Thursday, 14 May 2015

Continuity equation

Consider a control volume of fixed mass $\delta M$ that moves with the fluid. Letting $\delta V = \delta x \delta y \delta z$ be the volume, we can write: $$ \frac{1}{\delta M} \frac{D}{Dt} (\delta M) = \frac{1}{\rho \delta V} \frac{D}{Dt} (\rho \delta V) = \frac{1}{\delta \rho} \frac{D \rho}{Dt} + \frac{1}{\delta V} \frac{D \delta V}{Dt} = 0 $$

So, following the motion of a control volume, the mass is conserved, and the fractional change in density is equal and opposite to the fractional change in volume.
 

 We can rewrite the volume term: $$ \frac{1}{\delta V} \frac{D \delta V}{Dt} = \frac{1}{\delta x \delta y \delta z} \frac{D}{Dt} (\delta x \delta y \delta z) = \frac{1}{\delta x} \frac{D \delta x}{Dt} + \frac{1}{\delta y} \frac{D \delta y}{Dt} + \frac{1}{\delta z} \frac{D \delta z}{Dt} $$ What does $\frac{D \delta x}{Dt}$ represent? Consider the $u$ velocity at the left ($u1$) and the right ($u2$) of our control volume: $$ u1 = \frac{D x}{D t} \\ u2 = \frac{D (x + \delta x)}{D t} \\ u2 - u1 = \delta u = \frac{D \delta x}{D t} $$