4-1
Chapter 4 Continuity Equation and Reynolds Transport Theorem
4.1 Control Volume
4.2 The Continuity Equation for One-Dimensional Steady Flow
4.3 The Continuity Equation for Two-Dimensional Steady Flow
4.4 The Reynolds Transport Theorem
Objectives:
• apply the concept of the control volume to derive equations for the conservation of mass for steady one- and two-dimensional flows
• derive the Reynolds Transport Theorem for three-dimensional flow
• show that continuity equation can recovered by simplification of the Reynolds Transport Theorem
4-2
4.1 Control Volume▪ Physical system
~ is defined as a particular collection of matter or a region of space chosen for study
~ is identified as being separated from everything external to the system by closed boundary
• The boundary of a system: fixed vs. movable boundary
•Two types of system:
closed system (control mass) ~ consists of a fixed mass, no mass can cross its boundary
open system (control volume) ~ mass and energy can cross the boundary of a control volume
surroundings
4-3
→ A system-based analysis of fluid flow leads to the Lagrangian equations of motion in which particles of fluid are tracked.
A fluid system is mobile and very deformable.
A large number of engineering problems involve mass flow in and out of a system.
→ This suggests the need to define a convenient object for analysis. → control volume
• Control volume
~ a volume which is fixed in space and through whose boundary matter, mass, momentum, energy can flow
~ The boundary of control volume is a control surface.
~ The control volume can be any size (finite or infinitesimal), any space.
~ The control volume can be fixed in size and shape.
→ This approach is consistent with the Eulerian view of fluid motion, in which attention is focused on particular points in the space filled by him fluid rather than on the fluid particles.
4-4
4.2 The Continuity Equation for One-Dimensional Steady Flow
• Principle of conservation of mass
The application of principle of conservation of mass to a steady flow in a streamtube results in the continuity equation.
• Continuity equation
~ describes the continuity of flow from section to section of the streamtube
• One-dimensional steady flow
Fig. 4.1
Consider the element of a finite streamtube
- no net velocity normal to a streamline
- no fluid can leave or enter the stream tube except at the ends No flow
4-5
Now, define the control volume as marked by the control surface that bounds the region between sections 1 and 2.
→ To be consistent with the assumption of one-dimensional steady flow, the velocities at sections 1 and 2 are assumed to be uniform.
→ The control volume comprises volumes I and R.
→ The control volume is fixed in space, but in
dt
the system moves downstream.From the conservation of system mass
( m
I m
R t) ( m
R m
O t)
t (1)For steady flow, the fluid properties at points in space are not functions of time,
m 0 t
→
( m
R t) ( m
R t)
t (2)Substituting (2) into (1) yields
( ) m
I t ( m
O t)
t (3)Inflow Outflow
4-6
Express inflow and outflow in terms of the mass of fluid moving across the control surfce in time
dt
( ) m
I t
1 1A ds
12 2 2
( m
O t)
t A ds
(4)Substituting (4) into (3) yields
1 1
A ds
1 2A ds
2 2
Dividing by
dt
gives1 1 1 2 2 2
m A ds A ds
(4.1)→ Continuity equation
In steady flow, the mass flow rate,
m
passing all sections of a stream tube is constant.m AV
= constant (kg/sec)( ) 0
d AV
(4.2.a)→
d ( AV ) dA V ( ) dV ( A ) 0
(5)4-7
Dividing (a) by AV
results ind dA dV 0
A V
(4.2.b)→ 1-D steady compressible fluid flow
For incompressible fluid flow; constant density
→
d 0
t
From Eq. (4.2a)
( ) 0
d AV
d AV ( ) 0
(6)Set
Q
=volume flowrate (m3/s, cms)Then (6) becomes
1 1 2 2
const.
Q AV AV A V
(4.5)For 2-D flow, flowrate is usually quoted per unit distance normal to the plane of the flow,
b
→
q
flowrate per unit distance normal to the plane of flow m s m
3
4-8 Q AV
q hV
b b
hV
1 1 h V
2 2 (4.7)[re] For unsteady flow
+ inflow outflow
t t t
mass
mass
( m
R t)
t ( m
R t) ( ) m
I t ( m
O t)
tDivide by
dt
( ) ( )
( ) ( )
R t t R t
I t O t t
m m
m m
dt
Define
(
R t)
t(
R t) ( )
m m m vol
t dt t
Then
( )
( )
I t(
O t)
tvol m m
t
4-9
• Non-uniform velocity distribution through flow cross section
Eq. (4.5) can be applied. However, velocity in Eq. (4.5) should be the mean velocity.
V Q
A
A A
Q dQ vdA
1
V
AvdA
A
•The product
AV
remains constant along a streamline in a fluid of constant density.→ As the cross-sectional area of stream tube increases, the velocity must decrease.
