 ### Network # Improving power factor

#### cells

Aug 18, 2013
1
Hello. I have a devise (an ice cream machine but just think of it as a big motor) which uses 1100 watts at 240 volts. So it should draw 4.6 amps if the power factor was good however the manufacturers seem to have built it in a way where the power factor is terrible. It actually draws closer to 8.5 amps and has a power factor of around 0.55

This high amp draw is causing me problems as I have one line powering three of these devises so the amp draw is around 24 amps while the line has a trip at I think 20 amps so it is tripping. As such I can only use two of these machines at once.

So I have two choices. Install an additional new line (cost and time) or try to improve the very poor PF of the devises.

So any ideas on how I can improve the PF without opening up the machines and soldering on an appropriate capacitor? I am not confident to do that as I am no electrician and only know a little about electronics from A-level classes a long time ago.

Is there any off the shelf solution which are plug and play for this type of problem?

Any help appreciated

#### john monks

Mar 9, 2012
693
I calculated that you need about 79.2uF in parallel with each motor. This is assuming you have 60htz. The voltage rating needs to be in excess of 320 volts.
If you place such a capacitor in parallel with each motor I would expect your switch would have a rather big surge current therefor may not be practical.

I would be surprised if your PF is as bad as you say. Maybe you can check it again.

#### Rleo6965

Jan 22, 2012
585
AC outlet with zero crossing detector might help.

#### Y2KEDDIE

Sep 23, 2012
259
Power factor question

I calculated that you need about 79.2uF in parallel with each motor. This is assuming you have 60htz. The voltage rating needs to be in excess of 320 volts.
If you place such a capacitor in parallel with each motor I would expect your switch would have a rather big surge current therefor may not be practical.

I would be surprised if your PF is as bad as you say. Maybe you can check it again.

John,

I'm curious as to how you came up with the numbers, what formula did you use?

Eddie

As far as his initial surge, I would suggest bringing each motor on line in a delayed sequence. Next I would suggest soft-start motor starters.

#### john monks

Mar 9, 2012
693
The real current you machine draws is P = 1100 watts divided by 230 volts or 4.58
amps.
The current you are measuring is 8.5 amps. Some of this is real and some is
imaginary in terms of power.
The real current is at a phase angle of zero degrees.
The imaginary current is at 90 degrees inductive because this is a motor.
So you have a right triangle with one leg at 4.58 amps and a hypotenuse of 8.5
amps.
So the imaginary current is going to be the square root of (8.5^2-4.58^2) or 7.16
amps.
This current is inductive so you need to place a capacitor in parallel with the
motor that has the opposite current or current that is 180 degrees out of phase
with the imaginary current.
That is the capacitive reactance must be equal and 180 degrees out from the
inductive reactance of the motor.
The inductive reactance of the motor is 230 volts divided by the imaginary current
of 7.16 amps.
The minimum voltage of the capacitor will be the square root of 2 times the line voltage. That is 230 * 1.414 = 325 volts.
And that is 33.5 ohms.
The capacitance of the capacitor is 1/(2*pi*F*Xc) or 1/(6.28*60Hz*33.5ohms) or

79.2uF
Here is a short program I wrote to verify my math.

10 I=4.583334
20 II=SQR(8.5^2-I^2)
30 PRINT"Imag current ="II
40 XC=240/II
50 PRINT"Xc ="XC
60 C=1/(6.28*60*XC)
70 PRINT C

Last edited by a moderator:

Sep 23, 2012
259
Thanks John.

#### Y2KEDDIE

Sep 23, 2012
259
I was reading in a text that when you put capacitors in series with a start winding you want to avoid resonance because of high voltage across the said components. I realise this is series resonance and purpose is for starting.

Placing equal and opposite reactances in parrallel is used in the run windings circuit. I would think one would have the same problem with hih voltages devloped.

Am I'm missing something here? What are your thougts?

