chp_6.1-chp_6.7.4
TRANSCRIPT
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6. Differential Analysis of
Fluid Flow
6.1 Motion of Fluid Element
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Translation
y
x
Rotation
y
x
Angular Deformation
y
x
Linear Deformation
y
x
Figure 6.1 Pictorial representation of the components of fluid motion
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] ]
TT T T T
T
aDV
Dtu
V
xv
V
yw
V
z
V
tp
Total accelerationof a particle
Convective acceleration Localacceleration
! ! x
x
x
x
x
x
x
x
The acceleration of a fluid particle can be expressed as
!!
!!
!!
t
w
z
ww
y
wv
x
wu
Dt
Dwa
tv
zvw
yvv
xvu
DtDva
t
u
z
uw
y
uv
x
uu
Dt
Dua
zp
yp
xp
xx
xx
xx
xx
xxxxxxxx
xx
xx
xx
xx
(6.1)
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6.1.1 Fluid Rotation
y
x
(y
(xaa
b
b
(F
O
(\
(L(E
T T T T[ [ [ [! i j kx y z
Figure 6.2 Rotation of a fluid element in a two-dimensional flow field
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using a Taylor series expansion, as
Y Y xYx! o xx(
The angular velocity of line oa is given by
tx
tw
ttoa
(((!
(!
p(p(/limlim
00LHE
since
( ( (L
xY
x! x x t
xt
xtxxw
toa
x
xYxxY!
(
(((!
p(
//lim
0
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u uu
yyo!
x
x(
The angular velocity of line ob is given by
t
y
tw
ttob
(
((!
(
(!
p(p(
/limlim
00
\F
( ( (\x
x!
u
yy t
y
u
t
ytyyuw
tob x
xxx!
(
(((!
p(
//lim
0
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The rotation of the fluid element about the z axis isthe average angular velocity of the two mutually perpendicular
line elements, oa and ob, in the xy plane.
-
!
y
u
x
vwz
x
x
x
x
2
1
By considering the rotation of two mutuallyperpendicular lines in the yz and xz planes, one can show that
-
!z
v
y
wwx x
xxx
2
1
- !
x
w
z
uwyx
x
x
x
2
1
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Then,
-
!!
yu
xvk
xw
zuj
zv
ywiwkwjwiw zyx
xx
xx
xx
xx
xx
xx
TTTTTT
2
1
We recognize the term in the square brackets as
xVcurl !
Then, in vector notation, we can write
xVw !2
1(6.3)
(6.2)
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VxwTT
!! 2\
The vorticity, in cylindrical coordinates, is
!
xU
x
x
x
x
x
x
x
x
x
xU
xU
U rzz
zrzr
V
rr
rV
ri
r
V
z
Vi
z
VV
riV
111 TTTT
The Vorticity is a measure of the rotation of a fluid elementas it moves in the flow field. The circulation
is defined as the line integral of the tangential velocitycomponent about a closed curve fixed in the flow.
+
!+c
Vds (6.6)
(6.5)
(6.4)
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velocity components on the boundaries of a fluid elementshown in figure (6.3). For the closed curve oabc
d u xx
x y uu
yy x y+ ( ( ( ( (!
-
Y
xYx
xx
Y(
dx
uy
x y+ ( (!
xY
xxx
yxwd z ((!+ 2uv
O
b
a
c
(x
(y
Figure 6.3 Velocity components onthe boundaries of a fluid element
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dAVxdAwVdsA
z
A
z
C
!!!+T
2
Then,
(6.7)
Equation (6.7) is a statement of Stokes theorem in two dimensions.
Thus the circulation around a closed contour is the sum of
the vorticity enclosed within it [1].
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6.1.2 Angular Deformation
A,A
D
B
C
x
y
D
B
E
E
d
dt
v
x
d
dt
u
y
AB ADE x
x
E x
x! !and
tyx
vHH
x
x
x
v
y
u
dtdtdt
ADABxy
x
x
x
xEHEHHE!! (6.8)
Figure 6.4 Combined effects of rotation and angulardeformation on face ABCD of the infinitesimal fluid element
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x
w
z
u
dt
xz
x
x
x
xEH!
y
w
z
v
dt
yz
xx
xxEH
!
(6.9)
(6.10)
[z xt
[z xt
AAd
B
B
d
C
Cd
DDd
xExyxt/2
xExy xt/2
AAd
B
B
d
C
Cd
DDd
y
x
Fig.6.5 Rotation of face ABCD of the
infinitesimal fluid element
Fig.6.6 Angular deformation of face ABCD
of the infinitesimal fluid element
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-
!!
