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 2 years ago
Ok here is a question I was working on today but I cannot seem to visualize it properly.
A circular disk of radius r is used in an evaporator and is rotated in a vertical plane. If it is to be partially submerged in the liquid so as to maximize the exposed wetted area of the disk, show that the center of the disk should be positioned at a height r / sqrt(1 + pi^2) above the surface of the liquid.
What is the "wetted" area of the disk? It cannot be the part of the disk that is submerged because then the maximum area would be the entire disk submerged.
 2 years ago
Ok here is a question I was working on today but I cannot seem to visualize it properly. A circular disk of radius r is used in an evaporator and is rotated in a vertical plane. If it is to be partially submerged in the liquid so as to maximize the exposed wetted area of the disk, show that the center of the disk should be positioned at a height r / sqrt(1 + pi^2) above the surface of the liquid. What is the "wetted" area of the disk? It cannot be the part of the disk that is submerged because then the maximum area would be the entire disk submerged.

This Question is Closed

Edutopia
 2 years ago
Best ResponseYou've already chosen the best response.0It may just be a poorly worded question...

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0That's true. I found a solution here (http://answers.yahoo.com/question/index;_ylt=AjP6YX4F6.ERQSVCCXH8zhHty6IX;_ylv=3?qid=20091128095615AA1AuqI&show=7#profileinfo91HhPVl0aa) but that did not really help much. This particular question was posted in a chapter on integration yet it is an optimization problem. I am not sure how integration can be used here.

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0So this disk is actually an annulus? Is this bit of information assumed in the question?

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0Oh wait..scratch that. I see.

Mathmuse
 2 years ago
Best ResponseYou've already chosen the best response.1vertical plane was the give away....up/down. just trying to work out the integration set up

Mathmuse
 2 years ago
Best ResponseYou've already chosen the best response.1the integration involved is to determine the area of the submerged disc section of the circle

Mathmuse
 2 years ago
Best ResponseYou've already chosen the best response.1although, to find a max/min you will have to set a Area equation and take the derivative.

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, I agree. I do not believe integration is required to solve this problem. Thank you so much for the drawing! I see now the "wetted" portion of the disk is the portion exposed to the air ready to evaporate. I understand now how the person from Yahoo Answers set this question up. I will modify your drawing and add the geometry involveddw:1356733401864:dw I supposed now my question is how do we write the area of the triangle AOB and the area of the inner circle in terms of theta (angle AOB)? If we can do this the rest is a matter of optimization. Attached is my version of the region that needs to be maximized.

Mathmuse
 2 years ago
Best ResponseYou've already chosen the best response.1see attached, hopefully its not too messy. I've broken the drawing into 4 parts each can be solved relatively easily. summed they form the area of the 'wetted' area. 1. Derivative will find angle of maximum wetness 2. integration may be used to find area but you can sub your solved angle into that.

phi
 2 years ago
Best ResponseYou've already chosen the best response.1for A3, you should have a factor r^2

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0Yes that is true. Thank you for catching that, phi.

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks again for all the assistance. This one is going to close now !

Mathmuse
 2 years ago
Best ResponseYou've already chosen the best response.1@phi, right! the sides of the triangle in the lower left should be rcos@, and rsin@
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