Ghosh - 550
Page 1
2/13/2016
Viscous Flows
This chapter discusses all real fluid flow applications in the realm of incompressible
flows. Due to the limited time, compressible flow applications are covered in other
courses such as Ideal Flows or Aerodynamics. Incompressible viscous flow applications
may range from low to high speeds and internal and external flow configurations.
Internal flows occur when the fluid medium is surrounded by solid walls, which may
be stationary or moving. On the other hand, for solids surrounded by fluid particles,
we get the external flow configurations. A critical parameter in the flow study will be
the non-dimensional quantity Reynolds number, Re. This number is the ratio of inertia
and viscous forces in fluid flow. It is expressed as:
Re 
UL UL Inertia Force



 Viscous Force
In the above expression,  and  are absolute and kinematic viscosities, U =
Characteristic velocity, and L = Characteristic length. For example, for a steady external
flow over a circular cylinder, U = Free stream velocity and L = Diameter of the cylinder.
However, for steady internal pipe flow configurations,U= V = Average velocity of the
flow through a cross section.
We shall discuss flow based upon Re rather than speed. The advantage in doing that is
high or low speed flows can be discussed at the same time in the context of low viscosity
or highly viscous behavior. Another measure of flow speed can be by the amount of
turbulence. For very high-speed flows, Re is high enough to call the flow fully
turbulent. On the other hand, very low speed flows will be categorized invariably as
laminar.
Implications of Viscosity
(Parallel vs. Fully-Developed Flows)
We earlier saw that viscosity manifests itself through creation of both shear and rotation.
One of the most important characteristics of a real fluid is it satisfies “no slip” condition
on a solid surface. In other words, whenever fluid flows past a solid surface, the layer of
fluid in contact with the surface cannot slip against the surface, with the result that all
components of fluid velocity will be zero at the solid, impermeable surface. This
condition creates the highest shear on the fluid by a solid non-porous wall. As fluid
particles adjacent to the wall try to stop the next layer of fluid, the shear gradually
loses its strength as we move away from the wall. This is the cause of the boundary
layer formation on a solid surface. We shall investigate this in much more depth in the
next chapter on external flows. The internal flows discussed in this section will be
beyond the entrance length, which means that boundary layer growth from each
wall has already met at the center of the channel.
Ghosh - 550
Page 2
2/13/2016
x
x
Entrance Length
Velocity Profiles don’t change
with x  fully developed
Beyond the entrance length, which is typically 138-140 D for laminar pipe flows and 2540 D for turbulent pipe flows, we call the flow fully developed. That is:
u
0
x
Parallel Flows:
Instead of assuming fully developed flow, if we assume a parallel
flow, it means all fluid streamlines are parallel. In such a case of parallel flow along x,
u  0, v  w  0
Since all fluid media must satisfy the mass, momentum and energy equation, we find by
the application of continuity equation for incompressible flows,
v  0
w  0
u v w
u
 
 0,
0
x y z
x
We therefore find that a parallel flow is indeed fully developed.
With the introduction of parallel flows, simplification of the governing equations
becomes much simpler. We now develop the applications for some specific types of
internal flows.
Plane Poiseuille Flow
This is the case of fully developed incompressible flow between two infinitely large
parallel plates. We seek the velocity profile and shear flow field for such flows. As
before, if the flow is assumed parallel in the x-direction, u  0, v  w  0 . Therefore, the
u
 0 , which satisfies the fully developed condition. Let
continuity equation reduces to
x
Ghosh - 550
Page 3
2/13/2016
us investigate the y- and z-momentum equations for such a flow. Also, we assume that
the body forces are negligible. Therefore:
0
0
0
0
0
0
0
0
  v  v  v
 v
v
v
v 
p
y :   u  v  w     B y   2  2  2 
x
y
z 
y
y
z 
 t
 x
2
0
0
0
2
0
0
2
0
0
0
 w  w  w
 w
w
w
w 
p
z : 
u
v
w
    Bz   2  2  2 
x
y
z 
z
y
z 
 t
 x
2
2
2
p p

 0 from above, which means that pressure is a function of x only. Now we
y z
simplify the x-momentum equation:

