Help with Magnetic Stirrer

please anyone help me regarding the mechanical calculations for selecting a motor for magnetic stirrer.

1. The forces acting in the magnetic stirrer rod while in motion
2. from the force applied in the magnetic stirrer rod - how to select the motor based on torque(power)?

i am using two similar magnetic rods one affixed in motor and another inside liquid.
Planned to use a CPU fan motor. but i need calculations to prove the selection.

Have you considered frictional losses from the magnet in the vessel and viscosity of whatever is being stirred? Thus, there is no one answer based on the information you have given.

John

I used a motor from a cassette tape deck.
The mechanical load on the motor is more likely to be friction losses, than load from the stirrer. Due to the distance apart of the magnets, if the stirrer load was too great, the stirrer bar would stall.
I made up a stirrer with a tape deck motor because these have an inbuilt dc speed controller. They also have an internal adjustment pot and this can be moved to an outside control to give a variable speed. I find this important because with some liquids, the vortex is too great. For magnet, I got one out of a single cylinder petrol engine magneto. loudspeaker magnet would be OK too b ut you probably need to think about pole pieces.
good luck.
Too easy.

Could u please suggest me a link to calculate frictional losses in vessel by magnetic rod

Is this something you are trying to design for school or just something you want to get to work?

If it is the latter, then consider that the stirrer motor needs to overcome its own turning friction or it wouldn't turn. I assumed that was a given. It is then coupled to the bar in the vessel ("stirred bar") by magnetic force. Forces acting on the stirred bar will impede its spinning and require power from the stirrer. At some point, the forces on the stirred bar may exceed the strength of the magnetic couple and the stirrer will continue to turn, but the stirred bar will not.

So, as a rough estimate, you might say the motor needs to be at least as powerful as the magnetic couple. Take the two magnets you intend on using, separate them by an appropriate distance to represent what you will be doing, and see how much torque it takes to turn one while the other is stationary. So basically (first approximation), it is how much force does it take to separate two magnets of given strengths that are a fixed distance apart.

As a simple experiment, you could put a magnet on your PC fan and then take another magnet and see how close you can get without stopping the fan, or more appropriately, prevent the PC fan from starting. Obviously, there are other factors, such as the effect of what else is between the stirrer and the stirred magnets.

A practical design will also have to consider other factors besides motor power. For example egg-shaped magnets behave differently than rectangular magnets in the stirred vessel. Round-bottom vessels stir differently than flat-bottom vessels. Erlenmeyer flasks stir differently than beakers. Magnet size compared to the size of the vessel is yet another factor. Some configurations produce a more stable vortex than others. Variable speed control is essential.

John

Also, you would need to consider the speed of rotation of the bar in the liquid, since the drag on the bar will be "somewhat" proportional to the square of it's angular velocity...at least until the liquid starts moving. Many, many variables...but none critical, since you're just trying to stir a liquid.

Ken

i have to make it. i dunno how to write for report?? motor calculations. i am taking 2 cylindrical magnets. motor should be controlled for 5 rpm and 10 rpm. based ont his i have to analyse motor selection.

For those very slow, controllable speeds (never seen any commercials stirrers run that slow) you will need either a geared or stepper motor. With either one the torque will be pretty high. So, I would not worry about that.

In all the stirrers I've repaired, a strong, flattened bar or U-shaped magnet was attached to the motor shaft. The stirring bars were all just iron/steel rods embedded in a Teflon shell, not magnets.

But...what are the requirements of the report? Do you need to show calculations in magnetic coupling?...fluid dynamics?...motor control?...? Or do you just need to build a working model and explain how it works? Are we talking junior high?... high school?... university?

Ken

KMoffett said:

For those very slow, controllable speeds (never seen any commercials stirrers run that slow) you will need either a geared or stepper motor. With either one the torque will be pretty high. So, I would not worry about that.

Click to expand...

Agree. For very low speeds, an overhead stirrer is usually used, in my experience.

In all the stirrers I've repaired, a strong, flattened bar or U-shaped magnet was attached to the motor shaft. The stirring bars were all just iron/steel rods embedded in a Teflon shell, not magnets.

Click to expand...

Every commercial stir bar I have seen in the past 50+ years was magnetic. There were occasions when Teflon wouldn't work, and I had to make glass encased stir bars. After sealing the glass, I tried to re-magnitize them, but they typically were very weak. They still worked.

John

yes i am a university student. i have to prove that the motor i am planning to select has the force to rotate the magnetic rod inside the fluid. (along with fluid).
how do proceed?

Sounds like you are asking for the answer to a problem involving drag coefficients with several unknowns: size and shape of stirrer, density and viscosity of fluid, and speed. The drag changes with speed and geometry of the stirrer, dense fluids require more torque, etc. The smartest way to proceed would seem to be with an experiment.

Use jpanhault's method to determine the inch/oz needed to break the coupling. The motor magnet fixed, and the stirrer bar centered and attached to a pivot, the intended distance from the magnet. A lever arm perpendicular to the pivot. Measure the force at the arm's length necessary to break the coupling...you now have the maximum torque it takes to move the stir bar in the most viscous liquid. The fluid dynamics of the drag of the fluid on the bar, and the container on the fluid, to get the minimum torque are beyond me.

Ken

seems to me you have had some very useful advice.
All the advice seems to be; "suck it and see"
Sometimes in life we have a problem which is amenable to calculation; sometimes not. In this case the problem is not related to the power source but is related to the torque required to spin a magnet of unknown dimensions in a fluid of unknown viscosity in a magnetic field of unknown strength.
There is NO answer.
Your suggestion to use a computer fan is interesting in that there is NO information on its' rotational speed. My understanding of these fans is that they spin at as high a speed as possible. Such a fan is UNSUITABLE for a strirrer. You need to be able to ADJUST the speed. The required stirring speed is dependent on the liquid being stirred AND the conditions.
I suggested a cassette tape deck motor and this has a gearing arrangement and adjustable speed.
Your fan motor would be useless. You need to be able to adjust the speed for start. Can you do this with a brushless fan.
If you want to assess the problem you need to examine the stall speed against viscosity and produce a series of points of stall speed. In a practical situation to produce a device, you need to concentrate on maximising flux density in the gap between the rotating magnet and the stirrer. Fundamental calculations of the kind you advocate are a waste of time. You need to establish by experiment the performance of your system and then go back and formalise your system so that in a manufacturing environment you can establish tolerances for the magnets and the separation distances.
One lecturer I heard of, when conducting a mechanical engineering course, was discussing strength of materials for drive shafts. His lecture started; "we calculate for this shaft we need a diameter of 25 mm to transmit the required torque; so we take a shaft of 50mm".
This sort of states the way engineering decisions are taken.
enough said!.