Ferranti Effect in Transmission Line

Ferranti Effect in Transmission Line

THEORY

A long transmission line draws a substantial quantity of charging current. If such a line is open circuited or very lightly loaded at the receiving end the voltage at receiving end may become greater than voltage at sending end. This is known as Ferranti Effect and is due to the voltage drop across the line inductance being in phase with the sending end voltages. Therefore both capacitane and inductance is responsible to produce this phenomenon.

The capacitance and charging current is negligible in short line but significant in medium line and appreciable in in long line by equivalent π model. It is proportional to the square of lengths of lines, that is, ΔV α kx2, where x is the length of line and k is a constant for all voltage].

Ferranti Effect can be explained by considering a nominal π model of the line. Figure 1(b) shows the phasor diagram of Figure 1(a).Here OE represents the receiving end voltage Vr.OH represents the current Ic1

through the capacitor C/2 at the receiving end.The voltage drop Ic1R across the resistanc R is shown by EF.it is in phase with Ic1. The voltage drop across X is Ic1X.It is represented by the phasor FG which leads the phasor Ic1R by 90°.The phasor OG represents the sending end voltage Vs under no-load condition.It is

seen from the phasor diagram that Vs<Vr.In other words, the voltage at the receiving end is greater than the voltage at the sending end when the line is at no load.

Figure 1. (a) Nominal π model of the line at no load

Figure 1b) Phasor diagram

 

In practice, the capacitance of the line is not concentrated at some definite points.It is distributed uniformly along the whole length of the line.Therefore the voltage will increase from sending end to receiving end.At no load or light load the voltage at the receiving endis quite large as compared to the constant voltage at the sending end.

For a nominal π model of a line

Vs = (1+ ZY/2) Vr + ZIr

At no load, Ir = 0

Vs = (1+ ZY/2) Vr

Vs – Vr = (ZY/2) Vr

Z = ( r + jωl)S, Y = (jωc)S

If the resistance of the line is neglected,

Z = jωl S

And

Vs – Vr = ½( jωl S) (jωcS) Vr = - ½ (ω2s2)lc Vr

For overhead lines

1/ √lc = velocity of propagation of electromagnetic waves on the line = 3x 108 m/s

Vs – Vr = - ½ (2 π f) 2 S2 . 1/ (3x 108)2 Vr

Vs – Vr = - (4 π2/ 18 x 1016) f2S2Vr

This equation shows that (Vs – Vr) is negative. That is, Vr > Vs. This equation also shows that

Ferranti effect depends on frequency and electrical length of the line. The conductor diameter and spacing have no bearing on Ferranti effect.

In general, for any line

Vs = AVr + BIr

At no load,

Ir = 0, Vr = Vrnl

So, Vs = AVrnl, |Vrnl | = |Vs| / |A|

For a long line A is less than unity and it decreases with the increase in length of line. Hence Vrnl > Vs. As the line length increases the rise in the voltage at the receiving end at no load becomes more predominant