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REE 307Fluid Mechanics II

Lecture 5November 8, 2017

Dr./ Ahmed Mohamed Nagib Elmekawy

Zewail City for Science and Technology

DIFFERENTIAL ANALYSIS

OF FLUID FLOW

2

2

The fundamental differential equations of fluid motion are derived in

this chapter, and we show how to solve them analytically for some

simple flows. More complicated flows, such as the air flow induced by

a tornado shown here, cannot be solved exactly.

3

3

Objectives

• Understand how the differential equation of

conservation of mass and the differential linear

momentum equation are derived and applied

• Obtain analytical solutions of the equations of

motion for simple flow fields

4

4

LAGRANGIAN AND EULERIAN DESCRIPTIONS

Kinematics: The study of motion.

Fluid kinematics: The study of how fluids flow and how to describe fluidmotion.

With a small number of objects, such

as billiard balls on a pool table,

individual objects can be tracked.

In the Lagrangian description, one

must keep track of the position and

velocity of individual particles.

There are two distinct ways to describe motion: Lagrangian and Eulerian

Lagrangian description: To follow the path of individual objects.

This method requires us to track the position and velocity of each individual

fluid parcel (fluid particle) and take to be a parcel of fixedidentity.

5

• A more common method is Eulerian description of fluid motion.

• In the Eulerian description of fluid flow, a finite volume called a flow domain

or control volume is defined, through which fluid flows in and out.

• Instead of tracking individual fluid particles, we define field variables,

functions of space and time, within the control volume.

• The field variable at a particular location at a particular time is the value of

the variable for whichever fluid particle happens to occupy that location at

that time.

• For example, the pressure field is a scalar field variable. We define the

velocity field as a vector field variable.

Collectively, these (and other) field variables define the flow field. The

velocity field can be expanded in Cartesian coordinates as

6

6

In the Eulerian description, one

defines field variables, such as

the pressure field and the

velocity field, at any location

and instant in time.

7

• In the Eulerian description we

don’t really care what happens to

individual fluid particles; rather we

are concerned with the pressure,

velocity, acceleration, etc., of

whichever fluid particle happens

to be at the location of interest at

the time of interest.

• While there are many occasions in

which the Lagrangian description

is useful, the Eulerian description

is often more convenient for fluid

mechanics applications.

• Experimental measurements are

generally more suited to the

Eulerian description.

7

CONSERVATION OF MASS—THE CONTINUITYEQUATION

To derive a differential

conservation equation, we

imagine shrinking a control

volume to infinitesimal size.

8

The net rate of change of mass within the

control volume is equal to the rate at

which mass flows into the control volume

minus the rate at which mass flows out of

the control volume.

8

8

Derivation Using an Infinitesimal Control Volume

A small box-shaped control

volume centered at point P

is used for derivation of the

differential equation for

conservation of mass in

Cartesian coordinates; the

blue dots indicate the center

of each face.

At locations away from the center of the

box, we use a Taylor ser ies expansion

about the center of the box.

9

10

10

Conservation of Mass: Alternative forms

• Use product rule on divergence term

kz

jy

ix

kwjviuV

12

Continuity Equation in Cylindrical Coordinates

Velocity components and unit vectors in cylindrical coordinates: (a) two-

dimensional flow in the xy- or r𝜃-plane, (b) three-dimensional flow.

13

Conservation of Mass: Cylindrical coordinates

The divergence

operation in Cartesian

and cylindrical

coordinates.

15

Special C a s e s of the Continuity Equation

Special Case 1: Steady Compressible Flow

16

Special Case 2: Incompressible Flow

17

If the differential fluid

element is a material

element, it moves with the

flow and Newton’s second

law applies directly.

21

THE DIFFERENTIAL LINEAR MOMENTUM EQUATION

Derivation Using Newton’s Second Law

Acceleration Field

Newton’s second law applied to a fluid

particle; the acceleration vector (gray arrow)

is in the same direction as the force vector

(black arrow), but the velocity vector (red

arrow) may act in a different direction.

