KUNDU Fluid Mechanics 2 Episode 12 ppsx
KUNDU Fluid Mechanics 2 Episode 12 ppsx
...
2. 38)
Eq.
(1
2. 38).
and
( 12. 40),
the cigcnvalue problem for determining the marginal stale
(a
=
0)
is
(
D2
-
k2)’i,.
=
(1
+
ax)&,
(
D2
-
k2)2he
=
-Fd
k%,:
(1
2. 87) ... Eq. ( 12. 72) : the iirst and third
of
Eq. ( 12. 72)
are
added; the rest are simply transformed. The result is
iki
+
i,,
=
0.
These equations
are
exactly the sanc
as
Eq.
(...
KUNDU Fluid Mechanics 2 Episode 6 ppsx
... root of
Eq.
(7. 120 ) is
w2
=
gk,
(7. 121 )
which
is
the
same as that for a dcep water gravity wave. Equation (7.1 19) shows that
in this case
b
=
ne-'"!
(7. 122 )
It
.follows ...
the flow.
To
see
this,
consider the mechanical energy of
a
fluid particle at the surrace.
E
=
u2 /2
+
gH
=
Q2/2H2
+
gH.
Eliminatjng
Q
by
Eq.
(7.88)
we obtain, af...
KUNDU Fluid Mechanics 2 Episode 10 ppsx
... nonzero ones
are
pupcd
as
element matrices and
vectors,
At
(1
1 .22 1)
(I
1 .22 2)
(
11 .22 3)
(1
1 .22 5)
(
1
1 .22 6)
Thc prohlern can be nondirnensionalized using the djameter
of
the ... approximations
(
1
1. 12
1
)
and (1 1. 122 ), produccs
an
equation
for
the pressure correction
a:jP;j
+
=
-~~(u;+I/z.j
-
uT-l/z.j)
-
A-r(v;j+i/*
-
~zj-1j2).
(1 1...
KUNDU Fluid Mechanics 2 Episode 1 ppt
...
2
.
Analogy between Heat and Vorticity Diffusion
3
.
Pressure
Change Due
to
Dynamic Effects
20 9
21 3
21 6
21 8
22 1
22 5
22 7
23 0
23 2
23 4
23 8
24 0
24 2
24 5
24 a
25 0
25 4
25 5 ...
22 5
22 7
23 0
23 2
23 4
23 8
24 0
24 2
24 5
24 a
25 0
25 4
25 5
25 6
25 7
26 1
26 2
26 4
26 6
26 8
27 0
27 0
27 0
27 1
27 3
27 3
theory is also given...
KUNDU Fluid Mechanics 2 Episode 2 potx
...
(Figure 2. 7). Using
Eq.
(2. 12) , the components of he
stress
tensor
in
the rotated frame are
431
I&
-43
.;I
=
CllC21t 12
+
C2ICllt21
=
TZU
+
TlU
-
TU,
ti2
=
CllCBt 12
+
~21 ~12t21 ...
whcrc the two are related as
(2. 26)
(2. 27)
As
a check, Eq.
(2. 27)
givcs
R11
=
0
and
R 12
=
-~ 123 ~3
=
-w,
which is in agree-
ment with Eq.
(2...
KUNDU Fluid Mechanics 2 Episode 3 ppt
... vectors
il,
i2, and
i3
changc with
time.
To
this
observer the time derivative
of
P
is
d
(
$)F
=
Z(plil+
P2i2
+
-
.
dPi
.
dP2
.
dP3
dir di2
di3
-
11
-
+
12-
+
1.3 ...
the rotating frame by
n
Figure
4. 12
Coordinate
liamc
(XI.
x2.
x3)
rotating
at
angular
velocity
S2
with
respect
to
a
fixcd
frame
(XI
3
x2.
X:d.
x,
Figure
4.7
Surfa...
KUNDU Fluid Mechanics 2 Episode 4 pdf
...
(5 .22 )
can be written as
-V&nqi&ijkWk,jq
=
-V(&jSqk
-
&~haqj)wk.jq
=
-vWk,nk
+
vOn,jj
=
v%,jj.
(5 .25 )
If
we use
Eqs.
(5 .23 H5 .25 ),
vorticity equation
(5 .22 ) ...
'Itucsdcll,
C.
A.
(19 52) .
Stokes' principle of viscosity.
JoumZ
oJRationol Mechanics
ud
Analysi.s
1:
physical
Journul131:
4 424 47.
Univemity
Press.
22 8 -23 1...
KUNDU Fluid Mechanics 2 Episode 5 potx
...
22
1
Logcr
of
(
hsiant
Depth
22 3
Laycr
of
Vnriahle
Depth
H
(.r)
22 4
Yotilir!fnr Stwpeniirg it1
CI
.Voriili'pmii
.v
Mi~iiinti
22 5
H!rlruulir
Jiunp
22 7 ...
Water
21 2
It~hiem~
of.Su!$iri.
TrLsirm
2
1
3
Slctrrrling
Ilitrw
21 6
Cmup
1
hloixly rurd
I.,'rictg-
Fhx
2
1
8
CNJI.~)
Irlocic!
.
cmd
1Iiri.v
Dispmiirri...
KUNDU Fluid Mechanics 2 Episode 7 pdf
...
(7.168)
whcrc we have
usccl
p”
=
6’
N4p ,2/ 2w2g2,
found from
Eq.
(7.166)
after taking its
real
part. Use of thc dispersion rclation
w2
=
k2N2/(k2
+
m’)
shows that
Ek
=
Ep.
(7.169) ...
26 8
Intcrd
Fmiidt:
Number
26 8
RiclmnLqm
Aimher.
26 9
Mwh
Niuihs
27 0
I’ra~idtl
R’i~nhrx.
27 0
ficrc&es
27 0
Litemlure
Cited
27 0
Siipplementul
Rea...
KUNDU Fluid Mechanics 2 Episode 8 pot
... derivatives from
Eq.
(10 .22 ):
rlll
(10 .26 )
(10 .27 )
(10 .28 )
(10 .29 )
(1
0.30)
dS
3
Uqf"
dS
-
- -
u-
-[f
-
fq]
=
a2+
a+
-
=
Uf',
ay
i 12+
Uf"
a$
6
' ...
=
p(a2+/ay2)oY
where
the
subscript
zero
stands
for
y
=
0.
Ushg
a2$/ay2
=
Uf"/S
given in
Fq.
(20 .29 ).
we
obtain
to
=
pVf
"(0)/6,
and
finally
0.3...