Chapter 5
Example 5.7-15 ---------------------------------------------------------------------------------Consider the flow through a rectangular duct whose width is very large in the z direction
when compared to the gap (2h) in the y direction. Such a situation could occur in a die when
a polymer is being extruded at the exit into a sheet, which is subsequently cooled and
solidified. Determine the relationship between the flow rate and the pressure drop between
the inlet and exit, together with other relevant quantities.
Large
width
Inlet
y
Exit
x
2h
z
L
Figure 5.7-6 Flow through a rectangular duct.
Solution -----------------------------------------------------------------------------------------We can analyze the problem by referring to a cross section of the duct, shown in Figure 5.77, taken at any fixed value of z.
Exit
Inlet
y
P1
h
vx
x
P2
h
L
Figure 5.7-7 A cross section of the duct with flowing fluid.
We make the following assumptions regarding the flow through the duct
1. The flow is steady with Newtonian fluid and constant physical properties.
2. There is only one nonzero velocity component in the x direction so that vy = vz = 0.
5
Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 274
5-47
3. The entrance and exit effects are negligible and there is no variation of velocity in the z
v
direction because of the large width so that x = 0.
z
4. Gravity force is in the negative y direction; hence, gy =  g and gx = gz = 0.
5. There is no slip at the boundary (the wall), so that vx = 0 at y =  h.
The differential mass balance or continuity equation (

=  v) is simplified to v = 0
t
for constant density
v =
v x v y v z
+
+
=0
x
z
y
v x
= 0. Therefore vx = vx(y) is a function of y only. We now
x
consider the momentum equation in the x direction
Since vy = vz = 0, we have
  2v
v
v
v 
 v x
 2vx  2vx 
P
+   2x 
 v x x  v y x  v z x  = 
 2  + gx
x
x
y
z 
y 2
z 
 t
 x
 
This equation is simplified to
0=
P
 2v
+  2x
x
y
(E-1)
If the dependence of pressure on y is neglected, then P = P(x) is a function of x only.
Equation (E-1) becomes

P  P1
dP
d 2vx
=
= 2
2
dx
L
dy
(E-2)
A first integration gives
dv x
1 dP
=
y + C1
dy
 dx
(E-3)
Since the velocity is a maximum at y = 0, we have
0=
1 dP
(0) + C1  C1 = 0
 dx
A second integration of equation (E-3) with C1 = 0 gives
5-48
vx =
1 dP 2
y + C2
2  dx
The constant of integration C2 can be obtained from the boundary condition that vx = 0 at y =
h.
0=
1 dP 2
1 dP 2
h + C2  C2 = 
h
2  dx
2  dx
The velocity profile is finally
vx =
1 dP 2
(y  h2)
2  dx
The volumetric flow rate Q for the system with a width W can be obtained by first consider
the differential flow rate through an element Wdy
dQ = vxWdy
Hence
h
Q = 2  v xWdy = 2W
0
W dP
Q=
 dx
1 dP
2  dx

h
0
( y 2  h 2 )dy
h
 y3

2Wh 3

 h 2 y  =
3
 3
0
 dP 


 dx 
The maximum velocity occurs at the centerline y = 0
vx,max =
h 2  dP 


2   dx 
The average velocity is just the volumetric flow rate Q divided by the area of flow
Therefore
vx,ave =
Q
h 2  dP 
=


2Wh
3  dx 
vx,ave =
2
vx,max
3
The shear stress at any location within the fluid is given by
v 
 v
v
dP
yx =   x  y  =  x =
y
dx
y
x 
 y
5-49
Shell Balance
We can also apply the momentum balance directly to a differential fluid element to obtain the
differential equation required for the calculation. Consider a fluid element with dimensions
of x and y in the plane of diagram as shown in Figure 5.7-8. The element has a depth of W
in the direction normal to the plane of the diagram.
yx|y+y
y
P|x
x
y
P|x+x
yx|y
x
Figure 5.7-8 Momentum balance on a differential fluid element xyW.
The convective momentum transfers through the left-hand and right-hand face are
vxvxyW|x  vxvxyW|x+x = 0
We have assume that vx = vx(y) is a function of y only. Applying the momentum balance on
the fluid element xyW yields
P|xyW  P|x+xyW + yx|y+yxW  yx|yxW = 0
Dividing the equation by the control volume xyW gives

