Transmission Line Model and Performance
Dr Audih alfaoury
2014
Dr Audih alfaoury
Introduction
We shall represent lines as a "pi" section with
lumped parameters. There are three cases: short,
medium and long lines.
Z = R+jX
IS
+
VS
_
IR
+
VR
_
IL
In the short line model (under 80km), we
Y/2
Y/2
assume the shunt capacitance (the legs of the π )
are so small that they are open circuit (i.e. neglected). This leaves only the series RL
branch.
In the medium line model (80 to 240km), we assume that the capacitance may be
represented as two capacitors (legs of the π ) each equal to half the line capacitance. This
is known as the nominal π model.
In the long line model (over 240 km), we assume the line has distributed parameters
instead of lumped parameters. This yields exact results. After finding the exact model,
an equivalent π is found. This is called the equivalent π model because it has lumped
parameters which are adjusted so that they are equivalent to the exact distributed
parameters model.
1: Short Line Model
A short line with a load is shown (per phase) in the
figure to the right. Using circuit theory we
S*
have: I R = R* . We also know VS = VR + ZI R . Note that
VR
I S = I R . Now we will find the transmission parameters
(or ABCD parameters) of the short line:
IS
Z = R+jX
+
+
VS
VR
_
_
For a 2-port shown as "ABCD" the equations are as
IS
V   A B  VR 
follows:  S  = 
.
Comparing
with
the
 
+
 I S  C D   I R 
ABCD
VS
_
equations for the short line above, we see that: A = 1 ,
B = Z , C = 0 and D = 1 . This completes the equations
needed to represent the short line. The voltage regulation in percent is given by:
VR ( NL) VR ( FL )
Percent VR =
× 100
VR ( FL )
IR
SR
IR
+
VR
_
At no-load I R = 0 and thus VS = AVR ( NL ) , and since A = 1 , VR ( NL) = VS . This conclusion is
easily made by observation of the circuit above.
Dr Audih alfaoury
Example:Find the ABCD constants of a π circuit having a 600Ωresistor for shunt branch al sending end, a 1k Ω resistor for the shunt branch at the receiving end and an 80 Ω resistor for series branch. Solution: IL
IR
Is
80Ω
Vs
VR
600Ω
1000Ω
Vs  VR  V  VR  (80  I L )
V
I L  I R  I (1000  )  I R  ( R )
1000
V
Vs  VR  (80  I L )  Vs  VR  (80  ( I R  ( R )) 
1000
80VR
 1VR  80 I R 
 1VR  0.08VR  80 I R  1.08VR  80 I R
1000
from KCL
 Vs  1.08VR  80 I R and since Vs  AVR  BI R then;
A  1.08 and
B  80
And for C and D from KCL I s  I L  I  600 
VR
V
V
1.08VR  80 I R
)  ( s )  (I R  R )  (
)
1000
600
1000
600
 (0.001  0.0018) VR  (1  0.133) I R
I s  I L  I 600    ( I R 
And since I s  CVR  DI R  0.0028VR  1.133I R 
C  0.0028 and
D  1.133
Example
The ABCD constant of a three phase short length of TL are: A  D  0.936  j 0.016  0.9360.98o
B  33.5  j138  14276.4o 
C  (5.18  j 914) x106 S
Example
The load at the receiving end is 50MW at 220kV with a power factor of
0.9 lagging. Find the magnitude of the sending end voltage and the voltage regulation. Assume the magnitude of the sending end voltage remains constant. Dr Audih alfaoury
Solution:
P
50 106
IR 

145.8 25.84o [A]
3
3.VLL.cos 3  22010  0.9
cos1(0.9)  25.84o and since its lagging then the angle is negative,thus;
IR 145.8 25.84o
[ A]
VLL 220103
VR  
1270o kV and
3
3
Vs  AVR  BIR  (0.9360.98o  (1270o ) 103)  ((14276.4o )  (145.8 25.84o )) 
 (118,855  j2,033)  (13,153  j15,990) 1332337.77o V or133.237.77o kV
Vs 3  3 133.23  230.8 kV
VR(nL) 
Vs 3
A

