### 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 A_{1}) with a velocity v_{1} and leaves the other section (of cross-sectional area A_{2}) with a velocity v_{2}.

The quantity of fluid entering the left section during a time interval Δt is a cylindrical column of fluid of length v_{1}Δt and area of cross section A_{1}.

Thus, the volume of fluid entering the cross section of area A_{1} is

v_{1}A_{1}Δt = V_{1} ……..(i)

Within the same time interval Δt, the volume of fluid leaving the cross section of area A_{2} is

v_{2}A_{2}Δt = V_{2} ……..(ii)

If the density of fluid is ρ_{1} 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

Δm_{1} = v_{1}A_{1}ρ_{1}Δt ……..(iii)

and Δm_{2} = v_{2}A_{2}ρ_{2}Δt ……..(iv)

Since the flow is steady,

Δm_{1} = Δm_{2}

i.e. v_{1}A_{1}ρ_{1} = v_{2}A_{2}ρ_{2} ……..(v)

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

ρ_{1} = ρ_{2}

A_{1} v_{1} = A_{2} v_{2} ……..(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 cm^{2} and 25 mm^{2}respectively 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 :

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

$ \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 $