 |
 yx | y
P |x  x  P |x
+ yx y  y
=0
x
y
In the limit as xyW  0, we obtain
d yx
dy
=
dP
dx
(E-4)
The shear stress at any location within the fluid is given by
v 
 v
dv
yx =   x  y  =  x
dy
x 
 y
Equation (E-4) becomes
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d
dy
 dv x  dP
 
 =
dx
 dy 
For constant physical properties, we have
dP
d 2vx
 2 =
dx
dy
This equation is the same as the one derived from simplifying the following component of
the Navier Stokes equations.
  2v
v
v
v 
 v x
 2vx  2vx 
P
+   2x 
 v x x  v y x  v z x  = 
 2  + gx
2
x
x
y
z 

x

y
z 
 t

 
Example 5.7-2 ---------------------------------------------------------------------------------The geometry of angular drag flow between two concentric cylinders is shown in Figure 5.79. The inner and outer cylinders have radii of R and R respectively. The inner cylinder is
rotating with constant angular velocity  while the outer one is fixed. Find the velocity
distribution v(r) for the fluid between the cylinders and the torque required to rotate the
inner cylinder.

L
z
r
Figure 5.7-9 Flow between concentric cylinders.
Solution -----------------------------------------------------------------------------------------The r-component of the Navier Stokes equations for Newtonian isothermal incompressible
flow in cylindrical coordinates is given as
2
2
 vr
  1 
P
vr v v v2
vr 
 1  vr 2 v  vr 
 =
 
+   
 vr

  vz
( rv r )  2
 2
 2  + gr
2
r
r
r 
r
z 
r 
z 
 r 
 r  r r
 t
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For steady state flow with only one component of the velocity v(r), the equation is
simplified to

v2
P
=
r
r
We have a radial pressure gradient due to centrifugal acceleration. The -component of the
Navier Stokes equations is given as
2
  1 
v
v v
vv
v 
1 P
2 vr  2 v
 v
 1  v
 vr      r   v z   = 
+   
( rv )  2

 2
2
r 
r
r 
r
z 
r 2 
z
 t
 r 
 r  r r

Discarding the zero terms in the above equation yields
0=
d 1 d

( rv )

dr  r dr

Integrating the equation gives
1 d
( rv ) = a 
r dr
v =
d(rv) = ardr 
rv =
1 2
ar + b
2
1
b
ar +
2
r
The two constants of integrations a and b can be evaluated using the boundary conditions.
At r = R  v = R =
At r = R 
v = 0 =
1
b
aR +
2
R
1
b
aR +
2
R
2
R 2 
Solving two equations for two unknowns a and b: b =
and a = 
1/  
1/  
The velocity distribution is
v = 
2 r
R  R r 
R 2  1
+
=
  
1/   2
1/    r R 
1/   r
The shear stress r is given by
   v

 r  r
r =  r
d v 
 1 v r 
= r   


dr  r 
 r  
5-52

 + g

v
R  R 1 
=
  
1 /     r2 R 
r
Taking the derivative of the above expression gives
R 
d  v 
R
  =
 2 3 
1/   
dr  r 
r 
The shear stress is then
r = r
 R 2
d  v 
  =  2
1/    r2
dr  r 
At the surface of the inner cylinder r = R, therefore
r|R =  2

2 
1
=
1/   
1 2
The force applied on the inner surface of the cylinder is (2RL)r|R. The torque on the
cylinder is the product of this force time the moment arm of radius R.
 = (2RL)r|R(R) = (22R2L)
2 
4L 2 R 2 
=
1 2
1 2
We have assumed laminar flow that is achievable for concentric cylinders as long as the
rotational speed is below a value that satisfies
R 2 

<
40
 (1   ) 3 / 2
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