LearnChemE

#### Conservation of Mass: Summary

The answers to the ConcepTests are given below and will open in a separate window.
##### Key points from this module:
1. The mass of the system must always be constant because mass can neither be created nor destroyed. This means that $$\frac{DM_{sys}}{Dt}$$= 0.
2. If mass enters the control volume at a higher rate than it leaves, then the control volume is accumulating mass. It is not at steady state. This means that $$\frac{\partial}{\partial t} \int_{CV} \rho\,dV$$ > 0 and $$\int_{CS} \rho\,\vec{V} \cdotp \vec{n} \, dA$$ < 0.
3. If mass leaves the control volume at a higher rate than it enters, then the control volume is losing mass. It is not at steady state. This means that $$\frac{\partial}{\partial t} \int_{CV} \rho\,dV$$ < 0 and $$\int_{CS} \rho\,\vec{V} \cdotp \vec{n} \, dA$$ > 0.
4. At steady state, the mass flow rate of fluid leaving the control volume (in kg/s) minus the mass flow rate of fluid entering the control volume (in kg/s) must always equal zero. Because it is at steady state $$\frac{\partial}{\partial t} \int_{CV} \rho\,dV$$ = 0. This means that  $$\int_{CS} \rho\,\vec{V} \cdotp \vec{n} \, dA$$ is also zero.
##### From studying this module, you should now be able to:
• Draw and work with control volumes.
• Simplify the macroscopic continuity equation for a specified control volume.
• Calculate the average velocity of a fluid flowing through a constriction.
• Calculate the rate at which mass is accumulating or depleting within a control volume.

Prepared by: Jeffrey Knutsen, Department of Mechanical Engineering, University of Colorado Boulder