The equations of motion for fluid flow

(such as Newton’s second law) are

written for a fluid particle, which we

also call a material particle.

If we were to follow a particular fluid

particle as it moves around in the

flow, we would be employing the

Lagrangian description, and the

equations of motion would be directly

applicable.

For example, we would define the

particle’s location in space in terms

of a material position vector

(xparticle(t), yparticle(t), zparticle(t)).

22

Local

acce lera tion

Advective (convective)

acceleration

The components of the

acceleration vector in cartesian

coordinates:

Flow of water through the nozzle of a

garden hose illustrates that fluid

particles may accelerate, even in a

steady flow. In this example, the exit

speed of the water is much higher than

the water speed in the hose, implying

that fluid particles have accelerated

even though the flow is steady.

24

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 Va 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 ideal flow in a variable-area duct

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 and distributed over the volume of the fluid such as gravitational and electromagnetic.

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

Positive components of the stress

tensor in Cartesian coordinates on the

positive (right, top, and front) faces of

an infinitesimal rectangular control

volume. The blue dots indicate the

center of each face. Positive

components on the negative (left,

bottom, and back) faces are in the

opposite direction of those shown here.

Body Forces

27

For fluids at rest, the only

stress on a fluid element is

the hydrostatic pressure,

which always acts inward

and normal to any surface.

𝜏𝑖𝑗 called the

viscous s t ress

tensor

Surface Forces

Sketch illustrating the surface forces acting in the x-

direction due to the appropriate stress tensor component

on each face of the differential control volume; the blue

dots indicate the center of each face.

30

If the differential fluid

element is a material

element, it moves with the

flow and Newton’s second

law applies directly.

31

Newtonian versus Non-Newtonian Fluids

Rheological behavior of fluids—shear

stress as a function of shear strain rate.

32

Rheology: The study of the deformation of flowing fluids.

Newtonian fluids: Fluids for which the shear stress is linearly proportional to the shear strain rate.

Newtonian fluids: Fluids for which the shear stress is not linearly related to the shear strain rate.

Viscoelastic: A fluid that returns (either fully or partially) to its original shape after the applied stress is released.

Some non-Newtonian fluids are called

shear thinning fluids or

pseudoplas t ic fluids, because the

more the fluid is sheared, the less

viscous it becomes.

Plastic fluids are those in which the

shear thinning effect is extreme.

In some fluids a finite stress called the

yield s t ress is required before the

fluid begins to flow at all; such fluids

are called Bingham plastic fluids.

Derivation of the Navier–Stokes Equation for Incompressible, Isothermal Flow

The incompressible flow

approximation implies constant

density, and the isothermal

approximation implies constant

viscosity.

33

The Laplacian operator, shown

here in both Cartesian and

cylindrical coordinates, appears

in the viscous term of the

incompressible Navier–Stokes

equation.

34

28

The Navier–Stokes equation is the

cornerstone of fluid mechanics.

The Navier–Stokes equation is an

unsteady, nonlinear, second order, partial

differential equation.

Equation 9–60 has four unknowns (three

velocity components and pressure), yet it

represents only three equations (three

components since it is a vector equation).

Obviously we need another equation to

make the problem solvable. The fourth

equation is the incompressible continuity

equation (Eq. 9–16).

