**Ferranti Effect**is the rise in receiving-end voltage (V

_{R}) as compared to the sending-end voltage (V

_{S}) of a transmission line. It was first noticed by Sebastian Ziani de Ferranti on a project involving underground cables in a 10 kV distribution system in 1887, and was eventually named after him.

The ferranti effect normally occurs
on an energized long line (more than 80 km) with very light or worst, no load.
However, it can also happen with shorter lines composed of underground cables.

Looking at the power quality
perspective, long transmission lines and underground cable installations may
require protection for overvoltage such as surge arresters when loads are
suddenly disconnected.

**Parameters**

Ferranti effect is mainly due to the
charging current, which is associated with the line capacitance. In addition,
these basic points must be noted:

1. Capacitance is dependent on

- Composition –
Cables have more capacitance than bare conductors per unit length.
- Line length
(l) – Long lines have higher capacitance than short lines.

2. Charging current

- Becomes more significant
as load current decreases.
- Increases with
system voltage given the same capacitance value.

Accordingly, ferranti effect occurs
only for long lightly loaded or open-circuited energized lines. In addition, the
phenomenon becomes more evident with higher applied voltage and underground cables.

**Ferranti Voltage Rise Factor**

Using π -model, the voltage
equation for a transmission line is:

V

_{S}= (1 + YZ/2)V_{R}+ ZI_{L}
where: I

_{L}= load current, Y = line admittance and Z = line impedance
At very light or no load, I

_{L}can be neglected leaving,
V

_{R}= V_{S}/(1 + YZ/2)
With resistance neglected, the above
equation can then be detailed into:

V

_{R}= V_{S}/ (1 – ω^{2}l^{2}LC/2); where: ω = 2πf
The ferranti effect voltage rise
factor is the reciprocal of the term (1 – ω

^{2}l^{2}LC/2), which should be greater than one. For example, this factor could be as high as 1.16 for a 500 km line. As a result, the receiving-end voltage becomes higher than the sending-end (i.e. V_{R}> V_{S}). Moreover, it is clear from the equation that the voltage rise factor is proportional to the square of the line length, and consequently the inductance and capacitance.
## 4 comments:

Good!

I dont understand where I squared comes from in the reciprocal term?

Z=ω.l.L

Y=ω.l.C

where l is the line length, L and C are inductance and capacitance per unit length.

Therefore,

YZ/2 = ω2.l2.L.C/2

Good Explaination

Thanks Very much

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