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ME 2134: Fluid Mechanics IPart 2

By

T. T. Lim

Office: E2-03-18

Recommended and Reference Texts:

(1) Cengel Y.A. and Cimbala J.M.:“Fluid Mechanics: Fundamental and Applications” (McGraw-Hill 2006)

(2) E.John Finnemore and Joseph B. Franzini:“Fluid Mechanics with Engineering Applications” (10th edition), McGraw Hill, 2002.

(3) R.W. Fox and A.T. McDonald:“Introduction to Fluid Mechanics” (4th edition), John Wiley & Sons, 1992.

(4) K.L. Kumar:“Engineering Fluid Mechanics”. Eurasian Publishing house

(5) B. Massey (revised by J Ward-Smith):“Mechanics of Fluids”. Stanley Thornes (Publishers) Ltd, c1998.

(6) B.R. Munson, D.F. Young and T.H. Okiishi:“Fundamentals of Fluid Mechanics” (3rd edition), John Wiley & Sons, 1998.

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(7) M.C. Potter and D.C. Wiggert:“Mechanics of Fluids” (2nd edition), Prentice-Hall International, 1997

(8) Irving H. Shames:“Mechanics of Fluids” (3rd Edition), McGraw Hill, 1992.

(9) Alexander J. Smits:“A Physical Introduction to Fluid Mechanics” (1st edition), John Wiley, 2000.

(10) V.L. Streeter, E.B. Wylie and K.W. Bedford: “Fluid Mechanics” (9th edition), McGraw Hill, 1998.

(11) F.M. White:“Fluid Mechanics” (7th Edition), McGraw Hill, 2011.

(12) A. Jeffrey:“Handbook of Mathematical Formulas and Integration” (2nd Edition), Academic Press, 2000.

Module Outline:

1. Equilibrium of Moving Fluids (in translation and rotation).

2. Momentum and its Applications.

3. Dimensional Analysis and Similitude.

4. Analysis of Pipe Flow

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Equilibrium of Moving Fluids

Stationary Tank

p= pa + ρgz

Pa

z

p

V=ConstantMoving Tank

z

Pa

p

p= ??

(A) Statics of a Moving System

V=Constantz

Pa

p

p= pa + ρgz

When the entire continuum is in uniform RECTILINER motion, the governing principle of the statics of a fluid in the gravity field remains the same.

A container moving with a CONSTANT VELOCITY

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(B) Uniform Rectilinear Acceleration

a=Constant accelerationAccelerating Tankz

Pa

p

What does the fluid/air interface look like ?

p= ?

Questions:

Under constant acceleration

ax

aaz

z

x

dx

dz

2dx

xpp∂∂

−

2dz

zpp∂∂

−

2dx

xpp∂∂

+

2dz

zpp∂∂

+

p ax

az

wt

dz

dxA

B C

D

Fluid element has a unit depth

5

2dx

xpp∂∂

−

2dz

zpp∂∂

−

2dx

xpp∂∂

+

2dz

zpp∂∂

+

p ax

az

wtdz

dxA

B C

D

Fluid element has unit depth

( ) xa1.dz.dxdz2

dx.xppdz

2dx.

xpp ρ=⎟

⎠⎞

⎜⎝⎛

∂∂

+−⎟⎠⎞

⎜⎝⎛

∂∂

−

xaxp

ρ−=∂∂

The equation of motion in the x-direction is

Simplifying the above equation gives

(B1)

⎟⎠⎞

⎜⎝⎛

∂∂

−2

dxxpp 1.dz

2dx

xpp ⎟

⎠⎞

⎜⎝⎛

∂∂

−

⎟⎠⎞

⎜⎝⎛

∂∂

+2

dxxpp 1.dz

2dx

xpp ⎟

⎠⎞

⎜⎝⎛

∂∂

+

⎟⎠⎞

⎜⎝⎛

∂∂

−2dz

zpp 1.dx

2dz

zpp ⎟

⎠⎞

⎜⎝⎛

∂∂

−

⎟⎠⎞

⎜⎝⎛

∂∂

+2dz

zpp 1.dx

2dz

zpp ⎟

⎠⎞

⎜⎝⎛

∂∂

+

dx × 1AD

dx × 1BC

dz × 1CD

dz × 1AB

ForcePressureAreaFacex

z

2dx

xpp∂∂

−

2dz

zpp∂∂

−

2dx

xpp∂∂

+

2dz

zpp∂∂

+

p ax

az

wtdz

dxA

B C

D

Fluid element has a unit depth

( ) ( ) za1.dz.dx1.dz.dxgdx2dz.

