Download Ch2Aug2009

058:0160
Professor Fred Stern
Chapter 2
1
Fall 2009
Chapter 2: Pressure Distribution in a Fluid
Pressure and pressure gradient
In fluid statics, as well as in fluid
dynamics, the forces acting on a
portion of fluid (C.V.) bounded by
a C.S. are of two kinds: body
forces and surface forces.
Body Forces: act on the entire body of the fluid (force
per unit volume).
Surface Forces: act at the C.S. and are due to the
surrounding medium (force/unit areastress).
In general the surface forces can be resolved into two
components: one normal and one tangential to the surface.
Considering a cubical fluid element, we see that the stress
in a moving fluid comprises a 2nd order tensor.
y
σxy
σxz
x
z
Face
σxx
Direction
058:0160
Professor Fred Stern
Chapter 2
2
Fall 2009

  

ij
xx
yx
zx






xy
yy
zy
xz
yz
zz




Since, by definition, a fluid cannot withstand a shear
stress without moving, (deformation) a stationary fluid
must necessarily be completely free of shear stress (σij=0,
i ≠ j). The only non-zero stress is the normal stress,
which is referred to as pressure:
σii=-p
σn = -p, which is compressive, as it should be since fluid cannot
withstand tension. (Sign convention based on the fact that p>0 and in
the direction of –n)
n
Or
(one value at a point,
independent of
direction, p is a scalar)
px = py = pz = pn = p
i.e. normal stress (pressure) is isotropic. This can be easily
seen by considering the equilibrium of a wedge shaped fluid
element
z
F
x
F
: pn dA sin   px dA sin   0
pn  p x
pndA
α
dl
:  pn dA cos   pz dA cos   W  0
pn  p z
1
Where W  gdAcos dl sin 
2
= O  dl 
pxdAsinα
α
dA=dldy
z
W=ρgV
pzdAcosα
058:0160
Professor Fred Stern
Chapter 2
3
Fall 2009
Note: For a fluid in motion, the normal stress is different
on each face and not equal to p.
σxx ≠ σyy ≠ σzz ≠ -p
By convention p is defined as the average of the normal
stresses
p
1
1
 xx   yy   zz     ii

3
3
The fluid element experiences a force on it as a result of
the fluid pressure distribution if it varies spatially.
Consider the net force in the x direction due to p(x,t).
p 

dFx  pdydz   p  dx dydz
x 

p
=
dxdydz
x
net
dy
pdydz
p 

 p  dx dydz
x 

dz
dx
The result will be similar for dFy and dFz; consequently,
we conclude:
 p
p ˆ p ˆ 
dFpress    iˆ 
j  k  

x

y
z 

Or:
f  p
force per unit volume due to p(x,t).
Note: if p=constant, f  0 .
058:0160
Professor Fred Stern
Chapter 2
4
Fall 2009
Equilibrium of a fluid element
Consider now a fluid element which is acted upon by both
surface forces and a body force due to gravity
or f   g (per unit volume)
dF   gd
grav
grav
Application of Newton’s law yields: ma   F
da   f d
a   f  f
f
per unit d
body
surface
f body   g and g   gkˆ  f body   g kˆ
f surface  f
pressure
 f
z
g
viscous
(includes f viscous , since in general  ij   p ij   ij )
f
pressure
Viscous part
 p
  2V  2V  2V 
f viscous    2  2  2    2 V
y
z 
 x
For ρ, μ=constant, the viscous force will have this form (chapter 4).
 a  p   g  2V
inertial
pressure
gradient
gravity
with a 
viscous
V
 V  V
t
This is called the Navier-Stokes equation and will be
discussed further in Chapter 4. Consider solving the N-S
equation for p when a and V are known.


p   g  a   2 V  B( x, t )
058:0160
Professor Fred Stern
Chapter 2
5
Fall 2009
This is simply a first order p.d.e. for p and can be solved
readily. For the general case (V and p unknown), one
must solve the N-S and continuity equations, which is a
formidable task since the N-S equations are a system of
2nd order nonlinear p.d.e.’s.
We now consider the following special cases :
1) Hydrostatics ( a  V  0 )
2) Rigid body translation or rotation (  V  0 )
2
3) Irrotational motion (  V  0 )
  (  a)  (  a)   2 a



vector identity
if  cons tan t
 V  0 

 2 V  0  Euler equation 

 Bernoulli eqn.
also,
V  0  V  & if   const. 2  0
058:0160
Professor Fred Stern
Chapter 2
6
Fall 2009
Case (1) Hydrostatic Pressure Distribution
p   g    g k
p p
i.e.