230.8
 246.5 kV
0.9361
(%)regulation 
VR(nL)  VR(FL)
VR(FL)
.% 
246.5  220
100 12%
220
Example
What is the maximum length in km for a 1-phase transmission line having
2
copper conductor of 0·775 cm cross-section over which 200 kW at unity power factor and at 3300V
are to be delivered ? The efficiecny of transmission is 90%. Take specific resistance as 1.725 µ Ω
cm.
Solution.
Receiving end power = 200 kW = 2,00,000 W
Transmission efficiency = 0·9
∴
Sending end power = 2,00,000 = 2,22,222 W
0⋅9
∴
Line losses = 2,22,222 − 2,00,000 = 22,222 W
Line current, I =
or
200 × 103
= 60·6 A
3300 × 1
,
Let R Ω be the resistance of one conductor then for 2 conductors the total resistance is 2R.
2
Line losses = I 2R
2
22,222 = 2 (60·6) × R
22,222
∴
R =
2 = 3·025 Ω
2 × 60 ⋅ 6
Now,
R = ρ l/a
Ra = 3 ⋅ 025 × 0 ⋅ 775
6
∴
l =
−6 = 1·36×10 cm = 13·6 km
ρ
1 ⋅ 725 × 10
a f
Dr Audih alfaoury
Example
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.
Solution.
Load power factor, cos φR = 0·8 lagging
Receiving end voltage/phase, VR = 22,000
3 = 12,700 V
→
Impedance/phase,
Z = 4+j6
Line current,
5000 × 103
= 164 A
3 × 12700 × 0 ⋅ 8
∴ sin φR = 0·6
I =
As cos φR = 0·8
(note: we use here the phase voltage not line voltage)
Taking VR as the reference phasor
VR = VR + j 0 = 12700 V
→
I = I (cos φR − j sin φR) = 164 (0·8 − j 0·6) = 131·2 − j 98·4
(i) Sending end voltage per phase is
→→
VS = VR + I Z = 12700 + (131·2 − j 98·4) (4 + j 6)
= 12700 + 524·8 + j 787·2 − j 393·6 + 590·4
= 13815·2 + j 393·6
Magnitude of VS =
a13815 ⋅ 2f + a393 ⋅ 6f
2
2
= 13820·8 V
3 × 13820·8 = 23938 V = 23·938 kV
13820 ⋅ 8 − 12700
VS − VR
(ii)
% age Regulation =
× 100 =
× 100 = 8·825%
12700
VR
2
2
(iii)
Line losses = 3I R = 3 × (164) × 4 = 3,22,752 W = 322·752 kW
5000
∴
Transmission efficiency =
× 100 = 93·94%
5000 + 322 ⋅ 752
Line value of VS =
Example :
Estimate the distance over which a load of 15000 kW at a p.f. 0·8 lagging can be
delivered by a 3-phase transmission line having conductors each of resistance 1 Ω per kilometre.
The voltage at the receiving end is to be 132 kV and the loss in the transmission is to be 5%.
Dr Audih alfaoury
Solution.
Power delivered
15000 × 103
= 82 A
=
3 × line voltage × power factor
3 × 132 × 103 × 0 ⋅ 8
Line losses = 5% of power delivered = 0·05 × 15000 = 750 kW
Let R Ω be the resistance of one conductor ,for three conductors 3R.
2
Line losses = 3 I R
or
750 × 103 = 3 × (82)2 × R
Line current,
I =
750 × 103
= 37·18 Ω
3 × 82 2
Resistance of each conductor per km is 1 Ω (given).
∴
Length of line = 37·18 km
Example:
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.
Solution.
Resistance of each conductor, R = 5·31 Ω
Reactance of each conductor, XL = 5·54 Ω
Load power factor,
cos φR = 0·8 (lagging)
∴
R =
Sending end voltage/phase,
a f
VS = 33,000
3 = 19,052 V
Let VR be the phase voltage at the receiving end.
3
I = Power delivered / phase = 1200 × 10
VR × cos φ R
VR × 0 ⋅ 8
Line current,
=
150 × 10
VR
5
...(i)
(i) Using approximate expression for VS, we get,
VS = VR + I R cos φR + I XL sin φR
15 × 10
15 × 10
× 5·31 × 0·8 +
× 5·54 × 0·6
VR
VR
5
5
or
19,052 = VR +
or
VR2 − 19,052 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
15 × 10
15 × 10
=
= 81·36 A
18,435
VR
2
2
(iii) Line losses,
= 3 I R = 3 × (81·36) × 5·31 = 1,05,447 W = 105·447 kW
3600
∴ Transmission efficiency =
× 100 = 97·15%
3600 + 105 ⋅ 447
Example.
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.
5
(ii) Line current,
5
I =
Solution.
Resistance of each conductor, R = 6 Ω
Reactance of each conductor, X L = 8 Ω
Load power factor, cos φR = 0·9 lagging
Receiving end voltage/phase,
VR = 110 × 103
3 = 63508 V
Sending end voltage/phase,
VS = 120 × 10
3 = 69282 V
3
Let I be the load current. Using approximate expression for V S , we get,
Dr Audih alfaoury
Ps  PR  P  and I R  I s , P  RI 2
V .I cos   V .I cos   I R
V R + I R cos φR + I XL sin φR
V .I cos   I R
or
63508 + I × 6 × 0·9 + I × 8 × 0·435
cos  

