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EEL2026 Tut-2A Transmission Line Solutions

Power Transmission & Distribution (Multimedia University)

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EEL2026 Power Transmission and Distribution

Tutorial#2A Transmission line calculations

EET 2026 Power Transmission and Distribution

Tutorial # 2A
Transmission Line Calculation Solution-Some hints
1. A 69-kV, three-phase short transmission line is 16 km long. The line has a per phase
series impedance of 0.125 + j0.4375 ohm/km. Determine the sending end voltage,
voltage regulation, the sending end power, and the transmission efficiency when the
line delivers
(a) 70 MVA, 0.8 lagging power factor at 64 kV.
(b) 120 MW, unity power factor at 64 kV.

Solution
The line impedance is Z = (0.125 + j 0.4375)(16) = 2 + j 7 Ω
640
The receiving end voltage per phase is V R  36.95040 kV
3
(a) SR = 70/3 = 23.33 MVA/ph
PR = 23.33 x 0.8 = 18.67 MW/ph
IR =
23.33( MW )
(cos  j sin  ) 0.6315(0.8  j 0.6) (0.5052  j 0.3789) kA
36.9504(kV )

The sending end voltage is

VS VR  ZI R 36.95040 (kV )  (2  j 7)(0.5052  j 0.3789)(kA)

(40.61  j 2.78) 40.7083.910 kV

The sending end line-to-line voltage is V S ( L  L )  3 VS 70.508 kV

The sending end power PS = PR + IR2 R = 19.467 MW/ph

PS(3phase) = 58.4 MW

70.508  64
% Voltage Regulation = 100 10.169%
64

PR ( 3 ) 56
Transmission line efficiency =    100 95.90%
PS ( 3 ) 58.393

(b) The per phase power, PR = 120/3 = 40 MW

40( MW )
IR = (cos   j sin  ) (1.0825  j 0) kA
36.9504( kV )

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EEL2026 Power Transmission and Distribution

Tutorial#2A Transmission line calculations

The sending voltage is

VS VR  ZI R 36.95040  (2  j 7)(1.08250 ) (35.12  j 7.58) kV
VS 39.842710.9639 kV/ph

The sending end line-to-line voltage VS ( L  L )  3 VS 69.0096 kV

The sending end power = PS = PR + IR2 R = 40 (MW)+ 1.0825(kV)2 x 2 = 42.33 MW

PS(3phase) = 42.33 x 3 = 127 MW

69.0096  64
% Voltage Regulation = 100 7.8275%
64
PR ( 3 ) 120
Transmission line efficiency =    100 94.465%
PS ( 3 ) 127.031

2. A thee-phase short transmission line is supplying a load of 250 MW at 0.8 power

factor lagging. The voltage at the receiving end is kept constant at 230 kV. The
resistance and reactance per phase of the line are 5 Ω and 15 Ω respectively. (i)
Calculate the voltage regulation. (ii) At what value of the power factor is the voltage
regulation zero? Derive the expression used.

Solution
Z = 5 + j15 Ω
VR = 230/ 3 = 132.79 kV/ph
PR = 250/3 = 83.33 MW/ph
IR = (83.33/(0.8 x 132.79)) x (0.8 – j0.6) = 0.6275 - j0.4706 = 0.7844   36.87 0 kA
VS = VR + Z IR = 132.79 + (5 + j15)x (0.6275 - j0.4706)
= 143 + j7.06 = 143.16 2.83 0 kV
143.16  132.79
Voltage regulation = 100 = 7.81%
132.79

The leading power factor angle at which the voltage regulation is zero is given by
X IZ
  / 2  tan  1  sin  1 21.10
R 2VR
The power factor at which the voltage regulation will be zero is
cos (21.1o ) = 0.9329 leading

Derivation of the formula

We know that
VS = VR + (R + jX)(I cosφ + jI sinφ)
= VR + RI cosφ – XI sinφ + j(XI cosφ + RI sinφ)
|VS| = (VR + RI cosφ – XI sinφ)2 + (XI cosφ + RI sinφ)2
2

