Intro Fluid Dynamics homework question

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Discussion Overview

The discussion revolves around a homework problem related to fluid dynamics, specifically calculating the force required to lift a ring placed on the surface of water. The problem involves concepts such as surface tension, shearing stress, and compressibility, but lacks certain information, leading to uncertainty in the approach to find a solution.

Discussion Character

  • Homework-related
  • Exploratory
  • Technical explanation

Main Points Raised

  • Scott presents the problem and expresses confusion about the missing mass of the ring, suggesting it may relate to surface tension, compressibility, or shearing stress.
  • Scott attempts to manipulate the shearing stress equation but finds insufficient information regarding velocity and height.
  • Another participant relates the problem to the Du Noüy ring method, indicating they arrived at the correct answer through an ad hoc approach, ignoring the weight of the ring.
  • Scott inquires about the energy equations related to the problem, seeking clarification on how to apply them.
  • A later reply suggests that energy from surface tension could be calculated as tension times area, proposing a differentiation approach to find the force.

Areas of Agreement / Disagreement

Participants express uncertainty and differing approaches to solving the problem, with no consensus on a definitive method or solution. Some focus on surface tension while others consider different equations from the textbook.

Contextual Notes

The discussion highlights limitations in the problem statement, particularly the absence of the ring's mass and specific parameters needed for certain equations. Participants also note their varying levels of experience with fluid dynamics, which may affect their proposed solutions.

sc0tt
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How much force is necessary to lift a ring, diameter 20mm, made of fine wire, and placed on the surface of water at 20 degrees celsius?
All the equations from the chapter:
Shearing stress = F/A = Viscosity*(velocity parallel/h)
Kinematic Viscosity = viscosity/density
Change in pressure = 4(surface tension/diameter)
Capillarity height = 4Tcos(theta)/(g*diameter*density)
Bulk modulus = -V*(change in pressure/change in V)
compressibility = 1/Bulk modulus
PV=RT
PV^n=constant

I'm learning fluid dynamics on my own for possible research so I don't have any professor or guidance, just a book. This makes solving problems much more difficult. With this problem specifically, I'm lost because I feel like the mass of the ring must be known as part of the problem, but no other information is given. This leads me to believe it has something to do with surface tension, compressibility, or shearing stress (this equation contains force).
My initial reaction was to solve the first equation for force and substitute the second equality in for shear stress. This leaves:
F=A*(viscosity*velocity parallel/h)
but there is no velocity parallel or required height.
The other equations don't seem to be of any help.
The answer the book gives is 9.15*10^-3 N
Thanks for any help,
Scott
 
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Sounds interesting, what book are you reading?

I'm quite lost myself when it comes to fluid dynamics, but I'd relate this problem to surface tension. More specifically, the Du Noüy ring: http://en.wikipedia.org/wiki/Du_Noüy_ring_method.

I got the right answer, but using a bit ad hoc methods (well, maybe not, but had you not supplied the answer, I would probably not be writing this reply). I thought of the thing as creating two surfaces (inner, outer) when you begin to lift the ring. I wrote an equation for the energy, and thus obtained the force.

EDIT: And yeah, I ignored the weight of the ring thing, as I suppose if you do these things experimentally, you add a counterbalance to get rid of it + I suppose it's supposed to be very, very light.
 
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The book is "Introduction to Fluid Mechanics" by Y. Nakayama. so are you saying you don't use any of the equations the book gives? If possible could you please list out the energy equations, my physics is better but i don't see a solution using energy either. Gravitational is mgh, energy from surface tension would be... ?
Thanks
 
Well, that's how I would do it, but again, having little experience with fluid dynamics, there might be other, better, solutions.

Energy from surface tension would be the tension times the area, i.e. gamma * 2 * 2*pi*r*h. Differentiate w.r.t. h, and you've got the force, I suppose.
 

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