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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
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THE UNIVERSITY OF TEXAS AT DALLAS, MECHANICAL ENGINEERING DEPARTMENT
Incompresible Fluid Mechanics, MECH-6370,HW 3
Edgardo Javier García Cartagena
October 21, 2015
PROBLEM 1
The streamfunction for flow over a circular cylinder is ψ=Ur si nθ(1− r 2
0
r 2
)Evaluate the pres-
sure distribution on the cylinder.
vr = 1
r
∂ψ
∂θ=U
(1− r 2
0
r 2
)cos(θ)
vθ =−∂ψ∂r
=−U
(1+ r 2
0
r 2
)si n(θ)
On the surface of the cylinder (r = r0) vr = 0 and vθ =−2Usi nθ. Using the Bernoulli equationwe obtain:
P0 + 1
2ρU 2 = Ps + 1
2ρv2
θs
Ps = P0 + 1
2ρU 2 − 1
2ρ
(−U
(1+ r 2
0
r 2
)si n(θ)
)2
Ps = P0 + 1
2ρU 2 − 1
2ρU 24si n2θ
Ps = P0 + 1
2ρU 2(1−4si n2θ)
1
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5
θ
−3.0
−2.5
−2.0
−1.5
−1.0
−0.5
0.0
0.5
1.0
Ps−P
01 2ρU
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[1−exp(−yV0/ν)], 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 ψ.
v =−∂ψ∂x
−V0 =−∂ψ∂x
ψ(x, y) =V0x + f (y)
u = ∂ψ
∂y
U0[1−exp(−yV0/ν)] = ∂ψ
∂y
ψ(x, y) =U0 y +U0ν
V0e−y
V0ν + f (x)
∴ ψ(x, y) =V0x +U0 y +U0ν
V0e−y
V0ν
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: φ(r,Θ) = (Γ/2π)Θ− (m/2π)l og (r ), where Γ and m are constants. Show that the angle, α , between the velocity vector, u =ur ir +Uθiθ and the radial direction, r , remains constant throughout the flow field.
ur = ∂φ∂r = −m
2πr
uθ = 1r∂φ∂θ = Γ
2πr
t an(α) = uθur
= Γ/2πr
m/2πr= −Γ
m
α= t an−1(−Γ
m
)= const ant
PROBLEM 4
Velocity potential for flow impinging on a flat plate is described by the streamfunction, ψ =Ax y . Using MatLab (or some other application), draw streamlines in equal increments ofψ 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. ∇2φ = 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).
Ψ= Ax y + m
2πθ = A
2r 2si n(2θ)+ m
2πθ
Stagnation point is at x = 0, y = h (θ = π2 ,r = h) where h is the height of the bump.
vr = 1
r
∂ψ
∂Θ= Ar cos(2θ)+ m
2πr
vθ =−∂ψ∂r
= Ar si n(2θ)
At this stagnation point the velocity is zero meaning ur = 0 and vθ = 0, therefore from previ-ous equations the following relation is obtained:
0 = Ahcos(π)+ m
2πhor
h2 = m
2πA
4
−1.0 −0.5 0.0 0.5 1.0
X
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 1.0
X
0.0
0.2
0.4
0.6
0.8
1.0
Y
(b)
Figure: (A = 1,m = 1,h = 1/2π) (a) Stremfunction, (b) Velocity Vector Field
−1.0 −0.5 0.0 0.5 1.0
X
