PSC motor behavior when changing run capacitor value.

[xchcui]

Hi.

I was browsing and read alot of articles and forums around the web
and i didn't find a full explanation about the behavior of the current
and the voltage in the PSC motor.
Here are two situations that i would like to understand(and please,
correct me if i wrong about something).
1)if i will connect an higher value capacitor than the original rated,
the impedance in the series capacitor-start winding branch will be lower,so
i assume that the current and the torque,when the motor will be turn-on,will be higher
by using this higher value cap and the current will be also higher(of course,
compare to the original cap value)during the normal operation of the motor.
In that case,what will happen in the main winding?
Does the current will be the same/higher/lower,when we start the motor?
Does the current will be the same/higher/lower,during the normal operation of the motor?
2)In the other hand,if i will connect a lower value capacitor than the original rated,the impedance
in the series capacitor-start winding branch will be higher,so i assume,that the
current and the torque,when the motor will be turn-on,will be lower and
the current will be lower,also,during the normal operation of the motor.(along the start winding).
As in the first question:what will happen in the main winding?
Does the current will be the same/higher/lower,when we start the motor?
Does the current will be the same/higher/lower,during the normal operation of the motor?

Thanks in advance.

 

[waross]

First effect:
Within reason, increasing the capacitor value will probably increase the starting torque. There may be exceptions, it depends on the motor design.
Increasing the capacitor value will probably increase the running torque. There may be exceptions, it depends on the motor design.
Second effect:
Changing the value of the capacitor may upset the division of current between the windings. At rated load, one winding may be overloaded even though the motor is not nominally overloaded.
Added thoughts:
The OEM installed capacitor may not be the optimum value.
Motors seldom run at exactly 100% of their rated load. At less than full load you may be able to overload one winding slightly without motor burnout.

Anecdote:
Years ago I was faced with the challenge of opening and closing two large and heavy steel, rolling garage doors.
3/4 HP openers were ordered. 1/2 HP openers were supplied.
Note; These were capacitor start motors and the capacitor is only in circuit during starting but the capacitor action during starting is similar to a PSC motor.
The 1/2 HP motors did not have enough breakaway torque to start the doors moving. If we started the doors moving by nudging them with a pry bar just as the motor was energized, the motors could easily overcome the rolling friction and running current was much less than full load current.
We doubled the value of the starting capacitors and the motors easily developed enough torque to overcome the breakaway friction and the gates have been working well for years now.

The relationships between the torque developed by each winding, and the current in each winding are interacting with each other and as the speed increases from locked rotor and the slip frequency drops, the current sharing between the windings may be changing.
The answer to your question may be more grey scale than black and white.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jraef]

Bill is right, messing with the cap rating will change the starting torque, but affect the running efficiency. I've heard tales of someone doubling the capacitance and getting 200% torque, but dropping the running efficiency to just 18%! There is basically a compromise in the design of a PSC motor between starting torque and running efficiency, but is done to increase reliability by eliminating the need for a centrifugal switch (or Potential Relay) to take out the Start capacitor. So there are still two caps, but they are in the circuit the entire time. Messing with the Start cap can get you in trouble really fast.

If you need more starting torque, change to a Cap Start Induction Run (CSIR) or Cap Start Cap Run motor (CSCR) which will have the centrifugal switch. CSIR is the highest starting torque, but low efficieny, CSCR has slightly less starting torque by better efficiency. PSC compromises both aspects.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

 

[xchcui]

Thanks for your response,but you didn't actally answer my questions.
My main interset is in the current aspect.
If i connect larger capacitor(+10%):what will be the current in the run winding
and what will be the current in the start winding at the moment that i start the motor
and what will be the current in the run winding and what will be the current in the start winding at the normal operation of the motor.(compare to the oem capacitor).

And in the other hand,if i connect smaller capacitor(-10%):what will be the current in the run winding and what will be the current in the start winding at the moment that i start the motor
and,also,what the current will be(on the start and run winding)at the normal operation of the motor.(compare to the oem capacitor).