0
x y
z
g
p
  g
z
and
dp   gdz
2
or
2
2
1
1
p  p    gdz   g   ( z )dz
2
1
r 
g  go  o 
r
 const. near
earth ' s surface ro
liquids  ρ = const. (for one liquid)
p = -ρgz + constant
gases
 ρ = ρ(p,t) which is known from the equation
of state: p = ρRT  ρ = p/RT
dp
g dz

p
R T (z )
which can be integrated if T =T(z) is
known as it is for the atmosphere.
058:0160
Professor Fred Stern
Fall 2009
Chapter 2
7
Manometry
Manometers are devices that use liquid columns for
measuring differences in pressure. A general procedure
may be followed in working all manometer problems:
1.) Start at one end (or a meniscus if the circuit is
continuous) and write the pressure there in an appropriate
unit or symbol if it is unknown.
2.) Add to this the change in pressure (in the same unit)
from one meniscus to the next (plus if the next meniscus
is lower, minus if higher).
3.) Continue until the other end of the gage (or starting
meniscus) is reached and equate the expression to the
pressure at that point, known or unknown.
058:0160
Professor Fred Stern
Chapter 2
8
Fall 2009
Hydrostatic forces on plane surfaces
The force on a body due to a pressure distribution is:
F    p n dA
A
where for a plane surface n = constant and we need only
consider |F| noting that its direction is always towards the
surface: | F |  p dA .
A
Consider a plane surface AB entirely submerged in a
liquid such that the plane of the surface intersects the freesurface with an angle α. The centroid of the surface is
denoted ( x, y ).
F   sin  yA  pA
058:0160
Professor Fred Stern
Chapter 2
9
Fall 2009
Where p is the pressure at the centroid.
To find the line of action of the force which we call the
center of pressure (xcp, ycp) we equate the moment of the
resultant force to that of the distributed force about any
arbitrary axis.
ycp F   ydF
Note: dF  y sin dA
A
  sin   y 2 dA
A
y
2
dA  I o  moment of Inertia about O  O
A
2
 y A I
= moment of inertia w.r.t horizontal centroidal axis
 F  pA   sin  yA
I


2
ycp sin  yA   sin  y A  I
 y  y
cp

I
yA
and similarly for xcp
x F   xdF
cp
A
I
x 
x
yA
xy
cp
where
I  product of inertia
xy
I  I  x yA
xy
xy
058:0160
Professor Fred Stern
Chapter 2
10
Fall 2009
Note that the coordinate system in the text has its origin at the centroid
and is related to the one just used by:
x  xx
text
and
y  y  y 
text
Hydrostatic Forces on Curved Surfaces
z
x
y
In general,
Horizontal Components:
Fx  F  i    p n  i dA
F    p n dA
dAx
A
Fy    p dAy
Ay
dAx = projection of n dA onto a plane perp. to x direction
That is, the horizontal component of force acting on a
curved surface is equal to the force acting on a vertical
projection of that surface (which includes both magnitude
and line of action) and can be determined by the methods
developed for plane surfaces.
Fz   pn  k dA    p dAz    h dAz  
Az
Az
058:0160
Professor Fred Stern
Chapter 2
11
Fall 2009
Where h is the depth to any element area dA of the
surface. That is, the vertical component of force acting on
a curved surface is equal to the net weight of the total
column of fluid directly above the curved surface and has
a line of action through the centroid of the fluid volume.
Example Drum Gate
p   h   R 1  cos  
h
n   sin  i  cos  k
dA  lRd



ˆ  cos kˆ lRd
F   R1  cos  
 sin

i

  
dA
0
n
p
F .iˆ  Fx  lR

2
 1  cos sin d
0
058:0160
Professor Fred Stern
Chapter 2
12
Fall 2009
1

 

