Conservation of Mass: Summary
The answers to the ConcepTests are given below and will open in a separate window.
Key points from this module:
- 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.
- 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.
- 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.
- 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