MECH 341

Fluid Mechanics II

Darko Matovic

, 2012

Queen's University

Dept. of Mechanical and Materials Engineering

Review of Fluid Mechanics I:

Control Volume Concept,

Reynolds Transport Theorem,

Conservation Laws

Objectives:

●

Recap key parts of the MECH241 course

●

Distinguish between Lagrangian and Eulerian

flow description

●

Derive Reynolds Transport Theorem

●

Apply it to different conserved properties

MECH 241 (Fluid Mechanics I)

Covered (Textbook sections):

●

Fluid Properties (Chapter 1)

●

Fluid Statics (Chapter 2)

●

Integral relations/ Reynolds Transport Theorem

(Chap. 3)

●

Bernoulli Equation (sec. 3.7 and 6.12)

●

Pipe flow/losses (most of Chap. 6)

Eulerian vs. Lagrangian Frame

●

Eulerian view

–

Control volume

linked to space

–

Fluid flows through

the CV boundaries

–

Becomes point in

space when sized

reduced to zero

–

“

Point” may move,

but is not linked to

fluid

●

Lagrangian view

–

Control volume

linked to the lump

of fluid

–

No material

exchange through

the boundaries

–

Becomes “fluid

particle” when size

reduced to zero.

An analogy with tracking a car:

●

Watching the surroundings pass by from the

driver's seat (or by the bumper mounted

camera) gives you Lagrangian view

●

Watching from the side as the car passes by

puts you in the Eulerian frame of reference

●

This

YouTube video

of a simulated car race

switches between different frames quickly

and often!

Flux through a Control Surface

d

dt

(

B

syst

)

=

d

dt

(

∫

CV

β

ρ

dV

)

+

∫

CS

β

ρ

(

⃗

u

r

⋅

⃗

n

)

+

∫

CV

S

β

dV

B

−

Extensive property ;

β

−

intensive property

Reynolds Transport Theorem

d

dt

(

B

syst

)

=

d

dt

(

∫

CV

β

ρ

dV

)

+

∫

CS

β

ρ

(

⃗

u

r

⋅

⃗

n

)

+

∫

CV

S

β

dV

B

−

Extensive property ;

β

−

intensive property

Mass Conservation

β

=

1

;

B

=

m

;

S

β

=

0

(

dm

dt

)

syst

=

0

=

d

dt

(

∫

CV

ρ

dV

)

+

∫

CS

ρ

(

⃗

u

r

⋅

⃗

n

)

dA

or

(

dm

dt

)

syst

=

0

=

d

dt

(

∫

CV

ρ

dV

)

+

∫

CV

ρ

div

u

r

dV

Special Cases:

Steady State

d

dt

(

⋅

)

=

0

:

∫

CS

ρ

(

⃗

u

r

⋅

⃗

n

)

dA

=

∫

CV

ρ

div

⃗

u

r

dV

=

0

One-Dimensional inlets and outlets

∫

CS

→

∑

i

=

1

N

(

⋅

)

:

∑

i

=

1

N

(

ρ

i

A

i

U

r

,

i

)

=

0

Linear Momentum Conservation

β

=

⃗

U

;

B

=

m

⃗

U

;

∫

CV

S

β

dV

=

−

⃗

F

external

d

dt

(

m

⃗

U

)

syst

=

∑

⃗

F

=

d

dt

(

∫

CV

⃗

U

ρ

dV

)

+

∫

CS

⃗

U

ρ

(

⃗

U

r

⋅

⃗

n

)

dA

or

d

dt

(

∫

CV

⃗

U

ρ

dV

)

+

∫

CV

⃗

U

ρ

div

⃗

U

r

dV

=

∑

⃗

F

This is vector equation, thus, say, along

x

coordinate, where

F

=

F

x

and

⃗

U

=

U

x

⃗

i

+

U

y

⃗

j

+

U

z

⃗

k

d

dt

(

∫

CV

U

x

ρ

dV

)

