Chapter 5
Conservation Laws: Control-Volume Approach
5.1 Macroscopic Mass Balance (Integral Relation)
The general balance equation can be written as
Accumulation = Input + Generation - Output - Consumption
The terms (Generation - Consumption) are usually combined to call Generation with
positive value for net generation and negative value for net consumption.
Let
M = total mass (kg) of A within the system at any time.
m in = rate (kg/s) at which A enters the system by crossing the boundaries.
m out = rate (kg/s) at which A leaves the system by crossing the boundaries.
rgen = rate (kg/s) of generation of A within the system by chemical reactions.
rcons = rate (kg/s) of consumption of A within the system by chemical reactions.
Then the mass balance on species A can be written as
dM
= m in + rgen  m out  rcons
dt
(5.1-1)
If there is no chemical reaction, the mass balance equation is simplified to
dM
= m in  m out
dt
(5.1-2)
Example 5.1-1. ---------------------------------------------------------------------------------A tank contains 2 m3 of pure water initially as shown in Figure 5.1-1. A stream of brine
containing 25 kg/m3 of salt is fed into the tank at a rate of 0.02 m3/s. Liquid flows from
the tank at a rate of 0.01 m3/s. If the tank is well mixed, what is the salt concentration
(kg/m3) in the tank when the tank contains 4 m3 of brine.
Fi ,  Ai
V(t = 0) = 2 cubic meter
Fi = 0.02 m3 /s
Ai = 25 kg/m3
A
F , A
Figure 5.1-1 A tank system with input and output.
5-1
Solution -----------------------------------------------------------------------------------------Step #1: Define the system.
System: salt and water in the tank at any time.
Step #2: Find equation that contains A, the salt concentration in the tank at any time.
The salt balance will contain A.
Step #3: Apply the salt balance around the system.
d (V A )
= FiAi - FA = 0.5 – 0.01A
dt
(E-1)
where V is the brine solution in the tank at any time. Need another equation to solve
for V and A.
dV
= Fi – F = 0.02 – 0.01 = 0.01 m3/s
dt
Step #4: Specify the boundary conditions for the differential equation.
At t = 0, V = 2 m2, A = 0, at the final time t, V = 4 m2
Step #5: Solve the resulting equations and verify the solution.
Integrate Eq. (E-2)
V

2
t
dV = 0.01  dt , to obtain V = 2 + 0.01t
0
When V = 4 m3, t = 200 sec
The LHS of Eq. (E-1) can be expanded to
V
dV
d A
+ A
= 0.5 – 0.01A
dt
dt
Hence
(0.01t + 2)
d A
+ 0.01A = 0.5 - 0.01A
dt
The above equation can be solved by separation of variables
5-2
(E-2)
dt
d A
=
0.5  0.02  A 0.01t  2
-
1
1
ln (0.5 – 0.02A) =
ln(0.01t + 2) + C1
0.02
0.01
-
1
ln (0.5 – 0.02A) = ln(0.01t + 2) + C
2
(E-3)
at t = 0, A = 0, hence
1
- ln(0.5) = ln(2) + C
2
(E-4)
Eq. (E-3) - Eq. (E-4)
-
1  0.5  .02  A 
 0.01t  2 
ln 
 = ln 

2 
.5
2



1 - 0.04A = (1 + 0.005 t)-2
Finally
A = 25 -
25
(1  0.005t ) 2
(E-5)
at t = 200 sec
A = 25 -
25
(1  0.005  200)
2
= 18.75 kg/m3
Verify the solution
At t = 0, from (E-5); A = 0, as t  , A = 25 kg/m3
---------------------------------------------------------------------------------------------
5-3
We now consider the general open system or control volume fixed in space and located in a
fluid flow field, as shown in Figure 5.1-2. The streamline of a fluid stream is the curve where
the velocity at any point is tangent to it. For a differential element of area dA on the control
surface, the rate of mass efflux from this element = (v)( dAcos), where ( dAcos) is the
area dA projected in a direction normal to the velocity vector v,  is the angle between the
velocity vector v and the outward-directed unit normal vector n to dA, and  is the density.
Control volume
dA

Streamlines of
fluid stream
v
n
Normal to surface dA
Control surface
Figure 5.1-2 Flow through a differential area dA on a control surface.
(v)(dAcos) is the scalar or dot product of (vn)dA. Since the normal vector n is pointing
outward, the mass (efflux) leaving the control volume is positive ( < 90o) and the mass
(influx) entering the control volume is negative ( > 90o). If we now integrate this quantity
over the entire control surface A, we have the net outflow of mass across the control surface
~
or the net mass efflux from the entire control volume V .
mass
efflux   rate of
 net

 = 
 from control volume  from
mass
control
mass
efflux 
 net

 =
 from control volume
output   rate of

volume   from
mass
input 

control volume
 v cos dA =   (vn)dA
A
Since the rate of mass accumulation in control volume is
A
dM

