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 7Ω
The receiving end voltage per phase is VR is
(a)
64∠0°
√3
= 36.9504∠0° kV
SR = 70/3 = 23.33 MVA/ph
PR = (23.33 )( 0.8) = 18.67 MW/ph
IR =
23.33(𝑀𝑊)
36.9504(𝑘𝑉)
X(𝑐𝑜𝑠𝜃 −jsin𝜃) =0.6315(0.8-j0.6)=(0.502-j0.3789)kA
The sending end voltage is
Vs
=Va +ZIR =36.9504Z0 *(kV) + (2 +j7)(0.5052 - j0.3789)(kA)
=(40.61- j2.78) =40.708z3.91° kV
The sending end line-to-line voltage is |VS(L-L)|=√3|VS| =70.508 kV
The sending end power PS = PR + IR2R = 19.467 MW/ph
PS(3-phase)= 58.4 MW
% Voltage Regulation=
70.508 −64
64
Transmission line efficiency = n =
X 100 = 10.169%
PR(3phase)
PS(3phase)
=
56
58.393
x100 =95.90%
(b) The per phase power, PR =120/3 = 40 MW
IR =
56
58.393
X(𝑐𝑜𝑠𝜃 −jsin𝜃)
= (1.0825 + j0) kA
The sending end voltage is
V,
=Va +ZIR =36.9504∠0° + (2 + j7)(1.0825 ∠0° )
=(35.12 + i7.58) kV =39.8427.c10.9639° kV/ph
The sending end line-to-line voltage is |VS(L-L)|=√3|VS| =69.0096 kV
The sending end power PS = PR + IR2R = 19.467 MW/ph
PS(3-phase)=42.33 x 3= 58.4 MW
% Voltage Regulation=
69.0096 −64
64
Transmission line efficiency = n =
X 100 = 7.8275%
PR(3phase)
PS(3phase)
=
120
127.031
x100 =94.465%
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 52 and 15 12 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
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° kA
Vs = VR + ZIR = 132.79 + (5+j15)x(0.6275 - j0.4706)
= 143 + j7.06 = 143.16 ∠2.83° kV
Voltage regulation =
143.16 −132.79
132.79
X 100 = 7.81%
The leading power factor angle at which the voltage regulation is zero is given by
Φ= π/2 = tan-1
𝑋
𝑅
+ sin-1
𝐼𝑍
VR
=21.1°
The power factor at which the voltage regulation will be zero is
cos (21.1°' )= 0.9329 leading
Derivation of the formula
We know that
Vs
|Vs|2
= VR + (R +jX)(Icosφ+ jIsinφ)
=VR + RIcosφ-- XIsinφ + j(XIcosφ+ RIsinφ)
= (VR + RIcosφ-- XIsinφ )2 + (XIcosφ+ RIsinφ)2
= VR2 +I2 (R2+X2) + 2VR IRcosφ-- 2VR XIsinφ
Since |Vs| = |VR|
0= I2Z2+ 2VRI R cos φ - 2VRX I sinφ
5. A 345-kV, three-phase transmission line is 130 km long. The series impedance z =
0.036 + j0.3 Man, and the shunt admittance y = j4.22. 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
A= 0.989 + j0.00128 = 0.989
B =4.68 + j39
C = -3.52 x 10' + 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° kV/ph
VR(LL)=√3 x 190.92 = 330.68 kV
IR= 441.8Z - 31.65° A
PR = 190.92 x 0.4418 x cos(27.45)= 74.85 MW /ph = 224.5MW
QR = 190.92 x 0.4418 x sin(27.45)= 38.88 MVAR /ph = 116.65MVAR
Voltage Regulation=
345/0.989 −330.68
330.68
X 100 = 5.46%
6.
A 500-kV, three-phase transmission line is 250 km long. The series impedance z =
0.045 + j0.4 52/ph/km and shunt admittance y = j4 x10-6 S/km. Evaluate the
equivalent - π model and the ABCD constants.
Solution
z = (0.045 + j0.4)52
; y = j4 x 104 S
y =√zy =0.0001 +j0.0013
ZC =𝑥 =
√𝑍
𝑦
=316.23- j1.778)Ω
yl = 0.0018 + j0.3162
A = cosh ɣℓ= 0.9504 + j0.0006
B = Z C sinh ɣℓ=1.088 + j98.34
C = (sinh ɣℓ)/ Z. = j9.83 x 104
D=A
Z = Z C sinh ɣℓ= 1.088 + j98.34Ω
Y/2= j5.042 x 104S
The equivalent - π circuit is as shown
7.
A three-phase transmission line is 370 km long. The series impedance of the line is
0.5242∠79.4' Ω/km and the susceptance is j3.17 x10-6 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
(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+ℓ21.6611 x10-6 ∠169.04°
= 1 - 1.6308 x10-6 ℓ2 xj0.3158 x10-6 ℓ2
Neglect :1z the imaginary part as it is very small compared to the real part)
|A|= 0.9302 = 1 - 1.6308 x10-6 ℓ2
ℓ2 = 4280
Therefore the length of the line 1= 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 µF/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.