 lR 2   cos  0  cos 2 0   2lR 2
4


 
R 2
Rl
Same force as that on projection of gate
A
p
onto vertical plane

Fz   lR2  1  cos   cos  d
0

 1


 lR  sin    sin 2   2lR 2
2 4

0
2
 R 2 
  
 lR
 l 
2
2


2

Net weight of water above surface
Another approach:
1


F1    2 R 2 L  R 2 L  / 2
2


1 

 LR 2  2    / 2
2 

1
1

F2    R 2 L  F1   F  F2  F1  R 2 L
2
2

058:0160
Professor Fred Stern
Chapter 2
13
Fall 2009
Hydrostatic Forces in Layered Fluids
See textbook 2.7 in page 86.
Buoyancy and Stability
Archimedes Principle
FB  FV ( 2 )  FV (1)
= fluid wt. above 2ABC –
fluid wt. above 1ADC
= wt. of fluid equivalent to
body volume
In general, FB = ρg  (  = submerged volume).
The line of action is through the centroid of the displaced
volume, which is called the center of buoyancy.
Example
L
y
y
G
ρb
h
d
B
b
058:0160
Professor Fred Stern
Chapter 2
14
Fall 2009
hg  mg  0 where 
Weight of the block W  b Lb
0
A
wp
is displaced water volume by the block and  is the
specific weight of the liquid.

W  B   b Lbhg   w Lbdg  d  b h


 


w
W
B
Instantaneous displaced water volume:
  0  yAwp
F
V
..
 m y  B  W    0
  Awp y
..
m y   Awp y  0
..
y
 Awp
m
y0
y  A cos nt  B sin nt
.
Use initial condition
and B:
.
y  y0 cos nt 
y0
n
Where
n 
.
( t  0, y  y0 y  y 0 )
 Awp
m
sin nt
to determine A
058:0160
Professor Fred Stern
period
Chapter 2
15
Fall 2009
T
2

 2
m
 Awp
Spar Buoy
We can increase period T by increasing block mass m
and/or decreasing waterline area Awp .
Stability: Immersed Bodies
Stable
Neutral
Unstable
Condition for static equilibrium: (1) ∑Fv=0 and (2) ∑M=0
Condition (2) is met only when C and G coincide,
otherwise we can have either a righting moment (stable)
or a heeling moment (unstable) when the body is heeled.
058:0160
Professor Fred Stern
Fall 2009
Chapter 2
16
For a floating body the situation is slightly more
complicated since the center of buoyancy will generally
shift when the body is rotated, depending upon the shape
of the body and the position in which it is floating.
The center of buoyancy (centroid of the displaced
volume) shifts laterally to the right for the case shown
because part of the original buoyant volume AOB is
transferred to a new buoyant volume EOD.
The point of intersection of the lines of action of the
buoyant force before and after heel is called the
metacenter M and the distance GM is called the
metacentric height. If GM is positive, that is, if M is
above G, then the ship is stable; however, if GM is
negative, then the ship is unstable.
Consider a ship which has taken a small angle of heel α
058:0160
Professor Fred Stern
Chapter 2
17
Fall 2009
1. evaluate the lateral displacement of the center of
buoyancy, CC
2. then from trigonometry, we can solve for GM and
evaluate the stability of the ship
Recall that the center of buoyancy is at the centroid of the
displaced volume of fluid (moment of volume about yaxis – ship centerplane)
x   x d   xi i
This can be evaluated conveniently as follows:
moment of  before heel (goes to zero due to
x =
symmetry of original buoyant volume AKKD
about centerplane)
- moment of  AOB
+ moment of  EOD
x     x  d   ( x) d
AOB
tan   y
x
DOE
d  y dA  x tan  dA
058:0160
Professor Fred Stern
Chapter 2
18
Fall 2009
x   x tan  dA   x tan  dA
2
2
OA
OD
 tan  
ship  waterplane  area
x 2 dA
(moment of inertia of ship waterplane = I00, i.e, Izz)
x  I 00 tan 
CC '  x 
I 00 tan 

CC '  CM tan  (from Section View)
CM 
I 00

GM  CM  CG
GM 
I 00

 CG
058:0160
Professor Fred Stern
Chapter 2
19
Fall 2009
This equation is used to determine the stability of floating
bodies:
 If GM is positive, the body is stable
 If GM is negative, the body is unstable
Roll:
The rotation of a ship about the longitudinal axis
through the center of gravity.
Consider symmetrical ship heeled to a very small angle θ.
Solve for the subsequent motion due only to hydrostatic
and gravitational forces.


