Water Rimming inside a drum

[Tagger]

How do you calculate the point where water inside a copper drum begins to "rim" (attaches to the ID of the drum). The drum starts rotating at 180 RPM to a top speed of 1300 RPM. The ID of the drum is 15.5 inches and 28 inches long. Water enters/exits using a dual (inlet/outlet) rotating joint along the axis of rotation. It is a brake where the water cools the drum. Does the mass of the water relate to the speed of the drum (i.e. the more mass of water the faster the drum needs to rotate for the water to rim). I guess it is dependent on the friction coefficient inside, etc. The drum ID wall does become dirty over time, and there is no exact answer.
Thanks

 

[JKEngineer]

I don't have a real answer, but this is a problem addressed in the paper industry where they use horizontal, steam heated drums to dry the paper. Look at the literature or sites of dryer or syphon joint manufacturers. Johnson Rotary syphons are a major player.

They also use spoiler bars to break the rim at intermediate speeds.

Johnson has a video taken inside a drum as the speed was increased that I saw perhaps 15 years ago. Pretty neat. Jack M. Kleinfeld, P.E. Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering

http://www.KleinfeldTechnical.com

 

[quark]

If the axis of the water cylinder formed is vertical you can balance centriugal force with that of weight of water.

Roughly it should start at 238 rpm. Manufacturers of liquid ring vacuum pumps may help you in case of horizntal axis water cylinder.

 

[quark]

Correction to my post. The speed comes out to be 56 RPM. As there are lot of other parameters to consider, I will try to perfect it in my next post.

 

[Tagger]

Thanks for the help guys. Quark how did you come up with 56 RPM? Is it F=ma, where a is the vector component normal to the wall of the drum and it =r*omega^2? Where r is the radius of the ID of the drum and omega^2 is the angular velocity. In this case since we have a drum that is constantly accelerating you will have a tangnetial component =r*alpha where alpha=angular acceleration. Of course at consant velocity this conponent goes away. Friction between the wall and water comes into play...or does it have a significant role? And yes KJEngineer this is a tension brake that uses a copper drum to brake on, and water circulates inside the drum to cool it. The customer states that when the system is running, the water is not coming out until they stop the drum and water then gushes out. The inlet/outlet pipe is located along the axis of the horizontally mounted drum. I am wondering even if we redesign the syphon system where the syphon is somehow mounted to the wall the customer does not have enough pressure to overcome the centrifugal forces to exit through the axis of the drum.

 

[JKEngineer]

Tagger -
If you are trying to deal with getting the water out becuase it needs to circulate to provide the cooling, then I more strongly suggest looking to the suppliers for the paper industry. They have solved this problem pretty well and a long time ago, EXCEPT they are dealing with steam systems where the steam is supplied to the rotary joint and condensate with a little blow through is removed. That gives them a differential pressure and a differential density to work with that your system may not have available. Nevertheless, there are several syphon designs, both rotating and stationary, and spoiler bar designs, and a lot of experience in the industry.
For reference, if I remember correctly, typical dryer drum diameter is about 5-6 feet and surface speeds can be as high as 3500 ft/min, which is about 200 rpm. Jack M. Kleinfeld, P.E. Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering

http://www.KleinfeldTechnical.com

 

[quark]

Tagger!

You have to equate F = mg to F = mrw² where w is angular velocity. The assumption is that after the drum accelerates, it attains a constant speed. No slip condition between water layers is considered in this context. One more error is I have considered the ID of the drum, but you have to consider the ID of the water ring. If you know the quantity of water inside the drum you can do it by the equation V = 3.142(D1²-D2²H/4 where D1 is inside diameter of the water ring and D2 is inside diameter of the drum. If you consider inside diameter of the water ring, the RPM comes out to be higer (that is the right thing).

I couldn't get any equation on this topic but some books specified the same method.For circulation of water through the system, you better calculate the pressure drop across the water ring. Refer McCabe & Smith(Unit Operations in Chemical Engineering)or let me know. Centrifuges operate on the same principle (particularly continuous centrifuges).

Regards,