V .I
or
5774
V .cos   IR
cos  
V
or
5774/8·88 = 650·2 A
3 VR I cos φ R
3 × 63508 × 650 ⋅ 2 × 0 ⋅ 9
(i)
Power output =
kW =
1000
1000
= 1,11,490 kW
V cos φ R + I R 63508 × 0 ⋅ 9 + 650 ⋅ 2 × 6
=
(ii) Sending end p.f., cos φS = R
= 0·88 lag
VS
69282
Example: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.
Solution.
Resistance of each conductor,
R = 1·5 Ω
Reactance of each conductor,
XL = 4 Ω
VS
69282
8·88 I
I
=
=
=
=
2
s
s
s
R
R
R
2
R
R
R
s
s
R
s
R
s
s
11 × 103
= 6351 V
3
Load power factor,
cos φR = 0·8 lagging
Power delivered in kVA × 1000
Load current,
I =
3 × VR
×
5000
1000
=
= 262·43A
3 × 6351
Using the approximate expression for V S (sending end voltage per phase), we get,
VS = V R + I R cos φR + I XL sin φR
= 6351 + 262·43 × 1·5 × 0·8 + 262·43 × 4 × 0·6 = 7295·8 V
V − VR
7295 ⋅ 8 − 6351
% regulation = S
× 100 =
× 100 = 14·88%
6351
VR
2
2
3
Line losses = 3 I R = 3 × (262·43) × 1·5 = 310 × 10 W = 310 kW
×
Input power = Ouput power + line losses = 4000 + 310 = 4310 kW
Receiving end voltage/phase,
Transmission efficiency =
VR =
Output power
4000
× 100 =
× 100 = 92·8%
4310
Input power
Example :-- 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.
Solution.
Resistance of each conductor, R = 0·03 × 16 = 0·48 Ω
−3
Reactance of each conductor, XL = 2π f L × 16 = 2π × 50 × 0·7 × 10 × 16 = 3·52 Ω
3
VR = 11 × 10 = 6351 V
3
cos φR = 0·8 lagging
Receiving end voltage/phase,
Load power factor,
1000 × 10
1000 × 10
=
= 65·6A
3 × VR × cos φ 3 × 6351 × 0 ⋅ 8
V R + I R cos φR + I XL sin φR
6351 + 65·6 × 0·48 × 0·8 + 65·6 × 3·52 × 0·6 = 6515 V
VS − VR
6515 − 6351
× 100 =
× 100 = 2·58%
VR
6351
2
2
3
3 I R = 3 × (65·6) × 0·48 = 6·2 × 10 W = 6·2 kW
Output power + Line losses = 1000 + 6·2 = 1006·2 kW
1000
Output power
× 100 =
× 100 = 99·38%
1006 ⋅ 2
Input power
3
Line current, I =
Sending end voltage/phase, V S =
=
∴
%age Voltage regulation =
Line losses =
Input power =
∴
Transmission efficiency =
3
Dr Audih alfaoury
Example .
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.
Solution. Figure below shows the single diagram of the transmission system.The voltage drop
will be due to the impedance of transmission line and also due to the impedance of transformer.
Resistance of each conductor = 20 × 0·4 = 8 Ω
Reactance of each conductor = 20 × 0·5 = 10 Ω
Let us transfer the impedance of transformer secondary to high tension side i.e., 33 kV side.
Equivalent resistance of transformer referred to 33 kV side
2
R(traf.)=R(primery)+ratio sequare.R(secondry)= Primary resistance + 0·35 (33/6·6)
= 7·5 + 8·75 = 16·25 Ω
Equivalent reactance of transformer referred to 33 kV side
2
= Primary reactance + 0·65 (33/6·6)
= 13·2 + 16·25 = 29·45 Ω
Total resistance of line and transformer is
R = 8 + 16·25 = 24·25 Ω
Total reactance of line and transformer is
XL = 10 + 29·45 = 39·45 Ω
Receiving end voltage per phase is
VR = 33,000
3 = 19052 V
2000 × 10
= 35 A
3 × 33000
Using the approximate expression for sending end voltage V S per phase,
VS = V R + I R cos φR + I XL sin φR
= 19052 + 35 × 24·25 × 0·8 + 35 × 39·45 × 0·6
= 19052 + 679 + 828 = 20559 V = 20·559 kV
3
Line current,
I =
Sending end line voltage
Sending end p.f.,
cos φS
3 × 20·559 kV = 35·6 kV
V cos φ R + I R 19052 × 0 ⋅ 8 + 35 × 24 ⋅ 25
=
= R
= 0·7826 lag
VS
20559
=
3I R
3 × (35) × 24 ⋅ 25
kW =
= 89·12 kW
1000
1000
2
2
Line losses
=
Output power
= 2000 kVA × 0·8 = 1600 kW
1600
=
× 100 = 94·72%
1600 + 89 ⋅ 12
∴ Transmission efficiency
Dr Audih alfaoury
Short Transmission Line: Phasor Diagram
AC voltages are usually expressed as phasors.
Load with lagging power factor.
Load with unity power factor.
Load with leading power factor.
For a given source voltage V S and magnitude of the line current, the received voltage is lower for
lagging loads and higher for leading loads.
Dr Audih alfaoury
Medium Line Model
As the length of the line is increased to over 80km (but less than 1240km) the
medium line model is used. This is also known as the nominal π model as shown below.
The admittance Y and impedance Z shown are shown for the whole line, per-phase (and
not per km). For the "pi" model shown we can
(2) I R
I S (1) Z = R+jX
I I I 0
write the following equations for node (2):
+
+
Y
IL
VS = VR + ZI L . Using the first
VR
I L = IR + V
VR
VS
_
_
2
Y/2
Y/2
 ZY 
V + ZI R
equation in the second we have: VS =  1 +
2  R