- !!
-
!!
y
w
z
v
dt
x
w
z
u
dt
x
v
y
u
dt
yz
yz
xzxz
xy
xy
x
x
x
xQ
HEQX
xx
xxQHEQX
xx
xx
QHE
QX
(6.11)
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Example 6.1
For a certain two-dimensional flow field the velocity
is given by the equation
V= 4xyi+ 2 (x2-y2)j
Is this flow irrotational?
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Solution
For an irrotational flow the rotation vector, w, must be zero.
For the prescribed velocity field
u = 4xy v= 2 (x2-y2) w= 0
and therefore 02
1
!
-
! zv
y
wwx x
xxx
02
1!
-
!x
w
z
uwy x
xxx
0)44(2
1
2
1!!
-
! xx
y
u
x
vwz x
xxx
Thus, flow is irrotational
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Shear Deformation
Angular Motion and Deformation
Fluid elements located in a moving fluid move
with the fluid and generally undergo a change
in shape (angular deformation). A small
rectangular fluid element is located in the
space between concentric cylinders. Th
e innerwall is fixed. As the outer wall moves, the fluid
element undergoes an angular deformation.
The rate at which the corner angles change
(rate of angular deformation) is related to the
shear stress causing the deformation.
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6.2 Conservation of Mass
Conservation of mass requires the mass M, of a system
remains constant as the system moves through the flow fields
0!Dt
DMsys(6.12)
The control volume representation of the conservation of masscan be written as
!CV CS
AdnVdt
0.TT
VVxx
(6.13)
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6.2.1 Continuity Equation
The continuity equations are developed from
the general principle of conservation of mass as;
0 ! x
xV V
td VdA
CSCV
T T(6.14)
Assuming that the flow is steady the equation 6.14 will become;
!CS
AdV 0TT
V
Applying the equation to the control volume shown in the figure 6.7
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V V1 1 1 2 2 2V A V A!
m V A V A! !V V1 1 1 2 2 2(6.15)
m Q Q
Q Q A V A V
! !
! ! ! !
V V
V V
1 1 2 2
1 2 1 2 1 1 2 2
dA11
1
V2dA2
V1
controlvolume
Figure 6.7 Steady flow through a streamtube
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The average velocity over a cross-section is given by
VA
udA! 1
(6.16)
Vu HyHz +
Vv HxHz +
Vv HxHz
Vu HyHz
y
x V[ HxHy
V[ HxHy +yz)x( HHHV
x
xv
yzy)x( HHHV[
x
x
z
xz)y( HHHVxx
ux
Figure 6.8 Differential control volume in rectangular coordinates
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The net mass flow in x,y,z directions
zyxux
xzyux
zyuzyu HHHVx
xHHHV
x
xHHVHHV !
-
zyxvy
yzxvy
zxvzxv HHHVx
xHHHV
x
xHHVHHV !
-
zyxwz
zyxwz
yxwyxw HHHVx
xHHHV
x
xHHVHHV !
-
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The total net mass flow through the control volume will be
zyxwz
vy
ux
HHHVxxV
xxV
xx
- (6.17)
The rate of change of mass inside the control volume is given by;
zyxt
HHHx
xV(6.18)
This rate of change of mass is equal to the net mass which flowsthrough the control volume. Hence, from the equations (6.17) and (6.18);
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x
x
x
x
x
x
u
x
v
y
w
z ! 0
(6.19)
Equations 6.18 and 6.19 may be compactly written in vector notation.
By using fixed unit vectors in x,y,z directions,TTTi j k, , respectively, the operator
TV is defined as;
! T T Ti
xj
yk
z
x
x
x
x
x
x
and the velocity vector q is given by
(6.20)
wkvjuiVTTT
!p
(6.21)
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wkvjuiz
ky
jx
iV VVVxx
xx
xx
VTTTTTT
!
p
..
! x
xV
x
xV
x
xV
xu
yv
zw
Then;
because
TT TTi i i j. , .! !1 0 etc.
tV x
xVV !