= VR2 + I2(R2 + X2) + 2VR I R cosφ – 2VR X I sinφ

Since |VS| = |VR|
0 = I2 Z2 + 2VR I R cosφ – 2VR X I sinφ

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EEL2026 Power Transmission and Distribution

Tutorial#2A Transmission line calculations

(X/Z) sinφ – (R/Z) cosφ = I Z/(2VR)

sinθ sinφ – cosθ cosφ = I Z/(2VR) where θ is the angle of the line impedance.
sin(θ + φ – π/2) = I Z/(2VR)
Therefore φ = π/2 – θ + sin-1(I Z/(2VR))

3. A three-phase voltage of 17.32 kV is applied to a line having R = 10 Ω and X = 12 Ω

per conductor. At the end of the line is a balanced load of P MW at 0.707 power
factor leading. At what value of P is the voltage regulation zero.

Solution
VR = 17.32/1.732 = 10 kV
Z = 10 + j12 Ω
We have derived in the previous problem that
VS2 = VR2 + I2(R2 + X2) + 2VR I R cosφ – 2VR X I sinφ
For zero regulation |VS| = |VR|
0 = I2(244) + 2 x10 x I x10 x 0.707 – 2 x 10 x I x 12 x 0.707
0 = 244 I2 – 28.28 I
I = 28.28/244 = 0.1159 kA
PR = 3 x 10 x 0.1159 x 0.707 = 2.458 MW

4. A 345-kV, 50-Hz, three-phase transmission line is 130 km long. The resistance per
phase is 0.036 Ω per km and the inductance per phase is 0.96 mH per km. The shunt
capacitance is 0.0134 μF per km. The receiving end load is 270 MVA with 0.8 power
factor lagging at 325 kV. Use medium line model to find the voltage, current and
power at the sending end and the voltage regulation.

Solution
z = r + jωL = 0.036 + j 314 x 0.96 x10-3 = 0.036 + j 0.3016 Ω/km
Z = z l = (0.036 + j 0.3016)130 = 4.68 + j39.207 Ω
Y = j314 x 0.0134x10-6 x 130 = 0.000547 S
The BCD constants are:
A = 0.989 + j0.00128 = 0.989 0.07 0
B = 4.68 + j39.207 Ω
C = -3.525 x 10-7 + j0.000546 S
D=A
VR = 325/ 3 = 187.6 kV/ph
SR = 270/3 = 90 MVA/ph
90 ( MVA )
IR = (0.8  j 0.6) 0.3838 - j0.2878 0.4797 - 36.87 0 kA
187.6( kV )
VS = A VR + B IR = 199.41 4.014 0 kV/ph; VS(LL) = 199.41 x 3 = 345 kV
IS = C VR + D IR = 0.4211   26.010 kA
PS = 199.41 x 0.4211 x cos(30.0250) = 72.9 MW/ph = 218.7 MW
QS = 199.41 x 0.4211 x sin(30.0250) = 42.01 MVAR/ph = 124.03 MVAR
345 / 0.989  325
Voltage regulation = 100 = 7.33%
325

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EEL2026 Power Transmission and Distribution

Tutorial#2A Transmission line calculations

5. A 345-kV, three-phase transmission line is 130 km long. The series impedance z =

0.036 + j0.3 Ω/km, and the shunt admittance y = j4.22x10-6 S/km. The sending end
voltage is 345 kV and the sending end current is 400 A at 0.95 power factor lagging.
Use medium line model to find the voltage, current and power at the receiving end
and the voltage regulation.

Solution
0
A = 0.989 + j0.00128 = 0.989 0.07
B = 4.68 + j39 Ω
C = -3.52 x 10-7 + j0.000546 S
D=A
VS = 345/ 3 = 199.19 kV/ph
IS = 400x(0.95 – j0.3122) = 380 – j124.9 A
= 190.92   4.2 0 kV/ph
VR(LL) = 3 x 190.92 = 330.68 kV
IR = 441.8   31.65 0 A
PR = 190.92 x 0.4418 x cos(27.45) = 74.85 MW/ph = 224.56 MW
QR = 190.92 x 0.4418 x sin(27.45) = 38.88 MVAR/ph = 116.65 MVAR
345 / 0.989  330.68
Voltage regulation = 100 = 5.46 %
330.68

6. A 500-kV, three-phase transmission line is 250 km long. The series impedance z =

0.045 + j0.4 Ω/ph/km and shunt admittance y = j4x10-6 S/ph/km. Evaluate the
equivalent-π model and the ABCD constants.