Y
Y  ZY 

Y
We also see that I S = I L + VS and thus we have: I S = I R + 2 VR + 2 1 + 2 .  VR + ZI R 



2
 ZY 
 ZY 
= Y 1 +
VR + 1 +
 IR
4 
2 


L
R
c
Thus we now have
: the ABCD parameters:
 ZY 
A = 1 +
B=Z

2 

 ZY 
 ZY 
C = Y 1 +
D = 1 +


4 
2 


In general for a symmetrical 2-port A = D and AD − BC = 1 (the determinant of the
ABCD matrix in this case is unity). Thus it is easy to find the inverse relationship:
VR   D − B  VS 
 =
 
 I R   −C A   I S 
Note also that f is frequency in Hz and g is conductance to ground per unit length (zero
in the specific case of the medium line model). Note also that r , g, z, and y are all per
unit length of line. The series impedance is Z = ( r + jω L ) l = R + jX and the shunt
admittance is Y = ( g + jωC ) l .
*
Note the construction of phasor diagram. The load current I R lags behind VR by φR . The capacitive
current IC leads VR by 90º as shown. The phasor sum of IC and I R is the sending end current IS . The
drop in the line resistance is IS R ( AB) in phase with IS whereas inductive drop IS X L ( BC) leads IS by 90º.
Therefore, OC represents the sending end voltage VS . The angle φS between the sending end voltage VS
and sending end current IS determines the sending end power factor cos φS.
Dr Audih alfaoury
Example A medium single phase transmission line 100 km long has the following
constants :
Resistance/km = 0·25 Ω ;
Reactance/km = 0·8 Ω
−6
Susceptance/km = 14 × 10 siemen ;
Receiving end line voltage = 66,000 V
Assuming that the total capacitance of the line is localised 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.
Solution. Figurs
( i) and (ii) show the circuit diagram and phasor diagram of the line
respectively.
Total reactance,
XL = 0·8 × 100 = 80 Ω
−6
−4
Total susceptance,
Y = 14 × 10 × 100 = 14 × 10 S
Receiving end voltage,
VR = 66,000 V
15,000 × 103
= 284 A
66,000 × 0 ⋅ 8
cos φR = 0·8 ;
sin φR = 0·6
Taking receiving end voltage as the reference phasor [see Figure ( ii)], we have,
∴ Load current,
IR =
VR = V R + j 0 = 66,000V
Load current,
I R = IR (cos φR − j sin φR) = 284 (0·8 − j 0·6) = 227 − j 170
Capacitive current,
IC = j Y × V R = j 14 × 10
(i) Sending end current,
IS = IR + IC = (227 − j 170) + j 92
−4
× 66000 = j 92
= 227 − j 78
Magnitude of IS =
(ii) Voltage drop
Sending end voltage,
... (i)
(227)2 + (78)2 = 240 A
= IS Z = IS ( R + j XL ) = (227 − j 78) ( 25 + j 80)
= 5,675 + j 18, 160 − j 1950 + 6240
= 11,915 + j 16,210
VS = VR + IS Z = 66,000 + 11,915 + j 16,210
= 77,915 + j 16,210
...(ii)
2
2
(77915) + (16210) = 79583V
V − VR
79,583 − 66,000
× 100 =
× 100 = 20. 58%
= S
VR
66,000
Magnitude of V S =
(iii) % Voltage regulation
(iv) Referring to exp. (i), phase angle between VR and I R is :
−1
−1
θ1 = tan − 78/227 = tan (− 0·3436) = − 18·96º
(iv) Referring to exp. (i), phase angle between VR and I R is :
−1
−1
θ1 = tan − 78/227 = tan (− 0·3436) = − 18·96º
Referring to exp. (ii), phase angle between V and V is :
R
S
θ2 = tan −1 16210 = tan −1 (0 ⋅ 2036) = 11⋅ 50º
77915
∴ Supply power factor angle, φS = 18·96º + 11·50º = 30·46º
∴
Supply p.f. = cos φS = cos 30·46º = 0·86 lag
Dr Audih alfaoury
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 as shown in Figure . Therefore, in this arrangement, full charging current flows over half the line. In Figure , one phase of 3phase transmission line is shown as it is advantageous to work in phase instead of line-to-line values.
Let
IR
XL
cos φR
V1
=
=
=
=
load current per phase ;
inductive reactance per phase ;
receiving end power factor (lagging) ;