0
0
0
  2u  2u  2u 
 u
u
u
u 
dp
x :   u  v  w     Bx   2  2  2 
x
y
z 
dx
y
z 
 t
 x
(Note that
p
dp
was modified to
from the y and z equation results)
x
dx
u
u
 2u
 0 . Also
0 
Let us further assume that the flow is steady. 
 0.
t
x
x 2
Furthermore, the flow can be assumed to be free from the end conditions since the plates
u
 2u
 0 , which means
are infinitely long and deep. 
 0 also. Thus the xz
z 2
equation simplifies to:
dp
 2u
0   2
dx
y
 2 u 1 dp
 2 
 dx
y
[The partial derivatives in velocity are no
u u u


 0]
longer needed since
t x z
We can integrate this equation twice in y to write:
Ghosh - 550
u ( y) 
Page 4
1  dp  2
  y  C1y  C2
2  dx 
Boundary Conditions:
y
2/13/2016
(C1 and C2 = Constants)
Since both plates are stationary, u(0) = 0, u(h) = 0
u(y)
h
x
The velocity profile u(y) may be evaluated with 0  C2 , and,
1  dp  2
1  dp 
0
 h  C1h  C1 
 h
2  dx 
2  dx 
 u ( y) 

1  dp  2
  y  hy
2  dx 

To be able to plot this velocity profile, let us assume
dp
h  1 m,   10 1 N  sec/ m 2 ,
 5 N / m 3
dx
y
Flow
h=1 m
Parabolic Velocity
Profile
6.25
m/s
x
Note that the velocity profile starts with a zero value on the wall, reaches a peak value of
6.25 m/s in the middle of the channel (h = 0.5 m) before reducing to zero on the upper
dp
 0 (instead of –5
wall (h = 1 m) symmetrically. Also, try to plot the function when
dx
N/m3). You will see an unrealistic curve (showing fluid bulges out along "-" x direction).
We can check the volumetric flow rate to claim this point.
Ghosh - 550
Page 5
2/13/2016
h



Q   V  dA   u w dy , where w = depth of the channel and dA  w dy î
A
y0
or,


Q h
1  dp  h 2
  u dy     y  hy dy
w y0
2  dx  y  0
h
1  dp  y 3 y 2 
1  dp  3

     
 h
2  dx  3
2  0
12  dx 
From this expression, it is easy to see that since Q, w, h, and  are all positive
dp
 0 . Thus, we make an important discovery
quantities, Q cannot be positive unless
dx
for Plane Poiseuille Flow: A Plane-Poiseuille flow cannot exist if the pressure
dp
gradient,
, is not negative. We also introduce a new definition of average velocity in
dx
this context. Average velocity through any area A is defined as the volumetric flow
rate per unit depth, i.e.
1  dp  3
 h
Q
1  dp  2
12  dx 
V 

 h
A
hw
12  dx 

If we evaluate the maximum velocity in this flow,
du
1 dp
2 y  h   0
0
dy
2 dx
h
 y  , which occurs at the center of the channel.
2
1 dp 2

 u ( y) y  h / 2  u max 
y  hy 
2 dx
 yh / 2


1  dp 
 
8  dx 
Therefore we notice that the maximum velocity

Ghosh - 550
Page 6
2/13/2016
1  dp 
 
u max 8  dx  3

1  dp  2
V

 
12  dx 
3
or, u max  V for this flow.
2

Shear Stress Distribution:
 u v 
1  dp 
 yx        2 y  h 
2  dx 
 y x 
1  dp 
h
   y  
2  dx 
2
If we plot this function along with the velocity profile, we notice a linear variation of
shear stress and shear force as follows:
y
yx
umax
h=1 m
Fyx
u(y)
x
These plots were made with
dp
 0 as stated before.
dx
Couette Flow
This type of flow is also between infinite parallel plates. However, the boundary
conditions are a little different from Plane Poiseuille Flows. Here one of the plates
remains stationary, whereas the other moves with a constant velocity, U. For
visualization, we assume the bottom plate stationary and the top plate moving.
U
Flow
h
y
x
u=0
Ghosh - 550
Page 7
2/13/2016
All the assumptions applicable to the derivation of Plane Poiseuille flows hold in the case
of Couette flows. Thus, we may skip part of the derivation and start with the velocity
profile.
u ( y) 
Now,
1  dp  2
  y  C1y  C2
2  dx 
u (0)  0  C2  0
u ( y)  U  U 
 C1 
1  dp  2
 h  C1h
2  dx 
U 1  dp 
  h
h 2  dx 
 u ( y) 


