2001, W. E. Haisler
Chapter 2: Conservation of Mass
1
Chapter 2
CONSERVATION of MASS
FOR A CONTINUUM
MASS going in - MASS going out
= change in MASS
(during some time period)
(for the system/control volume as a function of
space and time)
2001, W. E. Haisler
Chapter 2: Conservation of Mass
2
The definition of the continuum system/control
volume depends on your need. It could be the
universe,
this building,
a living organism, or
a differential volume of an object.
The Coordinate System is another choice.
2001, W. E. Haisler
Chapter 2: Conservation of Mass
Consider a tank with fluid flowing into and out of the tank:
Fluid
Figure 1.1: Fluid Flow Into and Out of a Tank
3
2001, W. E. Haisler
4
Chapter 2: Conservation of Mass
For the macro-view, the system might be:
1 , A1 , v1
4 , A 4 , v 4
2 , A 2 , v2
Fluid
System
Boundary
3 , A 3 , v3
Figure 1.2: Entire Tank Taken as the System
2001, W. E. Haisler
5
Chapter 2: Conservation of Mass
Or, the system boundary might be a smaller micro-view:
Smaller
System
1 , A1 , v1
Boundary
4 , A 4 , v4
2 , A 2 , v2
Smaller
System
Fluid
3 , A 3 , v3
Boundary
Figure 1.3: Possible “Systems” That Could be Chosen
2001, W. E. Haisler
Chapter 2: Conservation of Mass
Must consider flow into and out of the smaller "system":
Fluid
Flow in/out of smaller system
Figure 1.4: Smaller System Chosen Within the Tank
6
2001, W. E. Haisler
7
Chapter 2: Conservation of Mass
Choose a Cartesian coordinate system for convenience to
define the system boundary and the mass flow directions:
Fluid
y
Flow in/out of smaller system
z
x
Figure 1.5: Rectangular System (Possibly Differential in Size)
2001, W. E. Haisler
Chapter 2: Conservation of Mass
8
What is a continuum and at what length scale is a material a
continuum? Is it possible to have too small a volume
element?
A continuum is defined to be a material system for which
all quantities of interest (variables) can be defined at every
point as functions of space (r) and time (t). Consider
density:
m
Continuum Hypothesis :  (r, t)  lim
V0 V
The above definition implies that mass density can be
defined as a function of position and time and therefore at
every point in a continuum, provided that the limit shown
above exists as the volume V  0 . The mass density, , is
in general a function of position and time ( = (r, t)).
2001, W. E. Haisler
9
Chapter 2: Conservation of Mass
Length scales which define a continuum:
Steel plate
Steel
Crystallographic lattice
for a BCC crystal
Water
water molecules
Water pipe
Building –
length scale ~ 10 m
Brick –
length scale ~ 0.1 m
Microphotograph
of a Composite length scale ~ 10-6 m
2001, W. E. Haisler
Chapter 2: Conservation of Mass
10
The length scale must be chosen such that variables of
interest (for example, density) are observed as being
constant with respect to the volume element chosen (at some
position and time within the material).
density 
Atomistic Length
Scales
Continuum Length
Scales
continuum limit
volume element (V)
2001, W. E. Haisler
11
Chapter 2: Conservation of Mass
Consider Conservation of Mass for a differential volume
element. Consider a Cartesian coordinate system for a
volume element located at position x,y,z and with
dimensions x, y, z:
y
y
(x,y,z)
z
j
i
k
z
x
x
2001, W. E. Haisler
Chapter 2: Conservation of Mass
12
In conservation of mass, if we consider mass going in (or
out); this implies the following:
1. there is mass or mass/volume ().
2. there must be a velocity of the fluid ( v  vxi vy j vzk ).
3. the mass must be flowing into something (a volume) and
through some surface area (the “in and out” area).
4. For a differential volume x y z, the mass flow rate
in the x direction through some area yz is given by
(vx)yz
and the total mass flow for a time increment t is
[(vx)yz]t
2001, W. E. Haisler
Chapter 2: Conservation of Mass
13
So where did the last result come from? How do we determine
the appropriate terms to put into the conservation equation? We need
the amount of mass entering the system during a certain period. The
mass is entering through a certain area at a certain velocity. We can
think of it this way. Define the mass flux to be mass flow rate per
unit area, or,
mass flow rate
mass flow

area
(area )(time)
mass length

 ( density )(velocity )   vx
volume time
mass flux 
Then the mass entering for a certain time period could be written as
(mass flux) x (area of entrance) x (time period of observation) or
 mass entering or 
 leaving a system    mass flux  area passing through  time interval 


