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

in Chapter2:Mechanical Properties of Fluids by | 178 views

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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 v1 at point P and v2 at point. Q. If A1, and A2, are the cross-sectional areas of the tube at these two points,

the volume flux across A1,  \(\frac{d}{dt}(V1) = A1v1\)

and that across A2\(\frac{d}{dt}(V2) =A2v2\)

By the equation of continuity of flow for a fluid,

A1v= A2v

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

If p1 and p2, are the densities of the fluid at P and Q, respectively, the mass flux across A1,

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

and that across A2,

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

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.

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

A1p1v1 = A2p2v2

i.e. Apv = constant

which is the required expression.

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