lecture 2 0 fluid dynamics

8/18/2019 Lecture 2 0 Fluid Dynamics

1/35

Fluid Dynamics

Dr. Hamdy A. Kandil

Review: Fluid Kinematics -Acceleration Field

Consider a fluid particle and Newton's second law,

The acceleration of the particle is the time derivative of theparticle's velocity.

However, particle velocity at a point is the same as the fluidvelocity,

To take the time derivative of V, chain rule must be used because.

particle part icle particleF m a

par ticl e

par ticl e

dV a

dt

, , particle particle particle particleV V x t y t z t

particle particle particle

particle

dx dy dzV dt V V V a

t dt x dt y dt z dt


2/35

Acceleration Field Since

In vector form, the acceleration can be written as

First term is called the local acceleration and is nonzero only forunsteady flows.

Second term is called the advective acceleration and accountsfor the effect of the fluid particle moving to a new location in theflow, where the velocity is different.

particle

V V V V

a u v wt x y z

, , particle particle particledx dy dz

u v wdt dt dt

V V t

V

dt

V d t z y xa

)(),,,(

then

k

z

j

y

i

x

k w jviuV

Where:

Acceleration Components

z

ww

y

wv

x

wu

t

wa

z

vw

y

vv

x

vu

t

va

z

uw

y

uv

x

uu

t

ua

z

y

x

The components of the acceleration are:

particle

V V V V a u v w

t x y z

Vector equation:

x- component

Y-component

z-component

Question: Give examples of steady flows with acceleration

Incompressible Steady idealflow in a variable-area duct


3/35

Streamlines A Streamline is a curve thatis everywhere tangent to theinstantaneous local velocityvector.

Consider an arc length

must be parallel to thelocal velocity vector

Geometric arguments resultsin the equation for astreamline

dr dxi dyj dzk

dr

V ui vj wk

dr dx dy dz

V u v w

u

v

dx

dy

Properties of Streamlines A streamline can not have a sharp corner

Two streamlines can not intersect or meet at a sharp corner

No flow across a tube formed by streamlines (steam tube).


4/35

Streaklines A Streakline is the locus of fluid

particles that have passedsequentially through aprescribed point in the flow.

Easy to generate inexperiments: dye in a waterflow, or smoke in an airflow.

Pathlines A Pathline is the actual pathtraveled (traced) by anindividual fluid particle oversome time period.

Same as the fluid particle'smaterial position vector

Particle location at time t:

Particle Image Velocimetry(PIV) is a modern experimentaltechnique to measure velocityfield over a plane in the flowfield.

, , par ticle particle par ticle x t y t z t

start

t

start

t

x x Vdt

For steady flow, streamlines, pathlines, andstreaklines are identical.


5/35

Which type of lines?

Conservation of Mass Differential CV

Infinitesimal control volumeof dimensions dx, dy, dz Area of right

face = dy dz

Mass flow rate throughthe right face of thecontrol volume


6/35


Now, sum up the mass flow rates into and out of the 6 faces of the CV

Plug into integral conservation of mass equation

Net mass flow rate into CV:

Net mass flow rate out of CV:

Rate of change of the control volume mass:


After substitution,

Dividing through by volume dxdydz

Or, if we apply the definition of the divergence of a vector

k z

j y

i x

k w jviuV

where

0)()()(

z

w

y

v

x

u

t


7/35

Conservation of Mass: Alternative forms

Use product rule on divergence term

k z

j y

i x

k w jviuV

Conservation of Mass: Cylindrical coordinates

There are many problems which are simpler to solve if theequations are written in cylindrical-polar coordinates

Easiest way to convert from Cartesian is to use vector form anddefinition of divergence operator in cylindrical coordinates


