Solution MidTermExam AH 1388 89

8/13/2019 Solution MidTermExam AH 1388 89

1/13

AmirKabir University of Technology

Faculty of Marine Technology

Advanced Hydrodynamics

Midterm Exam

First and Last Name:

Take home 9.00 a.m. Azar 28, 1388 Due time 9:00 a.m. Azar 29, 1388

Seyed Hossein Mousavizadegan

Email: hmousavi @ aut.ac.irTel: 6454-3111

Instruction:

1. It is supposed that you will be working independently without any consultation except for thecourse instructor ([email protected]).

2. You are allowed to consult any books and journals.

3. The above due time is sharp with no extension.

4. You should return the exam booklet.

5. The value of each question is given at the end of it.

Please return the exam booklet and the answer sheets to room 3 in the department main building.

Please send your computer program to ([email protected]) during the exam period.

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2/13

1. A hydraulic safety clutch should be decoupled at certain angular velocity. The frequencyadjustments occur by way of the filling H. The required coupling force Fc is produced by thefluid pressure acting on an annular piston (radius R and width s). Assuming s


3/13

2. Show by direct calculation that the two-dimensional mass conservation equation

t +

(u)

x +

(v)

y = 0

has the same form on variables appropriate to new axes rotated counterclockwise through anangle with respect to the original axes. (20)

Hint: You may follow the following steps:

i. Denote the original unit vector i and j and the rotated ones i and j and find the relation-ships between them.

ii. If the coordinates in original coordinate system are x, y and the coordinates in rotated oneis x, y, find the relationship between them.

iii. Denote the velocity components in the original coordinate system u, v and the velocitycomponents in the rotated coordinate system u, v. Find the relationships between thevelocity components in one coordinate system and the velocity components in the anothercoordinate system.

iv. By application of the chain rule of differentiation, show that the continuity equation remainsunchanged.

Solution a:

i i = cos , i j = sin j i = sin , j j = cos

r= xi+yj = xi +y j =

Multiply by i = x= x cos y sin Multiply byj = y = x sin +y cos Multiply byi = x =x cos +y sin Multiply by j = y = x sin +y cos

v= ui+vj = ui +v j =

Multiply byi = u= u

cos v

sin Multiply byj = v= u sin +v cos Multiply byi = u =u cos +v sin Multiply by j = v = u sin +v cos

Since is a scalar, then = (x, y) = (x, y)

x

y

x

y

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5/13

(b) Sketch the streamlines passing through two different your selected points at times t =t0, 2t0, 3t0, 4t0 and 5t0. You may assume a t0 and plot the curves. (5)

Solution b:

We may consider that x10 = 1, x20 = 1.

1 1.5 2 2.5 3 3.5 4 4.5 51

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3

x1

x2

t t0

t = 2 t0

t = 3 t0

t = 4 t0

t = 5 t0

(c) Obtain the pathline of the particle that is initially, t = 0, at A = (A1, A2). (5)

Solution c:

dx1

dt =v

1=

x1

t0+t =

ln x1

= ln (t+t0

) +c1

dx2dt

=v2 = x1t0+ 2t

= ln x2 = 12

ln(2t+t0) +c2

x= (A1, A2) at t= 0 =

c1 = ln A1 ln t0c2 = ln A2 12ln t0

ln x1= ln (t+t0) + ln A1 ln t0 = x1 = A1 t+t0t0

ln x2=1

2ln(2t+t0) + ln A2 1

2ln t0 = x2 = A2

2t+t0

t0

1/2

(d) Using the continuity equation give the density in material coordinates for the case that:

i. The initial density is:

= 0A1A2

at t= 0 (5)

ii. The initial density is:

= 0= const. at t= 0 (5)

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Solution d:

= 1J

( , , )

(,,)

1

J=

x1

A1

x2

A2x2A1

x2A2

= t+t0t0

0

0t+2t0t0

1/2

= t+t0

t02t+t0

t01/2

= 1

t+t0

t0

2t+t0

t0

1/21

(i.) = 0A1A2

t=0

= = 0 A1A2

t+t0

t0

2t+t0

t0

1/21

0 =

A1A2

t0

t0(t+t0)

2t+t0

(ii.) = 0t=0

= = 0 t+t0t0 2t+t0

t01/2

1

0 =

t0

t0(t+t0)

2t+t0

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(e) Determine the density field for (i) and (ii). (5)

Solution e:

(i.)

= 0A1

A2 t+t0

t0 t+2t0

t01/2

1

A1 = x1t+t0t0

A2 = x2t+2t0t0

1/2

= 0 =x1x2

t0

t+t0

2

(ii.)

0 = t0

t0

(t+t0)

2t+t0

7


8/13

4. Newtonian fluid (, = const.) passes between two parallel plates (h > Qh/so that the inertial terms in the equationof motion may be neglected. It also assumed that the body forces are also negligible.

hr

r

z

z

ri

ro

(a) Find the average radial velocityVr(r) overh. Consider that the volume flux is Q. (5)

Solution a:

Vr(r) = Q

2rh

(b) Write down the governing equations and simplify them. Show that the pressure is only afunction ofr. (5)

Solution b:

Continuity equation =

v= 1r (rvr)r + 1r v + vzz = 0

v = vz = 0

=

1

r

rvr

r

= 0

Momentum equation =

vrt + (v )vr

v2r = 1 pr +

2vr vrr2 2r2 v vt + (v )v+ vrvr = 1r p +

2v+ 2r2 vr vr2 vzt + (v )vz = 1 pz +2vz

The momentum equation is simplified according to:

t= 0, vz =v = 0

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and also for the assumption that the inertia terms are negligible. Therefore,