voltage across capacitor C
R = resistance per phase
C = capacitance per phase
V S = sending end voltage/phase
The *phasor diagram for the circuit is shown in Figure . Taking the receiving end voltage
as the reference phasor, we have,
*
Receiving end voltage,
VR = V R + j 0
Load current,
I R = IR (cos φ R − j sin φ R )
VR
Note the construction of phasor diagram. VR is taken as the reference phasor represented by OA. The load
current I R lags behind VR by φR. The drop AB = IR R/2 is in phase with I R and BC = IR·X L /2 leads I R by
90º. The phasor OC represents the voltage V1 across condenser C. The capacitor current IC leads V1 by
90º as shown. The phasor sum of I R and IC gives IS . Now CD = IS R/2 is in phase with IS while DE =
IS X L /2 leads IS by 90º. Then, OE represents the sending end voltage VS .
V1 = VR + I R Z / 2
Voltage across C,
= VR + IR (cos φR − j sin φR)
Capacitive current,
IC = j ω C V1 = j 2π f C V1
Sending end current,
IS = IR + IC
FG
H
FR + j X I
H2 2 K
L
IJ
K
Z = V + I R + j XL
VS = V1 + I S
S
1
2
2
2
A 3-phase, 50-Hz overhead transmission line 100 km long has the following
Sending end voltage,
Example :--constants :
Resistance/km/phase
= 0.1 Ω
Inductive reactance/km/phase
= 0·2 Ω
−4
Capacitive susceptance/km/phase = 0·04 × 10 siemen
Dr Audih alfaoury
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.
Solution. Figures ( i) and
( ii) show the circuit diagram and phasor diagram of the line
respectively.
Total resistance/phase,
Total reactance/phase.
Capacitive susceptance,
Receiving end voltage/phase,
Load current,
R
XL
Y
VR
=
=
=
=
0·1 × 100 = 10 Ω
0·2 × 100 = 20 Ω
−4
−4
0·04 × 10 × 100 = 4 × 10 S
66,000/√ 3 = 38105 V
10,000 × 103
= 109 A
3 × 66 × 103 × 0 ⋅ 8
= 0·8 ; sin φR = 0·6
IR =
cos φR
→
Impedance per phase,
Z = R + j XL = 10 + j 20
(i) Taking receiving end voltage as the reference phasor [see Figure
( ii)], we have,
Receiving end voltage,
VR = VR + j 0 = 38,105 V
Load current,
I R = IR (cos φR − j sin φR) = 109 (0·8 − j 0·6) = 87·2 − j 65·4
Voltage across C,
Charging current,
V1 = VR + I R Z/2 = 38, 105 + (87·2 − j 65·4) (5 + j 10)
= 38,105 + 436 + j 872 − j 327 + 654 = 39,195 + j 545
−4
IC = j Y V1 = j 4 × 10 (39,195 + j 545) = − 0 ⋅ 218 + j 15 ⋅ 6
Sending end current,
IS = IR + IC = (87·2 − j 65·4) + (− 0·218 + j 15·6)
∴ Sending end current
= 87·0 − j 49·8 = 100 É − 29º47′ A
= 100 A
(ii) Sending end voltage, VS = V1 + IS Z / 2 = (39,195 + j 545) + (87·0 − j 49·8) (5 + j 10)
= 39,195 + j 545 + 434·9 + j 870 − j 249 + 498
= 40128 + j 1170 = 40145 É1º40′ V
∴ Line value of sending end voltage
= 40145 × √ 3 = 69 533 V = 69·533 kV
(iii) Referring to phasor diagram in Fig. 10.14,
θ1 = angle between VR and VS = 1º40′
θ2 = angle between VR and IS = 29º 47′
∴
φS = angle between VS and IS
= θ1 + θ2 = 1º40′ + 29º47′ = 31º27′
∴ Sending end power factor, cos φS = cos 31º27′ = 0·853 lag
(iv)
Sending end power = 3 V S IS cos φS = 3 × 40,145 × 100 × 0·853
= 10273105 W = 10273·105 kW
Power delivered = 10,000 kW
∴
Transmission efficiency =
10,000 ×
100 = 97·34%
10273 ⋅ 105
Dr Audih alfaoury
Example :--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.
Solution.:-- Figurs ( i)
and ( ii) show the circuit diagram and phasor diagram respectively.
Total resistance/phase, R = 0·2 × 100 = 20 Ω
Total reactance/phase, X L = 0·4 × 100 = 40 Ω
Total capacitance admittance/phase, Y = 2·5 × 10− 6 × 100 = 2·5 × 10− 4 S
→
Phase impedance, Z = 20 + j40
Receiving end voltage/phase, V R = 110 × 103
3 = 63508 V
20 × 106
= 116.6 A
3 × 110 × 103 × 0 ⋅ 9
cos φR = 0·9 ; sin φR = 0·435