8/35

Conservation of Mass: Cylindricalcoordinates

Conservation of Mass: Special Cases Steady compressible flow

Cartesian

Cylindrical

0)()()(

z

w

y

v

x

u


9/35

Conservation of Mass: Special Cases Incompressible flow

Cartesian

Cylindrical

= constant, and hence

0

z

w

y

v

x

u

k z

j y

i x

k w jviuV

Conservation of Mass In general, continuity equation cannot be used by itself to

solve for flow field, however it can be used to

1. Determine if a velocity field represents a flow.

2. Find missing velocity component

equation. continuity hesatisfy t torequired , w:Determine?

flow ibleincompressanFor

Example

222

w

z yz xyv

z y xu

),(2

3:Solution2

y xc z

xzw


10/35

Conservation of MomentumTypes of forces:1. Surface forces: include all forces acting on the boundaries of a

medium though direct contact such as pressure, friction,…etc.

2. Body forces are developed without physical contact anddistributed over the volume of the fluid such as gravitationaland electromagnetic.

The force F acting on A may be resolved into two components,one normal and the other tangential to the area.

Stresses (forces per unit area)

Double subscript notation for stresses.• First subscript refers to the surface• Second subscript refers to the direction• Use for normal stresses and for tangential stresses

Surface ofconstant xSurface of

constant -x


11/35

Momentum Conservation From Newton’s second law: Force = mass * acceleration

Consider a small element x y z as shown

The stresses at the center of the element are presented by thestress tensor:

x

y

z

The element experiences an acceleration

DVm ( )

Dt

as it is under the action of various forces:

normal stresses, shear stresses, and gravitational force.

V V V V x y z u v w

t x y z

zz zy zx

yz yy yx

xz xy xx

T

Surface forces in the x direction acting on afluid element.

Consider only the forces in the x direction acting on all six surfaces

zz zy zx

yz yy yx

xz xy xx

T


12/35

Momentum Balance (cont.)

yxxx zx

Net force acting along the x-direction:

x x x x x y z x y z x y z g x y z

Normal stress Shear stressesBody force

z y xg y x z

z

y x z

z

z x y

y

z x y

y z y

x

x z y

x

xF

x zx

zx zx

zx

yx

yx

yx

yx xx

xx xx

xx x

)

2

()

2

()

2

(

)2

()2

()2

(

bxsx x F F F

)()( a z

uw

y

uv

x

uu

t

u

z y xg zx

yx xx x

The differential equation of motion in the x-direction is: xsxbx x x maF F maF

Similar equations can be obtained for the other two directions

Forces acting on the elementSurface forces

Body forces


13/35

Equation of Motion (momentum eq.s)

Velocities stresses ----- Unknowns

fluid. aformotionof equationaldifferenti General

)()(

)()(

)()(

c z

w

w y

w

v x

w

ut

w

z y xg

b z

vw

y

vv

x

vu

t

v

z y xg

a z

uw

y

uv

x

uu

t

u

z y xg

zz yz xz z

zy yy xy

y

zx yx xx x

zszbz z z

ysyby x x

xsxbx x x

maF F maF

maF F maF

maF F maF

Conservation of Linear Momentum

Unfortunately, this equation is not very useful

10 unknowns

Stress tensor, : 6 independent components

Note:

Density

Velocity, V : 3 independent components

4 equations (continuity + 3 momentum) 6 more equations required to close problem!

zz zy zx

yz yy yx

xz xy xx

T

zy yz zx xz yx xy and ,


14/35

Navier-Stokes Equations For Newtonian fluids the stress tensor components have some

definitions.

Substituting the stress tensor components into the equation ofmotion with produce the well-known Navier-Stokes equationsthat are suitable only for Newtonian Fluids:

Stress tensor components:

z

wV p

y

vV p

x

uV p

zz

yy

xx

2)3

2(

2)3

2(

2)3

2(

)(

)(

)(

y

w

z

v

z

u

x

w

y

u

x

v

zy yz

zx xz

yx xy

&

Navier-Stokes Equations

)()]([

)]([)].3

22([

a z

u

x

w

z

y

u

x

v

yV

x

u

x x

pg

Dt

Du x

)()]([

)].3

22([)]([

b y

w

z

v

z

V y

v

y y

u

x

v

x y

pg

Dt

Dv y

)()].3

22([

)]([)]([

cV z

w

z

yw

zv

y zu

xw

x z pg

Dt Dw z

x-momentum

y-momentum

z-momentum


15/35

Navier-Stokes Equations For incompressible fluids, constant µ:

Continuity equation: .V = 0

u z

u

y

u

x

u

z

w

y

v

x

u

x z

u

y

u

x

u

z x

w

y x

v

x

u

z

u

y

u

x

u

z

u

x

w

z y

u

x

v

y x

u

x

z

u

x

w

z y

u

x

v

yV

x

u

x

2

2

2

2

2

2

2

2

2

2

2

2

2

22

2

2

2

2

2

2

2

2

)(

)()(

)()(

)]}[()][()]2[({

)]([)]([)].3

22([

Navier-Stokes Equations For incompressible flow with constant dynamic viscosity:

x- momentum

Y- momentum

z-momentum

In vector form, the three equations are given by:

)()(2

2

2

2

2

2

a z

u

y

u

x

u

x

pg

Dt

Du x

)()(2

2

2

2

2

2

b z

v

y

v

x

v

y

pg

Dt

Dv y

)()(2

2

2

2

2

2

c z

w

y

w

x

w

z

pg

Dt

Dw z

V pg Dt

V D

2 Incompressible NSEwritten in vector form


16/35

Navier-Stokes Equation The Navier-Stokes equations for incompressible flow in vector

form:

This results in a closed system of equations!

4 equations (continuity and 3 momentum equations)

4 unknowns (U, V, W, p)

In addition to vector form, incompressible N-S equation can bewritten in several other forms including:

Cartesian coordinates

Cylindrical coordinates

Tensor notation

Incompressible NSEwritten in vector form

Euler Equations For inviscid flow (µ = 0) the momentum equations are given by:

x- momentum

Y- momentum

z-momentum

In vector form, the three equations are given by:

)()( a x

pg

z

uw

y

uv

x

uu

t

u x

pg Dt

V D

Euler equationswritten in vector form

)()( b y

pg

z

vw

y

vv

x

vu

t

v y

)()( c z

pg

z

ww

y

wv

x

wu

t

w z


17/35

Differential Analysis of Fluid FlowProblems Now that we have a set of governing partial differential

equations, there are 2 problems we can solve

1. Calculate pressure (P ) for a known velocity field2. Calculate velocity (U, V, W ) and pressure (P ) for known

geometry, boundary conditions (BC), and initial conditions(IC)

There are about 80 known exact solutions to the NSE

Solutions can be classified by type or geometry, for example:

1. Couette shear flows

2. Steady duct/pipe flows (Poisseulle flow)

Exact Solutions of the NSE

1. Set up the problem and geometry, identifying all relevantdimensions and parameters

2. List all appropriate assumptions, approximations, simplifications,

and boundary conditions3. Simplify the differential equations as much as possible4. Integrate the equations5. Apply BCs to solve for constants of integration6. Verify results

Boundary conditions are critical to exact, approximate, andcomputational solutions.

BC’s used in analytical solutions are No-slip boundary condition

Interface boundary condition

Procedure for solving continuity and NSE


18/35

boundary conditions

For a fluid in contact with asolid wall, the velocity of thefluid must equal that of the

wall

When two fluids meet at aninterface, the velocity andshear stress must be thesame on both sides

If surface tension effectsare negligible and thesurface is nearly flat

Interface boundary condition

No-slip boundary condition

Interface boundary condition Degenerate case of the interface BC occurs at the free surface

of a liquid. Same conditions hold

Since air


19/35

Example exact solution

Fully Developed Couette Flow For the given geometry and BC’s, calculate the velocity and

pressure fields, and estimate the shear force per unit area

acting on the bottom plate

Step 1: Geometry, dimensions, and properties

Fully Developed Couette Flow

Step 2: Assumptions and BC’s

Assumptions

1. Plates are infinite in x and z

2. Flow is steady, /t = 03. Parallel flow, v = 0

4. Incompressible, Newtonian, laminar, constant properties

5. No pressure gradient

6. 2D, w = 0, /z = 0

7. Gravity acts in the -y direction,

Boundary conditions1. Bottom plate (y=0) – no slip condition: u=0, v=0, w=0

2. Top plate (y=h) : no slip condition: u=V, v=0, w=0

gg jgg y ,ˆ


20/35


Step 3: Simplify 3 6

Note: these numbers refer

to the assumptions on the

previous slide

This means the flow is “fully developed” or not changing in the direction of flow