2001, W. E. Haisler
Chapter 2: Conservation of Mass
14
So what is the mass flux in the x direction? Must be connected to
density and the velocity in x direction. So we write vx . This is the
amount of mass per unit time that flows in the x direction though an
area whose normal points in the x direction. Only the component of
flow normal to the area enters the area.
In terms of mass flux, we could conveniently write conservation of
mass in the following way:
(mass flux in) (area of entrance) (time period of observation)
- (mass flux out) (area of exit) (time period of observation)
= change in mass within system during time period of observation.
2001, W. E. Haisler
Chapter 2: Conservation of Mass
15
A mass flow example: Consider each “ball” to be a mass
particle weighing 1 g and 1 mm in diameter, and that the balls are
one layer deep (plane of the paper). The mass particles are
flowing uniformly to the right with a velocity such that it takes
each ball 1 second to pass through the opening.
Note that 6 balls (6 g) will pass through the opening each second.
Hence the mass flow rate is 6 g/sec. In 3 seconds, 18 g has
passed through the opening.
2001, W. E. Haisler
Chapter 2: Conservation of Mass
16
Suppose now that the mass particles are flowing to the right with a
velocity that is inclined   60o from the horizontal.
Note that in 3 sec, a row of 3 balls will have moved up an imaginary
plane oriented at 60 (balls are moving at 1 ball per sec through the plane
but their velocity is oriented at 60). Note that there is room for only 3
rows of balls to get through the projected area of the opening. Hence in 3
seconds, 9 balls will pass through the opening, i.e., a mass flow rate of 3
g/sec, or 1/2 of the previous case. Note that the perpendicular component
of the velocity is v  v cos 60o  0.5 v . Hence, m   Av .
2001, W. E. Haisler
17
Chapter 2: Conservation of Mass
You can consider the mass flow rate through an area in two different
ways:
v
1. Only the perpendicular velocity of the balls
carries mass through the opening so that
A
m   Av   A v cos
v  v cos
v
2. The mass can flow through the projected
area that it “sees,” i.e., the area
perpendicular to the velocity vector, so
that m   ( A cos ) v .
Either way, you get exactly the same result.
A cos
A

2001, W. E. Haisler
18
Chapter 2: Conservation of Mass
Another mass flow example: Consider a faucet.
If water flows from a faucet with velocity v , and
the cross-sectional area of the faucet exit is A,
what is the mass flow rate?
Consider a time t . In this time, a cylindrical
volume of water, V, will be discharged of length
L and area A: V= A L
L
v
A
If the water velocity is v , then in time t a water
particle will have traveled: s= v t . Hence, over
time t , the length of cylindrical volume of water discharged is
L= v t . Hence, the volume of water discharged in t is: V=A v t .
The volumetric flow rate is the volume of water per unit time:
V  V / t  ( Avt ) / t  Av
The mass flow rate is mass per unit time, or the volumetric flow rate
times the density of the water:
m  m / t  V / t  (  Avt /)t   Av
2001, W. E. Haisler
19
Chapter 2: Conservation of Mass
Consider mass flow in the x direction with density  and
velocity vx flowing into the volume xyz through the
surface yz during a time increment of t.
y
y
(  vx ) x
( x, y , z )
(  vx ) x x
z
x
x
z
The mass entering the left boundary is: (  vx )yz t and
exiting the right boundary is: (  vx )yz
x  x
t .
x
2001, W. E. Haisler
20
Chapter 2: Conservation of Mass
For the entire volume, mass in - mass out:
x direction:
( vx ) yzt (vx )
yzt
x
xx
y direction: + ( v y )
y
z direction: + (vz )
xyt (vz )
z
xzt (  v y )
y y
z z
xzt
xyt
For the volume, change in mass (for time t):
= ( t t )xyz ( t )xyz
2001, W. E. Haisler
21
Chapter 2: Conservation of Mass
Conservation of mass requires that change in mass = mass in
- mass out (for a given time interval). Hence we have:
(
)xyz ( )xyz
t t
t
= (vx ) x yzt (vx ) xx yzt
xzt
+ ( v y ) xzt (v y )
y
y y
+ ( vz )
z
xyt (vz )
Divide by xyzt to obtain
z z
xyt
2001, W. E. Haisler
22
Chapter 2: Conservation of Mass
( v y ) ( v y )
(

v
)

