**Obtain an expression for conservation of mass starting from the equation of continuity.**

Consider a fluid in steady or streamline flow, that is its density is constant. The velocity of the fluid within a flow tube, while everywhere parallel to the tube, may change its magnitude.

Suppose the velocity is v_{1} at point P and v_{2} at point. Q. If A_{1}, and A_{2}, are the cross-sectional areas of the tube at these two points,

the volume flux across A_{1}, \(\frac{d}{dt}(V1) = A1v1\)

and that across A_{2}, \(\frac{d}{dt}(V2) =A2v2\)

By the equation of continuity of flow for a fluid,

A_{1}v_{1 }= A_{2}v_{2 }

i.e. \( \frac{d}{dt}(V1) = \frac{d}{dt}(V2)\)

If p_{1} and p_{2}, are the densities of the fluid at P and Q, respectively, the mass flux across A_{1},

\(\frac{d}{dt}(m1) = \frac{d}{dt}(p1v1) = A1p1v1\)

and that across A_{2},

\(\frac{d}{dt}(m2) = \frac{d}{dt}(p2v2) = A2p2v2\)

Since no fluid can enter or leave through the boundary of the tube, the conservation of mass requires the mass fluxes to be equal, i.e.

\(\frac{d}{dt}(m1) = \frac{d}{dt}(m2)\)

A_{1}p_{1}v_{1} = A_{2}p_{2}v_{2}

i.e. Apv = constant

which is the required expression.