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Viscosity (Fluid mechanics)

  • #1
162
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Homework Statement



Attached.

I am working on part (i). What is the resistive force?

Shear stress= μ*(du/dy) = μU/l = F/A

Since the force is up and down, it causes a shear stress, thus area is parallel to the force and thus it is the 4 faces B*L

Thus, A=4*B*L

→ F=μ*U*A/l = 4*μ*U*L*B/l

Where = viscosity
l=clearance gap
L=length of block
F=Resistive force
A=Area

I'm not sure if that is the correct method. Can someone please verify?
 

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Answers and Replies

  • #2
20,106
4,187
Looks good. The fluid in the gap exerts a vertical shear force on the 4 vertical sides of the block. Looks like the analysis is right on target.
 
  • #3
162
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Okay thanks, the second part asks me to find the "terminal velocity"

This is what I did

F=4*μ*U*L*B/l=ma
m=ρV
Where ρ is the density of the solid, not the fluid
and V is the volume of the block = (B^2)*L

Plugging that in and simplifying I got U=ρ*B*g*l / (4*μ)

I'm a bit hesitant on that answer because it did not involve the density of the fluid. Is this the correct method?
 
  • #4
20,106
4,187
No. There is an upward (buoyant) force from the surrounding liquid acting on the block, equal to weight of the displaced liquid ρliqVg. So the net downward force from the weight of the block combined with the upward (buoyant) force from the surrounding liquid is equal to....?
 
  • #5
162
0
No. There is an upward (buoyant) force from the surrounding liquid acting on the block, equal to weight of the displaced liquid ρliqVg. So the net downward force from the weight of the block combined with the upward (buoyant) force from the surrounding liquid is equal to....?
Oh, I think I'm supposed to ignore that. The problem stated to neglect buoyancy. Is that still necessary?
 

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