hes2340 fluid mechanics 1, semester 1, 2012, assignment 1 - basic concepts of fluid flow

SWINBURNE UNIVERSITY OF TECHNOLOGY (SARAWAK CAMPUS)

FACULTY OF ENGINEERING AND INDUSTRIAL SCIENCE

HES2340 Fluid Mechanics 1 Semester 1, 2012

Assignment 1: Basic Concepts of Fluid Flow

By

Stephen, P. Y. Bong (4209168)

Lecturer: Professor Alexander Gorin

Due Date: 4th

April 2012 (Wednesday)

Assignment 1: Basic Concepts of Fluid Flow Stephen, P. Y. Bong (4209168)

HES2340 Fluid Mechanics 1, Semester 1, 2012 of 10

ASSIGNMENT PROBLEM

A belt moves upward at velocity, V, dragging a film of viscous liquid of

thickness, h, as shown in the figure on the right. Near the belt, the film

moves upward due to no-slip. At its outer edge, the film moves downward

due to gravity.

Assume that the only non-zero velocity is u(y), with zero shear stress at the

outer film edge, using the general equations of viscous flow (Navier-

Stokes Equations) and stating clearly assumptions and reasons by which

you neglect or have to keep relevant terms in the Navier-Stokes Equations,

derive a formula for:

(a) u(y);

(b) The average velocity, Vavg, in the film; and

(c) The wall velocity, VC, for which there is no net flow either up or down.

(d) Sketch the non-dimensional velocity distribution in the film (velocity

distribution normalized by VC) for case (c).

(e) Calculate the volumetric flow rate of water film and oil (SAE 30) film

if h = 2 mm. Physical properties take from any relevant texts or

handbooks at temperature 20 °C (indicate in your report the source of

the information used).

Your report should include as well detailed calculations and graphs and

illustrations.

Deliverable:

Word processing report detailing results of your investigation to meet the above criteria, with graphs and

illustrations prepared by your own (hand written text and graphs and illustrations prepared by hand will

not be accepted).



SOLUTION

Problem Statement: For a given geometry and set of boundary conditions, an equation which governs the

fluid flow, u(y), the average velocity, Vavg, and the wall velocity, VC, are to be derived by using Navier-

Stokes Equations. The non-dimensional velocity distribution in the film is to be sketched and the

volumetric flow rate for both water film and oil (SAE 30) are to be determined.

Schematic Diagram:

Based on the schematic diagram above, the boundary conditions are:

At y = 0, v = V; at y = h, 0==dy

duµτ .

Assumptions:

1. The belt is infinite in x and z directions so that the flow is fully developed.

2. The flow is steady, i. e., 0=∂

∂=

∂

∂=

∂

∂

t

w

t

v

t

u.

3. This is a parallel flow (assume the velocities in y- and z-components are zero).

4. The fluid is incompressible and Newtonian with constant properties.

5. Pressure, p = Const. with respect to x (there is no applied pressure gradient pushing the flow in the x-

direction; the flow establishes itself due to viscous stresses caused by the moving belt.

6. The velocity field is purely one-dimensional, meaning here that v = w = 0, andy∂

∂;

z∂

∂of any velocity

component is zero.

7. Gravity acts in the negative x-direction and it can be expressed mathematically as g = – gi.



Physical Laws:

1. Conservation of Mass Principle – The Continuity Equation:

0=⋅∇ V

where ∇ is the Vector Operator, kjizyx ∂

∂+

∂

∂+

∂

∂=∇

V is the Velocity Vector, kjiV wvu ++=

( ) 000 =∂

∂+

∂

∂+

∂

∂⇒=++⋅

∂

∂+

∂

∂+

∂

∂⇒=⋅∇∴

z

w

y

v

x

uwvu

zyxkjikjiV

2. Newton’s Second Law of Motion – The Navier Stokes Equations

The General Navier Stokes Equations in x, y and z directions are as follows:

x-direction:

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂+

∂

∂+

∂

∂+

z

uw

y

uv

x

uu

t

u

zyxg zxyxxx

x ρττσ

ρ

y-direction:

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂+

∂

∂+

∂

∂+

z

vw

y

vv

x

vu

t

v

zyxg

zyyyxy

y ρτστ

ρ

z-direction:

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂+

∂

∂+

∂

∂+

z

ww

y

wv

x

wu

t

w

zyxg zzyzxz

z ρσττ

ρ

The viscous stress tensor for an incompressible Newtonian fluid with constant properties is given by:

ijij µετ 2=

where εij is the strain rate tensor and is given by:

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂

=

=

z

w

z

v

y

w

z

u

x

w

y

w

z

v

y

v

y

u

x

v

x

w

z

u

x

v

y

u

x

u

zzzyzx

yzyyyx

xzxyxx

ij

2

1

2

1

2

1

2

1

2

1

2

1

εεε

εεε

εεε

ε

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂

=

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂

=

z

w

z

v

y

w

z

u

x

w

y

w

z

v

y

v

y

u

x

v

x

w

z

u

x

v

y

u

x

u

z

w

z

v

y

w

z

u

x

w

y

w

z

v

y

v

y

u

x

v

x

w

z

u

x

v

y

u

x

u

ij

µµµ

µµµ

µµµ

µτ

2

2

2

2

1

2

1

2

1

2

1

2

1

2

1

2



When a fluid is moving, pressure acts inwards in the normal directions, but there is the existence of

viscous stresses as well. Thus, the stress tensor for an incompressible and Newtonian fluid can be

expressed as:

∂

∂+−

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂+−

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂+−

=

+

−

−

−

=

z

wp

z

v

y

w

z

u

x

w

y

w

z

v

y

vp

y

u

x

v

x

w

z

u

x

v

y

u

x

up

p

p

p

zzzyzx

yzyyyx

xzxyxx

ij

µµµ

µµµ

µµµ

τττ

τττ

τττ

σ

2

2

2

00

00

00

By comparisons, the normal and shear stresses for incompressible Newtonian fluids in Cartesian

coordinates can be expressed as:

x

upxx

∂

∂+−= µσ 2

y

vpyy

∂

∂+−= µσ 2

z

wpzz

∂

∂+−= µσ 2

∂

∂+

∂

∂==

x

v

y

uxyxy µττ

∂

∂+

∂

∂==

z

u

x

wzxxz µττ

∂

∂+

∂

∂==

z

v

y

wyzzy µττ

Substitute the normal and shear stresses above into the General Navier-Stokes Equation in x-direction

yields:

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂+

∂

∂+

∂

∂+

∂

∂+

∂

∂+

∂

∂

∂

∂+

∂

∂−

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂⋅

∂

∂+

∂

∂+

∂

∂⋅

∂

∂+

∂

∂+

∂

∂⋅

∂

∂+

∂

∂+

∂

∂−

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂⋅

∂

∂+

∂

∂+

∂

∂⋅

∂

∂+

∂

∂+

∂

∂+

∂

∂−

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂+

∂

∂

∂

∂+

∂

∂+

∂

∂

∂

∂+

∂

∂+

∂

∂−

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂+

∂

∂

∂

∂+

∂

∂+

∂

∂

∂

∂+

∂

∂+−

∂

∂+

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂+

∂

∂+

∂

∂+

z

uw

y

uv

x

uu

t

u

z

u

y

u

x

u

z

w

y

v

x

u

xx

pg

z

uw

y

uv

x

uu

t

u

z

w

xz

u

y

v

xy

u

x

u

xx

u

x

pg

z

uw

y

uv

x

uu

t

u

x

w

zz

u

x

v

yy

u

x

u

x

pg

z

uw

y

uv

x

uu

t

u

z

u

x

w

zx

v

y

u

yx

u

x

pg

z

uw

y

uv

x

uu

t

u

z

u

x

w

zx

v

y

u

yx

up

xg

z

uw

y

uv

x

uu

t

u

zyxg

x

x

x

x

x

zxyxxx

x

ρµρ

ρµρ

ρµρ

ρµµµρ

ρµµµρ

ρττσ

ρ

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

The term in the parentheses on the L. H. S. of the equation is zero 0=∂

∂+

∂

∂+

∂

∂

z

w

y

v

x

u due to the

equation of continuity for incompressible flow. Thus, the Navier-Stokes Equation in x-direction

becomes:

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂+

∂

∂+

∂

∂+

∂

∂−

z

uw

y

uv

x

uu

t

u

z

u

y

u

x

u

x

pg x ρµρ

2

2

2

2

2

2



Likewise, the Navier-Stokes Equations in y and z directions are:

y-direction:

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂+

∂

∂+

∂

∂+

∂

∂−

z

uw

y

uv

x

uu

t

u

z

v

y

v

x

v

y

pg y ρµρ

2

2

2

2

2

2

z-direction:

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂+

∂

∂+

∂

∂+

∂

∂−

z

uw

y

uv

x

uu

t

u

z

w

y

w

x

w

z

pg z ρµρ

2

2

2

2

2

2

Therefore, the Navier-Stokes Equation for incompressible flow in Cartesian coordinates can be

expressed as:

VgV 2∇++−∇= µρρ p

Dt

D

where DV/Dt is the material acceleration

2∇ is the Laplacian Operator, 2

2

2

2

2

22

zyx ∂

∂+

∂

∂+

∂

∂=∇

Analysis:

(a) The boundary conditions come from imposing the no-slip condition:

(1) At the free surface (y = h), τ = µ(du/dy) = 0

(2) At y = 0, v = V

The analysis is commenced with the continuity equation for incompressible flow in order to simplify

the partial differential equations.

0=∂

∂+

∂

∂+

∂

∂

z

w

y

v

x

u

Assumption 3 and 6 stress that the flow is parallel (velocities in y- and z- components are zero) and

the velocity field is purely one-dimensional. Which implies that ∂v/∂y = ∂w/∂z = 0, thus,

0=∂

∂

x

u Eq. (1)

Eq. (1) above tells that the velocity, u, is not a function of x. Therefore, it doesn’t matter where the

origin is placed as the flow will remains unchanged at any x-location. The phrase “fully developed”

in Assumption 1 also indicates that there is nothing special about any x-location since the belt is

infinite in length. Apart from that, based on Assumption 2, it can be concluded that the velocity, u, is

not a function of time as well. Therefore, the velocity, u, is at most a function of y.

Result of continuity: u = u(y) Eq. (2)

The y- and z-components of the incompressible Navier-Stokes Equations become unconcerned due to

parallel flow. Hence, the x-component of the incompressible Navier-Stokes Equation has been

utilized.



(a) (Continued)

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂+

∂

∂+

∂

∂+

∂

∂−

z

uw

y

uv

x

uu

t

u

z

u

y

u

x

u

x

pg x ρµρ

2

2

2

2

2

2

By applying the assumptions stated,

( ){ { { { { { {

∂

∂+

∂

∂+

∂

∂+

∂

∂=

∂

∂+

∂

∂+

∂

∂+

∂

∂−−

6 Assumption3 AssumptionContinuity2 Assumption6 Assumption

2

2

2

2

Continuity

2

2

5 Assumption

z

uw

y

uv

x

uu

t

u

z

u

y

u

x

u

x

pg ρµρ

Through term-by-term analysis, the Navier-Stokes Equation becomes a second-order linear

differential equation.

( )µ

ρµρ

g

dy

ud

dy

udg =⇔=+−

2

2

2

2

0

Integrate the second-order linear differential equation with respect to y gives:

12

2

Kyg

dy

dudy

gdy

dy

ud+=⇒= ∫∫ µ

ρ

µ

ρ Eq. (3)

Integrate again gives:

( ) 21

2

12

KyKyg

yudyCyg

dydy

du++=⇒

+= ∫∫ µ

ρ

µ

ρ Eq. (4)

By applying the 1st boundary condition, 0=

=hydy

du to Eq. (3) yields:

( )µ

ρ

µ

ρ ghKKh

g−=⇒+= 110

By applying the 2nd

boundary condition, u(0) = V to Eq. (4) gives:

( ) ( ) VKKKg

V =⇒++= 221

200

2µ

ρ

Substitute µ

ρghK −=1 and K2 = V into Eq. (4) gives the formula which describes the velocity

distribution of the fluid flow:

( ) Vygh

yg

yu +−=µ

ρ

µ

ρ 2

2 Ans.



(b) Based on the knowledge of Calculus, the average value on an interval [a, b] of an integrable function

f is given by:

( ) ( )dxxfab

f

b

a

1

avg ∫−=

Likewise, the average velocity, Vavg, of the film can be obtained by integrating the function of

velocity distribution, u(y), which derived in part (a) across the film:

( )

µ

ρ

µ

ρ

µ

ρ

µ

ρ

µ

ρ

µ

ρ

µ

ρ

30

26

1

26

1

2

1

0

1

223

0

23

0

2

0

avg

ghVVhh

ghh

g

h

Vyygh

yg

hdyVy

ghy

g

hdyyu

hV

hhh

−=

−

+−=

+−=

+−=

−= ∫∫

µ

ρ

3

2

avg

ghVV −=∴ Ans.

(c) With the relationship of the velocity distribution obtained in part (a), the equation of flow-rate per

unit width, Q, can be formulated from the relation

( )µ

ρ

µ

ρ

µ

ρ

µ

ρ

µ

ρ

32

2

3

0

2

0

2

0

ghVhVy

ghy

gdyVy

ghy

gdyyuQ

hhh

−=

+−=

+−== ∫∫

Since there has no net flow either up or down, therefore, µ

ρ

3

3gh

VC = . Ans.



(d) The equation which describe the velocity distribution, u(y), which derived in part (a) can be

expressed in dimensionless form by dividing both sides of the equation by the velocity V.

12

1222

12

12

2

222

2

2

22

2

+

−

=

+

⋅⋅−

⋅=

+⋅−⋅=

+−=

h

yA

h

yA

V

u

h

y

V

gh

h

y

V

gh

V

u

h

y

V

gh

h

y

V

gh

V

u

yV

ghy

V

g

V

u

µ

ρ

µ

ρ

µ

ρ

µ

ρ

µ

ρ

µ

ρ

where V

ghA

µ

ρ

2

2

=

The velocity distribution of the film at various y-locations which has the domain ranges from y/h = 0

to y/h = 1 are computed by introducing the variable A from A = 0 to A = 2. The computations are

carried out by using Microsoft Excel and the results are tabulated in the table below.

y/h u/V

A = 0 A = 0.5 A = 1 A = 1.5 A = 2

0 1 1 1 1 1

0.05 1 0.95125 0.9025 0.85375 0.805

0.1 1 0.905 0.81 0.715 0.62

0.15 1 0.86125 0.7225 0.58375 0.445

0.2 1 0.82 0.64 0.46 0.28

0.25 1 0.78125 0.5625 0.34375 0.125

0.3 1 0.745 0.49 0.235 -0.02

0.35 1 0.71125 0.4225 0.13375 -0.155

0.4 1 0.68 0.36 0.04 -0.28

0.45 1 0.65125 0.3025 -0.04625 -0.395

0.5 1 0.625 0.25 -0.125 -0.5

0.55 1 0.60125 0.2025 -0.19625 -0.595

0.6 1 0.58 0.16 -0.26 -0.68

0.65 1 0.56125 0.1225 -0.31625 -0.755

0.7 1 0.545 0.09 -0.365 -0.82

0.75 1 0.53125 0.0625 -0.40625 -0.875

0.8 1 0.52 0.04 -0.44 -0.92

0.85 1 0.51125 0.0225 -0.46625 -0.955

0.9 1 0.505 0.01 -0.485 -0.98

0.95 1 0.50125 0.0025 -0.49625 -0.995

1 1 0.5 0 -0.5 -1

The dimensionless velocity distribution in the film (velocity profile) shown in the figure below is

plotted based on the results above by using Microsoft Excel.



Figure: Plot of velocity distribution in the film (velocity profile) at various y-locations

As illustrated on the plot of velocity profile of the fluid flow shown above, there are still some

portions of the fluid flow downward (u/V < 0) although the belt is moving upward with a velocity V

as a result of small viscosity or slow belt speed. Apart from that, there will be a net upward flow of

fluid (positive V) if V > ρgh2/3µ. Thus, it can be concluded that in order to lift a fluid with small

viscosity, a relatively large belt speed is required.

hes2340 fluid mechanics 1, semester 1, 2012, assignment 1 - basic concepts of fluid flow

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