Chapter 9: Surface Resistance

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57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
1
Chapter 9: Surface Resistance
9.1 Introduction: drag and lift on immersed bodies
CD 

p  p  n  îdA    w t  îdA  = drag coefficient


1 2 S

S
V A
2
Cform (form drag)
Cf (skin friction drag)
1


p  p  n  ĵdA  = lift coefficient
CL 
1 2 

V A
2
1
For inviscid fluid, CD = 0 since both the form and skin
friction components are zero, which is known as
D’Alembert paradox. However, CL  0 and can often be
predicted accurately with ideal-flow theory.
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
In general,
low Re < 1: Cf > Cform Stoke’s flow (lab 1)
med & high Re: 1. Cf >> Cform
t/c << 1 i.e., streamlined body
2. Cform >> Cf
t/c  1 i.e., bluff body
1. = subject of this chapter
2. = subject of chapter 11 along with CL
Topics of Chapter 9
9.2 Surface Resistance with Uniform Laminar Flow
1. parallel plates (internal flow)
2. flow down an inclined plane (open channel flow)
3. parallel plates with pressure gradient (internal flow)
In this class:
1. parallel plates
2. extend as per 3. including inclined flow
3. flow down an inclined plane
9.3
external 9.4
9.5
flow
9.6
Boundary Layer Flow
Laminar
Turbulent
Transition control (brief comments)
Chapter 9
2
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
3
9.2 Surface Resistance with Uniform Laminar Flow
We now discuss a couple of exact solutions to the NavierStokes equations. Although all known exact solutions
(about 80) are for highly simplified geometries and flow
conditions, they are very valuable as an aid to our
understanding of the character of the NS equations and
their solutions. Actually the examples to be discussed are
for internal flow (Chapter 10) and open channel flow
(Chapter 15), but they serve to underscore and display
viscous flow. Finally, the derivations to follow utilize
differential analysis. See the text for derivations using CV
analysis.
1. Couette Flow
boundary conditions
First, consider flow due to the relative motion of two
parallel plates
Continuity
u
0
x
Momentum
d2u
0 2
dy
u = u(y)
v=o
p p

0
x y
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
or by CV continuity and momentum equations:
u1y  u 2 y
u1 = u2
 Fx  uV  dA  Qu 2  u1   0

dp 
d 

 py   p  x y  x     dy x = 0
dx 

 dy 
d
0
dy
d  du 
i.e.
   0
dy  dy 
d 2u
 2 0
dy
from momentum equation
du
 C
dy
C
u  yD

u(0) = 0  D = 0
U
u(t) = U  C = 
t
U
u y
t
du U
 
 constant
dy
t
Chapter 9
4
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
5
2. Generalization for inclined flow with a constant pressure
gradient
Continutity
u
0
x
Momentum

d2u
0   p  z    2
x
dy
i.e.,
d 2u
dh
 2 
dx
dy
u = u(y)
v=o
p
0
y
h = p/ +z = constant
plates horizontal
plates vertical
which can be integrated twice to yield

du
dh
  yA
dy
dx
dh y 2
u  
 Ay  B
dx 2
dz
0
dx
dz
=-1
dx
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
6
now apply boundary conditions to determine A and B
u(y = 0) = 0  B = 0
u(y = t) = U
dh t 2
U
dh t
U  
 At  A 

dx 2
t
dx 2
 dh y 2 1  U
dh t 
u ( y) 
 

 dx 2   t
dx 2 
=


 dh
U
ty  y 2  y
2 dx
t
This equation can be put in non-dimensional form:
u
t 2 dh  y  y y

1   
U
2U dx  t  t t
define: P = non-dimensional pressure gradient
t 2 dh
=
2U dx
Y = y/t
u
 P  Y(1  Y)  Y
U
parabolic velocity profile

p
z

z 2  1 dp dz 


2U   dx dx 
h
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
u Py Py 2 y

 2 
U
t
t
t
t
q   udy
0
t
 U dy
q 0
u 
t
t
tu t P
P
y
   y  2 y 2   dy
U 0 t
t
t
=
Pt Pt t
 
2 3 2
Chapter 9
7
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
8
u P 1
t 2  dh  U
  u
   