Continuity

X-momentum

2 Cont. 3 6 5 7 Cont. 6

02

2

y

U


Step 3: Simplify, cont.Y-momentum

2,3 3 3 3,6 3 33

Z-momentum

2,6 6 6 6 7 6 66

yg y

p

p = p(y)0 z

p yg

dy

dp

y

p

p = p(y)


21/35


Step 4: Integrate

y-momentum

X-momentumintegrate integrate

integrate

gdy

dp 3)( C gy y p


Step 5: Apply BC’s

Y = 0, u = 0 = C1(0) + C2 C2 = 0

Y = h, u =V =C1h C1 = V/h

This gives

For pressure, no explicit BC, therefore C3 can remain anarbitrary constant (recall only P appears in NSE).

Let p = p0 at y = 0 (C3 renamed p0)

1. Hydrostatic pressure

2. Pressure acts independently of flowgy p y p 0)(


22/35


Step 6: Verify solution by back-substituting into differentialequations

Given the solution (u,v,w)=(Vy/h, 0, 0)

Continuity is satisfied

0 + 0 + 0 = 0

X-momentum is satisfied


Finally, calculate shear force on bottom plate

Shear force per unit area acting on the wall

Note that w is equal and opposite to theshear stress acting on the fluid yx(Newton’s third law).

h

V

y

u

x

v yx xy

)(u=V/h


23/35

Parallel Plates (Poiseuille Flow) Given: A steady, fully developed, laminar flow of a Newtonian

fluid in a rectangular channel of two parallel plates where thewidth of the channel is much larger than the height, h, betweenthe plates.

Find: The velocity profile and shear stress due to the flow.

Assumptions: Entrance Effects Neglected No-Slip Condition No vorticity/turbulence

Additional and Highlighted ImportantAssumptions

The width is very large compared to the height of the plate. No entrance or exit effects.

Fully developed flow. THEREFORE…

Velocity can only be dependent on vertical location in theflow (u )

v = w = 0 The pressure drop is constant and in the x-direction only.

.inlengthaiswhere, p

Constant p

x L L x


24/35

Boundary Conditions No Slip Condition Applies

Therefore, at y = -a and y = +a, u=v=w = 0

The bounding walls in the z direction are often ignored. If wedon’t ignore them we also need: z = -W/2 and z = +W/2, u=v=w = 0, where W is the width of

the channel.

Fixed plate

Fixed plate

Fluid flow direction2ax

y

a y

a y

Incompressible Newtonian StressTensor

z

w

z

v

y

w

z

u

x

w

zv

yw

yv

yu

xv

z

u

x

w

y

u

x

v

x

u

2

2

2

τ

Now, we cancel terms out based

on our assumptions.This results in our new tensor:

000

00

00

y

u

y

u

T


25/35

gvvvv

2 p

t

Navier-Stokes EquationsIn Vector Form for incompressible flow:

z

y

x

g z

w

y

w

x

w

z z

ww

y

wv

x

wu

t

w

g

z

v

y

v

x

v

y z

vw

y

vv

x

vu

t

v

g z

u

y

u

x

u

x z

uw

y

uv

x

uu

t

u

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

p

:component-z

p

:component-y

p

:component-x

Which we expand to component form:

z

y

x

g z

w y

w x

w z z

ww ywv

xwu

t w

g z

v

y

v

x

v

y z

vw

y

vv

x

vu

t

v

g

z

u

y

u

x

u

x z

uw

y

uv

x

uu

t

u

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

p

:component-z

p

:component-y

p

:component-x

Reducing Navier-Stokes


26/35

Reducing Navier-Stokes This reduces to:

Including our constant pressure drop:

2

2

Pressure Modified

p0 y

u

x

2

2 p0

y

u

L

2

2 p0

: becomesequationtheTherefore b

dy

ud

L

yuuut

N-S equation therefore reduces to

02

2

xg y

u

x

p

0

yg y

p

0

zg z p

Ignoring gravitational effects, we get

02

2

y

u

x

p

0

y

p

0

z

pp is not a function of z

x - component:

y - component:

z - component:

x - component:

y - component:

z - component:

p is not a function of y

p is a functionof x only


27/35

Rewriting (4), we get

(5)

u is a function of y only and therefore LHS is afunction of y only

p is a function of x only and is a constant andtherefore RHS is a function of x only

LHS = left hand side of the equationRHS = right hand side of the equation

function of (y) = function of (x)Therefore (5) gives,

x p

L p

x

p

y

u

12

2

= constant

It means = constant

That is, pressure gradient in the x -direction is a constant.

Rewriting (5), we get

where is the constant pressure gradient in the

x-direction

Since u is only a function of y, the partial derivative becomes anordinary derivative.

(6)

Therefore, (5) becomes

(7)

L

p

x

p

y

u

1

2

2

x

p

L

p

L

p

dy

ud

2

2


28/35

Integrating (6), we get

(8)

where C1 and C2 are constants to bedetermined using the boundary

conditions given below:

Substituting the boundary conditions in (8), we get

Fixed plate

Fixed plate

Fluid flow direction

2ax

y

a yu at0a yu at0

a y

a y

21

2

2C yC

y

L

pu

01

C 2

2

2

a

L

pC

and

Therefore, (8) reduces to 222

ya L

pu

Parabolicvelocity profile

}no-slipboundarycondition

R. Shanthini15 March 2012

Fixed plate

Moving plateat velocity U


2ax

y

Steady, incompressible flow of Newtonian fluid in an infinite channelwith one plate moving at uniform velocity- fully developed plane Couette-Poiseuille flow

where C1 and C2 are constants to bedetermined using the boundary conditionsgiven below:

a y

a y

21

2

2C yC

y

L

pu

a yu at0

a yU u at

Substituting the boundary conditions in (8), we get

and

Therefore, (8) reduces toa

U C

2

1

22

2

2

a

L

pU C

yaa

U ya

L

pu

22

22

(8)

}no-slipboundarycondition

Parabolic velocity profileis super imposed on alinear velocity profile


29/35

R. Shanthini15 March 2012

Fixed plate



2ax

y

Steady, incompressible flow of Newtonian fluid in an infinite channelwith one plate moving at uniform velocity- fully developed plane Couette-Poiseuille flow

a y

a y

yaa

U ya

L

pu

22

22

If zero pressure gradient is maintained inthe flow direction, then ∆p = 0.

Therefore, we get

yaa

U u

2Linear velocity profile

Points to remember:- Pressure gradient gives parabolic profile- Moving wall gives linear profile

Starting from plane Coutte-Poiseuille flow

Let us get back to the velocity profile offully developed plane Couette-Poiseuille flow

yaa

U ya

L

pu

22

22

Let us non-dimensionalize it as follows:

a

yU

a

y

L

pau 1

21

2 2

22

Rearranging gives

Dividing by U gives

Introducing non-dimensional variables and , we get

a

y y

y y LU

pau

1

2

11

2

22

U

uu

a

y

a

y

LU

pa

U

u1

2

11

2 2

22


30/35

R. Shanthini

15 March 2012-1

0

1

-0.5 0 0.5 1 1.5

1

0.25

0

-0.25

-1

y y LU

pau

1

2

11

2

22

2/2 a LU

p

a

y y

U

uu

Plot the non-dimensionalized velocity profile of thefully developed plane Couette-Poiseuille flow

