Principle of Continuity in Fluid dynamics

Principle of Continuity :

Consider the streamline motion of a fluid in a tube. The fluid enters the left section of the tube (with cross-sectional area A1) with a velocity v1 and leaves the other section (of cross-sectional area A2) with a velocity v2.

The quantity of fluid entering the left section during a time interval Δt is a cylindrical column of fluid of length v1Δt and area of cross section A1.

Thus, the volume of fluid entering the cross section of area A1 is

v1A1Δt = V1     ……..(1)

Within the same time interval Δt, the volume of fluid leaving the cross section of area A2 is

v2A2Δt = V2     ……..(2)

If the density of fluid is ρ1 at the section X and the fluid density is ρ2 at section Y, then the mass of fluid entering X and leaving Y are respectively

Δm1 = v1A1ρ1Δt     ……..(3)

and Δm2 = v2A2ρ2Δt     ……..(4)

Since the flow is steady,

Δm1 = Δm2

i.e. v1A1ρ1 = v2A2ρ2       ……..(5)

As we have considered the fluid to be incompressible therefore, density at all cross sections will be same

ρ1 = ρ2

A1 v1 = A2 v2    ……..(6)

A v = Constant

This Equations  is known  as Equation of continuity.

Solved Example : A liquid is flowing through a tube of non-uniform cross section and having its axis horizontal. If two points X and Y on the axis of tube have cross-sectional areas 2.0 cm2 and 25 mm2respectively then find the flow velocity at Y when the flow velocity at X is 10 m/s.

Solution: According to the principle of continuity,

vx Ax = vy Ay

Therefore,

$ \displaystyle v_y = \frac{v_x A_x}{A_y} $

$ \displaystyle v_y = \frac{10\times 2}{25\times 10^{-2}} $

vy = 80 m/s

Therefore, the flow velocity at y is 80 m/s.

Solved Example : Water enters into a smooth horizontal tappering tube with a speed 2 m/sec and emerges out of the tube with a speed 8 m/sec. Find the ratio of the radii of cross-section of the tube at these two ends.

Solution: Equation of continuity :

A1 v1 = A2 v2

$ \displaystyle \pi r_1^2 v_1 = \pi r_2^2 v_2 $

$ \displaystyle v_2 = (\frac{r_1}{r_2})^2 v_1 $

$\displaystyle \frac{r_1}{r_2} = \sqrt{\frac{v_2}{v_1}} $

$ \displaystyle \frac{r_1}{r_2} = \sqrt{\frac{8}{2}} = 2 $

Also Read:

Nature of Flow
Bernoulli’s Theorem
Applications Of Bernoulli’s Theorem : Venturimeter
Velocity of Efflux
Surface Tension
Surface Energy
Excess Pressure inside a soap bubble
Angle of contact
Capillarity & Ascent Formula
Viscosity , Stoke’s Law & Terminal Velocity
Poiseuille’s formula

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