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Transcript
```058:0160
Professor Fred Stern
Chapter 2
1
Fall 2016
Chapter 2: Pressure Distribution in a Fluid
In fluid statics, as well as in fluid
dynamics, the forces acting on a
portion of fluid (CV) bounded by a
CS 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 CS 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 2016

  

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
(one value at a point,
independent of
direction; p is a scalar)
i.e. normal stress (pressure) is isotropic. This
can be easily seen by considering the
equilibrium of a wedge shaped fluid element
Or
p x = p y = p z = pn = p
058:0160
Professor Fred Stern
Chapter 2
3
Fall 2016
 F : p dA sin   p dA sin   0
x
n
x
p p
n
x
pndA α
dl
 F :  p dA cos  p dA cos  W  0
z
n
z
z
y
x
pxdAsinα
α
dA=dldy
W=ρgV=γV
Where:
pzdAcosα
W  V
V  y
1
xz
2
x  l cos 
W  dA cos 
z  l sin 
1
yl  dA  y  dA / dl
dl sin 
2
1
  p dA cos   p dA cos   dA cos  dl sin   0
n
z
2

 p  p  dl sin   0
n
z 2
p  p for dl  0 i.e. p  p  p  p
n
z
n
x
y
z
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
058:0160
Professor Fred Stern
Chapter 2
4
Fall 2016
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).
dy
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
5
Fall 2016
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
f   g (per unit volume)
dF
  gd or
grav
grav
Application of Newton’s law yields: ma   F
da   f d
 a   f  f body  f surface
f
body
  g and g   gkˆ  f
per unit d
body
  g kˆ
z
g
 f
f
pressure
viscous
surface
(includes f
, since in general  ij   p ij   ij )
viscous
f
f
pressure
Viscous part
 p
  2V  2V  2V 
   2V
f
 


viscous
 x 2 y 2 z 2 
For ρ, μ=constant, the viscous force will have this form (chapter 4).
 a  p   g  2V
inertial
pressure
gravity
viscous
with
a
V
 V  V
t
058:0160
Professor Fred Stern
Chapter 2
6
Fall 2016
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   2V  B( x, t )
This is simply a first order PDE for p and can be solved
readily. For the general case (V and p unknown), one
must solve the NS and continuity equations, which is a
formidable task since the NS equations are a system of 2nd
order nonlinear PDEs.
We now consider the following special cases:
1) Hydrostatics ( a  V  0 )
2) Rigid body translation or rotation ( 2V  0 )
3) Irrotational motion (   V
 0)
  (  a)  (  a)  2 a



vector identity
if  
constant
 V  0 
 2V
 0  Euler equation  
also,
V  0  V  & if   const. 2  0
 Bernoulli equation
058:0160
Professor Fred Stern
Chapter 2
7
Fall 2016
Case (1) Hydrostatic Pressure Distribution
p   g    g k
i.e.
or
p p

0
x y
and
z
p
  g
z
2
2
p  p    gdz   g   ( z )dz
2 1
1
1
g
dp   gdz
2
r 
g  g  0
0 r 
 
 constant near earth' s surface r
0
liquids  ρ = constant (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 2016
Chapter 2
8
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
9
Fall 2016
Hydrostatic forces on plane surfaces
The force on a body due to a pressure distribution is:
F    pn 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
Where p is the pressure at the centroid.
058:0160
Professor Fred Stern
Chapter 2
10
Fall 2016
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.
y F   ydF
Note: dF  y sin dA
  sin   y dA
cp
A
2
A
2
 y dA  I  moment of Inertia about O  O
o
A
2
 y A I
= moment of inertia WRT 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
I  product of inertia
where
cp
xy
A
I  I  x yA
I
x 
x
yA
xy
xy
xy
cp
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
058:0160
Professor Fred Stern
Chapter 2
11
Fall 2016
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 perpendicular to x direction
dAy = projection of n dA onto a plane perpendicular to y direction
The horizontal component of force acting on a curved
surface is equal to the force acting on a vertical projection
of that surface including 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
Where h is the depth to any element area dA of the surface.
