Question 5.5: A rigid tank of volume ∀ contains air at an absolute pressur...

Choose a fixed control volume as shown by the dashed line in Fig. 5.7(a).

From conservation of mass for the control volume, we get

$0=\frac{\partial}{\partial t} \int_{C V} \rho d \forall+\int_{C S}(\rho V \cdot \hat{n}) d A$                            (5.12)

Assuming that the properties in the tank are uniform, but time–dependent, the above equation can be written in the form

$\frac{\partial}{\partial t}\left[\rho \int_{C V} d \forall\right]+\int_{C S} \rho(V \cdot \hat{n}) d A=0$                              (5.12a)

Now,           $\int_{C V} d \forall=\forall$ .   Hence,

The only place where mass crosses the boundary of the control volume is at surface (1). Hence,

The flow is assumed uniform over surface (1), so that

Since the volume, $\forall$, of the tank is not a function of time,