zppdx

2dz.

zpp ρ=ρ−⎟

⎠⎞

⎜⎝⎛

∂∂

+−⎟⎠⎞

⎜⎝⎛

∂∂

−

( )gazp

z +ρ=∂∂

−

Similarly, the equation of motion in the z-direction is

Expanding and simplifying the above equation gives

(B2)

⎟⎠⎞

⎜⎝⎛

∂∂

−2

dxxpp 1.dz

2dx

xpp ⎟

⎠⎞

⎜⎝⎛

∂∂

−

⎟⎠⎞

⎜⎝⎛

∂∂

+2

dxxpp 1.dz

2dx

xpp ⎟

⎠⎞

⎜⎝⎛

∂∂

+

⎟⎠⎞

⎜⎝⎛

∂∂

−2dz

zpp 1.dx

2dz

zpp ⎟

⎠⎞

⎜⎝⎛

∂∂

−

⎟⎠⎞

⎜⎝⎛

∂∂

+2dz

zpp 1.dx

2dz

zpp ⎟

⎠⎞

⎜⎝⎛

∂∂

+

dx × 1AD

dx × 1BC

dz × 1CD

dz × 1AB

ForcePressureAreaFace

z

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Integrate equation (B1) gives, xaxp

ρ−=∂∂

(B1)

Therefore equation (B5) becomes ( )[ ] ozx pzgaxap ++ρ+ρ−= (B6)

Similarly, integrating equation (B2) gives

where f(x) is a term containing only x, and c2 is a constant.

( ) 2z c)x(fzgap +++ρ−= (B4) ( )ga

zp

z +ρ=∂∂

− (B2)

z

x

po

0

Substituting the boundary condition (at x=0, z=0, p=po.) into equation (B5) gives

c3=po

1x c)z(fxap ++ρ−= (B3)

where f(z) is a term containing only z, and c1 is a constant.

( )[ ] 3zx czgaxap ++ρ+ρ−= (B5)

Now equations (B3) and (B4) are compatible only if

where c3 is a new constant which depends on boundary condition.

To find the slope of the free surface, we substitute p=po into equation (B6) which leads to

gaa

dxdztan

z

x

+−==θ (B7)

It can be shown that the lines of constant pressure, also called ISOBARS are parallel to the free surface.

( )[ ] ozx pzgaxap ++ρ+ρ−= (B6)

aI SO BA R S

θ

A Large Open Cylinder A Small Closed Cylinder

θI SO BA R S

a

Po

Po

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[ ] opgzaxp +ρ+ρ−=

θ

For a simple case of acceleration along the x-axis only, substitute ax=a, az=0 into equation (B6), and we obtain

gatan −=θ

and the inclination of the free surface to the direction of the acceleration is given by

( )[ ] ozx pzgaxap ++ρ+ρ−= (B6)

ax = a

Applications: This type of analysis is important in the design of the tank-truck where brakes may be applied abruptly or a tank car for a railway. It is also important in the design of a fuel system of an airplane

Tank truckAcknowledgement: Wikimedia

Tank car of railwayAcknowledgement: China Railway

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Worked Example

Equilibrium of Fluid in ROTATION

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(C) Uniform Spin of Liquid in a Container

When a body of fluid rotates uniformly without relative motion between different elements of the fluid in a container, each particle moves in a circle. Under this condition, the fluid is said to undergo a SOLID-BODY ROTAION

Because an external torque is required to start the motion, the term “Forced Vortex” has also been used. Once steady conditions are established, there is no relative motion between fluid particles and thus no shear forces exist, even in a real fluid.