(i) Taking receiving end voltage as the reference phasor [see phasor diagram
have,
Load current, IR =
( ii)], we
VR = V R + j0 = 63508 V
Load current,
I R = IR (cos φR − j sin φR) = 116·6 (0·9 − j 0·435) = 105 − j50·7
Voltage across C,
V1 = VR + I R Z 2 = 63508 + (105 − j 50·7) (10 + j 20)
= 63508 + (2064 + j1593) = 65572 + j1593
Charging current,
Sending end current,
∴ Sending end current
Sending end voltage,
−4
IC = j YV1 = j 2·5 × 10 (65572 + j1593) = − 0·4 + j 16·4
IS = IR + IC = (105 − j 50·7) + (− 0·4 + j16·4)
= (104·6 − j 34·3) = 110 ∠− 18º9′ A
= 110 A
VS = V1 + IS Z / 2
= (65572 + j1593) + (104·6 − j34·3) (10 + j20)
= 67304 + j 3342
∴
Magnitude of V S = (67304)2 + (3342)2 = 67387 V
∴ Line value of sending end voltage
= 67387 × 3 = 116717 V = 116·717 kV
(ii) Total line losses for the three phases
2
2
= 3 IS R/2 + 3IR R/2
2
2
= 3 × (110) × 10 + 3 × (116·6) × 10
6
= 0·770 × 10 W = 0·770 MW
20
× 100 = 96·29%
∴ Transmission efficiency
=
20 + 0 ⋅ 770
Dr Audih alfaoury
Nominal π Method
In this method, capacitance of each conductor (i.e., line to neutral) is divided into two halves; one half
being lumped at the sending end and the other half at the receiving end as shown in Figure. 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
cos φR = receiving end power factor (lagging)
V S = sending end voltage per phase
The *phasor diagram for the circuit is shown in Figure. Taking the receiving end voltage as
the reference phasor, we have,
VR = V R + j 0
Load current,
I R = IR (cos φR − j sin φR )
Charging current at load end is
IC1 = j ω (C 2) VR = j π f C VR
Line current,
I L = IR + IC1
VS = VR + I L Z = VR + I L ( R + jXL )
Charging current at the sending end is
Sending end voltage,
IC2 = j ω (C 2) VS = j π f C VS
∴ Sending end current,
*
IS = I L + IC2
Note the construction of phasor diagram. VR is taken as the reference phasor represented by O A. The
current I R lags behind VR by φR . The charging current IC1 leads VR by 90º. The line current I L is the
phasor sum of I R and IC1 . The drop A B = IL R is in phase with I L whereas drop BC = IL X L leads I L by
90º. Then OC represents the sending end voltage VS . The charging current IC2 leads VS by 90º. Therefore, sending end current IS is the phasor sum of the IC2 and I L . The angle φS between sending end
voltage V S and sending end current IS determines the sending end p.f. cos φS.
Dr Audih alfaoury
Example
A 3-phase, 50Hz, 150 km line has a resistance, inductive reactance and ca−6
pacitive shunt admittance of 0·1 Ω, 0·5 Ω and 3 × 10 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.
Solution. Figure shows the circuit diagram for the line.
Total resistance/phase,
R = 0·1 × 150 = 15 Ω
Total reactance/phase,
XL = 0·5 × 150 = 75 Ω
−6
−5
Capacitive admittance/phase, Y = 3 × 10 × 150 = 45 × 10 S
Receiving end voltage/phase, VR = 110 × 10
Load current,
3
3 = 63,508 V
50 × 106
= 328 A
3 × 110 × 103 × 0 ⋅ 8
cos φR = 0·8 ; sin φR = 0·6
IR =
Taking receiving end voltage as the reference phasor, we have,
VR = V R + j 0 = 63,508 V
Load current,
IR =
IR (cos φR − j sin φR) = 328 (0·8 − j0·6) = 262·4 − j196·8
Charging current at the load end is
Line current,
−5
45 × 10 = j 14·3
IC1 = VR j Y = 63,508 × j
2
2
I L = IR + IC1 = (262·4 − j 196·8) + j 14·3 = 262·4 – j 182·5
VS = VR + I L Z = VR + I L ( R + j X L )
= 63, 508 + (262·4 − j 182·5) (15 + j 75)
= 63,508 + 3936 + j 19,680 − j 2737·5 + 13,687
= 81,131 + j 16,942·5 = 82,881 ∠ 11º47′ V
∴ Line to line sending end voltage = 82,881 × √ 3 = 1,43,550 V = 143·55 kV