35

Navier-Stokes Equations

)()]([)]([

)].3

22([)(

az

u

x

w

zy

u

x

v

y

Vx

u

xx

pg

z

uw

y

uv

x

uu

t

ux

)()]([)].3

22([

)]([)(

by

w

z

v

zV

y

v

y

y

u

x

v

xy

pg

z

vw

y

vv

x

vu

t

vy

)()].3

22([)]([

)]([)(

cVz

w

zy

w

z

v

y

z

u

x

w

xz

pg

z

ww

y

wv

x

wu

t

wz

x-momentum

y-momentum

z-momentum

Navier-Stokes Equations• For incompressible fluids, constant µ:

• Continuity equation: .V = 0

uz

u

y

u

x

u

z

w

y

v

x

u

xz

u

y

u

x

u

zx

w

yx

v

x

u

z

u

y

u

x

u

z

u

x

w

zy

u

x

v

yx

u

x

z

u

x

w

zy

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

az

u

y

u

x

u

x

pg

Dt

Dux

)()(2

2

2

2

2

2

bz

v

y

v

x

v

y

pg

Dt

Dvy

)()(2

2

2

2

2

2

cz

w

y

w

x

w

z

pg

Dt

Dwz

VpgDt

VD

2 Incompressible NSEwritten in vector form

Continuity and Navier–Stokes Equations in Cartesian Coordinates

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 be written 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:

)()( ax

pg

z

uw

y

uv

x

uu

t

ux

pgDt

VD

Euler equationswritten in vector form

)()( by

pg

z

vw

y

vv

x

vu

t

vy

)()( cz

pg

z

ww

y

wv

x

wu

t

wz

Continuity and Navier–Stokes Equations in Cylindrical Coordinates

Differential Analysis of Fluid Flow Problems

• 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 field

2. 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)

DIFFERENTIAL ANALYSIS OF FLUID FLOW PROBLEMS

•Calculating both the velocity and pressure fields for a flow of known geometry

and known boundary conditions

A general three-dimensional

incompressible flow field with

constant properties requires

four equations to solve for

four unknowns.

44

Exact Solutions of the NSE

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

2. List all appropriate assumptions, approximations, simplifications, and boundary conditions

3. Simplify the differential equations as much as possible

4. Integrate the equations

5. Apply BCs to solve for constants of integration

6. Verify results

• Boundary conditions are critical to exact, approximate, and computational solutions.▪ BC’s used in analytical solutions are

• No-slip boundary condition

• Interface boundary condition

Procedure for solving continuity and NSE

32

Exact Solutions of the Continuity

and Navier–Stokes Equations

Procedure for solving the

incompressible continuity and

Navier–Stokes equations.

Boundary Conditions

A piston moving at speed VP in a cylinder.

A thin film of oil is sheared between the

piston and the cylinder; a magnified view of

the oil film is shown. The no-slip boundary

condition requires that the velocity of fluid

adjacent to a wall equal that of the wall.

46

33

At an interface between two

fluids, the velocity of the two

fluids must be equal. In addition,

the shear stress parallel to the

interface must be the same in

both fluids.

Along a horizontal free surface of

water and air, the water and air

velocities must be equal and the

shear stresses must match.

However, since air << water, a

good approximation is that the

shear stress at the water surface is

negligibly small.

47

34

Boundary conditions along a plane of

symmetry are defined so as to ensure

that the flow field on one side of the

symmetry plane is a mirror image of

that on the other side, as shown here

for a horizontal symmetry plane.

Other boundary conditions arise

depending on the problem setup.

For example, we often need to define

inlet boundary conditions at a

boundary of a flow domain where fluid

enters the domain.

Likewise, we define outlet boundary

conditions at an outflow.

Symmetry boundary conditions are

useful along an axis or plane of

symmetry.

For unsteady flow problems we also

need to define initial conditions (at

the starting time, usually t = 0).

