REPRESENTATION OF
TRANSMISSION LINES
PERFORMANCE OF TRANSMISSION LINES
EE-PC 4111: POWER SYSTEM ANALYSIS
Constants of Transmission Lines
• A transmission line has three constants: R, L and C. These constants
are distributed uniformly along the whole length of the line.
• R and L form the SERIES IMPEDANCE.
• While the capacitance C () existing between the conductors forms the
SHUNT PATH.
Classification of Overhead Transmission Lines
• Depending upon the matter in which capacitance is taken into
account, overhead transmission lines are classified as:
1. Short Transmission Lines
2. Medium Transmission Lines
3. Long Transmission Lines
Short Transmission Lines
• When the length of an overhead transmission line is upto about 50 mi
and the line voltage is comparatively low (< 20 kV), it is usually
considered as a short transmission line.
• Due to smaller length and lower voltage, the capacitance effects are
small and hence can be neglected.
• Therefore, while studying the performance of a short transmission line,
only resistance and inductance of the line are taken into account.
Medium Transmission Lines
• When the length of an overhead transmission line is about 50- 150 mi
and the line voltage is moderately high (>20 kV < 100 kV), it is
considered as a medium transmission line.
• Due to sufficient length and voltage of the line, the capacitance
effects are taken into account.
• For purposes of calculations, the distributed capacitance of the line is
divided and lumped in the form of condensers shunted across the
line at one or more points.
Long Transmission Lines
• When the length of an overhead transmission line is more than 150
km and line voltage is very high (> 100 kV), it is considered as a long
transmission line.
• For the treatment of such a line, the line constants are considered
uniformly distributed over the whole length of the line and rigorous
methods are employed for solution.
Important Terms
• VOLTAGE REGULATION. When a transmission line is carrying
current, there is a voltage drop in the line due to resistance and
inductance of the line. The result is that receiving end voltage (ππ
) of
the line is generally less than the sending end voltage (ππ ). This voltage
drop (ππ − ππ
) in the line is expressed as a percentage of receiving end
voltage ππ
and is called voltage regulation.
Percent voltage regulation:
π½ πΊ − π½πΉ
%π½πΉ =
× πππ%
π½πΉ
Important Terms
• TRANSMISSION EFFICIENCY. The power obtained at the receiving
end of a transmission line is generally less than the sending end power
due to losses in the line resistance. The ratio of receiving end power to
the sending end power of a transmission line is known as the
transmission efficiency of the line.
Percent transmission efficiency:
π·πΉ
%πΌπ» =
× πππ%
π·πΊ
π½πΉ π°πΉ cos π½πΉ
%πΌπ» =
× πππ%
π½πΊ π°πΊ cos π½πΊ
Single-Phase Short Transmission Lines
• For short transmission lines, only resistance and inductance of the line
are taken into account.
• Here, the total line resistance and inductance are shown as
concentrated or lumped instead of being distributed.
Analysis of Single-Phase Short Transmission Lines
By KVL,
ππ − πΌ π
+ πππΏ − ππ
= 0
Rearranging,
ππ = ππ
+ πΌ π
+ πππΏ
For short transmission lines,
πΌ = πΌπ = πΌπ
Since π
+ πππΏ is represented as the series impedance of the line, then
π½πΊ = π½πΉ + π°πΉ π
In proper phasor notation,
π½πΊ ∠πΆ = π½πΉ ∠π + π°πΉ π
Three-Phase Short Transmission Lines
• This system may be regarded as consisting of three single phase units,
each wire transmitting one-third of the total power. As a matter of
convenience, we generally analyze 3-phase system by considering ONE
PHASE ONLY.
Analysis of Three-Phase Short Transmission Lines
• Since we consider one phase only, then we represent it by
one of its phases to the neutral.
• The equations we use in the analysis of single phase lines
can be used in the per-phase calculations for three-phase
transmission lines.
• However, we represent the quantities as their per phase
values. Hence,
π½πΊ(πππ πππππ) = π½πΉ(πππ πππππ) + π°πΉ(πππ πππππ) π(πππ πππππ)
• Interpreting these equations in their simple phasor forms, if ππ
is the reference
phasor then
π½πΊ ∠πΆ =
π½πΉ ∠π°
π
+ π°πΉ πππ πππππ π
πππ πππππ
[π/πβππ π]
Example Problem No. 1
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 β¦ and 15 β¦ respectively.
Determine:
(i) sending end voltage
(ii) sending end power factor and
(iii) transmission efficiency.
Solution:
First we solve for πΌ,
ππ
1100 × 103
πΌ = πΌπ
=
=
ππ
cos ππ
33 × 103 0.8
πΌ = πΌπ
= 41.67 π΄ πππππππ
In phasor form,
πΌ = 41.67∠ − 36.87° π΄
Example Problem No. 1
(i) sending end voltage
The sending end voltage is given by ππ = ππ
∠0° + πΌπ
Computing its value,
ππ = 33000∠0° + 41.67∠ − 36.87° 10 + π15
π½πΊ = πππππ. ππππ∠π. ππ° π½
(ii) sending end power factor
Since πΌ = 0.42° and ππ
= 36.87°, the sending pf angle is
ππ = πΌ + ππ
= 0.42° + 36.87°
ππ = 37.29°
So cos ππ = cos 37.29° = π. ππππ πππππππ
Example Problem No. 1
(iii) transmission efficiency
We know that
ππ
%π π =
× 100%
ππ
where ππ = ππ πΌπ cos ππ . Plugging in the values, we get
ππ
%π π =
× 100%
ππ πΌπ cos ππ
11000 × 103
%π π =
× 100%
33709.3176 41.67 0.7956
%πΌπ» = ππ. ππ%
Example Problem No. 2
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 β¦ and 6 β¦ respectively. Determine:
(i) percentage regulation and
(ii) transmission efficiency.
Solution:
Solving for ππ , we get
ππ =
ππ =
ππ
∠0°
3
22000∠0°
3
+
+
πππππ
∠ − cos −1 πππ
3ππ
cos ππ
5000 × 103
3 × 22000 × 0.8
π
+ ππΏ
∠ − cos −1 0.8
π½πΊ = πππππ. ππ∠π. π° π½/πππππ
4 + π6
Example Problem No. 2
(i) percentage regulation
Solving the percent voltage regulation can either be in phase or line quantities, in this
solution, let us use per phase quantities
ππ − ππ
%ππ
=
× 100%
ππ
22000
13822.65 −
3
%ππ
=
× 100%
22000
3
%π½πΉ = π. πππ%
(ii) transmission efficiency
Solving for πΌ,
πΌ = πΌππππ =
5000 × 103
3 × 22000 × 0.8
= 164.02 π΄/πβππ π
Example Problem No. 2
(ii) transmission efficiency
πππππ
5000 × 103
% ππ =
× 100% =
2
πππππ + 3πΌ π
5000 × 103 + 3 164.02
% πΌπ» = ππ. ππ%
2
4
× 100%
That’s all. Thank you!