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 1.0
X
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 1.0
X
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 1.0
X
0.0
0.2
0.4
0.6
0.8
1.0
Y
(b)
Figure: (A = 1,m = 2,h = (1/π)1/2) (a) Stremfunction, (b) Velocity Vector Field
5
−1.0 −0.5 0.0 0.5 1.0
X
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 1.0
X
0.0
0.2
0.4
0.6
0.8
1.0
Y
(b)
Figure: (A = 2,m = 1,h = (1/(4π))1/2 (a) Stremfunction, (b) Velocity Vector Field
−1.0 −0.5 0.0 0.5 1.0
X
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 1.0
X
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
dVAR
d t=
∫ni wi dS
the volume of an arbitrary region can be wrriten by the integral
VAR =∫
R(t )1d∀
using Leibnitz theorem
dVAR
d t= d
d t
∫R(t )
1d∀=∫∂1
∂td∀+
∫ni wi ·1dS
∫∂1
∂td∀= 0
∴dVAR
d t=
∫ni wi ·1dS
7
PROBLEM 6
Prob. 6.1 Panton Stoke’s flow over a sphere has velocity components
vr =Ucos(θ)
[1+ 1
2
(r0
r
)3− 3
2
(r0
r
)]
vθ =Usi n(θ)
[−1+ 1
4
(r0
r
)3+ 3
2
(r0
r
)]Compute all components of the viscous stress tensor in r,θ,φ coordinates.The viscous stress tensor is defined by
τ=−2
3µδ∇·v+2µS
if incompressibleτ= 2µS
τr r = 2µ∂vr
∂r
= 2µUcos(θ)
[−3
2
r 30
r 4 + 3
2
r0
r 2
]
τθθ = 2µ
[1
r
∂vθ∂θ
+ vr
r
]= 2µ
[Ucos(θ)
r
[−1+ 1
4
(r0
r
)3+ 3
2
(r0
r
)]+ Ucos(θ)
r
[1+ 1
2
(r0
r
)3− 3
2
(r0
r
)]]= 3
2
µUcos(θ)
r
(r0
r
)3
τφφ = 2µ
[1
r si n(θ)
∂vφ∂φ
+ vr
r+ vθcot (θ)
r
]= 2µ
[vr
r+ vθcot (θ)
r
]= 2µ
[Ucos(θ)
r
[1+ 1
2
(r0
r
)3− 3
2
(r0
r
)]+ Ucos(θ)
r
[−1+ 1
4
(r0
r
)3+ 3
2
(r0
r
)]]= 3
2
µUcos(θ)
r
(r0
r
)3
8
τθr = τrθ =µ[
r∂
∂r
( vθr
)+ 1
r
∂vr
∂θ
]=µ
[r∂
∂r
[Usi n(θ)
[−1
r+ 1
4
r 30
r 4 + 3
2
r0
r 2
]]+ 1
r
∂
∂θ
[Ucos(θ)
[1+ 1
2
(r0
r
)3− 3
2
(r0
r
)]]]
=µ[
Usi n(θ)
(1
r− r 3
0
r 4 − 6
2
r0
r 2
)−Usi n(θ)
(1
r+ 1
2
r 30
r 4 − 3
2
r0
r 2
)]
=µUsi n(θ)
[−3
2
r 30
r 4 + 3
2
r0
r 2
]
τφθ = τθφ =µ[
si n(θ)
r
∂
∂θ
(vφ
si n(θ)
)+ 1
r si n(θ)
∂vθ∂φ
]= 0
τrφ = τφr =µ[
1
r si n(θ)
∂vr
∂φ+ r
∂
∂r
( vφr
)]= 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 ∇·τ. Why isthis flow called inviscid?
vr =−Ucos(θ)
[1−
(r0
r
)2]
vθ =Usi n(θ)
[1+
(r0
r
)2]
p = 1
2ρU 2
[2(r0
r
)2(1−2si n2(θ))−
(r0
r
)4]
the viscous stress tensor is defined as
τ=−2
3µδ(∇·v)+2µS
first the divergence of the velocity is computed in cylindrical coordinates
∇·v = 1
r
∂
∂r(r vr )+ 1
r
∂vθ∂θ
+ ∂vz
∂z
expanding and setting vz = 0
∇·v = 1
r
[r
vr
∂r+ vr
]+ 1
r
∂vθ∂θ
∂vr
∂r= 2Ucos(θ)
r 20
r 3
∂vθ∂θ
=Ucos(θ)
(1+ r 2
0
r 2
)
substituting back into the divergence equation of the velocity