+

∫

CS

U

x

ρ

(

⃗

U

r

⋅

⃗

n

)

dA

=

∑

F

x

or

d

dt

(

∫

CV

U

x

ρ

dV

)

+

∫

CV

U

x

ρ

div

⃗

U

r

dV

=

∑

F

x

Linear Momentum in Cartesian

Coordinates

●

Linear momentum in 3D coordinates:

F

x

=

d

dt

[

∫

CV

U

x

dV

]

∫

CS

U

x

U

r

⋅

n

dA

F

y

=

d

dt

[

∫

CV

U

y

dV

]

∫

CS

U

y

U

r

⋅

n

dA

F

z

=

d

dt

[

∫

CV

U

z

dV

]

∫

CS

U

z

U

r

⋅

n

dA

Special Cases

●

Finite number of one-dimensional inlets/outlets

(can still be a 2D or 3D problem, but each

inlet/outlet is perpendicular to velocity vector)

●

Control volume fixed (non-moving, non-

deforming)

●

Control volume moves inertially (straight line,

constant velocity, non-deforming)

●

Control volume moves non-inertially (e.g.

accelerating or rotating).

Finite number of one-dimensional

inlets/outlets

F

x

=

∑

i

=

1

N

out

˙

m

i

U

i

,

x

−

∑

j

=

1

N

¿

˙

m

j

U

j

,

x

F

y

=

∑

i

=

1

N

out

˙

m

i

U

i

,

y

−

∑

j

=

1

N

¿

˙

m

j

U

j

,

y

F

z

=

∑

i

=

1

N

out

˙

m

i

U

i

,

z

−

∑

j

=

1

N

¿

˙

m

j

U

j

,

z

where

U

i

,

x

is the

x component of velocity

through the inlet/outlet

i,

etc

Source: F. White, Fluid Mechanics, McGraw Hill

Example – Fully Developed Pipe flow

●

Simple 1D flow

●

Constant area round pipe

●

Fully developed flow (i.e. velocity profile is

identical at each cross-section)

●

Steady

●

Incompressible flow (constant fluid density)

●

Constant fluid properties (viscosity constant)

One-dimensional case

Source, F. M. White, Fluid Mechanics 5

th

Ed.

Option 1: Control Surface Inside

F

friction

=

wall

A

wall

p

wall

x

p

wall

x

p

1

A

p

2

A

U

U

CV

CS

1D, steady, fully developed pipe flow

L

Option 2: Control Surface Outside

F

x

p

atm

p

atm

p

1

A

p

2

A

U

U

CV

CS

1D, steady, fully developed pipe flow

L

Where to draw CS

●

Draw CS inside pipe

–

To include Shear forces at CS, between the wall and

fluid (force acting on fluid by the wall)

–

Pressure forces vary along the wall, but cancel out at

each segment, due to symmetry

–

Flow crosses CS at inlet and outlet

●

Draw CS outside pipe

–

To find net force on pipe

–

At the cylindrical portion of the CS, atmospheric

pressure acts all around, cancells out again.

–

The shear force on fluid by the wall is hidden, the wall

is pushed forward by fluid friction, but is kept in place

by the reaction force

F

x

= F

friction

Momentum transport application

●

1D, steady flow, one inlet, one outlet:

∑

F

x

=

˙

m

U

out

−

U

in

p

1

A

−

p

2

A

−

w

A

w

=

˙

m

U

−

U

=

0

D

2

4

p

1

−

p

2

−

D

L

w

=

0

p

=

p

1

−

p

2

=

D

L

w

D

2

/

4

=

4

L

w

D

Friction Factor

●

Friction factor is non-dimensional parameter,

the ratio of relative pressure slope, and

the dynamic pressure, :

Δ

p

L

/

D

U

2

/

2

f

=

p

L

/

D

U

2

/

2

=

4

w

U

2

/

2

=

8

w

U

2

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