Fluid Conservation of Mass

system is defined as a collection of unchanging contents, so the conservation of mass principle for a system tells us that the time rate of change of the system mass is 0. In terms of the Reynolds Transport Theorem, we have

$$\begin{array}{rl}\frac{D{m}_{\text{sys}}}{Dt}& =0\\ \frac{\partial }{\partial t}{\int }_{\text{C.V.}}\rho d\forall +{\int }_{\text{C.S.}}\rho \mathbf{V}\cdot \text{n}dA& =0\end{array}$$

where

• $\frac{D{m}_{\text{sys}}}{Dt}$ is the time rate of change of mass in the system (equal to zero)
• $\frac{\partial }{\partial t}{\int }_{\text{C.V.}}\rho d\forall$ is the time rate of change of mass in the C.V.
• ${\int }_{\text{C.S.}}\rho \mathbf{V}\cdot \text{n}dA$ is the net mass flow rate through the C.S.

We can find an expression for the mean fluid velocity for an area $A$ by considering that the mass flow rate $\dot{m}$ can be written as

$$\begin{array}{rl}\dot{m}=\rho \bar{V}A& ={\int }_{\text{A}}\rho \mathbf{V}\cdot \mathbf{n}dA\\ \bar{V}& =\frac{{\int }_{\text{A}}\rho \mathbf{V}\cdot \mathbf{n}dA}{\rho A}\end{array}$$

Examples

Fixed, Non-deforming C.V.

Water flows steadily through a nozzle at the end of a fire hose. Determine the pumping capacity of the pump in $m^3/s$.

Apply the continuity equation:

$$\text{cancelto}0\frac{\partial }{\partial t}{\int }_{\text{C.V.}}\rho d\forall +{\int }_{\text{C.S.}}\rho \mathbf{V}\cdot \text{n}dA=0$$

We have steady flow, so the first term cancels to zero.

Then we have:

$$\begin{array}{rl}{\int }_{\text{C.S.}}\rho \mathbf{V}\cdot \mathbf{n}dA& ={\int }_{{\text{C.S.}}_{\text{in}}}\rho \mathbf{V}\cdot \mathbf{n}dA+{\int }_{{\text{C.S.}}_{\text{out}}}\rho \mathbf{V}\cdot \mathbf{n}dA\\ & ={\dot{m}}_2-{\dot{m}}_1\\ & =0\end{array}$$

Recall that the mass flowrate is $\dot{m}=\rho Q$, so we have

$${\rho }_2{Q}_2={\rho }_1{Q}_1$$

Assuming incompressibility, we have ${\rho }_2={\rho }_1$, which then gives

$$Q_1=Q_2=V_2{A}_2=0.0251{\text{ m}}^3\text{/s}$$

Moving C.V.

With a moving C.V., we have

$$\frac{D{B}_{\text{sys}}}{Dt}=\frac{\partial }{\partial t}{\int }_{\text{C.V.}}\rho bd\forall +{\int }_{\text{C.S.}}b\rho \mathbf{W}\cdot \mathbf{n}dA$$ $$\frac{\partial }{\partial t}{\int }_{\text{C.V.}}\rho d\forall +\underset{{\dot{m}}_{\text{out}}-{\dot{m}}_{\text{in}}}{\underbrace{{\int }_{\text{C.S.}}\rho \mathbf{W}\cdot \mathbf{n}dA}}$$

Deforming C.V.