→ Streamlines widely spaced indicate regions of low velocity, streamlines closely spaced indicate regions of high velocity.
1 1 2 2
:
1 2 1 2AV A V A A V V
4-10
[IP 4.3] p. 113The velocity in a cylindrical pipe of radius
R
is represented by an axisymmetric parabolic distribution. What isV
in terms of maximum velocity,v
c?[Sol]
2
1
2 cv v r
R
← equation of parabola2
2 0 2
1 1
1 2
R A c
Q r
V v dA v r dr
A A R R
3 2 4 2 2
2 0 2 2 2 2
0
2 2 2
2 4 2 4 2
R
c R c c c
v r v r r v R R v
r dr
R R R R R
→ Laminar flow[Cf] Turbulent flow
→ logarithmic velocity distribution
dA=2rdr dr
R
v
cr
4-11
4.3 The Continuity Equation for Two-Dimensional Steady Flow
(1) Finite control volume
Fig. 4.3
Consider a general control volume, and apply conservation of mass
( m
I m
R t) ( m
R m
O t)
t (a)For steady flow:
( m
R t) ( m
R t)
tThen (a) becomes
( ) m
I t ( m
O t)
t (b)i) Mass in
O
moving out through control surface(
O t)
t . .( cos )
C S out
m
ds dA
mass vol area 1 ds dAcos
4-12
• Displacement along a streamline;
ds vdt
(c)Substituting (c) into (b) gives
(
O t)
t . .( cos )
C S out
m
v dAdt
(d)
By the way,
v cos
= normal velocity component normal to C.S. atdA
Setn
= outward unit normal vector at
dA n 1
n
cos
v v n v
← scalar or dot product (e)
Substitute (e) into (d)
. . . .
(
O t)
tC S out C S out
m
dt v n dA dt v dA
where
dA n dA
=directed area element
[Cf] tangential component of velocity does not contribute to flow through the C.S.
→ Circulation
ii) Mass flow into I
(
I)
t . .( cos )
C S in
m ds dA
. .
( cos )
. .( )
C S in
v dAdt dt
C S in v n dA
90 cos 0
4-13
C S in. .
C S in. .
dt v ndA dt v dA
For steady flow, mass in = mass out
. . . .
C S out C S in
dt v dA dt v dA
Divide by
dt
. . . .
C S in
v dA
C S out v dA
. . . .
0
C S out
v dA
C S in v dA
(f)Combine c.s. in and c.s. out
. . . .
0
C S
v dA
C S v ndA
(4.9)where
. . C S
integral around the control surface in the counterclockwise direction→ Continuity equation for 2-D steady flow of compressible fluid
[Cf] For unsteady flow
( mass inside c v . .) t
= mass flowrate in – mass flowrate out4-14
(2) Infinitesimal control volumeApply (4.9) to control volume ABCD
AB
v ndA
BC v ndA
CD v ndA
DA v ndA 0
(f)By the way, to first-order accuracy
2 2
AB
dy v dy
v ndA v dx
y y
2 2
BC
dx u dx
v ndA u dy
x x
2 2
CD
dy v dy
v ndA v dx
y y
(g)2 2
DA
dx u dx
v ndA u dy
x x
2 u u dx
x
2
AB
dx x
2 v n v v dy
y
2
u u dx x
4-15
Substitute (g) to (f), and expand products, and then retain only terms of lowest order (largest order of magnitude)
( )
22 2 4
v dy dy v dy
vdx dx v dx dx
y y y y
( )
22 2 4
v dy dy v dy
vdx dx v dx dx
y y y y
( )
22 2 4
u dx dx u dx
udy dy u dy dy
x x x x
( )
22 2 4 0
u dx dx dx
udy dy u dy dy
x x x x
v u 0
dxdy v dxdy dxdy u dxdy
y y x x
v u 0
v u
y y x x
( u ) ( v ) 0
x y
(4.10)→ Continuity equation for 2-D steady flow of compressible fluid
• Continuity equation of incompressible flow for both steady and unsteady flow (
const.)u v 0
x y
(4.11)4-16
[Cf] Continuity equation for unsteady 3-D flow of compressible fluid
( u ) ( v ) ( w ) 0
t x y z
For steady 3-D flow of incompressible fluid
u v w 0
x y z
• Continuity equation for polar coordinates
Apply (4.9) to control volume ABCD
AB
v ndA
BC V ndA
CD V ndA
DA V ndA 0
2 2
t AB t
d v d
V ndA v dr
( )
2 2
r BC r
dr v dr
V ndA v r dr d
r r
2 2
t CD t
d v d
V ndA v dr
B D C
A
4-17
2 2
r DA r
dr v dr
V ndA v rd
r r
