A feeder booster is an important equipment for electrical energy transmission lines, especially for long feeder of transmission lines where the voltage dropped a the end point due to some technical limitation of transmission lines that not easy to illiminate as required.

Basically a booster is a generator whose function is to add to or inject into a circuit, a certain voltage that is sufficient to compensate for the IR drop in the feeders etc. IR drop means the voltage drop due to the resistance (R) of transmission conductor when flowing current (I) through the conductors.

In an electricl transmission network system it not possible to keep the feeder length equal or uniform load distribution for all; for that case load and voltage becomes unbalance in the system network. In a d.c. system, it may sometimes happen that a certain feeder is much longer as compared to others and the power supplied by it is also larger. In that case, the voltage drop in this particular feeder will exceed the allowable drop of 6% from the declared voltage. This can be remedied in two ways:

- by increasing the cross-section of the feeder conductor, so that its resistance and hence IR drop is decreased;
- or by increasing the voltage of the station bus-bars.

The difficulties for both cases that the increasing the conductor means the cost of transmission lines will increase few times and it is not be economy for utility company without increasing the per unit electricity cost; on the other hand the bas-bar voltage increasing is not easy technically for a certain feeder.

The second method is not practicable because it will disturb the voltage of other feeders, whereas the first method will involve a large initial investment towards the cost of increased conductor material.

To avoid all these difficulties, the usual practice is to install a booster in series with this longer feeder as shown in Figure above. Since it is used for compensating drop in a feeder, it is known as feeder booster. It is a (series) generator connected in series with the feeder and driven at a constant speed by a shunt-motor working from the bus-bars.

The drop in a feeder is proportional to the load current, hence the voltage injected into the feeder by the booster must also be proportional to the load current, if exact compensation is required. In other words, the booster must work on the straight or linear portion of its voltage characteristic.

Example:A 2-wire system has the voltage at the supply end maintained at 500. The line is 3 km long. If the full-load current is 120 A, what must be the booster voltage and output in order that the far end voltage may also be 500 V.Take the resistance of the cable at the working temperature as 0.5 ohm/kilometre.

Solution:Total resistance of the line is = 0.5 × 3 = 1.5 ΩFull-load drop in the line is = 1.5 × 120 = 180 VHence, the terminal potential difference of the booster is 180 V (i.e. 180/120 = 1.5 volt per ampere of line current).Booster-output = 120 × 180/1000 = 21.6 kW

A Feeder Booster add voltage to the feeder and compensate the voltage drop. Hence it increase the efficiency and reliability of the system. That is why it is one of the important device of power system.

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