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.
058:0160
Professor Fred Stern
Chapter 2
12
Fall 2016
Example Drum Gate
h=R-Rcosθ=R(1-cosθ)
p   h   R 1  cos  
h
n   sin  i  cos  k
dA  lRd



F    R1  cos 
 sin iˆ  coskˆ lRd 
  
dA
0
n
p
F .iˆ  Fx  lR

2
 1  cos sin d
0
1

 

 lR 2   cos  0  cos 2 0   2lR 2
4


Same force as that on projection of gate
 
R 2
Rl
p
A
onto vertical plane perpendicular
direction
058:0160
Professor Fred Stern
Fz   lR
2
Chapter 2
13
Fall 2016

 1  cos  cos d
0

 1


 lR 2  sin    sin 2   2lR 2
2 4

0
 R 2 
  
 lR
 l 
2
 2 
2

Net weight of water above curved surface
Another approach:
1


F1    2 R 2l  R 2l  / 2
2


1 

 lR 2  2    / 2
2 

1

F2    R 2l  F1 
2

1
 F  F2  F1  R 2l
2
058:0160
Professor Fred Stern
Fall 2016
Chapter 2
14
Hydrostatic Forces in Layered Fluids
See textbook 2.7
Buoyancy and Stability
Archimedes Principle
F F
F
B
V (2) V (1)
= fluid weight above 2ABC –
= weight of fluid equivalent to the 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.
058:0160
Professor Fred Stern
Chapter 2
15
Fall 2016
Example: Floating body in heave motion
L
y
y
G
ρb
h
d=draft
B
b
hg  mg  0 where  is
Weight of the block W  b Lb
0
A
wp
displaced water volume by the block and  is the specific
weight of the liquid.

W  B  b Lbhg   w Lbdg  d  b h  Sb h


 


w
W
B
b   w : d  h
b   w : d  h sink
b   w : d  h floating
Instantaneous displaced water volume:
  0  yAwp
F
V
..
 m y  B  W    0
  Awp y
058:0160
Professor Fred Stern
Chapter 2
16
Fall 2016
..
m y   Awp y  0
..
y
 Awp
m
y0
y  A cos nt  B sin nt
.
Use initial condition ( t  0,
and B:
y  y0 cos nt 
.
y  y0 y  y 0 )
to determine A
.
y0
n
sin nt
Where
n 
period
T
2

 Awp
m
 2
m
 Awp
Spar Buoy
T is tuned to decrease response to ambient waves: we can
increase T by increasing block mass m and/or decreasing
waterline area Awp .
058:0160
Professor Fred Stern
Chapter 2
17
Fall 2016
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.
Stability: Floating Bodies
For a floating body the situation is 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.
058:0160
Professor Fred Stern
Fall 2016
Chapter 2
18
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.
058:0160
Professor Fred Stern
Fall 2016
Chapter 2
19
Consider a ship which has taken a small angle of heel α
1. α=small heel angle
2. evaluate the lateral displacement of the center of
buoyancy, x = CC
3. then from trigonometry, we can solve for GM and
evaluate the stability of the ship: x =centroid;
CM=GM+CG
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   x 
i i
This can be evaluated conveniently as follows:
moment of  before heel (goes to zero due to
x =
symmetry of original buoyant volume AKKD
- moment of  AOB
+ moment of  EOD
058:0160
Professor Fred Stern
Chapter 2
20
Fall 2016
x     x  d   ( x) d
AOB
tan   y
DOE
d  y dA  x tan  dA
x
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

058:0160
Professor Fred Stern
Chapter 2
21
Fall 2016
GM  CM  CG , i.e. CM=GM+CG
GM 
I 00

 CG
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
058:0160
Professor Fred Stern
Roll:
Chapter 2
22
Fall 2016
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)
M g   GCˆj  CC iˆ   cosˆj  sin iˆ
  GC sin   CC  cos kˆ
  GC  GM sin kˆ
 GM sin kˆ

Note: recall that M  F  d ,
o
where d is the perpendicular
distance from O to the line of
d
action of F .
