This document contains 3 short problems related to power transmission lines: 1. A 15kVA line operates at 33kV and calculates voltage, power factor, regulation, and efficiency. 2. A 5MW, 11kV line with 10% power loss calculates sending end voltage and power factor. 3. A 110kV, 150km line calculates the receiving end load current needed for unity power factor.

1. Short and Medium Lines
1. 15000 KVA is received at 33KV at 0.85 pf lagging over an 8km 3 phase OH line. Each line has R =
0.29 Ω/km, and X = 0.65 Ω/km. calculate (b)the voltage at the sending end (b) the power factor
at the sending end, (c) the regulation, and (d) the efficiency of the transmission line.
[20.31|_2.37°KV, 0.8275 lagging, 6.58%, 96.38%]
Radius?and
find Now you will get the correct answer (power is
inductance? 15000KVA)
2. A 3 phase line, 10km long, delivers 5MW at 11KV, 50Hz, 0.8pf lagging. The power loss in the line
is 10% of the power delivered. The line conductors are situated at the corners of an equilateral
triangle of 2m side. Calculate the voltage and pf at the sending end. Assume resistivity of
copper to be 1.774µΩcm. [7.535|_5.266°KV, 0.7416 lagging]
Solution:
Vr = 6350V
Ir = 328A
It si short line
The power loss in the line is 3Ir2R
Power loss = 10% of power delivered
3Ir2R = 10% 5MW; R = 1.549 ohm = ρl/a
l= 10km; a = 1.147cm2; Hence r = 0.6042cm; r’ = 0.7788r
Spacing between the conductors, D = 2m = 200cm
Inductance L = 2x10-7 ln [200/(0.6042x0.7788)] = 1.21 x 10-6 H/m
Therefore total inductance L = 10,000 X 1.21 x 10-6 H
Reactance X = 3.8 ohm
Vr = 6350|_0°V
Rec. pf angle = cos-1(0.8) = 36.87°
Z = 1.549 + j3.8 = 4.1|_67.8226° ohm
Vs = Vr + IrZ = 7535.13|_5.266°V

2. The phase difference between Vs and Is will be
5.266° – (-36.87°) = 42.136°
Sneding end pf = cos42.136° = 0.7416(lagging)
3. A 3 phase 50Hz, 150km line operates at 110KV between the lines at the sending end. The total
inductance and capacitance per phaseare 0.2H and 1.5µF. Neglecting losses calculate the value
???
of receiving –end load having a pf of unity for which the voltage at the receiving end will be the
same as that at the sending end. Assume one-half of the total capacitance of the line to e
concentrated at each end. [173.275A]
Solution :
Vr = 110,000/sqrt(3) = 63508.53|_0°V
XL =2 πf(0.2) = 62.832 ohms
Z = j62.832 ohms
Y = j2 πfC = j4.712 x 10-4 S = 4.712 x 10-4|_90° mho or Siemens or S
Iab =( Y/2 )Vr = j14.963 A
Let the load current be Ir, since the load pf is unity,
Ir = Ir |_0°
Hence I = Ir + Iab = Ir + j 14.963
Vs = Vr + IZ = Vr|_0° + (Ir + j 14.963)( j62.832) = (Vr – 940.155) + j62.832(Ir)

3. Vs2 = (Vr – 940.155)2 + [62.832(Ir)]2
63508.532 = (63508.53– 940.155)2 + [62.832(Ir)]2
Ir = 173.275A