(6.22)
and equation (6.19) 0. ! V
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dU
U
VU Vr
Vz
r
Typical (r,U,z) point
dr
dzr
Cylindrical Axisz
Typical Infinitesimal Element
Figure 6.9 Definition sketch for the cylindrical coordinate system
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The net rate of mass flow through the control surface
VxV
x
xV
xU
xV
xU
UV
r V
r
r V r V
zdrd dzr
r z
-
(6.23)
The rate of change of mass inside the control volume
xV
xU
trdrd dz (6.24)
Hence, the continuity equation in cylindrical coordinates;
VxV
x
xV
xU
xV
x
xV
xUV
r V
r
r V r V
zr
tr
r z ! 0
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dividing by r gives
V xV
xxV
xUxV
xxVx
UVV
r r
V V
z tr
r z !1
0
1 1 0r
r Vr r
V Vz t
r zx Vx
xVxU
xVx
xVx
U !
or
(6.25)
If the flow is steady x x/ t ! 0
1 1
0r r
r Vr
Vz
Vr zx
xV
x
xUV
x
xVU ! (6.26)
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If the flow is incompressible;
xV x/ t $ 0
whether flow is steady or unsteady the equation (6.26) will be;
1 1 0r r
rVr
Vz
Vr zxxx
xUxxU
! (6.27)
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Example 6.2
The velocity components for a certain incompressible, steady flow field are
u = x2 + y2 + z2
v= xy + yz + z
w= ?
Determine the form of the z component,
w, required to satisfy the continuity equation.
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Solution
Any physically possible velocity distribution must
for an incompressible fluid satisfy conservation
of mass as expressed by the continuity equation
xx
xx
xx
ux
vy
wz
! 0
For the given velocity distribution
xx
u2!
xx zx
y
v!
xx
and
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so that the required expression forxw/xz is
zxzxxz
w!! 3)(2
x
x
Integration with respect to z yields
),(23
2
yxf
z
xzw !
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6.3 Stream Function for Two-Dimensional Incompressible Flow
For a two-dimensional incompressible flow in the x-y planeconservation of mass can be written as
xx
xx
u
x
v
y ! 0 (6.28a)
If a continuous function, (x,y,t), called the stream function,is defined such that
u
y
and v
x
! ! x]
x
x]
x
(6.28b)
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then the continuity equation is satisfied exactly, since
x
x
x
x
x ]
x x
x ]
x x
u
x
v
y x y y x !
2 2
(6.29)
As long as the velocity vectorT T T TV iu jv kw! is parallel to
dr idx jdy kdzT T T T
! , then on the streamline
T TV x dr! 0
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Instantaneous Streamline
y
x
z
X Tr dr
Tr
drT
TV
Figure 6.10. Relation between the velocity vectorand the instantaneous streamline
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Hence the equation of the streamline is
dxu
dyv
dzw
! ! (6.30)
since, the flow is assumed two-dimensional. Therefore
udy vdx ! 0 (6.31)
equation (6.31) is the equation of a streamline in a two-dimensional flow.
Substituting for the velocity components u and v in terms of the stream
function, , from equation (6.28), then along a streamline
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x]
x
x]
xxdx
ydy ! 0 (6.32)
since = (x,y,t), then at an instant,
to, = (x,y,t); at this instant, a change in may be evaluated
as though =(x,y). Thus, at any instant,
dxdx
ydy] x]
xx]x
! (6.33)
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comparing eqs. (6.38) and (6.39), we see that along an instantaneous
streamline, d =0; is a exact, the integral of d between any two points
in a flow field, 2- 1, depends only on the end points of integration
vu
vA
BC D
FE
]3
]2
]1
x
y
Figure 6.11 Instantaneous streamlines in a two-dimensional flow
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from the definition of a streamline, we recognize that there can be no
flow across a streamline. As seen from figure 6.11 that the rate of
flow between streamlines 1 and 2 across the lines AB, BC, DE, and
DF must be equal.
The volume flow rate, Q, between streamlines 1 and 2 can be evaluated
by considering the flow across AB or across BC. For a unit depth, the
flow rate across AB is [1];
Q udyy
dyy
y
y
y
! ! 1
2
1
2 x]
x(6.34)
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Along AB, x=constant, and d] x] x! y dy , therefore,
Qy
dy dy
y
y
y
! ! ! x]
x] ] ]
1
2
1
2
2 1(6.35)
For a unit depth, the flow rate across BC is
Q vdxx
dxx
x
x
x
! ! x]x
1
2
1
2
(6.35)
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Along BC, y=constant, and d x dx] x] x! / . Therefore, ?1A;
Qx
dx dx
x
! ! ! x]
x] ] ]
]
]
1
2
1
2
2 1(6.36)
In the r plane of the cylindrical coordinate system, the incompressible
continuity equation can be written as
xx
xxU
UrV
r
Vr ! 0 (6.37)
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y
xO
o
P
Pd
vu
dx
dy
O x
r
U
PPd
Vr
VUrdU
drrx
v
ry
u
x
x]
x
x]xU
x]
x
x]
U !!
