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Class 2 (2nd Engineer) MET 📅 Oct 2024

Exam Question

(a) Explain the process of voltage buildup in a self-excited shunt generator. (6)

(b) A shunt-wound generator has a magnetisation-curve given by the figures below. The total resistance in the field circuit is 20 Ohm and the armature resistance is 0.02 Ohm.

With the machine on load, estimate the e.m.f. generated and the armature current when the terminal voltage of the machine is 140V.

(10)

Field current (i) amperes 1.2 2.8 5.0 7.0 7.7 9.0 11.0

Generated e.m.f. (e)_ Volts 46 88 126 149 154 162 168

Exam question diagram

Reference Answer

(a) The basic principles of a self-excited generator:
When the generator is first started, there is a small amount of residual magnetism in the rotor winding. This residual magnetism is due to the fact that the rotor winding is made of ferromagnetic material, which retains a small amount of magnetism even when there is no current flowing through it. The prime mover (usually a diesel engine or a turbine) rotates the rotor of the generator. As the rotor rotates, the residual magnetism in the rotor winding induces an electromotive force (EMF) in the stator winding. The EMF induced in the stator winding is proportional to the speed of rotation of the rotor and the strength of the residual magnetism.
The EMF induced in the stator winding is fed to the automatic voltage regulator (AVR). The AVR rectifies the AC voltage from the stator winding and uses it to excite the rotor winding. The DC current from the AVR flows through the rotor winding, creating a magnetic field. This magnetic field interacts with the residual magnetism in the rotor winding, creating a stronger magnetic field. The stronger magnetic field induces a higher EMF in the stator winding.
The process of self-excitation continues until the generator reaches its rated voltage. At this point, the AVR reduces the excitation current to maintain the generator voltage at the desired level.
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