U 6 2
12  dx  2
For laminar flow
ut
 1000

Recrit  1000
The maximum velocity occurs at the value of y for which:
d u
P 2P
1
du

0


y

0
 
dy  U 
t t2
t
dy
y
t
P  1  t  t @ umax
2P
2 2P
 u max  u y max  
note: if U = 0:
u
u max

UP U U
 
4
2 4P
P P 2

6 4 3
for U = 0, y = t/2
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
9
The shape of the velocity profile u(y) depends on P:
dh
 0 the pressure decreases in the
dx
direction of flow (favorable pressure gradient) and the
velocity is positive over the entire width
1. If P > 0, i.e.,

dh
d  p  dp
    z 
  sin 
dx
dx  
 dx
a)
dp
0
dx
b)
dp
  sin 
dx
dh
 0 the pressure increases in the
dx
direction of flow (adverse pressure gradient) and the
velocity over a portion of the width can become
negative (backflow) near the stationary wall. In this
case the dragging action of the faster layers exerted on
the fluid particles near the stationary wall is insufficient
to over come the influence of the adverse pressure
gradient
2. If P < 0, i.e.,
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
10
dp
  sin   0
dx
dp
  sin 
dx
3. If P = 0, i.e.,
u
a)
b)
or
 sin  
dp
dx
dh
 0 the velocity profile is linear
dx
U
y
t
dp
 0 and  = 0
dx
dp
  sin 
dx
For U = 0 the form
Note: we derived
this special case
u
 PY1  Y   Y is not appropriate
U
u = UPY(1-Y)+UY
t 2 dh
Y1  Y   UY
=
2 dx
Now let U = 0:
t 2 dh
u
Y1  Y 
2 dx
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
11
Flow down an inclined plane
uniform flow  velocity and depth do not
change in x-direction
du
0
dx

d2u
x-momentum 0   p  z    2
x
dy

y-momentum 0   p  z   hydrostatic pressure variation
y
dp

0
dx
Continuity
d 2u
 2    sin 
dy
du

  sin y  c
dy

57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
12

y2
u   sin   Cy  D

2
du


 0   sin d  c  c   sin d
dy yd


u(0) = 0  D = 0

y2 
u   sin   sin  dy

2 
=
u(y) =

sin  y2d  y 
2
g sin 
y2d  y 
2
d
 2 y3 

q   udy  sin dy  
2
3 0
0

d
=
V avg
1 3
d sin 
3
q 1 2
gd 2
 
d sin  
sin 
d 3
3
discharge per
unit width
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
13
in terms of the slope So = tan   sin 
gd 2So
V
3
Exp. show Recrit  500, i.e., for Re > 500 the flow will
become turbulent
p
   cos 
y
Re crit 
Vd
 500

p    cos  y  C
pd   p o   cos  d  C
i.e.,
p   cos  d  y  p o
* p(d) > po
* if  = 0
if  = /2
p = (d  y) + po
entire weight of fluid imposed
p = po
no pressure change through the fluid
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
14
9.3 Qualitative Description of the Boundary Layer
Recall our previous description of the flow-field regions for
high Re flow about slender bodies
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
15
w = shear stress
w  rate of strain (velocity gradient)
=
u
y
y 0
large near the surface where
fluid undergoes large changes to
satisfy the no-slip condition
Boundary layer theory is a simplified form of the complete
NS equations and provides w as well as a means of
estimating Cform. Formally, boundary-layer theory
represents the asymptotic form of the Navier-Stokes
equations for high Re flow about slender bodies. As
mentioned before, the NS equations are 2nd order nonlinear
PDE and their solutions represent a formidable challenge.
Thus, simplified forms have proven to be very useful.
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
16
Near the turn of the century (1904), Prandtl put forth
boundary-layer theory, which resolved D’Alembert’s
paradox. As mentioned previously, boundary-layer theory
represents the asymptotic form of the NS equations for high
Re flow about slender bodies. The latter requirement is
necessary since the theory is restricted to unseparated flow.
In fact, the boundary-layer equations are singular at
separation, and thus, provide no information at or beyond
separation. However, the requirements of the theory are
met in many practical situations and the theory has many
times over proven to be invaluable to modern engineering.
The assumptions of the theory are as follows:
Variable
u
v

