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2001, W. E. Haisler Chapter 3: Conservation of Linear Momentum 45

Plane Couette Flow between two large parallel plates (one stationary and one moving)

Couette flow is the case with two parallel plates separated by a distance d. The bottom plate is stationary and the top plate move horizontally at a velocity uo. There is no pressure differential along the length of the plate. The


driving force is the movement of the top plate. Assume the flow is steady and incompressible.

a) Assume the following boundary conditions:steady state incompressible no flow in x or y direction flow in x direction is prevented by plates if no pressure gradient in y direction and plates are

long in y direction, then flow in middle (y direction) can be considered to have only a y component, i.e., ignore edge effects of plates in y direction).

body force (gravity) is zero b) Conservation of mass is given by


and reduces to or

for plane motion, i.e., we consider only a slice of the fluid in the middle of the plate (ignore edge effects), then

Thus only!c) Conservation of linear momentum equations are:


Noting that , , ignoring gravity effects so that , and noting that no pressure gradients exist ( , etc.), then the COLM equations reduce to:


x component of linear momentum

y component of momentum

z component of momentum

d) Assume the flow produces no normal stresses. Since flow is in x-z plane, there is no shear in the x-y or y-z plane (only shear in x-z plane)

(fluid property assumption)


The last assumption is observed from experiment. The traction (shear stress) on the wall is proportional to velocity

gradient normal to wall:


For a 3-D viscous flow field, we can show that the complete set of constitutive equations relating stresses and velocity gradients will be given by:

From the above, it is clear that if and , then (all normal deviatoric stresses are


zero) and (only non-zero shear stresses are in the x-z plane).


e) The 3 linear momentum equations reduce to

Note that since , the third momentum equation is actually an ordinary differential equation:

We assume a “no slip” Boundary Condition for the fluid at each wall. Since the lower wall is stationary, then


. The upper wall moves at a velocity so that . Integrating the third momentum equation and assuming is a constant, we obtain .

Substituting boundary conditions for velocity yields

at x=0: at x=d:

or

Thus, the velocity profile is a straight line for the plane Couette flow problem.


The shear stress is obtained by substituting the velocity into the constitutive equation:

which is constant from bottom to top. Thus, as uo increases, the shear stress increases. As d increases, the shear stress decreases. At the wall, the shear stress acting on the wall must be equal and opposite to the shear stress acting on the fluid!

BE CAREFUL! THE RESULTS ABOVE APPLY ONLY FOR COUETTE FLOW!


Some real world examples of Couette flow:a) Wing moving through calm air at speed uo. At some

distance far away from the wing (normal to direction wing is moving), the air is motionless – think of this point as a fixed boundary where the fluid velocity is zero. At the surface of the wing, the fluid velocity is uo if we assume a no-slip condition – think of this as the moving boundary. So, looks just like Couette flow.

b) Piston moving up and down in the cylinder of an engine. Between the piston and cylinder wall is lubrication oil with a thickness of d. The cylinder wall is the fixed boundary and the piston wall is the moving boundary.


Some viscosity coefficient values:Air at standard sea level conditions: =1.79 x 10-5 kg/(m s)Water: = 1.005 x 10-3 kg/(m s) at 20CMotor oil: = 1.07 kg/(m s) at 20CNote: 1 centipoise = 10-3 kg/(m s) = 6.72 x 10-4 lbm/(ft s)See: http://www.lmnoeng.com/fluids.htm

Some exercises1. A flat-bottomed boat with a wetted surface area of 25 sq.ft.

moves through the water. Assume the boat does not “push” any water in front of it. How much force in lbs. is required to propel the boat at 15 mph in water that is 3 ft. deep?

2. A piston with a diameter of 3 in. moves in a cylinder of diameter 3.01 in. Oil with viscosity of 1,000 centipoise fills the gap. How much force is required to move the cylinder if its velocity is 4 ft/sec?

http://www.lmmoeng.com/fluids.htm


Poiseulle Flow between two parallel plates (both stationary)

Poiseulle flow is the case of fluid flow between two fixed parallel plates separated by a distance d and a pressure gradient in the z direction. The driving force is the pressure

differential from left to right ( ). Assume the flow is

steady and incompressible.


a) Assume the following boundary conditions:steady state

incompressible

no flow in x or y direction flow in x direction is prevented by plates if no pressure gradient in y direction and plates are

long in y direction, then flow in middle (y direction) can be considered to have only a y component, i.e., ignore edge effects of plates in y direction).

body force (gravity) is zero in y and z directions so that

b) Conservation of mass is given by


and reduces to

or for plane motion, i.e., we consider only a slice

of the fluid in the middle of the plate (ignore edge effects), then

Thus only!c) Conservation of linear momentum equations are:


Noting that , , and considering gravity effects only in the x direction so that , the COLM equations reduce to:

2001, W. E. Haisler 62

x component of linear momentum

y component of momentum

z component of momentum

d) Assume the flow produces no normal stresses. Since flow is in y-z plane, their is no shear in the x-y or y-z plane (only shear in x-z plane)


The last assumption is observed from a Couette-type experiment. The traction (shear stress) on the wall is

proportional to the velocity gradient normal to wall,

(see Couette flow problem)


For a 3-D viscous flow field, we can show that the complete set of constitutive equations relating stresses and velocity gradients will be given by:

From the above, it is clear that if and , then (all normal deviatoric stresses are


zero) and (only non-zero shear stresses are in the x-z plane).

e) With all the assumptions and the constitutive equation, the 3 linear momentum equations reduce to

We assume a “no slip” Boundary Condition for the fluid at each wall. Since both walls are stationary, then


and .

We further take the pressure gradient to be a given value (Boundary Condition). Integrating the third (z) momentum equation and assuming is a constant, we obtain

. Substituting boundary

conditions for velocity yields:

Substituting and into the equation above gives the following result:


Thus, the velocity profile is a quadratic in x for Poiseulle Flow between two stationary parallel plates. The shear stress is given by substituting the velocity into the constitutive equation to obtain:

The shear stress is a maximum at either wall, and zero at the center. Interestingly, the shear stress does NOT depend on the viscosity coefficient, .


A photograph of velocity profiles of fluid starting from rest and flowing from left to right is shown below.

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