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
OneDimensional 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 onedimensional inlets/outlets
(can still be a 2D or 3D problem, but each
inlet/outlet is perpendicular to velocity vector)
●
Control volume fixed (nonmoving, non
deforming)
●
Control volume moves inertially (straight line,
constant velocity, nondeforming)
●
Control volume moves noninertially (e.g.
accelerating or rotating).
Finite number of onedimensional
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 crosssection)
●
Steady
●
Incompressible flow (constant fluid density)
●
Constant fluid properties (viscosity constant)
Onedimensional 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 nondimensional 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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