=
dV , the mass balance
dt
t V
with no chemical reaction is

dV = 
t V
  (vn)dA
(5.1-3)
A
dM
= m in - m out .
dt
However equation (5.1-3) is more general since it can account for the variation of density
~
over the control volume V and the variation of velocity v over the control surface A.
Equation (5.1-3) has the same physical meaning as equation (5.1-2)
5-4
Example 5.1-2. ---------------------------------------------------------------------------------Water is flowing through a large circular conduit with inside radius R and a velocity
profile given by the equation
  r 2 
v(fps) = 8 1    
  R  
Determine the mass flow rate through the pipe and the average water velocity in the 2.0 ft
pipe.
r
6 ft
2 ft
Solution -----------------------------------------------------------------------------------------Since v is a function of r, we first need to determine the mass flow rate through the
differential area dA = 2rdr
 = (v)(dAcos) = (v)( 2rdr) since cos = cos(0) = 1
dm
The mass flow rate through the area R2 is then obtained by an integration over the area
2
R 
r 
 = 2  81     rdr
m
0
  R  
Let z =
(E-1)
r
 dr = Rdz, equation (E-1) becomes
R
1
m = 16R2  1  z 2 
1
0
 z2 z4 
 = 16R2  2  4 
zdz  m

0
m = 4R2 = 462.432 = 7057 lb/s
The average velocity in the 2-ft pipe is
vave =
7057
m
=
= 36 ft/s
2
62.4    12
R
5-5
5.2 Microscopic Mass Balance (Differential Relation)
z
vy|y+y
z
vx | x
x
vx|x+x
y
x v |
y y
y
Figure 5.2-1 Illustration of a differential element in Cartesian coordinates.
Appling the conservation of mass to a 3-D control volume xyz in Cartesian coordinates
we obtain

dM
= xyz
= m in  m out
t
dt
The mass flow into and out of each surface are given by
m in - m out = yz[(vx)|x  (vx)|x+x] + xz[(vy)|y  (vy)|y+y]
+ xy[(vz)|z  (vz)|z+z]
Therefore
xyz

= yz[(vx)|x  (vx)|x+x] + xz[(vy)|y  (vy)|y+y]
t
+ xy[(vz)|z  (vz)|z+z]
Divide the equation by xyz and take the limit as xyz 0
In the limiting process of making x, y, z 0
5-6
(5.2-1)
x dx
y dy
z dz
lim
x  0 
( v x ) |x  x ( v x ) |x ( v y ) | y y ( v y ) | y ( v z ) |z  z ( v z ) |z
=


y  0 t
x
z
y
z  0
This limit process produces partial derivatives
limit ( v x ) |x  x ( v x ) |x
 ( v x )
=
x  0
x
x
limit
( v y ) | y y ( v y ) | y
y  0
y
limit
(  v z ) | z  z  (  v z ) | z
z  0
z
=–
=–
( v y )
y
 ( v z )
z
We then obtain the differential mass balance or the continuity equation. This equation must
be satisfied at all points within any flowing fluid.
 ( v x ) ( v y )  ( v z )

=


t
x
z
y
(5.2-2)
If the fluid is incompressible,  is a constant and equation (5.2-2) becomes
v x v y v z
+
+
=0
x
z
y
(5.2-3)
Equation (5.2-2) can be written in vector notation as

=  v
t
The vector differential or del operator  is defined in RCCS (Rectangular Cartesian
Coordinate System) as
 = ex



+ ey
+ ez
x
z
y
In this expression, ex, ey, and ez are the units vector in RCCs with the properties
5-7
ei ej = 0 for i  j and ei ei = 1 for i = j where i, j = x, y, or z
Therefore
v = (ex
v =
Since



+ ey
+ ez )( exvx + eyvy + ezvz)
x
z
y
 ( v x ) ( v y )  ( v z )
+
+
x
z
y
 ( v x )
v

=  x + vx
, v = v + v
x
x
x
v is the divergence of vector v and is given in RCCS as
v =
v x v y v z
+
+
x
z
y
 is the gradient of the scalar  and is given in RCCS as
 = ex



+ ey
+ ez
x
z
y
The gradient of  is a vector in the direction in which  increases most rapidly with distance.
The term v is the overall or net rate of mass loss. v is the rate of outflow of volume
(per unit volume). This situation could occur if the fluid were expanding due to a decrease in
pressure. This would result in an outflow of volume across the boundaries of a fixed unit
volume. Therefore v is the rate of mass loss due to expansion. v is the net loss of
mass due to flow if there is an increase in density in the gradient direction. There will be
more mass flow out of the system than the flow in.
Example 5.2-1. ---------------------------------------------------------------------------------Evaluate the divergence of the velocity vector v = vxex + vyey, where
vx =  ceky sin(kx  t), vy = ceky cos(kx  t); c, k, and  are constants
Solution -----------------------------------------------------------------------------------------v =
v x v y


+
=
[ ceky sin(kx  t)] + [ceky cos(kx  t)]
x
x
y
y
v =  ckeky cos(kx  t) + ckeky cos(kx  t) = 0
The flow is incompressible since v =
v x v y
+
= 0.
x
y
5-8