∇·v = 1
r
[2Ucos(θ)
r 20
r 3 −Ucos(θ)
[1−
(r0
r
)2]+Ucos(θ)
(1+ r 2
0
r 2
)]
= 1
rUcos(θ)
[−2
r 20
r 2 −1+1+ r 20
r 2 + r0
r 2
]= 0
the viscous stress tensor is reduced to
10
τ= 2µS
τr r = 2µ∂vr
∂r
=−4µUcos(θ)r 2
0
r 3
τθθ = 2µ
(1
r
∂vθ∂θ
+ vr
r
)= 2µ
(1
r
(Ucos(θ)
[1+ r 2
0
r 2
])− 1
r
(Ucos(θ)
[1− r0
r 2
]))
= 2µ
rUcos(θ)
[1+ r 2
0
r 2 −1+ r 20
r 2
]
= 4µUcos(θ)r 2
0
r 3
τzz = 0
τθr = τrθ =µ[
r∂
∂r
( vθr
)+ 1
r
∂vr
∂θ
]=µ
[∂vθ∂r
− vθr
+ 1
r
∂vr
∂θ
]=µ
[−2Usi n(θ)
r 20
r 3 − Usi n(θ)
r
(1+ r 2
0
r 2
)+ Usi n(θ)
r
(1− r 2
0
r 2
)]
=µUsi n(θ)
[−2
r 20
r 3 − 1
r− r 2
0
r 3 + 1
r− r 2
0
r 3
]
=−4µUsi n(θ)r 2
0
r 3
τzr = τr z = 0
τzθ = τθz = 0
[∇·τ]r = 1
r
∂
∂r(rτr r )+ 1
r
∂
∂θτθr −
τθθ
r
= 1
r8µUcos(θ)
r 20
r 3 − 1
r4µUcos(θ)
r 20
r 3 −4µUcosθr 2
0
r 4
= 0
[∇·τ]θ =1
r 2
∂
∂r(r 2τrθ)+ 1
r
∂
∂θτθθ
= 1
r 2 uµUsi n(θ)r 2
0
r 2 + 1
r4µU (−si n(θ))
r 20
r 3
= 0
11
[∇·τ]z = 0
∴ ∇·τ= 0
since 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
τ : ∇v =−2
3µ(∇·v)2 +2µS : S (0.1)
the viscous tress tensor is
τ=−2
3µδ∇·v+2µS (0.2)
τ : ∇v =(−2
3µδ∇·v+2µS
): ∇v
=(−2
3δi j
∂vk
∂xk+2µSi j
)∂vi
∂x j
=(−2
3δi j
∂vk
∂xk
∂vi
∂x j+2µSi j
∂vi
∂x j
)=−2
3
∂vk
∂xk
∂vi
∂xi+2µSi j
[S j i +W j i
]=−2
3
∂vk
∂xk
2
+2µSi j S j i +Si j W j i
=−2
3
∂vk
∂xk
2
+2µSi j S j i
=−2
3µ(∇·v)2 +2µS : S
where Si j is the symetric part of the velocity gradient, Wi j the antisymetric part of the velocitygradient, and the multiplication between a symetric and antisymetric tensor Si j W j i is equalto zero.
13
PROBLEM 9
Prob. 12.13 Panton Same as problem 1
PROBLEM 10
Prob. 12.14 Panton Find the pressure distribution on the surface of Hill’s spherical vortex.The streamfunction is given by
ψ= U R2
2
( r
R
)2[
1−( z
R
)2−
( r
R
)2]
for streamfuncion for axissymmetric flow in cylindrical coordinates the velocities are definedas
vr = 1
r
∂ψ
∂z
= U R2
r 2
( r
R
)2[
2
R2 z
]= U
r
( r
R
)2z
=Ur z
R2
vz =−1
r
∂ψ
∂r
=−U R2
r 2
[2r
R2 −( z
R
)2 2r
R2 − 4
R4 r 3]
=U
[1−
( z
R
)−2
( r
R
)2]
using Bernoulli equation
P0 + 1
2ρU 2 = Ps + 1
2ρ||v ||2s
where the magnitude of the velocity in the Hill’s spherical vortex at the surface and using thefact that ( r
R
)2+
( z
R
)2= 1
||v ||s =√
v2r + v2
z =Ur
R
P0 + 1
2ρU 2 = Ps + 1
2ρ
(U
r
R
)2
Ps = P0 + 1
2ρU 2
[1−
( r
R
)2]
14
0.0 0.2 0.4 0.6 0.8 1.0
r/R
0.0
0.2
0.4
0.6
0.8
1.0
Ps−P
01 2ρU
2
Figure: Pressure distribution on the surface of Hill’s spherical vortex
15
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