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When we increase the speed frequency will increase in genset.

what will happen when we increase the exitation?

how to improve power facor in generating station.( with out adding capacitor)

2007-09-27 03:31:32 · 2 answers · asked by sivaket2001 1 in Science & Mathematics Engineering

2 answers

Your question suggests that you are talking about Synchronous Machines and I'll assume so. There is a standard experiment in machines' labs called "V-curves of Synch. m/c's"...this V-curve is the relationship between p.f. and excitation...An underexcited machine is inductive and would have a lagging p.f....as excitation is increased, it approaches unity p.f. and finally crosses over to leading p.f., so leading p.f. implies overexcited machine. Therefore, load current at the same power level plotted against excitation current will have a minima and hence V-curve.

This is one of the strategies to control p.f....connect synch. machines to transmission lines and vary the excitation current to inject reactive power into power system (mostly we need capacitive compensation). Other methods involve active compensation using Power Electronic circuits that dynamically control the flow of reactive power. However, such systems are mostly used in conjunction with static compensation using passive elements (large capacitor banks).

2007-09-27 04:50:54 · answer #1 · answered by Saurabh T 2 · 0 1

When you increase the excitation, you put more energy into the excitation field so that more energy can be coupled from it into the load windings (assuming there is sufficient energy available to drive the rotor at the same field) Increasing the excitation field increases its available force and, as you should remember from physics, work done is equal to the integral of force timed distance.

Since power factor is largely controlled by the nature of the generators load, there really isn't much (other than adding reactance to the load to bring its phase angle closer to zero) one can do to change the power factor. If the load is largely reactive, it has to be balanced by a reactance of the opposite sign (so that the quadrature components of the total current add up to 0).

Doug

2007-09-27 03:58:16 · answer #2 · answered by doug_donaghue 7 · 0 1

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