F b  cosˆj  sin iˆ g
M g  r  Fb
( g = Δ)
058:0160
Professor Fred Stern
Chapter 2
20
Fall 2009

 
M g   GCˆj  CC iˆ   cosˆj  sin iˆ
  GC sin   CC  cos kˆ
  GC  GM sin kˆ
 GM sin kˆ
tan  
CC ' sin 

CM cos

Note: recall that M  F  d ,
o
where d is the perpendicular
distance from O to the line of
action of F .
M  Gz 
G
 GM sin  
..
 M  I 
I = mass moment of inertia about long axis through G
 = angular acceleration
..
..
I   GM sin   0
.. GM
for small  :  
 0
I
GM  g GM mgGM


I
I
I
k I
m
2
definition of radius of gyration
mk  I
2
m
O
F
G
k I
d
GM gGM

I
k
2
058:0160
Professor Fred Stern
Chapter 2
21
Fall 2009
The solution to this equation is,
.

 (t )   cos t  sin  t

0 for no initial
velocity
o
o
n
n
n
 = the initial heel angle
where
o

n
= natural frequency

gGM
k2

gGM
k
Simple (undamped) harmonic oscillation:
The period of the motion is
T
2

n
T
2k
gGM
Note that large GM decreases the period of roll,
which would make for an uncomfortable boat ride (high
frequency oscillation).
Earlier we found that GM should be positive if a ship
is to have transverse stability and, generally speaking, the
stability is increased for larger positive GM. However,
the present example shows that one encounters a “design
058:0160
Professor Fred Stern
Chapter 2
22
Fall 2009
tradeoff” since large GM decreases the period of roll,
which makes for an uncomfortable ride.
Parametric Roll:
2
2
e  2
 2
 T  2Te
Te
T
GM t    trough
 crest
Case (2) Rigid Body Translation or Rotation
In rigid body motion, all particles are in combined
translation and/or rotation and there is no relative motion
between particles; consequently, there are no strains or
strain rates and the viscous term drops out of the N-S
equation  V  0 .
2
p   g  a 
058:0160
Professor Fred Stern
Chapter 2
23
Fall 2009
from which we see that p acts in the direction of g  a ,
and lines of constant pressure must be perpendicular to
this direction (by definition, f is perpendicular to f =
constant).
Rigid body of
fluid translating
or rotating
The general case of rigid body translation/rotation is as
shown. If the center of rotation is at O where V  V , the
velocity of any arbitrary point P is:
0
V  V 0    r0
where  = the angular velocity vector
and the acceleration is:
dV0
dV
d
a
     r 0  
 r0
 dt
dt
dt




2
1
3
058:0160
Professor Fred Stern
Fall 2009
Chapter 2
24
1
=
acceleration of O
2
=
centripetal acceleration of P relative to O
3
=
linear acceleration of P due to Ω
Usually, all these terms are not present. In fact, fluids can
rarely move in rigid body motion unless restrained by
confining walls for a long time.
1.) Uniform Linear Acceleration
058:0160
Professor Fred Stern
Chapter 2
25
Fall 2009
p=constant
p   g  a   constant

  g  a  k  a i
^
^
z
x

p
   ax
x
1. ax  0
2. ax  0
p increase in +x
p decrease in +x
p
    g  az 
z
p decrease in +z
1. az  0
2. az  0 and az  g p decrease in +z but slower than g
3. az  0 and | az | g p increase in +z
unit vector in the direction of p :
s  p
| p |

 g  az  k  ax i
 g  a z   a x 


2
2
1
2
lines of constant pressure are perpendicular to p .
058:0160
Professor Fred Stern
Chapter 2
26
Fall 2009
n  s j 
ax k   g  az  i
 a   g  az  


2
2
x
1
2
unit vector in direction of p=constant
angle between n and x axes:
  tan
1
a
(g  a )
x
z
In general the pressure variation with depth is greater than
in ordinary hydrostatics; that is:
1
dp
2
2 2
 p  s    ax  ( g  az ) 
ds
G
which is > ρg
p  Gs  cons.
 Gs
gage pressure
2). Rigid Body Rotation
Consider a cylindrical tank of liquid rotating at a constant
rate Ω = Ω k :
p   g  a 
058:0160
Professor Fred Stern
Chapter 2
27
Fall 2009
a      r 
0
^
  r e
2

r

p   g  a   gkˆ  r2eˆr
p
  r2
r
i.e.
and
p
p

2

2
r   f ( z)  c
2
2
p
  g
z
p  f '   g
z
p   gz  f (r )  c
f ( z )   gz
r   gz  const.
2
2
058:0160
Professor Fred Stern
Chapter 2
28
Fall 2009
The constant is determined by specifying the pressure at
one point; say, p = p0 at (r,z) = (0,0).
1
p  p0   gz  r 2 2
2
(Note: Pressure is linear in z and parabolic in r)
Curves of constant pressure are given by:
p p r
z

