Conservation of Mass using Control Volumes
Conservation of Mass
In the Reynolds transport theorem for conservation of mass, let B = m, i.e. Bsys = msys
= mass of the system,
For our system, we know that
So, Reynold's Transport Theorem (R.T.T.) becomes
Incompressible Steady Flow
( density = constant)
- Let's analyze the last term in the equation:
- n = outward normal unit vector (perpendicular to surface
area element dA)
- V = velocity vector, which can be in any arbitrary direction
- Recall, from definition of dot product:
- Outlets: V is outward (mass leaving C.V.)
- Inlets: V is inward (mass entering C.V.)
- Now, it turns out that
flow rate outward through a surface
so, if we integrate over the entire surface,
and the conservation of mass equation becomes:
*this is the most useful form.
- e.g. Given: A rigid tank of volume V with p = p0
at t = 0. Air is pumped in at constant mass flow rate
Find: p in tank as a function of time
Solution: first, draw a C.V. inside the entire tank.
Now use our conservation of mass equation.
assume density is equal everywhere in the tank, and only varies with time.
There are no outlets and the only term remaining is the mass term
at the inlet:
This is a differential equation we must solve by separating the variables
and integrating from t=0 to some arbitrary t.
Finally, use the ideal gas law to get the pressure.
(NOTE: pressure varies linearly
Special Case: STEADY FLOW
For a steady flow, nothing is a function of time, the d/dt term
in the conservation of mass equation goes away.
in other words, "What goes in must come out"
If more is going in than out, mass will be accumulating with time inside
the control volume, and it would not be steady state!!
( and you need to include the unsteady volume term)
Since the density is constant,
outside of the integral
So the integral term
is equal to Q, the volume flow rate
i.e. where mass flow
rate = density X volume flow rate
So the conservation of mass equation becomes:
"volume flow rate it" = "volume flow
Another simplification: 1-D inlets & outlets
(for incompressible flow)
Consider an outlet:
thus, Qout = V A. Therefore, at an inlet
1-D outlet means velocity is parallel to n,
the unit normal vector.
V = constant across the outlet.
= constant across the outlet
we usually say:
.since the negative sign is accounted for in
the conservation of mass equation,
If an inlet or exit is not 1-D, we can still
use but VAV must be the average
EXAMPLE 3.14 in text:
A container of water,
Q3, D1, D2 and h equal constants.
V1 = 3 m/s; Q3 = 0.01 m3/s; h = constant;
D1 = 0.05m; D2= 0.07m
Find: Average exit velocity
Solution: Use conservation
of mass. First draw the C.V. shown.
Since it is steady and incompressible,
Using the above equations, solve for V2: