Question 3.2: The balloon in Fig. E3.2 is being filled through section 1, ...

• System sketch: Figure E3.2 shows one inlet, no exits. The control volume and system
expand together, hence the relative velocity V_{r}= 0 on the balloon surface.
• Assumptions: Unsteady flow (the control volume mass increases), deformable control surface, one-dimensional inlet conditions.
• Approach: Apply Eq. (3.16)

with V_{r}= 0 on the balloon surface and V_{r} =V_{1} at the inlet.
• Solution steps: The property being studied is mass, B =m and \beta = dm/dm = unity.
Apply Eq. (3.16). The volume integral is evaluated based on average density \rho_{b} , and
for the rate of change of system mass within the balloon at this instant.the surface integral term is negative (for an inlet):
\left(\frac{dm}{dt} \right)_{syst} =\frac{d}{dt} \left(\int_{CV}^{}{\rho d°v} \right)+\int_{CS}^{}{\rho (V_{r}\cdot n )dA}=\frac{d}{dt}\left(\rho_{b}\frac{4\pi }{3}R^3 \right)-\rho_{1}A_{1}V_{1}
• Comments: The relation given is the answer to the question that was asked. Actually, by the conservation law for mass, Eq. (3.1), (dm/dt)_{syst} = 0, and the answer could be rewritten as
\frac{d}{dt}(\rho_{b}R^3 )=\frac{3}{4\pi } \rho_{1}A_{1}V_{1}
This is a first-order ordinary differential equation relating gas density and balloon radius.
It could form part of an engineering analysis of balloon inflation. It cannot be solved without further use of mechanics and thermodynamics to relate the four unknowns \rho_{b}, \rho_{1}, V_{1}, and R. The pressure and temperature and the elastic properties of the balloon would also have to be brought into the analysis.