Equation of Continuity in Fluid Mechanics

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     ……..(i)

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

v₂A₂Δt = V₂     ……..(ii)

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     ……..(iii)

and Δm₂ = v₂A₂ρ₂Δt     ……..(iv)

Since the flow is steady,

Δm₁ = Δm₂

i.e. v₁A₁ρ₁ = v₂A₂ρ₂       ……..(v)

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

ρ₁ = ρ₂

A₁ v₁ = A₂ v₂    ……..(vi)

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,

v_x A_x = v_y A_y

Therefore,

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

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

v_y = 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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