# Viscosity of a Fluid!!Watch

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Thread starter 11 years ago
#1
Hello, I really need help in this experiment which is stressing me out a lot!

Basically I have tubes of different radius filled with a fluid (some type of oil) I drop different sized ball bearings in, and calculate the terminal velocity, then using stoke's law im able to calculate the viscosity of the fluid, What I dont understand is that, using different sized ball bearings and different sized radius tubes gives me different terminal velocity, which therefore gives a different value of viscosity. How can a different sized ball bearing affect the viscosity of the fluid its falling through??

Thanks a lot
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11 years ago
#2
look up Stoke's Law

it doesn't affect the viscosity, it does affect the drag force
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Thread starter 11 years ago
#3
Yeah i did, but if terminal velocity is proportional to radius of ball bearing multiplied by radius of tube filled with fluid, then clearly this will affect the viscosity of the fluid wouldnt it?, the thing is i keep the radius of the ball bearing constant and change the radius of the tube and terminal velocity changes which changes the viscosity...I need help in this please!

Thanks!
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11 years ago
#4
The radius of the tube will have no effect - providing its big enough for the ball to drop through without getting too close to the walls.

Terminal velocity is when weight of ball = viscous drag force

mass of ball x g = 6 x pi x (radius of ball) x viscosity x velocity

( The radius of the ball affects its mass too )

I dont think v will be prop to radius of ball! You should beable to work out the expected relationship from what I've given you.
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Thread starter 11 years ago
#5
The thing is I was using tubes of different diameter but with the same fluid, and when putting in ball bearings of the same size, it gave me different terminal velocities for each..therefore when put into stokes law equation it gave me a slightly different value for viscosity..

I also found out that

radius of ball bearing multiplied by diamater of tube is proportional to terminal velocities..
Is this the case..could the viscosity actually vary just slightly by changing the tube diameter?
I really dont understand why it varies then..Please help

Thanks!
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11 years ago
#6
There are various edge effects when a viscous fluid is forced through a small gap. I would expect that once you get to a certain diameter then increasing dia further would have no effect.
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11 years ago
#7
I don't think that the diameter of the tube containing the oil should have any effect on viscosity.

In the experiment you are describing, I assume that you are trying to keep the viscosity of oil constant.

I suspect that narrow/small tubes are affecting the streamline flow of the ball bearing (which is a requirement of Stokes' Law) and there may be friction between the ball bearing and the sides of the tube.
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11 years ago
#8
(Original post by teachercol)

Terminal velocity is when weight of ball = viscous drag force

mass of ball x g = 6 x pi x (radius of ball) x viscosity x velocity
Correct me if I'm wrong, but this equation ignores the buoyancy acting on the falling ball bearing: i.e. weight of ball = buoyancy + viscous drag.

The equation you stated would normally apply to objects moving horizontally, not being dropped vertically.

Thus, combining buoyancy to the equation, you get:
Velocity = (2/9 x r^2 (density ball - density liquid) x g )/viscosity
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11 years ago
#9
I am doing a similar experiment for A2 coursework, except I am investigating the effect of temperature on viscosity.

I wanted to know what actually causes the viscosity to decrease as temperature is increased.
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11 years ago
#10
(Original post by mumakil142)
I am doing a similar experiment for A2 coursework, except I am investigating the effect of temperature on viscosity.

I wanted to know what actually causes the viscosity to decrease as temperature is increased.
not always ,,some polymers's viscosity increase with the temperature MDI (eggs hardden with heat)
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11 years ago
#11
(Original post by mumakil142)
Correct me if I'm wrong, but this equation ignores the buoyancy acting on the falling ball bearing: i.e. weight of ball = buoyancy + viscous drag.
No youre right, of course. But I like to leave some things for the OP to find out
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11 years ago
#12
some1 in my physics class is doing something similar I believe

I'm dropping paper cones & working out how the nose angle effects time taken to reach T.V - which my teacher has said requires the use of Stokes law.
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