Homework: Ideal Flows
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Solution of a the velocity field around a cylinder, and pressure distribution on the cylinder. Stream-function of an asymptotic suction profile. Hill's spherical vortex
Transcript of Homework: Ideal Flows
THE UNIVERSITY OF TEXAS AT DALLAS, MECHANICAL ENGINEERING DEPARTMENT
Incompresible Fluid Mechanics, MECH-6370,HW 3
Edgardo Javier Garca Cartagena
October 21, 2015
PROBLEM 1
The streamfunction for flow over a circular cylinder is =Ur sin(1 r
20
r 2
)Evaluate the pres-
sure distribution on the cylinder.
vr = 1r
=U(
1 r20
r 2
)cos()
v =
r=U
(1+ r
20
r 2
)sin()
On the surface of the cylinder (r = r0) vr = 0 and v =2Usin. Using the Bernoulli equationwe obtain:
P0+ 12U2 = Ps + 1
2v2s
Ps = P0+ 12U2 1
2
(U(
1+ r20
r 2
)sin()
)2
Ps = P0+ 12U2 1
2U24sin2
Ps = P0+ 12U2(14sin2)
1
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
0.5
1.0
PsP
01 2U
2
Figure: Pressure distribution on the surface of a circular cylinder
2
PROBLEM 2
Find the streamfunction for the aymptotic suction pforile, u(y)=U0[1exp(yV0/)], whichoccurs when a streaming flow with outer velocity,U0, flows above a porous wall with suctionvelocity, V0. Use Matlab to plot streamlines with equal increments of .
v =x
V0 =x
(x, y)=V0x+ f (y)
u = y
U0[1exp(yV0/)]= y
(x, y)=U0y +U0 V0
eyV0 + f (x)
(x, y)=V0x+U0y +U0 V0
eyV0
0.0 0.2 0.4 0.6 0.8 1.0
X
0.0
0.2
0.4
0.6
0.8
1.0
Y
1.500
3.000
4.500
6.000
7.500
9.000
10.500
(a)
0.0 0.2 0.4 0.6 0.8 1.0
X
0.0
0.2
0.4
0.6
0.8
1.0
Y
(b)
Figure: (a) Streamfunction, (b) Velocity Vector Field
3
PROBLEM 3
The velocity potential for a spiral vortex flow is given by: (r,) = (/2pi) (m/2pi)log (r ), where and m are constants. Show that the angle, , between the velocity vector, u =ur ir +Ui and the radial direction, r , remains constant throughout the flow field.
ur = r = m2piru = 1r
= 2pir
tan()= uur
= /2pirm/2pir
= m
= tan1(m
)= constant
PROBLEM 4
Velocity potential for flow impinging on a flat plate is described by the streamfunction, =Axy . Using MatLab (or some other application), draw streamlines in equal increments of for this flow. The addition of a source of strength, m, at the flow origin, O, creates thepresence of a bump of height, h, protruding in the wall-normal direction. By conservation(i.e. 2 = 0), streamlines will flow around this bump. Add the source to your code, andexplore the relationship between h, A, and m (5 different combinations will be adequate).
= Axy + m2pi
= A2r 2sin(2)+ m
2pi
Stagnation point is at x = 0, y = h ( = pi2 ,r = h) where h is the height of the bump.
vr = 1r
= Arcos(2)+ m
2pir
v =
r= Ar sin(2)
At this stagnation point the velocity is zero meaning ur = 0 and v = 0, therefore from previ-ous equations the following relation is obtained:
0= Ahcos(pi)+ m2pih
or
h2 = m2piA
4
1.0 0.5 0.0 0.5 1.0X
0.0
0.2
0.4
0.6
0.8
1.0
Y
-0.5
00-0.
250
0.000
0.000
0.25
0
0.250
0.500 0.750
1.000
(a)
1.0 0.5 0.0 0.5 1.0X
0.0
0.2
0.4
0.6
0.8
1.0
Y
(b)
Figure: (A = 1,m = 1,h = 1/2pi) (a) Stremfunction, (b) Velocity Vector Field
1.0 0.5 0.0 0.5 1.0X
0.0
0.2
0.4
0.6
0.8
1.0
Y
-0.7
50
-0.5
00
-0.2
50
0.000
0.250
0.500
0.750
(a)
1.0 0.5 0.0 0.5 1.0X
0.0
0.2
0.4
0.6
0.8
1.0
Y
(b)
Figure: (A = 1,m = 0,h = 0) (a) Stremfunction, (b) Velocity Vector Field
1.0 0.5 0.0 0.5 1.0X
0.0
0.2
0.4
0.6
0.8
1.0
Y
0.000
0.000
0.250
0.250
0.50
00.