(

v
)
 t t   t
x x
x xx 
y
y y

t
x
y
( vz ) ( vz )
z
z z

z
Take the limit of each term; x, y, z, t  0,
 v
  v x 
t
x






















y  v z
y
z










2001, W. E. Haisler
23
Chapter 2: Conservation of Mass
CONSERVATION OF MASS
(The Continuity Equation)



















 v






y   v z
    v x 
t
x
y
z

































valid for any point x,y,z in a continuum and
any time t.
2001, W. E. Haisler
Chapter 2: Conservation of Mass
24
Recall that the vector operator  is called the
divergence and is defined by
  i   j   k
x y z
Hence, conservation of mass (continuity
equation) can be written as
 (v)
t
In vector notation, the conservation equation is
valid for any coordinate system.
2001, W. E. Haisler
25
Chapter 2: Conservation of Mass
Conservation of Mass (cylindrical
coordinate system)
 v  v


rv
  1

r 1
z
r
r
t
r

z




































































2001, W. E. Haisler
Chapter 2: Conservation of Mass
26
For the solution of most engineering physics
problems, we must have the appropriate
1. Governing equations (conservation of
mass, momentum, energy, etc.)
2. Boundary conditions (what is known or
assumed on the boundary of the system). This
could be known displacements, pressures,
temperature, heat flux, mass flow rate,
velocity, etc.
3. Initial conditions (values of system
variables which are known at some initial
time).
2001, W. E. Haisler
Chapter 2: Conservation of Mass
27
Some examples of conservation of mass
(continuity):
1. Steady state: variables are not a function of
time.
0 (v )
2. Incompressible:  = constant
2001, W. E. Haisler
28
Chapter 2: Conservation of Mass
3. Steady state, incompressible, 1-D flow between two
parallel plates (Poiseulle flow)
y
flow
x
z
vy  vz  0
 (vx)
0
x
Boundary Conditions:
Continuity gives:

vx  f ( y, z) C
1
for plane motion, vx independent of z 
vx  vx( y)
2001, W. E. Haisler
29
Chapter 2: Conservation of Mass
At this point, conservation of mass (continuity) tells us
that the velocity in the x direction is some function of y
as shown below:
d
y
flow
x
What else can we determine about the velocity distribution,
or other variables like pressure in the fluid?
Nothing! … Until we consider conservation of liner
momentum and define more information about the fluid
(constitutive equations) and kinematics for the fluid.
2001, W. E. Haisler
30
Chapter 2: Conservation of Mass
4. Laminar, steady state flow through a cylindrical tube:
z
flow
r
 (v ) 
t











 1r










 v
 v
 
r 1
z
r
r

z
  rv








Boundary Conditions: vr  v  0.




















Thus:
 vz  0 which implies v  f (r, ) C .
z
1
z
Assume symmetry with respect to  so that vz  vz (r) .
2001, W. E. Haisler
Chapter 2: Conservation of Mass
31
Lets take a look at what a velocity field looks like and what
conservation of mass implies. Suppose over a square planar
region, the velocity in Cartesian coordinates is given by:


 0  x  1, 0  y  1
2 
v y  1  2 y  y / 2 
 1
vx  2 x  (1  x) y
You should verify
that v x and v y satisfy
COM. Lets use
Maple to plot the
velocity field vx ( x, y )
and v y ( x, y ) over the
square region.
2001, W. E. Haisler
Chapter 2: Conservation of Mass
32
Notice that on the top and bottom boundaries, fluid is flowing
into the square region. On the left and right boundaries, fluid
is flowing out of the region. Velocity is greatest near the
lower right corner and almost zero near the center.
2001, W. E. Haisler
Here are two
localized plots
(zoomed in).
Note that at
some points,
the velocity
becomes zero.
Chapter 2: Conservation of Mass
33