!!
V
1V r
Figure 6.12 Selection of path to show relation of velocitycomponents to stream function
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To define a stream function as = (r,), then the velocity components
Vr
and Vr
r ! ! 1 x]
xU
x]
xU (6.38)
For rz plane of the cylindrical coordinate system,
the incompressible two-dimensional continuity equation
10
r
rV
r
V
z
r zx
x
x
x
! (6.39)
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The stream function =(r,z) and the velocity components
V
r z
and V
r r
r z! ! 1 1x]
x
x]
x
(6.40)
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The expressions for the body and surface forces can now be used todevelop the equations of motion.
xx amF HH !
yyamF HH !
zz amF HH ! ,
Figure 6.13 Surface forces in the x direction acting on a fluid element.
6.4 Equation Of Motion
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where zyxm HHHVH !and the acceleration components are given by Eq. 6.1. It now follows that
xx
xx
xx
xx!
xx
xx
xx
z
uw
y
uv
x
uu
t
u
zyxg zx
yxxxx V
XXWV (6.41a)
x
x
x
x
x
x
x
x!
x
x
x
x
x
x
z
vw
y
vv
x
vu
t
v
zyx
gzyyyyx
y VXWX
V (6.41b)
xx
xx
xx
xx
!x
x
x
x
xx
z
ww
y
wv
x
wu
t
w
zyxg zz
yzxzz V
WXXV (6.41c)
where the element volume zyx HHH cancels out.
Equations 6.41 are the general differential equations of motion for a fluid
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6.5 Equation Of Motion For Incompressible Flow
6.5.1 Momentum Equation (Equations of Motion) for Frictionless Flow:
Euler's Equations
Applying Newton's second law of motions to an infinitesimal fluidelement of, dm, with sides dx, dy, and dz, as shown in figure 6.14 gives
dF a dm
T T!
or
dFDV
Dt
dxdydzT
T
! V
(6.42)
(6.43)
There are in general two types of forces acting on a fluid.These are the body forces and the surface forces(the normal stress is the negative of the thermodynamic pressure ?3A).
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yP
dz
dx
O
z
y
x
PP
x
dx
dydz i
x
x 2 ( )T
PP
z
dzdxdy k
x
x 2( )
T
P
P
y
dydxdz j
x
x 2 ( )
T
PP
z
dzdxdy k
xx 2 ( )
T
PP
x
dxdydz i
x
x 2( )
T
P P
y
dydxdz j
x
x 2( )T
Figure 6.14 The forces acting on an infinitesimal inviscid fluid
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The evaluate the properties at each of the six faces of the control surface,we use a Taylor series expansion about point p. The pressure on
the left face of the infinitesimal fluid is
AP PP
y
dy P
y
dyy dy !
/ ! !. . .
2
2
2
21
1 2
1
2 2
xx
x
x
! PP
y
dyx
x 2
and at the right face of the infinitesimal fluid element
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P P
P
y
dy P
y
dyy dy !
/ ! ! . . .2
2
2
21
1 2
1
2 2
x
x
x
x
! PP
y
dyx
x2
The net surface force acting on the element is
dF PP
x
dxdydz i P
P
x
dxdydz i P
P
y
dydxdz js
T T T T!
x
x
x
x
x
x2 2 2
P
P
y
dydxdz j P
P
z
dzdxdy k P
P
z
dzdxdy k
x
x
x
x
x
x2 2 2
T T T
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or
dFP
xi P
yj P
zk dxdydz Pdxdydzs
TT T T!
-
! xx
xx
xx
The total force acting on the infinitesimal element is
PdxdydzdxdydzgFdFdFd sb !! TTTT V
Finally equation (6.43) becomes
V VDV
Dtg P
TT
! (6.44)
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That is known as the Euler's equation
Vx
xV
TT T TV
tV V g P
-
! . (6.45)
Equation (6.45) may be expressed in terms of three scalar equations as
aDu
Dt
u
tu
u
xv
u
yw
u
z
P
xgx x! ! !
x
x
x
x
x
x
x
x V
x
x
1
a DvDt
vt
u vx
v vy
w vz
P
ygy y! ! ! x
xxx
xx
xx V
xx
1
aDw
Dt
w
tu
w
xv
w
yw
w
z
P
zgz z! ! !
