# Review of Fluid Mechanics I: Control Volume Concept, Reynolds Transport Theorem, Conservation Laws

Mechanics

Oct 24, 2013 (4 years and 6 months ago)

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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

Fluid flows through
the CV boundaries

Becomes point in
space when sized
reduced to zero

Point” may move,
fluid

Lagrangian view

Control volume
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
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:
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)

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