Charging current at the sending end is
Sending end voltage,
IC 2 = jVS Y / 2 = (81131
, + j 16,942 ⋅ 5) j
45 × 10
2
−5
= − 3·81 + j 18·25
Sending end current,
∴ Sending end current
IS = I L + IC2 = (262·4 − j 182·5) + (− 3·81 + j 18·25)
= 258·6 − j 164·25 = 306·4 ∠ − 32·4º A
= 306·4 A
Example
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 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
Dr Audih alfaoury
Solution. The figure shows the circuit diagram for the line.
Total resistance/phase,
R = 0·1 × 100 = 10 Ω
Total reactance/phase,
XL = 0·5 × 100 = 50 Ω
−6
−4
Susceptance/phase,
Y = 10 × 10 × 100 = 10 × 10 S
3
Receiving end voltage/phase, VR = 66 × 10 /√3 = 38105 V
Load current,
IR =
20 × 106
= 195 A
3 × 66 × 103 × 0 ⋅ 9
cos φR = 0·9
;
sin φR = 0·435
Taking receiving end voltage as the reference phasor, we have,
VR =V R + j0 = 38105 V
I R = IR (cos φR – j sin φ R) = 195 (0·9 − j 0·435) = 176 − j 85
Charging current at the receiving end is
Load current,
10 × 10
Y
IC1 = VR j = 38105 × j
2
2
Line current,
−4
= j 19
I L = IR + IC1 = (176 − j 85) + j 19 = 176 − j 66
VS = VR + I L Z = VR + I L ( R + j X L )
= 38,105 + (176 − j 66) (10 + j 50)
= 38,105 + (5060 + j 8140)
= 43,165 + j 8140 = 43,925 ∠10·65º V
3
Sending end line to line voltage = 43, 925 × √ 3 = 76 × 10 V = 76 kV
Charging current at the sending end is
Sending end voltage ,
,
+ j 8140) j
IC2 = VS jY / 2 = (43165
= − 4·0 + j 21·6
∴ Sending end current,
10 × 10
2
−4
IS = IL + IC2 = (176 − j 66) + (− 4 ⋅ 0 + j 21 ⋅ 6)
= 172 − j 44·4 = 177·6 ∠ − 14·5º A
(i) Referring to phasor diagram in Figure,
θ1 = angle between V and V = 10·65º
R
S
θ2 = angle between VR and IS = − 14·5º
∴
φS = angle between VS and IS = θ2 + θ1
= 14·5º + 10·65º = 25·15º
∴ Sending end p.f., cos φS = cos 25·15º = 0·905 lag
V − VR
43925 − 38105
(ii) % Voltage regulation
= S
× 100 =
× 100 = 15·27 %
VR
38105
(iii) Sending end power
= 3 V S IS cos φS = 3 × 43925 × 177·6 × 0·905
6
= 21·18 × 10 W = 21·18 MW
Transmission efficiency = (20/21·18) × 100 = 94 %
Dr Audih alfaoury
Athree-phase,60-Hz,completelytransposed345-kV,200-km, and.the.following.positivesequence line constants:
EXAMPLE
z ¼ 0:032 þ j0:35 W=km
y ¼ j4:2 106
S=km
Full load at the receiving end of the line is 700 MW at 0.99 p.f. leading and at
95% of rated voltage. Assuming a medium-length line, determine the following:
a. ABCD parameters of the nominal p circuit
b. Sending-end voltage VS , current IS , and real power PS
c. Percent voltage regulation
d. Transmission-line e‰ciency at full load
SOLUTION
a. The total series impedance and shunt admittance values are
Z ¼ zl ¼ ð0:032 þ j0:35Þð200Þ ¼ 6:4 þ j70 ¼ 70:29 84:78
6
Y ¼ yl ¼ ð j4:2 10 Þð200Þ ¼ 8:4 10
4
90
A ¼ D ¼ 1 þ ð8:4 104 90 Þð70:29 84:78 Þ
W
S
1
2
¼ 1 þ 0:02952 174:78
¼ 0:9706 þ j0:00269 ¼ 0:9706 0:159
B ¼ Z ¼ 70:29 84:78
C ¼ ð8:4 104 90 Þð1 þ 0:01476 174:78 Þ
¼ ð8:4 104 90 Þð0:9853 þ j0:00134Þ
¼ 8:277 104 90:08
b. The receiving-end voltage and current quantities are
VR ¼ ð0:95Þð345Þ ¼ 327:8
kVLL
327:8
VR ¼ pffiffiffi 0 ¼ 189:2 0
3
kVLN
700 cos1 0:99
IR ¼ pffiffiffi
¼ 1:246 8:11
ð 3Þð0:95 345Þð0:99Þ
kA
The.sending-end.quantities.are:
VS ¼ ð0:9706 0:159 Þð189:2 0 Þ þ ð70:29 84:78 Þð1:246 8:11 Þ
¼ 183:6 0:159 þ 87:55 92:89
¼ 179:2 þ j87:95 ¼ 199:6 26:14
VS ¼ 199:6
kVLN
pffiffiffi
3 ¼345:8 kVLL
IS ¼ ð8:277 104 90:08 Þð189:2 0 Þ þ ð0:9706 0:159 Þð1:246 8:11 Þ
¼ 0:1566 90:08 þ 1:209 8:27
¼ 1:196 þ j0:331 ¼ 1:241 15:5
kA
Dr Audih alfaoury
and the real power delivered to the sending end is
pffiffiffi
PS ¼ ð 3Þð345:8Þð1:241Þ cosð26:14 15:5 Þ
¼ 730:5 MW
