1. Conservation of Mass
V
dz
dy
dx
A continuous fluid can be thought of as being made up of a
large group of elemental fluid particles.
These particles jostle along shoulder to shoulder as the fluid
moves.
2. Fluid Particle
V
dz
dy
dx
Each particle moves at a velocity (V) and
has a volume (dx, dy, dz) and density ().
Each has an individual mass ( m= .dx.dy.dz).
3. Conservation of Particle Mass
When moving with the flow, particles may be distorted
(stretched, compressed, sheared or rotated) due to the
pressure and temperature fields that they pass through.
However their individual masses remain unchanged.
4. Control Volume

For the control volume (), the
mass contained within will be
the sum of the masses of all the
fluid particles.
m

All Particles
in Volume
 dx dy dz 
  d
Volume
5. Flow through Control Volume
V
If the fluid is moving, then all
fluid particles can be tracked
over a small period of time (t).
Particle mass will be conserved.
The change in fluid mass in the
volume will equal the difference
in mass of particles entering
compared to mass of particles
leaving.
(t)
6. Inflow and Outflow
This can be written as
Mass Change in Volume () = Inflow (B) –Outflow mass(C)
Outflow
C
(t)
Inflow
B
V
7. Equations of Mass Flow
Mass Change in Volume () = Inflow (B) –Outflow mass(C)
m (t  t )  m (t )  mB (t  t )  mC (t  t )
m (t  t )  m (t )  mB (t  t )  mC (t  t )  0
m mB (t  t ) mC (t  t )


0
t
t
t
In limit as t 0
m
m
 B m
C 0
t
8. Inflow Rate
The amount of mass entering the
control volume through an elemental
surface, area dA, can be calculated
using the component of velocity
normal to the surface
mB   dl dA   (Vn) t dA
m B
 V cos() dA
t
Thus incoming flow rate is

mB  
 V cos() dA
inf low
surface
-Vn
dl
V
dA
n

B
9. Outflow Rate
The amount of mass exiting the control volume through surface,
dA, can also be calculated using the component of velocity
normal to this surface
m C   dl dA   ( Vn ) t dA
Vn

dl
V
mC
 V cos() dA
t
Thus the flow rate through the
exit surface is
dA

C
mC  
 V cos() dA
outflow
surface
10. Integral Equation Form
m
m
 B m
C 0
t
Substituting for the incoming and outgoing flow rates
in the mass conservation balance for volume () leads
to the result,
m
 (   V cos() dA )  (   V cos() dA )  0
t
inf low
outflow
surface
surface
m
 (  V cos() dA )  0
t
Complete
Surface
Integral Form