1. Fluid is contained between two parallel
Bottom plate is
stationary
Top plate moves with
velocity
vx = v H x y
vx = 0
y+y
x + x
Fluid of
constant , 
No pressure drop.
Fluid is contained between two parallel
plates. The top plate moves with
velocity V while the bottom plate is
stationary. Find the velocity profile
of the fluid. Where is the maximum
velocity? What is the force required to
move the top plate for length L and
width W?
We select rectangular coordinates for this problem with origin midway between the two plates.
The distance between the plates is H. We perform a force balance on a representative fluid
element located between x and x+x and between y and y+ y. The flow is assumed to be
steady and fully developed(What do these terms mean?). Fluid properties of density and
viscosity are constant. Under these conditions, the force or momentum balance becomes:
 Fx = (yx x z)y - (yx x z)y+ y = 0
Dividing by xy z,
(yx )y - (yx )y+ y = 0 y

2. Now we shrink the element down to infinitesimal size,
Lim { (yx )y - (yx )y+ y } = 0
x y
y0
You should recognize this as the definition for the derivative of . The shear stress is a
function of y only for this problem. Hence,
d yx = 0 dy
This is a separable differential equation:
dyx = 0 dy
Integrating,
yx = C1
We have no boundary condition to evaluate the constant, so we find the velocity profile
using Newton’s law of viscosity. Substituting(yx =  dvx/dy),
 dvx/dy = C1
This is also a separable differential equation that can be solved for the velocity.

3. Rearranging and integrating,
 dvx = C1  dy
  
Evaluating the integrals with viscosity constant, we find
vx = C1y + C2  Velocity profile is known to within 2 constants 
We use the boundary conditions to find the constants of integration. At y = - H/2, vx = 0.
vx = 0 = C1(-H/2) + C2 
and at y = H/2, vx = V:
V = C1(H/2) C2
These are two equations for the two unknown constants. Solving yields:
C1 = V/H and C2 = V  vx = V(y/H +1/2) 2

4. Where is the velocity a maximum? Look at dvx/dy = 0. dvx/dy = V/H = 0
There is no value of y that makes this true. That is because this is a linear velocity profile.
Let’s look at a diagram.
Fluid of
constant , 
vx = v
There is no local maximum or
minimum. The maximum occurs
at the boundary y=H/2. y H x
vx = 0
What is the force required to move the plate? The frictional force is related to the shear
stress (yx )H/2 at the surface of the top plate.
yx =  (dvx/dy)H/2 =  (V/H) H/2 = V/H (Force on fluid is +)
We must multiply by the area of the top plate of length L and width W to get the force:
Fx = VLW/H We must apply a force of this magnitude in the plus x-direction.