# Eqn of continuity

1. Sep 19, 2008

### chhitiz

does eqn of continuity apply to only incompressible fluids?is there an eqn for compressible fluids?

2. Sep 19, 2008

### ank_gl

equation of continuity is just another expression for conservation for mass.

rho*area*velocity = constant

rho cancels out while dealing with a incompressible flow, & stays for compressible flows

3. Sep 19, 2008

### minger

In pseudo form, the equation of continuity over a control volume is simply

$$\sum m_{in} = \sum m_{out} + \sum m_{accumulated}$$

So, if you have a box with marbles in it, and you put more marbles into it, either you accumulate marbles in the box, or if the box is full, marbles must come out.

For incompressible flow, pressure must remain constant, this means that the number of marbles (think molecules) must remain the same. That means for every marble that comes in, one must go out. However, for compressible flow, there can be an accumulation inside the control volume.

IIRC the actual equation in one of the 4 forms is something like:

$$\frac{\partial \rho}{\partial t} + \nabla (\vec{\rho V}) = 0$$

The second term is called Divergence of Velocity and ends up being a rather important term when deriving the N-S equations.

Last edited: Sep 19, 2008
4. Sep 19, 2008

### stewartcs

For the general case, the RHS should be the time rate change of the mass inside the control volume, not the summation.

$$\sum \dot{m}_{in} - \sum \dot{m}_{out} = \Delta \dot{m}_{system}$$

For a steady-state, steady-flow process the mass and energy of the control volume are constant with time. The RHS becomes:

$$\frac{dm_{cv}}{dt} = \Delta \dot{m}_{cv} = 0$$

Since the mass of the control volume is constant with time during the steady-state, steady-flow process, the conservation of mass principle becomes:

$$\sum \dot{m}_{in} = \sum \dot{m}_{out}$$

The mass flow rate is related to volume flow rate and fluid density by

$$\dot{m} = \rho \dot{V}$$

For one entrance, one exit steady-flow control volume, the mass flow rates are related by

$$\dot{m}_{in} = \dot{m}_{out}$$

Hence,

$$\rho_{in} \dot{V}_{in} = \rho_{out} \dot{V}_{out}$$

or

$$\rho_{in} \vec{V}_{in}A_{in} = \rho_{out} \vec{V}_{out}A_{out}$$

CS

5. Sep 19, 2008

### stewartcs

Actually, looking back at this I think we're saying the same thing since you're not using the mass flow rate.

CS

6. Sep 19, 2008

### minger

toe-mae-toe, toe-ma-toe :)