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Viscosity (2013)

Page history last edited by Joe Redish 7 years ago

Class content Kinds of Forces Resistive forces

 

Prerequisites

 

When an object moves in a fluid -- a liquid or gas -- it drags bits of the fluid with it along its surface.  This results in a layer of fluid sliding over a neighboring layer of fluid.  The interactions of the molecules in the fluid result in a kind of internal friction that acts to slow the relative motion of neighboring layers of fluid.  The full description of the mechanism of viscosity is rather complex, so we will treat it phenomenologically.  Let's start by looking at the simplest possible example.

 

Imagine a liquid that is sandwiched between two plates.  If the top plate is moved while the bottom is held fixed, the liquid will be sheared -- pulled so that the amount of deformation of the fluid changes perpendicular to the direction of flow.  Some of the fluid sticks to and moves with the top plate, while the parts of the liquid next to the bottom plate remain at rest.  This sets up a gradient (a rate of change with respect to space) in velocity across the liquid.  This requires the layers of the fluid to move past one another as shown in the lower picture.

 

An equation for the viscous force

To get an equation for the resistive force that the fluid exerts on the plates as a result of the internal sliding, consider the example shown in the figure.  Suppose the two plates have an area A and are separated by a thickness y of liquid.  Suppose we are holding the bottom plate fixed and are dragging the top plate with a constant velocity u.  If there were no internal resistance, any force on the top plate would continue to speed it up.  But if we apply a constant force, app, on the plate, we discover that it will speed up, but then its velocity will increase more slowly until it approaches a constant velocity, which we will call u.  What's happening is that the acceleration of the plate, P, is determined by Newton's second law for the plate:

 

 

 

When the plate reaches a steady speed, the acceleration is 0, so the force we are applying is equal (and opposite) to the viscous force. 

 

What we discover from our experiments is:

  • the viscous force is proportional to the speed of the plate
  • the viscous force is proportional to the area of the plate
  • the viscous force is inversely proportional to the distance between the moving plate and the fixed plate.

 

The proportionality constant, μ, is called the viscosity of the fluid and is defined by:

 

                    Formula    

 

The viscous force on a moving sphere

While the two-plate experiment is a good way to figure out what's going on in the phenomenon of viscosity, it's a bit tricky to apply to an object moving in a fluid.  From our experiments with the plates, we see that our viscosity coefficient has the dimensionality of

 

Now suppose we have a small sphere of radius R moving through a fluid with a velocity v.  We expect there to be a viscous force holding back the fluid, and we expect it to depend on the viscosity coefficient, μ, the velocity of the object, v, and the size of the object -- some function of R.

 

We want to construct a force, which has dimensionality the same as "ma", of ML/T2.  We expect it to be proportional to μ, which has dimensionality, M/LT.  We expect it to be proportional to velocity, v,  which has dimensionality L/T.  So μ*v has dimensionality (M/LT) * (L/T) = M/T2.  We're missing a factor with dimensionality "L".  The only one we have is the radius.  So we expect our force to look something like μ*v*R. This turns out to be right -- but with a dimensionless factor of 6π -- something we couldn't know by dimensional analysis.  The result is

 

Viscous force on a sphere of radius R moving in a fluid:

 

Viscosity of different fluids

Since viscosity has dimensions of M/LT, it will have units (in the SI system) of kg/m-s.  Sometimes it's convenient to express this unit in different forms.  For example, since we will typically be building a force with it, we might want to rearrange this so it looks like Newtons: 1 N = 1 kg-m/s2.  So we can make the units of viscosity include a Newton by multiplying by m2-s and dividing by the same factor.  The result is

The combination unit Newton/m2 is a pressure and is often called a Pascal.  So the units you will see for viscosity are typically "Pascal-seconds (Pa-s)".  The viscosity for air and water is:

 

μair = 1.81 10-5 Pa-s

μwater = 1.00 10-3 Pa-s.

μseawater = 1.07 10-3 Pa-s.

 

Follow-ons

 

 

Karen Carleton and Joe Redish 9/29/11

 

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