 a  br
g
2g
2
0
2
2
which are paraboloids of revolution, concave upward,
with their minimum points on the axis of rotation.
The position of the free surface is found, as it is for linear
acceleration, by conserving the volume of fluid.
058:0160
Professor Fred Stern
Fall 2009
Chapter 2
29
058:0160
Professor Fred Stern
Chapter 2
30
Fall 2009
The unit vector in the direction of p is:
 gkˆ  r2eˆr
sˆ 
1/ 2
See Fig 2.23
( g )2  ( r2 )2

on page 99

dz
tan  
g 2
r
dr
  z
i.e. r  C exp 

g


2
1
z
r
θ
slope of s
equation of p surfaces
s
Case (3) Pressure Distribution in Irrotational Flow
(Bernoulli Equation)
 a  ( p)  gkˆ   2 V  ( p  z )   2 V
 V

 V .V   ( p  z )   (.V )    (  V )
 t


Viscous term=0 for =constant and =0, i.e., Potential
flow solutions also solutions NS under such conditions!
1. Assuming inviscid flow: =0
 V

 1 / 2V .V  V  (  V )  ( p  z )
 t


Euler Equation
058:0160
Professor Fred Stern
Chapter 2
31
Fall 2009
2. Assuming incompressible flow: =constant
V 2 p

V
 
  gz   V  
t

2

V 2  V .V
3. Assuming steady flow: ddt=0
B  V  
2
V
p
B
  gz
2

Consider:
B perpendicular B= constant
V   perpendicular V and 
Therefore, B=constant contains streamlines and vortex
lines:
058:0160
Professor Fred Stern
Fall 2009
4. Assuming irrotational flow: =0
B  0 B= constant (everywhere same constant)
5. Unsteady irrotational flow
V  
V 2   .
  V 2 p



  gz   0

 t 2

 V
p

  gz  B (t )
t 2

2
B(t)= time dependent constant
Chapter 2
32
058:0160
Professor Fred Stern
Fall 2009
Chapter 2
33
058:0160
Professor Fred Stern
Fall 2009
Chapter 2
34
Flow Patterns: Streamlines, Streaklines, Pathlines
058:0160
Professor Fred Stern
Fall 2009
Chapter 2
35
1.)
A streamline is a line everywhere tangent to the
velocity vector at a given instant.
2.)
A pathline is the actual path traveled by a given
fluid particle.
3.)
A streakline is the locus of particles which have
earlier passed through a particular point.
Note:
058:0160
Professor Fred Stern
Chapter 2
36
Fall 2009
1. For steady flow, all 3 coincide.
2. For unsteady flow, ψ(t) pattern changes with time,
whereas pathlines and streaklines are generated as
the passage of time.
Streamline
By definition we must have V  dr  0 which upon
expansion yields the equation of the streamlines for a
given time t  t1
dx dy dz


 ds
u
v
w
s= integration parameter
So if (u,v,w) known, integrate with respect to s for t=t1
with I.C. (x0,y0,z0,t1) at s=0 and then eliminate s.
Pathline
The pathline is defined by integration of the relationship
between velocity and displacement.
dx
u
dt
dy
v
dt
dz
w
dt
Integrate u,v,w with respect to t using I.C. ( x , y , z , t )then
eliminate t.
0
Streakline
0
0
0
058:0160
Professor Fred Stern
Chapter 2
37
Fall 2009
To find the streakline, use the integrated result for the
pathline retaining time as a parameter. Now, find the
integration constant which causes the pathline to pass
through ( x0 , y0 , z0 ) for a sequence of times   t . Then
eliminate  .
Example:
by:
u
x
1 t
an idealized velocity distribution is given
v
y
1  2t
w0
calculate and plot: 1) the streamlines 2) the pathlines 3)
the streaklines which pass through ( x , y , z ) at t=0.
0
0
0
1.)
First, note that since w=0 there is no motion in
the z direction and the flow is 2-D
dx
x

ds 1  t
dy
y

ds 1  2t
s
s
x  C1 exp(
)
y  C2 exp(
)
1 t
1  2t
s  0 at ( x0 , y0 ) : C1  x0
C 2  y0
and eliminating s
058:0160
Professor Fred Stern
s  (1  t ) ln
y  y0 (
x n
)
x0
Chapter 2
38
Fall 2009
x
y
 (1  2t ) ln
x0
y0
where
n
1 t
1  2t
This is the equation of the streamlines which pass
through ( x0 , y0 ) for all times t.
2.)
To find the pathlines we integrate
dx
x