Solution
z = (0.045 + j0.4) Ω; y = j4 x 10-6 S
  zy 0.0001  j 0.0013
z
Zc  (316.23  j1.778) 
y
γl = 0.0018 + j0.3162

A = cosh γl = 0.9504 + j0.0006

B = Zc sinh γl = 1.088 + j98.34 Ω
C = (sinh γl) / Zc = j9.83 x 10-4
D=A

Z’ = Zc sinh γl= 1.088 + j98.34 Ω

Y’/2 = j5.042 x 10-4 S
The equivalent – π circuit is as shown
1.088 + j98.32

4
-4
j5.042x10 j5.042x10-4

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EEL2026 Power Transmission and Distribution

Tutorial#2A Transmission line calculations

7. A three-phase transmission line is 370 km long. The series impedance of the line is
0.524  79.4o Ω/km and the susceptance is j3.17x10-6 S/km. The voltage at the
sending end is 400 kV. (i) Find the sending end current and the receiving end voltage
when there is no load on the line. (ii) Determine the maximum permissible line length
if the receiving end no-load voltage is not to exceed 430 kV.

Solution
l  zy l 0.477284.52 0
z
Zc  406.4  5.48 0 
y
The A and C constants are:

A = 0.8904  1.340
C = j0.0011 S

(i) At no load:
The receiving end voltage VRNL = 400/|A| = 400/0.8904 = 449.24 kV (line-to-line)
The sending end current ISNL = C VSNL = j0.0011 x 449.24/ 3 = j0.2853 kA

(ii) The permissible VR = 430 kV

The corresponding value of |A| = |VS/VR| = 400/430 = 0.9302
We can write |A| = 1 + YZ/2 = 1 + l2 1.6611 x 10-6  169.040
= 1 – 1.6308 x 10-6 l2 + j0.3158 x 10-6 l2
Neglecting the imaginary part (as it is very small compared to the real part)
|A| = 0.9302 = 1 – 1.6308 x 10-6 l2
l2 = 4280.
Therefore the length of the line l = 206.9 km

8. A three-phase, 60-Hz, 765-kV transmission line is 400 km long. The line inductance
is 0.8885 mH/km per phase and its capacitance is 0.01268 F/km per phase. Assume
a lossless line. (a) Determine the phase constant, the surge impedance (characteristic
impedance), velocity of propagation, the line wave length, and the SIL (b) The
receiving end rated load is 2000 MW, 0.8 power factor lagging at 735 kV. Determine
the sending end quantities and the voltage regulation.
Solution
(a)
L = 0.8885 mH/km and C = 0.01268 μF/km

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EEL2026 Power Transmission and Distribution

Tutorial#2A Transmission line calculations

Propagation constant   LC 0.001265 rad/km

βl = 290
Wave length λ = 2π/β = 4967 km
1
Velocity of propagation = = 2.9793 x 10-5 km/s
LC
Characteristic impedance Zc = 263.7 Ω
SIL = 7652/264.7 = 2210.89 MW

(b)
It is a lossless line. The ABCD constants are computed as:
A = cosβl = 0.8746
B = jZc sinβl = j128.33 Ω
C = j(1/Zc) sinβl = j0.001832 S
D=A

VR = 735/ 3 = 424.35 kV
PR = 2000/3 = 666.67 MW/ph
IR = (666.67/424.35) x (0.8 – j0.6) = (1.2568 – j0.5428) kA
VS = A VR + B IR = 517.86  18.150 kV/ph
VSLL = 806.96 kV
IS = C VR + D IR = 1.1  -2.460 kA
Sending complex power = s VS IS* = 1600 (MW) + j601.5 (MVAR)
806.96 / 0.8746  735
Voltage regulation = 100 39.53%
735

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