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

Answer in Brief

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

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Solution

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 `vec"v"_1,` at point P and `vec"v"_2` at point Q. If A1 and A2 are the cross-sectional areas of the tube at these two points, the volume flux across A1, `"d"/"dt"("V"_1) = "A"_1"v"_1`
and that across A2,`"d"/"dt"("V"_2) = "A"_2"v"_2`

By the equation of continuity of flow for a fluid,

`"A"_1"v"_1 = "A"_2"v"_2`

i.e. `"d"/"dt"("V"_1) = "d"/"dt"("V"_2)`

If `rho_1  "and"  rho_2` are the densities of the fluid at P and Q, respectively, the mass flux across `"A"_1, "d"/"dt"("m"_1)`

`= "d"/"dt"(rho_1"V"_1) = "A"_1rho_1"v"_1`

and that across A2, `"d"/"dt"("m"_2) = "d"/"dt"(rho_2"V"_2) = "A"_2rho_2"v"_2`

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.,

`"d"/"dt" ("m"_1) = "d"/"dt"("m"_2)`

i.e. A1ρ1v1 = A2ρ2v2

i.e. Apv = constant which is the required expression.

Concept: Mechanical Properties of Fluids
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APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 2 Mechanical Properties of Fluids
Exercises | Q 7 | Page 54
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