48

Summary

• Introduction

• Conservation of mass-The continuity equation

Derivation Using an Infinitesimal Control Volume

Continuity Equation in Cylindrical Coordinates

Special Cases of the Continuity Equation

• The Navier-Stokes equation

Derivation of the Navier–Stokes Equation for Incompressible, Isothermal Flow

Continuity and Navier–Stokes Equations in CartesianCoordinates

Continuity and Navier–Stokes Equations in CylindricalCoordinates

• Differential analysis of fluid flow problems

Exact Solutions of the Continuity and Navier– Stokes Equations

49

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

• Assumptions1. Plates are infinite in x and z

2. Flow is steady, /t = 0

3. Parallel flow, v = 0

4. Incompressible, Newtonian, laminar, constant properties

5. No pressure gradient

6. 1D, 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

ggjgg y ,ˆ


• Step 3: Simplify3 6

Note: these numbers referto 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

0

z

w

y

v

x

u

0

x

u

)()(2

2

2

2

2

2

z

u

y

u

x

ug

x

p

z

uw

y

uv

x

uu

t

ux


• Step 3: Simplify, cont.y-momentum

2,3 3 3 3,6 3 33

z-momentum

2,6 6 6 6 7 6 66

ygy

p

p = p(y)0

z

pyg

dy

dp

y

p

p = p(y)

)()(2

2

2

2

2

2

z

v

y

v

x

vg

y

p

z

vw

y

vv

x

vu

t

vy

)()(2

2

2

2

2

2

z

w

y

w

x

wg

z

p

z

ww

y

wv

x

wu

t

wz


• Step 4: Integrate

y-momentum

X-momentum

integrate integrate

integrate

gdy

dp 3)( Cgyyp


• 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 an arbitrary constant (recall only P appears in NSE).• Let p = p0 at y = 0 (C3 renamed p0)

1. Hydrostatic pressure2. Pressure acts independently of flowgypyp 0)(


• Step 6: Verify solution by back-substituting into differential equations• Given the solution (u,v,w)=(Vy/h, 0, 0)

• Continuity is satisfied0 + 0 + 0 = 0

• X-momentum is satisfied

)()(2

2

2

2

2

2

z

u

y

u

x

ug

x

p

z

uw

y

uv

x

uu

t

ux


• Finally, calculate shear force on bottom plate

Shear force per unit area acting on the wall

Note that w is equal and opposite to the shear stress acting on the fluid yx

(Newton’s third law).

h

V

y

u

x

vyxxy

)(u=V/h

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

rectangular channel of two parallel plates where the width of the channel is much larger than the height, h, between the 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 Important Assumptions

• The width is very large compared to the height of the plate.

• No entrance or exit effects.

• Fully developed flow.

• Steady flow

• THEREFORE…• Velocity can only be dependent on vertical location in the flow (u)

• v = w = 0

• The pressure drop is constant and in the x-direction only.

.in length a is where,p

Constantp

xLLx

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 we don’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 direction

2ax

y

ay

ay

gvvvv

•

2pt

Navier-Stokes EquationsIn Vector Form for incompressible flow:

z

y

x

gz

w

y

w

x

w

zz

ww

y

wv

x

wu

t

w

gz

v

y

v

x

v

yz

vw

y

vv

x

vu

t

v

gz

u

y

u

x

u

xz

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

gz

w

y

w

x

w

zz

ww

y

wv

x

wu

t

w

gz

v

y

v

x

v

yz

vw

y

vv

x

vu

t

v

gz

u

y

u

x

u

xz

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

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 yp is a function of x only

Rewriting (4), we get

(5)

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

p is a function of x only and is a constant and therefore 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 an ordinary derivative.

(6)

Therefore, (5) becomes

(7)

L

p

x

p

y

u

12

2

x

p

L

p

L

p

dy

ud

2

2

1CyL

p

dy

du

integrate

Integrating again, we get

(8)

where C1 and C2 are constants to be determined using the boundary conditions given below:

Substituting the boundary conditions in (8), we get

Fixed plate

Fixed plate


2ax

y

ayu at0

ayu at0

ay

ay

21

2

2CyC

y

L

pu

01 C2

2

2

a

L

pC

and

Therefore, (8) reduces to 22

2ya

L

pu

Parabolic velocity profile

}no-slip boundarycondition

Fixed plate

Moving plate at velocity U


2ax

y

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

where C1 and C2 are constants to be determined using the boundary conditions given below:

ay

ay

21

2

2CyC

y

L

pu

ayu at0

ayUu at

Substituting the boundary conditions in (8), we get

and

Therefore, (8) reduces toa

UC

21

22

2

2

a

L

pUC

yaa

Uya

L

pu

22

22

(8)