500
0.750
0.750
1.000
(a)
1.0 0.5 0.0 0.5 1.0X
0.0
0.2
0.4
0.6
0.8
1.0
Y
(b)
Figure: (A = 1,m = 2,h = (1/pi)1/2) (a) Stremfunction, (b) Velocity Vector Field
5
1.0 0.5 0.0 0.5 1.0X
0.0
0.2
0.4
0.6
0.8
1.0
Y
-1.2
00
-0.6
00
0.000
0.00
0
0.600
1.200
1.800
(a)
1.0 0.5 0.0 0.5 1.0X
0.0
0.2
0.4
0.6
0.8
1.0
Y
(b)
Figure: (A = 2,m = 1,h = (1/(4pi))1/2 (a) Stremfunction, (b) Velocity Vector Field
1.0 0.5 0.0 0.5 1.0X
0.0
0.2
0.4
0.6
0.8
1.0
Y
0.080
0.16
0
0.24
0
0.320
0.400
0.480
(a)
1.0 0.5 0.0 0.5 1.0X
0.0
0.2
0.4
0.6
0.8
1.0
Y
(b)
Figure: (A = 0,m = 1,h =) (a) Stremfunction, (b) Velocity Vector Field
6
PROBLEM 5
Show that the volume of an arbitrary region is given by
dVARdt
=niwidS
the volume of an arbitrary region can be wrriten by the integral
VAR =R(t )
1d
using Leibnitz theorem
dVARdt
= ddt
R(t )
1d=1
td+
niwi 1dS
1
td= 0
dVARdt
=niwi 1dS
7
PROBLEM 6
Prob. 6.1 Panton Stokes flow over a sphere has velocity components
vr =Ucos()[
1+ 12
(r0r
)3 3
2
(r0r
)]
v =Usin()[1+ 1
4
(r0r
)3+ 3
2
(r0r
)]Compute all components of the viscous stress tensor in r,, coordinates.The viscous stress tensor is defined by
=23v+2S
if incompressible= 2S
r r = 2vrr
= 2Ucos()[3
2
r 30r 4+ 3
2
r0r 2
]
= 2[
1
r
v
+ vrr
]= 2
[Ucos()
r
[1+ 1
4
(r0r
)3+ 3
2
(r0r
)]+Ucos()
r
[1+ 1
2
(r0r
)3 3
2
(r0r
)]]= 3
2
Ucos()
r
(r0r
)3 = 2
[1
r sin()
v
+ vrr+ vcot ()
r
]= 2
[vrr+ vcot ()
r
]= 2
[Ucos()
r
[1+ 1
2
(r0r
)3 3
2
(r0r
)]+Ucos()
r
[1+ 1
4
(r0r
)3+ 3
2
(r0r
)]]= 3
2
Ucos()
r
(r0r
)3
8
r = r =[r
r
(vr
)+ 1r
vr
]=[r
r
[Usin()
[1r+ 1
4
r 30r 4+ 3
2
r0r 2
]]+ 1r
[Ucos()
[1+ 1
2
(r0r
)3 3
2
(r0r
)]]]
=[Usin()
(1
r r
30
r 4 6
2
r0r 2
)Usin()
(1
r+ 1
2
r 30r 4 3
2
r0r 2
)]
=Usin()[3
2
r 30r 4+ 3
2
r0r 2
]
= =[sin()
r
(v
sin()
)+ 1r sin()
v
]= 0
r = r =[
1
r sin()
vr
+ r r
(vr
)]= 0
9
PROBLEM 7
Prob. 6.3 Panton An ideal "inviscid" flow over a cylinder has the velocity components givenin problem 5.1. Compute all components of the viscous stress tensor. Compute . Why isthis flow called inviscid?
vr =Ucos()[
1(r0r
)2]v =Usin()
[1+(r0r
)2]p = 1
2U2
[2(r0r
)2(12sin2())
(r0r
)4]the viscous stress tensor is defined as
=23(v)+2S
first the divergence of the velocity is computed in cylindrical coordinates
v= 1r
r(r vr )+ 1
r
v
+ vzz
expanding and setting vz = 0
v= 1r
[rvrr+ vr]+ 1r
v
vrr
= 2Ucos()r20
r 3
v
=Ucos()(
1+ r20
r 2
)
substituting back into the divergence equation of the velocity
v= 1r
[2Ucos()
r 20r 3Ucos()
[1(r0r
)2]+Ucos()
(1+ r
20
r 2
)]
= 1rUcos()
[2r
20
r 21+1+ r
20
r 2+ r0r 2
]= 0
the viscous stress tensor is reduced to
10
= 2S
r r = 2vrr
=4Ucos()r20
r 3
= 2(
1
r
v
+ vrr
)= 2
(1
r
(Ucos()
[1+ r
20
r 2
]) 1r
(Ucos()
[1 r0
r 2
]))
= 2rUcos()
[1+ r
20
r 21+ r
20
r 2
]
= 4Ucos()r20
r 3
zz = 0
r = r =[r
r
(vr
)+ 1r
vr
]=[vr
vr+ 1r
vr
]=[2Usin()r
20
r 3Usin()
r
(1+ r
20
r 2
)+Usin()
r
(1 r
20
r 2
)]
=Usin()[2r
20
r 3 1r r
20
r 3+ 1r r
20
r 3
]
=4Usin()r20
r 3
zr = r z = 0z = z = 0
[]r = 1r
r(rr r )+ 1
r
r
r
= 1r
8Ucos()r 20r 3 1r
4Ucos()r 20r 34Ucos r
20
r 4
= 0
[] =1
r 2
r(r 2r)+
1
r
= 1r 2uUsin()
r 20r 2+ 1r
4U (sin())r20
r 3
= 0
11
[]z = 0
= 0since this is the term that accounts for viscous forces in the momentum equation , we can saythat the flow is "inviscid"
12
PROBLEM 8
Prob. 6.7 Panton For a Newtonian fluid, show that the dissipation is given by
:v=23(v)2+2S : S (0.1)
the viscous tress tensor is
=23v+2S (0.2)
:v=(2
3v+2S
):v
=(2
3i j
vkxk