x

y

order of magnitude
U
<<L
O(1)
O()
L
O(1)
1/
O(-1)
2
2
 = /L
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
17
The theory assumes that viscous effects are confined to a
thin layer close to the surface within which there is a
dominant flow direction (x) such that u  U and v << u.
However, gradients across  are very large in order to
satisfy the no slip condition.
Next, we apply the above order of magnitude estimates to
the NS equations.
  2u  2u 
u
u
p
u  v     2  2 
x
y
x
y 
 x
1 1  -1
2 1
-2
 2v 2v 
v
v
p
u  v     2  2 
x
y
y
x 
 x
1   1
2 1
-1
elliptic
u v

0
x y
1
1
Retaining terms of O(1) only results in the celebrated
boundary-layer equations
u
u
p
 2u
u  v    2
x
y
x
y
p
parabolic
0
y
u v

0
x y
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
18
Some important aspects of the boundary-layer equations:
1) the y-momentum equation reduces to
p
0
y
i.e.,
p = pe = constant across the boundary layer
from the Bernoulli equation:
1
p e  U e2  constant
2
p e
U e
i.e.,
 U e
x
x
edge value, i.e.,
inviscid flow value!
Thus, the boundary-layer equations are solved subject to
a specified inviscid pressure distribution
2) continuity equation is unaffected
3) Although NS equations are fully elliptic, the
boundary-layer equations are parabolic and can be
solved using marching techniques
4) Boundary conditions
u=v=0
y=0
u = Ue
y=
+ appropriate initial conditions @ xi
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
19
There are quite a few analytic solutions to the boundarylayer equations. Also numerical techniques are available
for arbitrary geometries, including both two- and threedimensional flows. Here, as an example, we consider the
simple, but extremely important case of the boundary layer
development over a flat plate.
9.4 Quantitative Relations for the Laminar Boundary Layer
Laminar boundary-layer over a flat plate: Blasius solution
(1908)
student of Prandtl
u v

0
x y
Note:
p
=0
x
for a flat plate
u
u
 2u
u v  2
x
y
y
u=v=0 @y=0
u = U
@y=
We now introduce a dimensionless transverse coordinate
and a stream function, i.e.,
 y
U y

x

  xU  f 
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
u
  

 U  f 
y  y
v
Chapter 9
20
f   u / U
 1 U 
f   f 

x 2
x
substitution into the boundary-layer equations yields
ff   2f   0
f  f  0 @  = 0
Blasius Equation
f  1 @  = 1
The Blasius equation is a 3rd order ODE which can be
solved by standard methods (Runge-Kutta). Also, series
solutions are possible. Interestingly, although simple in
appearance no analytic solution has yet been found.
Finally, it should be recognized that the Blasius solution is
a similarity solution, i.e., the non-dimensional velocity
profile f vs.  is independent of x. That is, by suitably
scaling all the velocity profiles have neatly collapsed onto a
single curve.
Now, lets consider the characteristics of the Blasius
solution:
u
vs. y
U
v
U
U
vs. y
V
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999

5x
Re
Chapter 9
21
value of y where u/U = .99
Re x 
U x

57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Typed by Stephanie Schrader Fall 1999
Chapter 9
22
U f (0)

w 
2x / U

cf 
i.e.,
2 w 0.664 


2
Re x x
U 
see below
1L
C f   c f dx  2c f (L)
L0
1.328
=
Re L
UL

Other:

u 
x
*
dy  1.7208
displacement thickness
   1 
U
Re
0

x
measure of displacement of inviscid flow to due
boundary layer

u  u
x

   1 
dy  0.664
U  U
Re x
0
momentum thickness
measure of loss of momentum due to boundary layer
*
H = shape parameter = =2.5916

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