Continuity equation = 1r

rvrr

= 0 = rvr =f(z)

Momentum equation =

1 pr +2vr vrr2 = 0

p

= 0

pz = 0

= 1 dpdr +

1r

r

r vrr +

2vrz2 vrr2

=

p= p(r)

rvr =f(z)

1 dpdr +1r

r

r vrr

+

2vrz2 vrr2

= 0

= 1dpdr

+2vrz2

= 0

(c) Compute and the velocityvr(r, z) = Vr(r)Z(z) from the Navier Stokes equation and theno-slip condition at the walls and plot the velocity profile. (5)

Solution c:

vr(r, z) = Vr(r)Z(z) = Q2rh

Z(z)

2vrz2

= Q

2rhZ(z)

1

dp

dr+

Q

2rhZ(z) = 0 = Z(z) = 2rh

Q

dp

dr =

Z(z) = = Z(z) = 2

z2 +c1z+c2

vr = Q

2rh

2z2 +c1z+c2

=

vr(r, z)z=0

= 0 = c2 = 0

vr(r, z)z=h

= 0 =

c1 =

2

h

vr = 2

Q2rh

z2 zh

(d) Compute the shear stresszr distribution and plot it. (5)

Solution d:

zr =vrz

= zr =2 Q2rh (2z h)

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10/13

(e) Compute the pressure distribution subject to the boundary condition p(ri) = pi andp(ro) = po and plot it. (5)

Solution e:

2rhQ

dpdr

= = dp= Q2h

drr

= p= Q2h

ln r+c p(ri) = pip(ro) = po

=

pi= Q2hln ri+c

po = Q2hln ro+c

pi po = Q2h

ln

riro

= = pi po

ln ri ln ro2h

Q

c= pi Q2h

ln ri = c= pi pi poln ri ln ro ln ri

p= pi+ (pi po) ln rln riln riln ro

(f) Calculate the volume flux,Q; (5)

Solution f:

vr =

2

Q2rh

z2 zh

= pipoln riln ro2hQ

= vr = 12

pi poln ri ln ro

1

r

z2 zh

Q=

vdA =

v= vr =

12

pipoln riln ro

1r

z2 zh

dA= rddz

Q= 2

0 h

0

1

2

pi poln ri

ln ro z

2 zh ddzQ=

pipoln roln ri

h3

6

10


11/13

(g) Show that in the limit (ro ri)/ri = /ri


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5. Consider two parallel uniform streams,U1 (upper) andU2 (lower) meeting atx = 0. The twofluids are identical (1 = 2 , 1 = 2).

(a) Write down the motion equations when Re1& Re2 >>1 for both streams. (5)

Solution a:

Since Re1 & Re2 >> 1, we can apply the boundary-layer equation for both layers. It isalso assumed a steady flow.

uix +

viy = 0

uiuix +vi

uiy =

2uiy2

=i= 1, 2i= 1 = y 0i= 2 = y 0

(b) Write down the boundary conditions for each layer. (5)

Solution b:

Inlet: =

ui(0, y), vi(0, y) = 0, are known

Patching to the outer layer: =

u1(x,) U1u2(x,) U2

At the interface, y = 0 =

u1(x, 0) = u2(x, 0)v1(x, 0) =v2(x, 0) = 0

1u1y =2

u2y

1 = 2= u1y = u2y

(c) Using the similarity method and the following similarity parameters:

i = y

U12xi

, fi = uiU1

i= 1, 2

to obtain the similarity equations for each layer as follows.

fi +fif

i = 0 i= 1, 2

Provide the boundary conditions for each layer of the flow.(10)

Solution b:We may write the momentum equation as follows using the definition of the streamfunction.

iy

2ixy

ix

2iy2

=i3iy3

Using the given similarity parameter i= y

U12xi

, we taken into account that:

i(x, y) = U1g(x)fi()

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13/13

where g(x) =

2xiU1

. It is also defined that bi(x) =

U12xi

, therefore, i = ybi(x).

According to the definition for , it can be written that:

vi= ix

= U1[g

i(x)fi(i) +gi(x)fii

ix

]

= U1[g

i(x)f(i) +

b i(x)

bi gi(x)if

i(i)]

2ixy

= U1[g

i(x)f

i(i)iy+b ibi

gi(x)f

i(i) +b ibi

gi(x)f

i (i)bi]

= U1[big

if

i(i) +b

igif

i(i) +b

igiif

i (i)]

ui= i

y = U1gif

i(i)iy = U1gibif

i(i)

2iy2

= U1b2igif

i(i)

3i

y3 = U1b3

igif

i (i).

Substituting these values into momentum equation and doing some manipulations, weobtain

U1ib2i

[big

i+b

igi]f2

i U1giibi

fif

i =f

i . (1)

If we take the relationships for bi(x) and gi(x) and substitute in the above equation, we

obtainedbigi+ b

igi = 0 and U1g

i

ibi= 1. Therefore, we get the following similarity equation.

fi +fif

i = 0 , i= 1, 2

The boundary conditions are:

Kinematic B.C. =

f1(0) =f2(0) = 0 f

1

(0) =f

2

(0)= 0

Dynamic B.C. =

u1y

y=0

=

u2y

y=0

= f1(0) =f2(0)

Far field B.C. = f1() = 1 f2() =U2U1

(d) Write a computer program to obtained the velocity profile in each layer. Please providea complete explanation of your applied schemes. (25)

(e) Compute the thickness of each layers.(10)

Good Luck

S. H. Mousavizadegan

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Solution MidTermExam AH 1388 89