B|
α)

The power component can be expressed as

|A||VR|2
PR +
cos(β − α)
|B|
=
|Vs||VR|
cos (β − δ
|B|
)

|A||VR|2
QR +
sin(β − α)
|B|
=
|Vs||VR|
sin (β − δ
|B|
)
Receiving End Power Circle
Diagram :
16

Squaring and adding these equations will give
{PR +
|A||V |2
|A||VR|2 cos(
β − α)}2 + {QR + |B|R sin(β − α)
|B|
}
2
2
= |Vs|2|VR| { cos2( β − δ )+ sin2 (β −
|B|
α)}
2
= |Vs|2|VR|
|B|




It is an equation of a circle. The coordinates of centre of a circle are:
|A||VR|2
X-coordinate of the circle = - |B| cos(β −
α)
|A||V |2
Y- coordinate of the circle = - |B|R sin(β −
α)
|Vs||V |
Radius of the circle = |B| R
Construction of circle diagram:
17

Plot the centre of the circle N on a suitable scale.

From N draw an arc of a circle with the calculated radius

From the origin O draw the load line OP inclined at angle ϕ R with the
VSVR
.
B
horizontal.

Let it cut the circle at P, then the receiving end true power and reactive
power will be represented by OP and PQ respectively.

If the voltages VS and VR are taken phase voltage in volts then the powers
indicated on X-axis and Y-axis will be in watts and VARs per phase
respectively.
Construction of circle diagram:
18

If the voltages VS and VR are taken line voltage in volts then the powers
indicated on X-axis and Y-axis will be in watts and VARs for all three phases
respectively.

If the VS and VR are taken from line to line and in kV then the power
indicated will be in MW and MVAR and for all the three phases.

To determine the maximum power a horizontal line is drawn from the centre
of the circle intersecting vertical axis at the point L and the circle at the
point M.

Distances LM represents the maximum power for the receiving end.
Receiving End Power Circle
Diagram :
19
Sending End Power Circle
Diagram :
20
We have seen…
21


Receiving end True power,
2
PR = |Vs||VR| cos (β − δ ) - |A||VR | cos(β −
|B|
|B|
α)
Receiving end Reactive power,
Q = |Vs||VR| sin (β − δ ) - |A||VR | sin(β − α)
2
R


|B|
|B|
|Vs||VR| |A||VR2|
cos(β −
PR(max) =
|B|
|B|
α)
|A||VR2|
QR(max) = sin(β − α)
|B|

Construction of circle diagram

Receiving End Power Circle Diagram

Sending End Power Circle Diagram :
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