Question 5.5: A rigid tank of volume ∀ contains air at an absolute pressur...
A rigid tank of volume \forall contains air at an absolute pressure of P and temperature T. At t = 0, air begins escaping from the tank through a valve with a flow area of A_{1}. The air passing through the valve has a speed of V_{1} and a density of \rho_{1}. Determine the instantaneous rate of change of density in the tank at t = 0, assuming it to be uniform within the tank.
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,
\frac{\partial(\rho \forall)_{C V}}{\partial t}+\int_{C S} \rho(\vec{V} \cdot \hat{n}) d A=0The only place where mass crosses the boundary of the control volume is at surface (1). Hence,
\int_{C S} \rho(\vec{V} \cdot \hat{n}) d A=\int_{A_{1}} \rho(\vec{V} \cdot \hat{n}) d AThe flow is assumed uniform over surface (1), so that
\frac{\partial(\rho \forall)}{\partial t}+\rho_{1} V_{1} A_{1}=0Since the volume, \forall, of the tank is not a function of time,
\forall \frac{\partial \rho}{\partial t}+\rho_{1} V_{1} A_{1}=0\frac{\partial \rho}{\partial t}=-\frac{\rho_{1} V_{1} A_{1}}{\forall}
