Poof power factor or what!

[Surefire]

I have a small 3 phase motor and decided to do a little power factor calculation. The motor is rated as follows: 0.18KW, delta amps=1.01A, Star amps=0.58A, full load power factor=0.76 and full load efficiency=58%.

Now power factor can be calculated as follows: Power (watts) / root3 x volts x amps x efficiency

For star connection, the PF comes out at 0.86. If the math is done when connected in delta, the PF comes out at 0.44!

To check this is true I displayed the current (using a current proble) and voltage on a scope and the shift between the voltage and the current was indeed big at 4mS on the same phase (bare in mind that 5mS is 90 degrees). 4mS equates to a lagging current of around 72 degrees. The cosine of 72 degrees gives a power factor of 0.3! (These values were recorded when the motor was running un-loaded)

Is this correct or am I doing something wrong. If it is correct it means that running a motor in delta is incredibly inefficient and will almost certainly need some kind of power factor correction.

Note: I just mechanically loaded up the motor and the shift between voltage and current reduced 3mS which give a PF of 0.58 so it just goes to show that a motor on no load has a poorer power factor than a motor on full load.

 

[Marvo]

If you used an oscilloscope and it showed this lag then it must be accurate. Very small motors are notoriously poor PF plus if the motor was without load it would be even worse. The motor efficiency increases the nearer you load it to its max.

 

[Knobhead]

Is this correct or am I doing something wrong. If it is correct it means that running a motor in delta is incredibly inefficient and will almost certainly need some kind of power factor correction.

Note: I just mechanically loaded up the motor and the shift between voltage and current reduced 3mS which give a PF of 0.58 so it just goes to show that a motor on no load has a poorer power factor than a motor on full load.

The only reliable way to load test a motor is using a dynamometer.

As for PF correction, is it worth it? Do you pay for KW or KVA?

 

[Lucien Nunes]

Oops, slight error in your calcs. The power factor does not depend on whether the motor is connected in star or delta, those are just two different wiring options to match the motor phase voltage to the voltages available from your supply. The product of line voltage and line current in either case will be the same, hence the apparent power will be the same. pf = real power /apparent power.

Yes, when unloaded, induction motor power factor falls dramatically, but this is not a black mark against the motor because the real power falls too. If the motor were perfectly efficient and lossless, when unloaded it would still draw current at zero pf, because a certain amount of wattless current (aka magnetising current) must flow all the time it is operation to provide the rotating magnetic field. This is exactly like a lossless inductor across the supply, which will pass a wattless current equal to V/Xl. If the motor is highly efficient this fixed wattless current will dwarf the loss current and the pf will be very low indeed. If the motor is inefficient and has excessive iron loss, for example, the real component of the current will be larger and the pf higher.

Therefore, in the extreme case of being unloaded, low pf is not always a bad sign. Note that an unloaded motor also has an efficiency of zero! Where the problem lies with low pf on small motors is that when you have large numbers of them idling, you can't ignore their load current just because the total work done is negligible. The sum of their magnetising currents in a large plant where machinery often runs light can impact the overall pf. Clearly, a motor with a lower magnetising current and higher pf improves the efficiency / capacity of the wiring, without improving the efficiency of the motor itself.

 

[telectrix]

read the thread title and thought it was about that daley bloke.

 

[danzor]

He looks sexy in trunks