Question 5.6: Compressed air exhausts from a small hole in a rigid spheric...

Choose a fixed control volume as shown by the dotted line in Fig. 5.8 (a).

From the conservation of mass for the fixed control volume, we get

0=\frac{\partial}{\partial t} \int_{C V} \rho d \forall+\int_{C S}(\rho V . \hat{n}) d A                       (5.13)

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 . \hat{n}) d A=0                           (5.13a)

Now,  \int_{C V} d \forall=\forall_{\operatorname{tank}} , is not a function of time. Hence,

\forall_{\operatorname{tank}} \frac{d \rho}{d t}+\dot{m}_{e}=0                             (5.14)

Since  \dot{m}_{e} \propto \rho , we assume \dot{m}_{e}=k \rho , where k is a proportionality constant. Thus,

\forall_{\operatorname{tank}} \frac{d \rho}{d t}+k \rho=0                 (5.14a)

Integrating,

\frac{\rho}{\rho_{0}}=\exp \left[-\frac{k}{\forall_{\operatorname{tank}}} t\right]                             (5.15)

Now,  p_{0}=400 kPa ,  and  T_{0}=400 K , then  \rho_{0}=\frac{p_{0}}{R T_{0}}=\frac{400}{0.287 \times 400}=3.48 kg / m ^{3}

Or                           k=\frac{0.02}{3.48}=0.005747 m ^{3} / s

The tank volume is  \forall_{ tank }=\frac{\pi}{6} D^{3}=\frac{\pi}{6}(0.6 m )^{3}=0.113 m ^{3}

For the given final conditions,

t = 18 s