x
x
x
x
x
x
x
x V
x
x
1
(6.46a)
(6.46b)
(6.46c)
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The momentum equation for inviscid flow also can be written incylindrical coordinates. The equations are
aV
tV
V
r
V
r
VV
V
z
V
r
P
rgr
rr
r rz
rr! !
xx
xx
xxU
xx V
xx
U U2
1
aV
tV
V
r
V
r
VV
V
z
V V
r r
Pgr z
rU
U U U U U UU
xx
xx
xxU
xx V
xxU
! ! 1
aV
tV
V
r
V
r
VV
V
z
P
zgz
zr
z zz
zz! !
xx
xx
xxU
xx V
xx
U 1
(6.47b)
(6.47c)
(6.47a)
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or
-
x
x! VV
t
Vg .VVV (6.48)
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6.5.2 Irrotational Flow
021 !
xx
xx!
yu
xv
z[ and, therforeyu
xv
xx!xx(6.49)
Similarly
z
u
y x
x!
x
x[(6.50) and
xz
u
x
x!
x
x [(6.51)
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Figure 6.15 Various regions of flow: (a) around bodies; (b) through channels
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6.5.3 Bernoulli Equation Applied to Irrotational Flow
The Eulers equation have been integrated along a streamline for steady,
incompressible, inviscid flow to obtain the Bernoulli equation ?1A;
P Vgz
V
2
2=constant (along a streamline) (6.52)
This equation (6.52) can be applied between any two points on the same streamline.The value of the constant will vary, in general, from streamline to streamline.
If, in addition to being inviscid, steady and incompressible, the flow field is alsoirrotational
(the velocity field is such that 02 !! VTTT
[ ).
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we see that for irrotational flow, since
0. ! VTT
, then ).(2
1
).( VVVV
TTTT
!
and Eulers equation for irrotational flow can be written as;
)(2
1).(
2
11 2VVVzgPTTT
!!V
(6.54)
During the interval, dt, a fluid particle moves from the vector positionTr
, to the position,T Tr dr
; the displacement, dTr, is an arbitrary infinitesimal displacement in any direction.
Taking the dot product of dr dxi dyj dzkT T T T
!
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with each of the terms in equation (6.54), we have
! 1 1
2
2
VPdr g zdr V dr
T T T T( )
and hence,
!dP
gdz d VV
1
2
2( )T
ordP
gdz d VV
!1
202( )
T
integrating this equation gives
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dPgz
V
V
2
2
=constant (6.55)
For incompressible flow, V= constant, and
Pgz
V
V
2
2=constant (6.56)
Since dTr
inviscid flow that is also irrotational, equation (6.56) is valid between anytwo points in the flow field ?1A.
was an arbitrary displacement, then for a steady, incompressible,
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6.6 Inviscid Incompressible Flow
6.6.1 Stream Function
The function is known as the stream function.
Figure 6.16 definition of a stream line
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can also be defined in terms of the velocity components. Thus
x
-=V
y
U
x
x
x
x! (6.57)
The continuity equation, in cartesian coordinates,for the flow field under consideration is
x
x
x
x
U
x
V
y ! 0
Now introduce a function which is defined as equation 6.57
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W
V
x
U
yz !
-
1
2
x
x
x
x
The condition of irrotationality is
^x
x
x
x! !
-
!2 0W
V
x
U
yz
(6.58)
(6.59)
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x
x
x
x
2
2
2
2
= =
x y
Substituting equation 6.57 into equation6.58 will let us have the following equation
That is called vorticity ().
(6.60)
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udyvdxdyydxxd !
=
=
!= x
x
x
x
Figure 6.17 Two streamlines showing the components of thevolumetric flow rate across an element of controlsurface joining the streamlines.
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Where equation 6.57 has been used.
Then the eq. of the line =constant will be
0=-vdx+udy ordy
dx
v
u
-
!
=
The total volume of fluid flowing between the streamlines per-unit timeper-unit depth of flow field will be
Q udy vdxA
B
A
B
!
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But it was observed earlier that
d= ! vdx udy
so that, integrating this expression between the two points A and B,it follows that (10)
!==B
A
B
A
vdyudx12
comparing these two expression confirms that 2-1=Q
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Example 6.3
The velocity components in a steady, incompressible, two-dimensional flow field are
xv
yu
4
2
!!