1  dp  2
Uy
  y  hy 
2  dx 
h
If we compare the above velocity profile with that obtained for Plane Poiseuille flows, we
Uy
find the right hand side has an additional term,
. The plot of just this term is a linear
h
velocity profile from y = 0, u = 0 to y = h, u = U. Thus, the Couette flow velocity profile
may be thought of as the superposition of the Plane Poiseuille flow’s velocity profile and
this additional linear profile. Because of this additional fluid momentum, Couette flows
dp
 0 ). Recall that the
can exist even with mild adverse pressure gradient (i.e.,
dx
existence of Q > 0 makes the flow possible.
y
U
For dp  0
For dp  0
dx
Flow
dx
h
x
Couette Flow Velocity Profiles
For dp  0
dx
Ghosh - 550
Page 8
2/13/2016


1  dp  2
Uy
, all the flow quantities
  y  hy 
2  dx 
h
such as volumetric flow rate, average velocity, maximum velocity, shear stress and shear
force distributions can be computed as before using their respective formulae.
Since we know the velocity profile u ( y) 
Hagen Poiseuille Flow (or, Pipe Flow)
Now we come to derive the most popular application of the internal flows, commonly
known as Hagen Poiseuille Flow or, simply pipe flows. Since pipes have cylindrical
geometry, we use the cylindrical form of the momentum equations. Let us assume an
incompressible, steady flow through a circular pipe without any appreciable body
forces. Assuming a parallel flow in the z-direction, Vz  0 , but Vr  V  0 .
0
Continuity equation
0


r Vr   1 V  Vz  0
r
r 
z

Vz
0
z
As in the case of Plane Poiseuille flow, writing out the momentum equations in  and r
p p

 0 . Therefore, let us focus on z-direction.
direction will simply result in
r 
0
0
0
0
V V V
V 
 V
z :  z  Vr z    z  Vz z 
r
r 
z 
 t
0
  2 Vz 1 Vz 1  2 Vz  2 Vz 
dp

   Bz   2  
 2

2
2 
dz
r

r

r
r



z


We can further assume
Vz
 0 because of the cylindrical symmetry.

  2 Vz 1 Vz 
dp

 0     2  
dz
r

r

r


  2 Vz 1 Vz 
dp

    2  
dz
r r 
 r
[
Vz Vz Vz


 0]
t
z

Ghosh - 550


Page 9
2/13/2016
dp    Vz 

r

dz r r  r 
d  dVz  r dp
r

dr  dr   dz
or, integrating twice over “r”, we get
r 2  dp 
Vz (r ) 
   C1 ln r  C 2
4  dz 
r

z
2R
(C1, C2 = Constants)
Since the pipe radius is R, the boundary
conditions may be written as
Vr (r  R)  0
dVz
and
(r  0)  0 .
dr
The second boundary condition is due to flow symmetry at r = 0, whereas the first one is
due to “no-slip” condition. Solving the constants C1 and C2 we get
R 2  dp 
r2 

Vz (r )  
 1 
4  dz  R 2 
As in the case of Plane Poiseuille flows,
dp
 0 for this flow to exist (i.e., Q > 0).
dz
Some additional results are:
 R 4  dp 
 R 2  dp 
Q
 , V 
  , VZ max  2V ,
8  dz 
8  dz 
dVz r  dp 
and  r  
  
z
dr
2  dz 

[Note: You must use an annular area element dA  2 r dr ê z to derive V and Q results.]
y
dr
x
Ghosh - 550
Page 10
2/13/2016
Velocity Profiles in Pipes
The laminar velocity profile that we just derived is parabolic in shape, given by

r2 
Vz (r )  Vmax 1  2 
 R 
(Vzmax = Constant)
However, as the flow Reynolds numbers increase beyond Recr = 2300, the fluid flow in
pipes can be considered turbulent. Turbulent velocity profiles are much flatter compared
to the laminar profiles because of more friction near the walls. These are given by a
“power law”:
1
n

Vz
r
1  
Vmax
R
where, 6 < n < 10. The larger the Reynolds number, the higher the value of n. The most
popularly used velocity profile for turbulent fully developed flows is for n = 7.
Laminar
2R
Turbulent
z
Laminar Vs. Turbulent
Velocity Profile
Examples
Also due to the flatter profiles
V
the turbulent flow
is
Vz max
closer to 1.0. The power law
profiles satisfy
V
Vz max

2n 2
(n  1)( 2n  1)
Continue