For example:
1)i know that if i install +10% larger capacitor,the current on the start winding will be higher when i will start the motor,but will the run winding,also,be higher at that starting moments?(compare to oem capacitor).
At the normal operation of the motor,with this 10% higher capacitor,the start winding current will be higher,but will the run winding,also,be higher?(compare to the oem capacitor).

2)If i install -10% smaller capacitor,the current on the start winding will be lower when i will start the motor,but will the run winding,also,be lower at that starting moments?(compare to the oem capacitor).
And what happens with the currents at the run winding and start winding at the normal operation of the motor with that -10% smaller capacitor,compare to the oem capacitor?

Thanks.

 

[waross]

The answer to your question may be more grey scale than black and white.
Try it and see.
Note: A motor burnout may be an indication that you have exceeded the safe operating envelope of the motor.
Your next question may be about strange current readings.
When a induction motor is started there is typically a very high but short transient current, followed by the normal starting surge. To make life more interesting, the value of the transient may vary greatly depending on the 'point on wave' that the motor is energized.
The transient may cause fast acting fuses or instantaneous breakers to clear.
The transient has little effect on the motor heating because of the very short nature of the transient.
The surge that follows the transient is what causes the I[sup]2[/sup]R losses in the motor and is most important.
Ideally the heating will be influenced by the RMS of the transient plus the surge, but the transient may be safely ignored in the case of small motors.
The early electronic clamp meters would lock onto the transient value and were of little use in trouble shooting.
Modern clamp meters may have algorithms that more accurately indicate the value of the surge. It's up to you to check your meter specs to see if your meter is suitable.
As the frequency changes, the impedance changes. As the impedance changes the current changes.
As the motor speed changes the rotor frequency changes over a range ( for a 50 Hz system) of 50 Hz at locked rotor to as low as less than 5 Hz at full load.
There may also be interaction between the running winding and the starting winding.
The answer to your question may be more grey scale than black and white.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[xchcui]

Waross,your explanations are very informative and i learn alot from them,
But i don't understand why in my case,you say that the answer to my question is grey.
There is only one variable,the capacitor,which in one time is 10% higher and
in the other time is 10% lower than the oem cap.
So,i am pretty sure that only one result can happen,am i wrong?
When the cap will be higher the current should be higher and the motor will run hotter
and when the cap is smaller,the current should be lower and the motor will run cooler.
What do you think?Is it right?

 

[waross]

The current drawn by the windings depends on several variables.
Changing the capacitor value may change several other variables.
There is a physical angle between the start and the run windings.
There is also an electrical angle between the start and the run windings due to the action of the capacitor.
In a well designed motor the physical angle and the electrical angle are close to the same.
Change the capacitor value and you change the electrical angle.
The winding with the most advanced angle (with allowance for the physical displacement between the windings) will hog the load.
Now changing the capacitor value changes the primary impedance of the winding. The angle also changes and the reflected impedance from the rotor may also change.
Now add that the capacitor value chosen by the OEM may be the most economical rather than the most efficient.
If you are serious about finding the black and white answer, sign up for a course in electrical engineering. You may have to progress to the Masters level before your are able to calculate the various interactions between the components of a motor.

and when the cap is smaller,the current should be lower and the motor will run cooler.

Maybe but probably not. If the other winding has to do 10% more than it's share of the work, the I[sup]2[/sup]R losses will be +21%. That will probably run hotter.



Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[xchcui]

If the other winding has to do 10% more than it's share of the work, the I2R losses will be +21%. That will probably run hotter.

Warros,can you expand on this case?
When you mentioned the power losses(I2R),Do you mean that the main winding will consume more current in order to balance the motor?

 

[waross]

Roughly:
110% work ≈ 110% current. 110% current[sup]2[/sup] =121%, R is constant.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[xchcui]

Thanks for your help