T=0

ω

T>T1

r

uθ

p

p

p

dθ

r dr

2dr.

rpp∂∂

− 2dr.

rpp∂∂

+

Top-view

ω

r

z

o

zo dzdrz

r 2dr.

rpp∂∂

+

2dz.

zpp∂∂

+

pdr/2dz

Wt2dr.

rpp∂∂

−

2dz.

zpp∂∂

−

Side-view

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( ) ( )drdzrdrdz2

dpdr2dzd)(drr2dr

rppdzrd

2dr

rpp 2 θρω−=

θ+θ+⎟

⎠⎞

⎜⎝⎛

∂∂

+−θ⎟⎠⎞

⎜⎝⎛

∂∂

−

( ) drrpdrdrr2dr

rppr

2dr

rpp 22ρω−=++⎟

⎠⎞

⎜⎝⎛

∂∂

+−⎟⎠⎞

⎜⎝⎛

∂∂

−

drrpdrdr2dr

rpr.

2dr

rppdrprr.

2dr

rppr 22ρω−=+⎟

⎠⎞

⎜⎝⎛

∂∂

+∂∂

++−∂∂

−

drrdr2dr

rpr.dr

rp 22ρω−=

∂∂

−∂∂

−

The velocity of a typical element of dimension δr, rδθ, dz at a radial distance r from the axis of rotation is

u=uθ=ωr

The acceleration of the same element is given by rω2 in a radially inward direction.

Equation of motion of the fluid element in the radial (r) direction may be written as

Simplifying the above equation leads to

ρω=∂∂ r

rp 2 (C1)

2dr.

rpp∂∂

+pdz

2dr.

rpp∂∂

−

p

p

p

dθ

r dr2dr.

rpp∂∂

− 2dr.

rpp∂∂

+

2dz.

zpp∂∂

−

2dz.

zpp∂∂

+

wt

Radial direction

rω ωr

2dr.

rpp∂∂

+pdz

2dr.

rpp∂∂

−

p

p

p

dθ

r dr2dr.

rpp∂∂

− 2dr.

rpp∂∂

+

2dz.

zpp∂∂

−

2dz.

zpp∂∂

+

wt

( ) ( ) ( ) 022

=θρ−θ⎟⎠⎞

⎜⎝⎛

∂∂

+−θ⎟⎠⎞

⎜⎝⎛

∂∂

− dz.rd.drgrd.drdzzpprd.drdz

zpp

In the z-direction

gzp

ρ−=∂∂ (C2)

Integrate equation (C1) gives

c)z(frp ++ρω=2

22 (C3)

ρω=∂∂ r

rp 2 (C1)

where f(z) is a term containing only z, and c is a constant

1c)r(fgzp ++ρ−=

Integrate equation (C2) gives

(C4)

where f(r) is a term containing only r, and c1 is a constant

2

22 c

2rgzp +ρω+ρ−=

Equations (C3) and (C4) are compatible only if

(C5)

where c2 is a new constant which depends on boundary condition

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At r=0, z=zo and p = po (reference pressure) (see figure),it can be shown that

c2= po+ρg zo

Therefore, equation (C5) becomes

( ) oo przzgp +ρω+−ρ−= 22

21

(C6)

2

22 c

2rgzp +ρω+ρ−= (C5)

ω

o

zoz

r

po

( ) 22

210 rzzg o ρω+−ρ−=

which is a PARABOLOID OF REVOLUTION (see figure below)

ozrg

z +ω= 22

21

(C7)

To find the equation of the free surface, we let p=po

Therefore, equation (C6) becomes ( ) oo przzgp +ρω+−ρ−= 22

21

(C6)

Zo (i.e. Zmin)

Zmax

R

ISOBARSOriginal fluid level before rotation

Z1

Isobars in Rotating Cylinder with Liquid

12

α=ω

=ω

= tang

rg

rdrdz 22

22

The slope of the liquid level at any radius r is given by

R

z

r

α

r

ozrg

z +ω= 22

21

(C7)

Application: A centrifugal pump and a centrifuge make use of this principle. Here, an enclosed mass of water is whirled rapidly to create the pressure difference between the inlet and outlet.

Centrifugal pump

Acknowledgement: Encyclopedia Britannica Acknowledgement: ITT Corporation

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InitialState

Increasing Rotational Speed (Open Container)

ω

1 2 3 4 5 6 7

Tall Cylinder: No spillage of fluid

Rotation stops

InitialState

Increasing Rotational Speed (Open Container)

ω

1 2 3 4 5 6 7Rotation

stops

Short cylinder: Spillage of fluid

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InitialState

Increasing Rotational Speed (Closed Container)

ω

1 2 3 4 5 6 7

Short cylinder: Spillage of fluid

Let us go through a simple example