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The performance of a transformer should be judged by all-day efficiency and this efficiency is always less than the commercial efficiency

The ordinary or commercial efficiency of a transformer is given by the ratio –(output in watts)/(Input in watts). But there are certain type of transformer whose performance can not judged by this efficiency. Distribution transformers have their primaries energized all the 24 hours,although their secondaries supply little or no load most of the time during the day except the   housing lighting period.It means that whereas core loss occurs only when the transformers are loaded.This is why the transformer’s all day efficiency is always less than the commercial efficiency.

What is meant by back e.m.f?

Back e.m.f.: When the motor armature rotates,the conductors also rotate and hence cut the flux.In accordance with the laws of electromagnetic induction, e.m.f.is induced in them whose direction,as found by Fleming , s Right-hand Rule,is in opposition to the applied voltage. Because of its opposite direction, it is referred to as counter e.m.f. or back e.m.f.E b.                                                                                                                                                                                Fig:1 Prove that the gross power developed by a DC motor is maximum when the back e.m.f. (E b ) equals to the half of the applied voltage(V):  Back e.m.f.E b =V t -I a R a   Where V t =applied terminal voltage,     I a =armature current,      R a =armature resistance                                                                                                                                                            T

Two shunt generators connected in parallel is inherently stable

Justify the statement: Stability of operation is an important factor in the parallel operation of generators. Shunt generators share the load proportionately. Should anything to occur to upset this proportionate division of load, it would be desirable for the generators to restore automatically the division of load. Assuming that the speed governor of one of the motors driving the generators is such that it will not hold the speed constant and because of this condition the speed of generator 2 will increase. The increase in speed will produce a large r induced voltage in generator 2,which in turn will cause generator 2 to deliver more load. Of course ,if generator 2 delivers more of the load, generator 1 delivers less. This larger load upon generator 2 will cause more current to flow in the armature of generator 2,producing a larger I a R a drop, a larger drop due to armature reaction, and a larger reduction in armature current, all of which tends to oppose the increas

How load is shifted from one generator to the other?

With two generators connected in parallel, adjustments can be made for one generator to deliver all or any part of the load. If the resistance of the field rheostat of a generator is lowered, the resulting increase in field current will cause the generated voltage to increase. This increase in generated voltage causes the load delivered by the generator to increase. Naturally if one generator takes more load, the other generator drops some load. Hence by adjusting the field rheostats of the generators, the load can be shifted from one generator to the other or divided in any proportion desired.