Continuity equations

Continuity equation is used to describe the transport of some quantities. It is based on the law of conservation of mass which states that mass can neither be created nor destroyed. For a flow process through a control volume where the stored mass does not change, fluid enters and leaves the controlled volume through its surface called controlled  surface. Inflow of fluid equals to outflow which is net mass flow out of control volume through surface S = time rate of decrease of mass inside control volume $v$. $$\frac{\partial }{{\partial t}}\mathop{{\int\!\!\!\!\!\int\!\!\!\!\!\int}\mkern-31.2mu \bigodot}\limits_v  {\rho dv + \mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}\limits_S  {\rho v.ds = 0} }$$

Continuity equation in the form of partial differential equation is  $$\frac{{\partial \rho }}{{\partial t}} + \nabla .\left( {\rho u} \right) = 0$$ For a steady flow, $$\frac{\partial }{{\partial t}} = 0$$ $$\nabla  \cdot \left( {\rho v} \right) = 0$$ For a flow with constant density on applying continuity equation, Area x velocity = constant. $${A_1}{V_1} = {A_2}{V_2}$$

Example: Water is flowing through a pipe, having inlet inside diameter of 80 mm at 8m/sec. Calculate the velocity at exit  if inside diameter is reduced to 30 mm.

Solution : From continuity equation, we know that $${A_1}{V_1} = {A_2}{V_2}$$   where 1 , 2 represents inlet and exit conditions $${A_1} = \pi {\left( {\frac{{80}}{2}} \right)^2}$$ $${V_1} = 8\,m/s$$ $${A_2} = \pi {\left( {\frac{{30}}{2}} \right)^2}$$ $${V_2} = \,\,?$$ $$\Rightarrow \pi {\left( {\frac{{80}}{2}} \right)^2} \times 8 = \pi {\left( {\frac{{30}}{2}} \right)^2} \times {V_2}$$ $$\Rightarrow {V_2} = \frac{{80 \times 80 \times 8}}{{30 \times 30}}$$ $$= 56.88\,m/s.$$ Therefore  water is flowing at exit of pipe at 56.88 m/s.