R. Shanthini

15 March 2012-1

0

1

-0.25 0.75

0

-0.5

y y LU

pau

1

2

11

2

22

2/2 a LU

p

a y y

U

uu

BACK FLOW



31/35

R. Shanthini

15 March 2012-1

0

1

-1 0 1

0

-0.25

-0.5

-1.5

y y LU

pau

1

2

11

2

22

2/2 a LU

p

a

y y

U

uu


yaa

U ya

L

pu

22

22

Let us determine the volumetric flow ratethrough one unit width fluid film along z -direction, using the velocity profile

Fixed plate



2a x

y

a y

a y

a

a

a

a

a

a

yay

a

U y ya

L

pdy ya

a

U ya

L

pudyQ

223222

23222

22322232

2

2

3

3

2

2

3

3 aaa

U aa L

paaa

U aa L

pQ

Ua L

paQ

3

2 3

Determine the volumetric flow rate of thefully developed plane Couette-Poiseuille flow


32/35

-2

-1

0

1

2

3

4

-4 -2 0 2 4

Plot the non-dimensionalized volumetric flow rate of thefully developed plane Couette-Poiseuille flow

LU

pa

Ua

Q

2

3

21

Ua

Q

U

pa

2

Why the flow rate becomes zero?

Non-dimensionalizing Q, we get

Steady, incompressible flow of Newtonian fluid in a pipe- fully developed pipe Poisuille flow

Fixed pipe

z

r

Fluid flow direction 2a 2a


33/35

Laminar Flow in pipes

.of linear ais pressure..

0

zeidz

dp

z

p

r vv

vv

z z

r

Assumptions:

Now look at the z -momentum equation.

2

2

2

2

2

11

)(

dz

vv

r r

vr

r r dz

dp

z

vv

v

r

v

r

vv

t

v

z z z

z z

z zr

z

Steady Laminar pipe flow.

The above “assumptions” can be obtained from the singleassumption of “fully developed” flow.

In fully developed pipe flow, all velocity components are assumed to

be unchanging along the axial direction, and axially symmetric i.e.:

Results from Mass/Momentum

Or rearranging, and substituting for the pressure gradient:

The removal of the indicated terms yields:

011

r

vr

r r dz

dp z

In a typical situation, we would have control over dp/dz . That is,we can induce a pressure gradient by altering the pressure at one

end of the pipe. We will therefore take it as the input to thesystem (similar to what an electrical engineer might do in testinga linear system).

dz

dp

dr

dvr

dr

d

r

z

11

dz

dpr

dr

dvr

dr

d z


34/35

Laminar flow in pipes

But at r = 0 velocity = max or dvz/dr = 0 C1 = 0

Integrating once gives:

Integrating one more time gives:

1

2

2 C dz

dpr

dr

dv

r

z

dz

dpr

dr

dvr z

2

2

dz

dpr

dr

dv z

2

Divide by r

2

2

4C

dz

dpr v z

at pipe wall r = a velocity = 02

2

40 C

dz

dpa

dz

dpaC

4

2

2

Laminar flow in pipesSubstituting in the velocity equation gives:

)4

(44

2222

r a

dz

dp

dz

dpa

dz

dpr v z

Volume flow rate is obtained from:

tioncross

z dAvQsec

ar

r z dr rvQ

02

ar

r dr r ar dz

dp

Q0

22

2

)(

dz

dpaQ

8

4

dz

dp DQ

128

4


35/35

Laminar flow in pipesAverage velocity, vav= Q/area

Maximum velocity at r = 0:

dzdp D D

dz

dp D

v ave z

324/128

224

,

)16

()4

(22

max,

D

dz

dpa

dz

dpv z

vavergae = ½ vmax

Shear stress:

)(8)4

()4

2)((

,

D

v D

dz

dpa

dz

dp

dr

dv ave zar

z

Laminar flow in pipes

Coefficient of friction: f = 4 Cf

eave

ave

ave

w f

R Dv

v

v

C 1616

2

1 22

Drag Coefficient:

e R f

64

lecture 2 0 fluid dynamics

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