c. The.no-load.receiving-end.voltageis
VRNL ¼
VS
345:8
¼
¼ 356:3 kVLL
A
0:9706
and
percent VR ¼
356:3 327:8
100 ¼ 8:7%
327:8
d.The full-load line losses are PS PR ¼ 730:5 700 ¼ 30:5 MW and the
full-load transmission e‰ciency is
percent EFF ¼
PR
700
100 ¼ 95:8%
100 ¼
PS
730:5
Long Line Model
This is a distributed parameters model of the line. It is very exact and gives the voltage
and current anywhere along the line. The distance x is measured from the receiving end
towards the sending end. If we let the x = l (the length of the line) the equations for the
long line model have the following ABCD parameters:
A = cosh γ l
C=
1
sinh γ l
Zc
B = Z c sinh γ l
D = cosh γ l
z
and γ = α + j β = zy .
y
It is now possible to find an accurate π model whose ABCD parameters are equal in
value to the ones shown above. Without proof the results are outlined below:
sinh γ l
Z ′ = Z c sinh γ l = Z
γl
Y′ 1
Y tanh γ l / 2
=
tanh γ l =
2 Zc
2 γl/2
where the primed quantities are for the exact π or equivalent π model, and the unprimed
quantities are those for the nominal π .
where Z c is the line characteristic impedance given by Z c =
Dr Audih alfaoury
EXAMPLE
A three-phase 765-kV, 60-Hz, 300-km, completely transposed line has the
following positive-sequence impedance and admittance:
z ¼ 0:0165 þ j0:3306 ¼ 0:3310 87:14
y ¼ j4:674 106
W=km
S=km
Assuming positive-sequence operation, calculate the exact ABCD parameters
of the line. Compare the exact B parameter with that of the nominal p circuit.
SOLUTION
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
0:3310 87:14
¼
7:082 10 4 2:86
Zc ¼
4:674 106 90
¼ 266:1 1:43
W
and
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð0:3310 87:14 Þð4:674 106 90 Þ ð300Þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 1:547 106 177:14 ð300Þ
gl ¼
¼ 0:3731 88:57 ¼ 0:00931 þ j0:3730
e gl ¼ e 0:00931 eþj0:3730 ¼ 1:0094 0:3730
per unit
radians
¼ 0:9400 þ j0:3678
and
egl ¼ e0:00931 ej0:3730 ¼ 0:9907 0:3730
radians
¼ 0:9226 j0:3610
Then,
coshðglÞ ¼
ð0:9400 þ j0:3678Þ þ ð0:9226 j0:3610Þ
2
¼ 0:9313 þ j0:0034 ¼ 0:9313 0:209
sinhðglÞ ¼
ð0:9400 þ j0:3678Þ ð0:9226 j0:3610Þ
2
¼ 0:0087 þ j0:3644 ¼ 0:3645 88:63
A ¼ D ¼ coshðglÞ ¼ 0:9313 0:209
per unit
B ¼ ð266:1 1:43 Þð0:3645 88:63 Þ ¼ 97:0 87:2
W
C¼
0:3645 88:63
¼ 1:37 103 90:06
266:1 1:43
S
Using,the B parameter for the nominal p circuit is
Bnominal p ¼ Z ¼ ð0:3310 87:14 Þð300Þ ¼ 99:3 87:14
W
which is 2% larger than the exact value.
Dr Audih alfaoury
: Consider a 500 km long line for which the per kilometer line impedance
Example
and admittance are given respectively by z = 0.1 + j0.5145 Ω and y = j3.1734 × 10−6 mho.
Therefore
ZC =
z
=
y
0.1 + j 0.5145
0.5241∠79°
0.5241
 79° − 90° 
=
=
∠

−6
−6
−6
j 3.1734 × 10
3.1734 × 10 ∠90°
3.1734 × 10
2


= 406.4024∠ − 5.5° Ω
and
 79° + 90° 

2


γl = yz × l = 0.5241 × 3.1734 × 10− 6 × 500∠
= 0.6448∠84.5° = 0.0618 + j 0.6419
We shall now use the following two formulas for evaluating the hyperbolic forms
cosh (α + jβ ) = cosh α cos β + j sinh α sin β
sinh (α + jβ ) = sinh α cos β + j cosh α sin β
Application of the above two equations results in the following values
cosh γl = 0.8025 + j 0.037 and sinh γl = 0.0495 + j 0.5998
Therefore the ABCD parameters of the system can be written as
A = D = 0.8025 + j 0.037
B = 43.4 + j 240.72 Ω
C = −2.01 × 10− 5 + j 0.0015
Dr Audih alfaoury
Dr Audih alfaoury
Dr Audih alfaoury
Dr Audih alfaoury
Dr Audih alfaoury
Dr Audih alfaoury
Dr Audih alfaoury
Dr Audih alfaoury
Dr Audih alfaoury
Dr Audih alfaoury
Dr Audih alfaoury
Dr Audih alfaoury
VED
APPRO