S
olution
From the definition of the stream function
yy
u 2x
=x! x
xv 4!
x
=x!and
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The first of these equations can be integrated to give
)(1
2xfy !=
where
)(1
xf is an arbitrary function of x. Similarly from the second equation
)(22
2 yfx !=where
)(2
xf is an arbitrary function ofy.It now follows that in order to satisfy both expressions for thestream function
cyx != 222
where C is an arbitrary constant.
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we set C = 0
the stream function is
222 yx !=
2220 yx ! or xy 2s!
12/
22
!=
=
xy
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6.6.2 Potential Function
The function is defined by
ux
! xJ
x
xJ
x, V = +
y
Now for irrotational flow
x
x
x
x
x
x
xJ
x
x
x
xJ
x
v
x
u
y y x!
-
!
-
x y
x J
x x
x J
x x
2 2
x y x y! (6.61)
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If the expression for u and v are substituted in the continuity equationthe following result is obtained
x
x
x
x
u
x
v
y = 0
x y
x
x
xJ
x
x
x
xJ
xx y
-
-
! 0
x J
x
x J
x
2
2
2
2x y = 0
p
!2
0Jor
This is the well-known Laplace equation.
(6.62)
xJ xJ
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dxdx
ydyJ
xJx
xJx
! ! 0 and for constant
dx
dxy
dy= = =! ! 0 xx
xx
from the first equation.
dy
dx
x
y
u
vconstJ
xJ xxJ x
!
! !
.
/
/and for constant
dy
dx
x
y
v
uconst=
=
=!
! !.
/
/
x x
x xThus
dy
dx
dy
dxconst const
-
-
!
! !J . .=
1 (6.63)
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Figure 6.18 shows such a network for irrotational, incompressible, two-dimensional flow in a reducing elbow (12)
Figure 6.18 Orthogonal flow not for the flow in a two-dimensional reducingelbow. The net is formed by lines of constant and constant .
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Example 6.4
The two-dimensional flow of a nonviscous, incompressible fluid in thevicinity of the corner of Fig. E6.4a is described by the stream function 90
U] 2sin2 2r!
where has units of m2/s when r is in meters.
(a)Determine, if possible, the corresponding velocity potential.
(b) If the pressure at point (1) on the wall is 30 kPa, what is the pressure atpoint (2)? Assume the fluid density is 10 kg/m3 and the xy plane ishorizontal that is, there is no difference in elevation between points(1)and (2).
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Figure E 6.4
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Solution
(a) The radial and tangential velocity components can be obtained from the
stream function as
UU
]2cos4
1r
rvr !
x
x! U
]U 2sin4r
rv !
xx
!and
since
rvr
x
x!
Jit follows that U
J2cos4r
r!
xx
and therefore by integration
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)(2cos21
2 UUJ fr !
where )(1 Ufis an arbitrary function of
U. Similarly
UU
JU 2sin4
1r
rv !
x
x!
and integration yields
)(2cos22
2 rfr ! UJ
(1)
(2)
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where
)(2
rf is an arbitrary function of r. To satisfy both Eqs. 1 and 2,the velocity potential must have the form
Cr ! UJ 2cos2 2 (ans.)
where C is an arbitrary constant. As is the case for stream functions,the specific value ofC is not important, and it is customary to let C = 0so that the velocity potential for this corner flow is
UJ 2cos2 2r! (ans.)
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6.6.3 Stream Function And Velocity Potential
For a two-dimensional, incompressible, irrotational flow we have
expression for the velocity components, u and v, in terms of both
the stream function , and the velocity potential, ,
Figure 6.19 Selection of path to show relation of velocity componentsto stream function.
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u y x x y! ! ! !
x]
x
xJ
x
x]
x
xJ
x, v
(6.64)
and in plane polar coordinates (r,) the velocity components, u and v, interms of both the stream function, , and the velocity potential, ,
Vr r r r r
! ! ! ! 1 1x]
xUxJx
xJxU
x]xU
, V
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Example 6.5
Consider the flow field given by =ax2-ay2, where a=1 sec-1
a) Show that the flow is irrotationalb) Determine the velocity potential for this flow
Solution:
If the flow is irrotational, then Wz=0
since 2WV
x
U
yz!
-
xx
xx
and uy
!x]
x
x]
x, v = -
x
then uy
ax ay ay! ! x
x( )2 2 2 and v
xax ay ax! !
xx
( )2 2 2
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Thus wv
x
u
y xax
yay a a
w
z
z
2 2 2 2 2 0
2 0
! ! ! !