dt 1  t
dy
y

dt 1  2t
x  C1 (1  t )
t0
y  C2 (1  2t )
1
2
( x, y )  ( x0 , y0 ) : C1  x0
C 2  y0
now eliminate t between the equations for (x, y)
y  y0 [1  2(
1
x
 1)] 2
x0
058:0160
Professor Fred Stern
Chapter 2
39
Fall 2009
This is the pathline through ( x , y ) at t=0 and does not
coincide with the streamline at t=0.
0
0
3.)
To find the streakline, we use the pathline
equations to find the family of particles that have
passed through the point ( x , y ) for all times   t .
0
x  C1 (1  t )
C1 
x0
1 
y  C2 (1  2t )
C2 
0
1
2
y0
(1  2 )
1
2

x0
1
y
 1  (1  2t )( 0 ) 2  1
x
2
y

y
1  2t
( )2 
x
y0
1  2[(1  t )( 0 )  1]
x
  (1  t )
t  0:
y 
x

 1  2 0  1
y0 
x


1
2
058:0160
Professor Fred Stern
Chapter 2
40
Fall 2009
The Stream Function
Powerful tool for 2-D flown in which V is obtained by
differentiation of a scalar  which automatically satisfies
the continuity equation.
cont.:
ux  vy  0
say:
u   y and v   x


( y )  ( x )   yx  xy  0
x
y
V   y iˆ  x ˆj
then:
by definition!
curl V  kˆ z  kˆ 2
DV
 ( p   z )   2 V)
Dt
 curl (V. V)   2curl V steady
Steady flow


 (u  v )(kˆ 2 )   2 ( kˆ 2 )
x
y
curl ( 
  

 
( 2 ) 
( 2 )    4
x y
 y x


ary conditions (4 required):
at infinity :
u  y  U
v   x  0
on body :
u  v  0   y   x
single scalar equation, but 4th order!
bound
058:0160
Professor Fred Stern
Chapter 2
41
Fall 2009
Irrotational Flow
 2  0 2nd order linear Laplace equation
on S :   U  y  const.
on S B :
  const.
Geometric Interpretation of 
Besides its importance mathematically
important geometric significance.

also has

= constant = streamline
Recall definition of a streamline:
V  dr  0
dx dy

u
v
udy  vdx  0
compare with
i.e. d  0
dr  dxiˆ  dyˆj
d   x dx   y dy  vdx  udy
along a streamline
Or  =constant along a streamline and curves of constant
 are the flow streamlines. If we know  (x, y) then we
can plot  = constant curves to show streamlines.
058:0160
Professor Fred Stern
Fall 2009
Chapter 2
42
Physical Interpretation
dQ  V .ndA
 ˆ  dy ˆ dx ˆ
 (iˆ
j
).( i 
j )  ds 1
y
x ds
ds
  y dy  x dx
 d
i.e. change in d is volume flux and across streamline
dQ  0 .
2
2
1
1
Q12   V .ndA   d  2   1
Consider flow between two streamlines:
058:0160
Professor Fred Stern
Fall 2009
Chapter 2
43
Incompressible Plane Flow in Polar Coordinates
cont.:
or
1 
1 
(rvr ) 
(v )  0
r r
z 


( rvr ) 
(v )  0
r

1 

v  
r 
r
 1 


then
(r
)
(
)0
r r 

r
as before d  0 along a streamline and dQ  d
volume flux  change in stream function
say:
vr 
058:0160
Professor Fred Stern
Chapter 2
44
Fall 2009
Incompressible axisymmetric flow
1 

(rvr )  (vz )  0
r r
t
1 
1 
say: vr  
vz 
r z
r r
1  1 
 1 
then
(r
) (
)0
r r
r z
z r r
as before d  0 along a streamline and dQ  d
cont.:
Generalization
Steady plane compressible flow:


( u )  ( v)  0
x
y


define : u 
v  
y
x
S.L.
udy  vdx  0
compare with
1
1
 dy   x dx  0
 y

d   x dx   y dy 
1

(d )  0 i.e. d  0 and   constant is a streamline
058:0160
Professor Fred Stern
Fall 2009
Now:
dm   (V .n)dA  d
2
m 12    (V .n)dA   2   1
1
Change in  is equivalent to the mass flux.
Chapter 2
45