conservation-of-mass-summary

Conservation of Mass: Summary

The answers to the ConcepTests are given below and will open in a separate window. ConcepTest 1 AnswerConcepTest 2 Answer

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

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