! !
x
x
x
x
x
x
x
x
^
( ) ( )
Therefore, the flow is irrotational. The velocity component can be
written in terms of the velocity potential as
ux
! xJ
xv
y!
xJ
xConsequently u
xay! !
xJx
2
Integrating with respect to x gives
J ! 2ayx f y( )
where f(y) is an arbitrary function of y.
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Then
y
fax
y
yfaxax
yfaxyyy
axv
x
x
x
x
xx
xxJ
!!
!!!
2)(
22
)(2(2
Sodf
dy! 0 and f is constant. Therefore
constant2 ! axyJ
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We also can show that lines of constant and are orthogonal.
] ! ax ay2 2
For =constant d axdx aydy] ! ! 0 2 2
Hence
dy
dx
x
yc
-
!
!]
dx
dxy
dy udx vdyJxJ
x
xJ
x! !
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6.6.4 Circulation And Vorticity
The rotation about the z axis of a fluid particle was shown to be the
following equation:
w vx
uyz
!
-
1
2xx
xx
This can be further extended to the more general concepts of circulation
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and vorticity all of which are interrelated. The general definition ofcirculation is
!+ dsEV cos (6.65)This is the line integral of the component of velocity along a lineelement around a closed or complete control (11).
The vorticity (xi) is defined as the circulation / units area, or moreexplicitly the vorticity at a point is
Limareap0
Area
+
It can easily be shown that =2
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The value of circulation is the strength of the vortex.
Figure 6.20 circulation
For example consider an infinitesimal rectangle of sides dx dy (see
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For example, consider an infinitesimal rectangle of sides, dx, dy (seefigure 6.20). Then, the circulation round the rectangular element is
x x xx
x x xx
x x x+ !
u x v
u
xx y u
u
yy x v y
Taking positive in the anticlockwise sense
xxx
xx
x x+ !
v
x
u
yx y or
x :x x [x x+ ! !x y x y2 (6.66)
6 6 5 El t Pl Fl S l ti
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6.6.5 Elementary Plane-Flow Solutions
Several potential-flow problems of interest can be constructed fromthree types of elementary solutions:
1) Uniform stream2) Source or sink3) Vortex
6.7.1 Uniform Stream
The simplest flow pattern are those in which the streamlines are allstraight lines parallel to each other. First let the streamline be in the xdirection and solve for the resulting and functions
u uy x
Cons t! ! ! !gx
x
x
x
= *tan
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yx
v
x
x
x
x *!
=!! 0 (6.67)
integrating we obtain
=uy+C1 =ux+C2 (6.68)
Now the integration constant C1 and C2 have no effect whatever onvelocities or pressure in the flow. Therefore we shall consistently ignoresuch irrelevant constants and for a uniform stream in the x direction (5)
=u y =u x
These are plotted in figure 6.21 and consist of a rectangular mesh ofstraight stream-lines normal to straight potential lines.
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Figure 6.21 Flow net for a uniform stream: (a) stream in the x direction;(b) stream at angle .
In terms of plane polar coordinates (r,) equation 6.69 becomes
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uniform stream =u rsin = u rcos
The plot of course is exactly the same as figure 6.21a if we generalize toa uniform stream at an angle to the x axis, as in figure 6.21b, werequire that
xy
vv
xyuu
x
x
x
xE
x
x
x
xE
*!
=!!
*!
=!!
g
g
sin
cos
(6.70)
(6.71a)
(6.71b)
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= -v.x + u.y = -u .Sinx + u .Cosy
= u.x + v.y = u .Cosx + u .Siny
integrated we obtain for a uniform stream at angle ( )
= u .(y. Cos - x. Sin)
= u .(x. Cos + y. Sin)
These are useful in airfoil angle of attack problems.