Substituting these terms yields
( )
22 2 4
t t
t t
v d d v d
v dr dr v dr dr
( )
22 2 4
t t
t t
v d d v d
v dr dr v dr dr
2 2
r r
r r
v dr v dr
v rd v drd rd drd
r r
2 2
2 2 2 2
r r
r r
dr dr dr v dr v
v rd v drd rd drd
r r r r r r
2
2 2 2 0
r r
r r
v dr dr v dr
v rd rd v rd rd
r r r r
t r
t r
v v
d dr v d dr rdrd v rdrd
r r
2 2 3
1 1 1
( ) ( ) ( ) 0
2 2 2
r r
r r
v v
v drd dr d v dr d dr d
r r r r
Divide by
drd
1 1 1
2 2 2 0
t r r r
t r r r
v v v v
v v r v dr v dr dr
r r r r r r
0
r t
r r t
v v
r v r v v
r r
4-18
Divide byr
0
r r t
r t
v v v
v v
r r r r r
( ) ( )
0
t
r r
v
v v
r r r
(4.12)For incompressible fluid
0
t
r r
v
v v
r r r
(4.13)4-19
n [IP 4.4] p. 117
A mixture of ethanol and gasoline, called "gasohol," is created by pumping the two liquids into the "wye" pipe junction. Find
Q
eth andV
eth691.1 kg m
3
mix
1.08 m s V
mix
3 3
30 / 30 10 m /s Q
gas l s
680.3 kg m
3
gas
788.6 kg m
3
eth
[Sol]
2 2
1
(0.2) 0.031m
A 4
;
A
2 0.0079m
2;A
3 0.031m
23
1
30 10 / 0.031 0.97 m/s
V
(4.4)1
v n dA
2 v n dA
3 v n dA 0
(4.9)1
v n dA 680.3 0.97 0.031 20.4 kg/s
2 2
2
v n dA 788.6 V 0.0079 6.23 V
3
v n dA 691.1 1.08 0.031 23.1 kg/s
.
20.4 6.23
223.1 0
c s
v ndA V
2
0.43 m/s V
3 3
2 2
(0.43)(0.0079) 3.4 10 m /s 3.4 /s
Q
ethV A
l
No flow in and out
4-20
4.4 The Reynolds Transport Theorem• Reynolds Transport Theorem (RTT)
→ a general relationship that converts the laws such as mass conservation and Newton’s 2nd law from the system to the control volume
→ Most principles of fluid mechanics are adopted from solid mechanics, where the physical laws dealing with the time rates of change of extensive properties are expressed for systems.
→ Thesre is a need to relate the changes in a control volume to the changes in a system.
• Two types of properties
Extensive properties (
E
): total system mass, momentum, energy Intensive properties (i
): mass, momentum, energy per unit massE i
system mass,
m
1system momentum,
mv
v
system energy,
m v
2( ) v
2system system
E i dm i dvol
(4.14)Osborne Reynolds (1842-1912);
English engineer
4-21
▪ Derivation of RTT
Consider time rate of change of a system property
(
0) ( )
t dt t R t dt R I t
E
E E E
E E
(a)0 . .
( )
t dtc s out
E
dt i v dA
(b.1)
. .
(
I)
tc s in
E dt i v dA
(b.2)
(
R t dt)
R t dt
E
i dvol
(b.3)
(
R t)
R t
E i dvol
(b.4)Substitute (b) into (a) and divide by
dt
t dt t
1
R t dt R t
E E
i dvol i dvol
dt dt
. . . .
c s out
i v dA
c s ini v dA
4-22
c v. .
c s. .dE i dvol i v dA
dt t
(4.15)
①
dE
dt
= time rate of change ofE
in the system②
c v. .i dvol
t
= time rate change within the control volume → unsteady term③
c s. .i v dA
= fluxes ofE
across the control surface▪ Application of RTT to conservation of mass For application of RTT to the conservation of mass,
in Eq. (4.15),
E m
,i 1
anddm 0
dt
because mass is conserved.
c v. .dvol
c s. .v dA
c s out. .v dA
c s in. .v dA
t
(4.16)For flow of uniform density or steady flow, (4.16) becomes
. . . .
0
c s out
v dA
c s in v dA
~ same as Eq. (4.9) Unsteady flow: mass within the control volume may change if the density changes4-23
For one-dimensional flow2 2 2 c s out. .
v dA V A
1 1 1 . .
c s in
v dA V A
1 1 1
AV
2A V
2 2
(4.1)• In Ch. 5 & 6, RTT is used to derive the work-energy, impulse-momentum, and moment of momentum principles.
4-24
Homework Assignment # 4 Due: 1 week from today
Prob. 4.9 Prob. 4.12 Prob. 4.14 Prob. 4.20 Prob. 4.31