Electrical Power

Electrical Power
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1

Question 1.2
Three-phase short transmission line has impedance,
Z= ( R+ jX )= ( 1+ j 4 ) Ω
phase
(i) Sketch the short-line equivalent circuit per phase and annotate it with the given data
(ii) For a constant receiving-end line voltage, V 2 L=11 kV ∠ 300and line current I =250 A
into a star connected load:
(a) Determine and tabulate for five leading and lagging power factor λ=0,0.8,1:
(i) The sending-end input line voltage (complex/rectangular format)
(ii) The power angle in degrees
(iii) Voltage regulation (%)
(b) Calculate:
(iv) The 5 approximate voltage regulations and tabulate the values using the
approximation
ε va ≈ RIcos φ−XI sin φ
V 2
SOLUTION
Part i
Transmission lines with a length less than 50 km are the short-line transmission lines.
The capacitance and leakage resistance are negligible in these transmission lines. The two-port
network analogy is implemented. The equivalent short transmission circuit is implemented as,
Z= (1+ j 4 ) Ω
phase
2

The phasor diagram of the short transmission line can be referenced to an inductive load or a
capacitive load such that,
Figure 1 Phasor diagram with reference to an inductive load
Figure 2 phasor diagram with reference to a capacitive load
Part ii
Using the V R as the reference, the equation can be written as,
V S =V R + ( I R cos φR ± j I R sin φR ) ( R+ j X L )
±−used for lagging∧leading power factor
V S = √ ( V R + IR cos φR + IX sin φR )2+ ( IX cos φ R ± IR sin φR )2
The load angle is given as,
δ=φs−φ R
Alternatively,
δ=tan−1
( IX cos φR ± IR sin φR
V R + IR cos φ R + IX sin φR )
3

[ V S
I S ] =[ A B
C D ] [ V R
I R ]
[ V S
I S ] =[ 1 Z
0 1 ] [ V R
I R ]
[ V R
I R ] = [ 1 −Z
0 1 ] [ V S
I S ]
Voltage regulation
%VR=
|V S|−|V R|
|V R| x 100
It is approximated as,
%VR= R cos φR ± X sin φR
V R
x 100
Using actual values,
The line current is given as,
I =250 A , Z= (1+ j 4 )
IZ=250∗( 1+ j 4 )=250+ j 1000 V
The sending-end input line voltage,
V R ( L− N ) =V R ( L− L )
√ 3 = 11∗103 ∠300
√ 3 =6350.853∠ 300 V
V S ( L−N ) =V R ( L−N ) + IZ
¿ 6350.853 ∠300 +250+ j 1000 V
V S ( L−N ) =5500+ j 3175.43+250+ j 1000
V S ( L−N ) =5750+ j 4175.43
In phasor notation,
V S ( L−N ) =7.106 kV ∠35.9850
The power degree is given as
δ=35.9850
Using the alternative method with line current and line voltage as the reference,
4