M G  GZ 
 GM sin  
O
F
058:0160
Professor Fred Stern
Chapter 2
23
Fall 2016
CC ' sin 

CM cos
..
 M  I 
tan  
G
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
k I
m
GM gGM

I
k
mk  I
2
2
m
2
The solution to this equation is,
.
 (t )   cos t 
o
n

sin  t

o
n
n
 = the initial heel angle
where
o

n
= natural frequency

gGM
k2

gGM
k
0 for no initial
velocity
058:0160
Professor Fred Stern
Chapter 2
24
Fall 2016
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
tradeoff” since large GM decreases the period of roll,
which makes for an uncomfortable ride.
058:0160
Professor Fred Stern
Chapter 2
25
Fall 2016
Parametric Roll:
The periodicity of the encounter wave causes variations of
the metacentric height i.e. GM=GM (t). Therefore:
..
I   GM (t )  0
Assuming GM (t )  GM 0  GM1 cos(t ) :
..
I     GM 0  GM1 cos(t )   0 
..
  n2  Cn2 cos(et )   0
where n 
gGM 0
k
; C
GM1
;   mg ; I  mk 2 ; and e  encounter wave freq.
GM 0
By changing of variables (   et
):
..
n2
 ( )   1  C cos  ( )  0 and   2
e
This ordinary 2nd order differential equation where the restoring moment varies
sinusoidally, is known as the Mathieu equation. This equation gives unbounded
solution (i.e. it is unstable) when
n2  2n  1 
  2 
n  0,1, 2,3,..
e  2 
For the principle parametric roll resonance, n=0 i.e.
2
e  2n
2
2
 2
 Tn  2Te
Te
Tn
058:0160
Professor Fred Stern
Chapter 2
26
Fall 2016
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 
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
058:0160
Professor Fred Stern
Chapter 2
27
Fall 2016
The general case of rigid body translation/rotation is as
shown. If the center of rotation is at O where V  V 0 , the
velocity of any arbitrary point P is:
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
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.
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Professor Fred Stern
Chapter 2
28
Fall 2016
1.) Uniform Linear Acceleration
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
058:0160
Professor Fred Stern
Chapter 2
29
Fall 2016
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 .
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  constant
 Gs
gage pressure
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Professor Fred Stern
Chapter 2
30
Fall 2016
2). Rigid Body Rotation
Consider a cylindrical tank of liquid rotating at a constant
rate Ω = Ω k :
p   g  a 
a      r 
0
^
  r e
2

r

p   g  a   gkˆ  r2eˆr
i.e.
p
  r2
r
p
  g
z
058:0160
Professor Fred Stern
and
p
p

2

2
Chapter 2
31
Fall 2016
p  f '   g
r   f ( z)  c
2
2
z
f ( z )   gz  C
r   gz  Constant
2
2
The constant is determined by specifying the pressure at
one point; say, p = p0 at (r,z) = (0,0).

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.
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Professor Fred Stern
Chapter 2
32
Fall 2016
The unit vector in the direction of p is:
 gkˆ  r2eˆr
sˆ 
1/ 2
( g )2  ( r2 )2


dz
tan  
g 2
r
dr
z
r
slope of s
θ
2
dr
2 z

dz   
 ln r
g
r
g
  z
i.e. r  C exp 

 g 
2
1
equation of p surfaces
s
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Professor Fred Stern
Chapter 2
33
Fall 2016
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 / 2V  V  V  (  V )  ( p  z )
 t


Euler Equation
2. Assuming incompressible flow: =constant
V 2 p

V
     gz   V  
t

2

V 2  V V
058:0160
Professor Fred Stern
Fall 2016
Chapter 2
34

3. Assuming steady flow: t  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 2016
4. Assuming irrotational flow: =0
B  0 B= constant (everywhere same constant)
V  
V 2    
     p



  gz   0
2

 t

    p

  gz  B(t )
t
2

B(t)= time dependent constant
Chapter 2
35
058:0160
Professor Fred Stern
Fall 2016
Chapter 2
36
058:0160
Professor Fred Stern
Fall 2016
Chapter 2
37
Larger speed/density or smaller R require larger pressure
058:0160
Professor Fred Stern
Fall 2016
Chapter 2
38
Flow Patterns: Streamlines, Streaklines, Pathlines
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.
