i have to submit this assignment tomorrow ... i am not very good at electrical machines .... can anybody help me out with this ... and tell me in what ways can we make the power factor of a machine equal to one ... i have to tell a number of ways and not just one way .. and i am clueless about the whole thing
thanks in advance
You are talking of asynchronous AC motors (or generators)?
They have an inductive impedance and a 0.6 to 0.7 power factor (cos phi) typically at nominal load.
- capacitor banks can be used for reactive power compensation. They can be included with the machine (individual compensation) or placed at a plant's power distribution (common compensation). The latter mostly uses automatic adaption by a power factor controller.
- electronical power circuits (involving PWM operated IGBT) can be used for compensation of reactive power and harmonics at the same time
- synchronous machines can be used to generate respecively consume reactive power by varying the excitation current.
its nothing like that in particular ..... we have any machine which has a stator ( field poles) and rotor (armature) and now we want to make its power factor one ... this is what the task is
That's the problem with unclear questions! Is it a DC or synchronous AC machine? Armature would generally mean DC machine, it's only used from AC supply for small tools. Power factor isn't a topic here.
we have to write about both Synchronous Motor (AC) & Asynchronous Induction Motor (AC) can you explain me the necessary steps in a little detail so i can understand them better ... or send me a related file or link
U can always use the capacitor banks to improve power factor, it should be connected with the circuit that supplies the magnetizing current to the motor.
And as far as the synchronous motor (it is rarely used to drive any heavy load) is concerned it is mostly used for power factor correction it self called "synchronous condenser", maybe if u r electrical engineering student u should have books of B.L.THERAJA
hi
for power factor correction,from Φ1 to Φ2 and P, you need Q reactive power.
Q=P[Tg(Φ1)-Tg(Φ2)]
where Φ1: target phase angle
Φ2: existing phase angle
P :active power
Q :reactive power