#### john monks

Mar 9, 2012
693
Yes. When an inductor and a capacitor of equal absolute reactances are in series with some input AC voltage the total reactance approaches 0 ohms and the current heads for infinity. This causes high voltages to occur across the individual components and can cause damage. The same components in a parallel circuit will only have the input voltage across both components.
Keep in mind that the current leads to voltage in a capacitor and the current lags or follows the voltage in an inductor. So in a series circuit the total reactance approaches 0 ohms.
Now in your ice-cream machines the situation is a little more complicated because the motors have the characteristics of a resistor and an inductor and therefore more complicated math is required such and the pathagorean theorem or trigonometry.
If you are confused let me know where and I will explain in greater detail.

#### Y2KEDDIE

Sep 23, 2012
259
Thank you for your explanation. I like hearing different explanations from different perspectives. I totally agree and like your explanation and math. I wonder if we can trust the nameplate data (1100 W @ 240). Obviously he's drawing 8.5 A (measured) , I wonder is that continuous under load or peak in-rush. It sounds mighty inefficient. (Maybe there are run capacitors but they are defective).

#### john monks

Mar 9, 2012
693
Capacitors in motors do go bad because they are typically two electrolytic capacitors wired back to back and they dry out especially if they are old. It might be worth wile replacing them.

#### Y2KEDDIE

Sep 23, 2012
259
John,

When you calculated the Capacitance needed to bring the power factor to unity, was that makining an assumption there were no capacitance in the circuit to start with?

I'm thinking what if there were some capacitance but the value reduced to a low level. Would it be best to disconnect any capacitor attached to the windings and measure full load current without? Would the motor eevn run without some capacitance?

Thanks for sharing,

Eddie

#### Y2KEDDIE

Sep 23, 2012
259
I believe I read some other post where it was suggested placing differnt values of capacitance in parallel and measring the voltage developed until it peaked.

Possibly another way would put a varialble resistor in series and measuring alternately across the winding and resistor until equal votage is achieved, then measure the resistance which would be equal to the reactance.

I don't know if either these are practical. I would think the motor would have to be running under load.

#### john monks

Mar 9, 2012
693
I looked at the rated power and divided that by the line voltage to figure out the current at 0 degrees relative to the voltage. Then from there, assuming that the motor looked inductive, I figured out the current that was shifted by 90 degrees from the voltage by using the current that you posted and using the pythagorean theorem. So any capacitance that already existed in the circuit would automatically be cancelled out.

The capacitor in the motor may be low and that may be the problem.

I am not sure but I assume that if you disconnected all capacitors from the motor it would not start, it would only buzz. I would know for sure if I was sure how the motor was wired.

The motor would run without a capacitor but you would have to mechanically start the motor turning and then the motor would not be very efficient.

Motors have the characteristics of a resistor and an inductor. Ideally a motor would only look like a resistor so that the input current would be as low as possible.

In placing a capacitor in parallel with a motor with a motor you would be looking for when the current from the line was at its minimum. Remember that in an inductor the current lags and a capacitor the current leads. So with the two in parallel the current would be going one way in the inductor and the exact opposite way in a capacitor. So if the currents were equal the current would null to about zero. So now the only thing left in a motor is the current that is in phase with the voltage. And this current times the voltage is the true power that the motor is drawing from the line.

If you wanted to you could place a very large 0.1ohm resistor in series with the motor while measuring the voltage across that resistor while trying various capacitors across the motor and looking for the minimum voltage across the resistor. You might be able to simply measure the voltage across one of the two wires feeding the motor with a voltmeter and trying various capacitors. Maybe this would be the most practical approach.

I just have to add one thing to all this. Understanding these kinds of things are not that difficult. It is not vary wise to hung up on the math. You need to understand what the motor is doing and why it is doing what it does and understanding the characteristics of capacitors. The trick is you want the voltage and the current to be feeding the circuit at the same time. That is you want the phase to be the same so you don't blow circuit breakers and cause your utility problems. Once you understand the physics the mathematics becomes obvious.

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