(6.72a)
(6.72b)
Example 6.6
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pObtain the flow field streamlines fora) u = 20 m/sec in the direction of +ve x valuesb) v = -10 m/sec in the direction of -ve y valuesc) A combined of (a) and (b)
Solution:
a) v = 0 , u = 20 m/sec
= - v. dx + u. dy taking the integration
= -v.x + u.y + c since v = 0 => = + u.y + c
Boundary condition 0 = 0 when x =0 ; y = 0 . Hence c = 0 , then ;
= +u.y
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Consider y = -0.2 m with u = 20 m/sec or y = -0.4 , -0.6, -0.8, -1 with
u = 20 m/sec
1 = -u.y = (+20).(-0.2) = -4
Figure E 6.6 Straight line flow
b) u = 0 , v = -10 m/sec
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= -v.x + u.y + c
Taking the boundary conditions as 0 = 0 when x = 0 and y = 0 then c = 0
since u = 0 and c = 0 the equation will be as ;
= -v.x Taking the value of x as x = 0.2, 0.4, 0.6, 0.8, 1 with v = -10 m/sec
Then = -v.x = -0.2 . (-10) = 2 => =2 so on,
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c) A combination a and b
Here = -v.x + u.y = -(-10).x + 20.y
or this may evaluated by summing the separate fields of (a) and (b) as
= a+ b
= -vx + uy = -10.x + 20.y
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y y
1 = -(-10 . 0.2) + 20. 0.2 = -2
= -(-10. 0.6) + 20.(-0.4) = -2'
1
9=
6 7 2 Line Source or Sink
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6.7.2 Line Source orSink
A line normal to the xy plane, from which fluid is imagined to flowuniformly in all directions at right angles to it, is a source. It appears as apoint in the customary two-dimensional flow diagram. The total flow perunit time per unit length of line is called the strength of the source (6).
In steady flow the amount of fluid crossing any given cylindricalsurface of radius r and length b is constant.
Q = Vr.(2 rb) = constant = 2 b.m
or
Vr,source = m/rWhere m is a convenient constant. This is called a LINE SOURCE if m ispositive and a LINE SINK if m is negative. Obviously the source streamlinesflow outward, as in fig.6.22.a and the tangential velocity V= 0. So, we cansolve for the plane polar version of and .
(6.73)
xxx
x 1
0V1m
V
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x!
x!!
x!
x!!
rr
0V
r
rrV r
Integrating, we obtain a line source or sink
= m. and = m.Lnr
Figure 6.22 Flow net for a line source and line vortex a) Line source, b) Line vortex
(6.74)
The equivalent cartesian forms are
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The equivalent cartesian forms are
= m.tan-1(y/x) and = m.Ln(x2 + y2)1/2 (6.75)
6.7.3 Line Vortex
Reversing the roles of and in equation 6.74 yields
= -K.Lnr and = K. (6.76)
By direct differentiation of either one we obtain the velocity pattern (seefig.6.22b)
Vr= 0 and V =K/r (6.77)
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The vortex strength K has the same dimensions as the source strengthm, namely, velocity times length (m/sec . m or v x L)
6.7.4 Circulation
The line-vortex flow is irrotational everywhere except at the origin,
where the vorticity p p
x V
This means that a certain line integral called the fluid circulation
does not vanish when taken around a vortex center.
is infinite.
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Figure. 6.23 Definition of the fluid circulation .
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!!!c c c
dzudyvdxds.Vdscosv
starting and ending at the same point will lead to have the circulation as
(6.78)
= 0
path enclosing a vortex = 2 K
(6.79)
(6.80)
Alternatively the calculation can be made by defining a circular path ofradius r around the vortex center, from equation 13.22
!
!!
2
0c
2rdrK.dsV J (6.81)
Example 6.7
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A liquid drains from a large tank through a small opening as illustrated inFig. E6.7. A vortex forms whose velocity distribution away from the tank
opening can be approximated as that of a free vortex having a velocitypotential
U
T
J2
+!
Determine an expression relating the surface shape to the strength ofthe vortex as specified by the circulation +
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Figure E.67
Since the free vortex represents an irrotational flow field the Bernoulli
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Since the free vortex represents an irrotational flow field, the Bernoulliequation
2
2
22
1
2
11
22z
g
VPz
g
VP!
KK
can be written between any two points. If the points are selected at thefree surface, P1 = P2= 0 so that
g
Vz
g
VS
22
2
2
2
1 ! (1)
where the free surface elevation, is measured relative to a datumpassing through point (1).
The velocity is given by the equation
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y g y q
rrv TU
JU
2
1 +!x
x!
We note that far from the origin at point (1) 01
}! UvV
so that Eq. 1 becomes
grzS 228T
+
! (ans)
Vortex in a Beaker
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Vortex
A flow field in which the streamlines
are concentric circles is called a vortex. A
vortex is easily created using a magnetic
stirrer. As the stir bar is rotated at the
bottom of a beaker containing water, the
fluid particles follow concentric circularpaths. A relatively high tangential velocity
is created near the center which decreases
to zero at the beaker wall. This velocity
distribution is similar to that of a free
vortex, and th
e observed surface profilecan be approximated using the Bernoulli
equation which relates velocity, pressure,
and elevation.