}no-slip boundarycondition

Parabolic velocity profile is super imposed on a linear velocity profile

Fixed plate



2ax

y

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

ay

ay

yaa

Uya

L

pu

22

22

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

Therefore, we get

yaa

Uu

2Linear velocity profile

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

Starting from plane Couette-Poiseuille flow

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

yaa

Uya

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

yy

yyLU

pau

1

2

11

2

22

U

uu

a

y

a

y

LU

pa

U

u1

2

11

2 2

22

R. Shanthini

15 March 2012

-1

0

1

-0.5 0 0.5 1 1.5

1

0.25

0

-0.25

-1

yyLU

pau

1

2

11

2

22

2/2 aLU

p

a

yy

U

uu

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

R. Shanthini

15 March 2012-1

0

1

-0.25 0.75

0

-0.5

yyLU

pau

1

2

11

2

22

2/2 aLU

p

a

yy

U

uu

BACK FLOW


R. Shanthini

15 March 2012

-1

0

1

-1 0 1

0

-0.25

-0.5

-1.5

yyLU

pau

1

2

11

2

22

2/2 aLU

p

a

yy

U

uu


yaa

Uya

L

pu

22

22

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

Fixed plate


Fluid flow direction2ax

y

ay

ay

a

a

a

a

a

a

yay

a

Uyya

L

pdyya

a

Uya

L

pudyQ

223222

23222

22322232

22

33

22

33 a

aa

Uaa

L

paa

a

Uaa

L

pQ

UaL

paQ

3

2 3

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

-2

-1

0

1

2

3

4

-4 -2 0 2 4

Plot the non-dimensionalized volumetric flow rate of the fully 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

Laminar Flow in pipes

.oflinearaispressure..

0

zeidz

dp

z

p

rvv

vv

zz

r

Assumptions:

Now look at the z-momentum equation.

2

2

2

2

2

11

)(

dz

vv

rr

vr

rrdz

dp

z

vv

v

r

v

r

vv

t

v

zzz

zz

zzr

z

Steady Laminar pipe flow.

The above “assumptions” can be obtained from the single assumption 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

rrdz

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 the system (similar to what an electrical engineer might do in testing a linear system).

dz

dp

dr

dvr

dr

d

r

z

11

dz

dpr

dr

dvr

dr

d z

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

2C

dz

dpr

dr

dvr z

dz

dpr

dr

dvr z

2

2

dz

dpr

dr

dvz

2

Divide by r

2

2

4C

dz

dprvz

at pipe wall r = a velocity = 0 2

2

40 C

dz

dpa

dz

dpaC

4

2

2

Laminar flow in pipesSubstituting in the velocity equation gives:

)4

(44

2222

ra

dz

dp

dz

dpa

dz

dprvz

Volume flow rate is obtained from:

tioncross

zdAvQsec

ar

rzdrrvQ

02

ar

rdrrardz

dp

Q0

22

2

)(

dz

dpaQ

8

4

dz

dpDQ

128

4

Laminar flow in pipesAverage velocity, vav= Q/area

Maximum velocity at r = 0:

dz

dpDD

dz

dpDv avez

324/

128

224

,

)16

()4

(22

max,

D

dz

dpa

dz

dpvz

vavergae = ½ vmax

Shear stress:

)(8)4

()4

2)((

,

D

vD

dz

dpa

dz

dp

dr

dv avez

ar

z

Laminar flow in pipes

Coefficient of friction: f = 4 Cf

eave

ave

ave

wf

RDv

v

v

C1616

2

1 22

Drag Coefficient:

eRf

64

lecture 5 - drahmednagib.com · the navier–stokes equation isthe cornerstone of fluid mechanics....

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