Integral continuity for a compressing duct

Integral form

For a fixed control volume (CV) with control surface (CS):

$$\frac{\partial}{\partial t}\int_{CV}\rho\,dV+\int_{CS}\rho\,(\vec V\cdot\hat n)\,dA=0$$

The first term is the rate of change of mass stored in the CV; the second is the net mass outflow through the surface. For steady flow the first term is zero, so total mass inflow equals total outflow. For one inlet and one outlet:

$$\rho_1A_1V_1=\rho_2A_2V_2=\dot m$$

The integral form is the basis of the finite volume method, since it is applied to each cell. The differential form follows by applying the divergence theorem to a very small volume.

Numerical

Mass flow rate:

$$\dot m=\rho_1A_1V_1=1.2\times0.020\times20=0.48\ \text{kg/s}$$

Outlet velocity:

$$V_2=\frac{\dot m}{\rho_2A_2}=\frac{0.48}{1.5\times0.010}=32\ \text{m/s}$$

Check: halving the area doubles the velocity, while the 1.25 times higher density reduces it by a factor of 1.25, giving a net factor of 1.6: $V_2=V_1\times\dfrac{\rho_1A_1}{\rho_2A_2}=20\times\dfrac{0.024}{0.015}=32$ m/s.

Answer: $\dot m=0.48$ kg/s; $V_2=32$ m/s.