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