Starting power factor of induction motor

At starting, most of the current is magnetising, hence low pf. As the load increases, pf improves. Please check out this link : http://electojects.com/motors/tesla-induction-motors-4.htm

Good site sridhar.Thanks for sharing Best Regards

At very start motor is so to say "overloaded" - overcoming the inertia of the rotor and the load on shaft - compared to its being "lunloaded" to "loaded" to even "fully loaded" while running. Actually, at the start the motor consumes the locked rotor current - the active power mostly, hence PF at start should be the highest, not low.

Be that as it may, the starting pf is low....

Check this link : http://www.lmphotonics.com/m_control.htm

Simplify the situation and it will become a little more clear. Take a coil and connect it to a battery. At t=0+ all the energy is going to create the magnetic field (e.g. you have a purely reactive load).

And how many cycles it takes for AC to create the field in the stator windings and rotor bars? 10? 50? But if it takes for the motor to be speeded up to the rated speed several seconds - and what, all this time the AC does no active work on your view - it is just creating the magnetic field with its low PF - as pertains to reactive powers?

It should be specified what period of time is meant by "starting."

Let's go back to the coil, but connected to AC this time... For the sake of discussion, let's take it as an idea coil (ie purely inductive).

In this case even though there is current and voltage this power is all reactive (ie a pf of 0, because there is no real power) and is only alternately absorbed and returned to the source. So yes you can measure volts and amps, but in the ideal case there are no watts and this is all regardless of the number of cycles.

Now back to the motor, again at t=0+ you only have the reactive load! Here's another way to look at it... Initially ALL components have NO current flow (thus no real, apparent, or reactive power). In an inductive load, the current lags the voltage (note in the ideal coil its a 90 degree shift), until there is current flow no voltage can be induced anywhere to provide the counter emf (which is where you're real work comes in). So I would expect the pf to be lowest until the current begins to flow (about a quarter cycle)

Anyone who can explain it better please chime in...

Actually just read the link above... here's an excerpt...

Power factor (PF) varies considerably with the motor mechanical load (Figure below). An unloaded motor is analogous to a transformer with no resistive load on the secondary. Little resistance is reflected from the secondary (rotor) to the primary (stator). Thus the power line sees a reactive load, as low as 10% PF. As the rotor is loaded an increasing resistive component is reflected from rotor to stator, increasing the power factor.

So I would expect the pf to be lowest until the current begins to flow (about a quarter cycle)

Will the picture I'm going to describe below be far-fetched on your view, I dont know.

Imagine starting up an unloaded, 2 poles 50 Hz motor having, for instance, rated PF 0.9. At very start its PF is lowest - within the first cycle of the AC wave (according to your citation above, with which I will agree) - 0.1 (motor rotation is 0 rpm) , but then, rises rapidly to 0,9 (let s say now the speed is 1 rpm), is 0.8 at 100 rpm, 0.7 at 1000 rpm, 0.6 at 2000 rpm, 0.5 at 2400 rpm, 0.3 at 2950 rpm. Loading up the motor, PF will now increase, and fully loaded, will reach 0.9 at 2850 rpm.

I would agree with your statement.

Because at zero speed, although it is taking current, it isn't doing any work!

It seemed we have agreed on a winding-resistance - the current is (doing the work of intensive) heating up the coil's wire (while magnetic saturation is low).

This is only one explanation I can find why inrush is low if switching on occurs in the pick of the sine of the wave - flux comes quicker in the coil - compared with a switching the coil on at the beginning of the rise of the wave (0 V point).

Dear Mr.polerz,

The Induction Motor CURRENT profile follows the CIRCLE DIAGRAM. It is NOT like DC Motor, where the current has Linear Relation with Load, in the form of y = mX +C

While starting the motor, the current drawn may go as high as 4 to 6 time of full load current,and the phase angle between the Voltage and the current will be very large as high as 72 Deg. and hence the Power Factor cos72 = 0.30.

Once the motor is started, the speed is increasing and reaches as high as 96 to 97% of Rated Speed, simultaneously current will also reduce to No-Load Current Level, but follow the Circle Diagram (not Linear) and again phase angle between the Voltage and the Current will be larger and hence again low power factor.

Pl. refer a Standard Text Book, how a CIRCLE DIAGRAM is to be drawn.

Thanks.

DHAYANANDHAN.S

Just 2 quotations.

"The magnetizing current is independent of the load"

"Induction motor stall reduces the motor power factor"