058:0160
Professor Fred Stern
Chapter 2
39
Fall 2016
Note:
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 IC (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 IC ( x0 , y0 , z0 , t0 ) then
eliminate t.
058:0160
Professor Fred Stern
Chapter 2
40
Fall 2016
Streakline
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 ( x0 , y0 , z0 ) at t=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  C exp(
)
y  C exp(
)
1
2
1 t
1  2t
s  0 at ( x , y ) : C  x
C y
0 0
1
0
2
0
and eliminating s
058:0160
Professor Fred Stern
Chapter 2
41
Fall 2016
x
y
 (1  2t ) ln
x
y
0
0
x
1 t
y  y ( )n
where n 
0 x
1  2t
0
s  (1  t ) ln
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
1
x  C (1  t ) y  C (1  2t ) 2
1
2
t  0 ( x, y )  ( x , y ) : C  x
0 0
1
0
C y
2
0
now eliminate t between the equations for (x, y)
1
x
y  y [1  2(  1)] 2
0
x
0
This is the pathline through ( x , y ) at t=0 and does not
coincide with the streamline at t=0.
0
0
058:0160
Professor Fred Stern
Chapter 2
42
Fall 2016
3.)
To find the streakline, we use the pathline
equations to find the family of particles that have
passed through the point ( x0 , y0 ) for all times   t .
x  C1 (1  t )
C1 
x0
1 
y  C2 (1  2t )
C2 
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
Fall 2016
Chapter 2
43
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.
Continuity
boundary conditions (4 required):
at infinity :
on body :
u 
U
v    0
y

x
u  v  0    
y
x
058:0160
Professor Fred Stern
Fall 2016
Irrotational Flow
 2  0 2nd order linear Laplace equation
on S :   U y  const.


on S :   const.
B
u 
y

x
v    
x
y
Ψ and φ are orthogonal.
d   dx   dy  udx  vdy
x
y
d   dx  dy  vdx  udy
x
y
dy
u
1
i.e.
 
dx   const
v dy
dx   const
Chapter 2
44
058:0160
Professor Fred Stern
Chapter 2
45
Fall 2016
Geometric Interpretation of 
Besides its importance mathematically
important geometric significance.

also has

= constant = streamline
Recall definition of a streamline:
V  dr  0
dr  dxiˆ  dyˆj
dx dy

u
v
udy  vdx  0
compare with d   dx   dy  vdx  udy
x
y
i.e. d  0 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 2016
Chapter 2
46
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
(note that ψ and Q have same dimensions: m3/s)
i.e. change in d is volume flux and across streamline
2
2
1
1
Q12   V .ndA   d  2   1
Consider flow between two streamlines:
dQ  0 .
058:0160
Professor Fred Stern
Fall 2016
Chapter 2
47
Incompressible Plane Flow in Polar Coordinates
1 
rv r   1  v   0
r r
r 

rv r    v   0
or :
r

continuity :
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 
Incompressible axisymmetric flow
   
1

rv 
v 0
r r r z z
1 
1 
say : v  
v 
r
z r r
r z
1    1     1  
then :
r
 
0
r r  r z  z  r r 
as before d  0 along a streamline and dQ  d
continuity :
058:0160
Professor Fred Stern
Chapter 2
48
Fall 2016
Generalization


( u )  ( v)  0
x
y


define : u 
v  
  compressib le flow stream function
y
x
SL
udy  vdx  0
1
1
compare with
 y dy   x dx  0

d   x dx   y dy 

1

(d )  0 i.e. d  0 and   constant is a streamline
Now:
dm   (V .n)dA  d
2
m 12    (V .n)dA   2   1
1
Change in  is equivalent to the mass flux.
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