As we learn about conservation of mass in chemistry and other branches , such a way equation of continuity is a mathematical expression of the law of conservative of mass in fluid dynamics .

This equation defines the steady flow of a fluid in a tube. It states that in steady flow mass of the liquid enter in one tube will be same as the mass of liquid leaving from other end .

**Derivation of equation of continuity.**

Fig. 1 shows a section of a tube in which at the ends , the cross sectional areas are A_{1} and A_{2} and the velocity of the fluid are v_{1} and v_{2} respectively .

According to the equation of continuity we can say that , if there is study flow then , mass of fluid entering at end A_{1} per second will be equal to mass of fluid leaving the end A_{2} per second .

In time dt the fluid enters a distance v_{1}dt at end A , so the volume entered in time dt is dV = A_{1}v_{1}dt and the volume entered per second is

dV/dt = A_{1}V_{1}

Hence mass entering per second at A_{1} is = A_{1}V_{1}ρ

Similarly mass leaving per second at A_{2} is = A_{2}V_{2}ρ

According to the definition of steady flow A_{1}V_{1}ρ = A_{2}V_{2}ρ

**A _{1}V_{1} = A_{2}V_{2}**

Therefore, the velocity of the liquid is smaller in the wider parts of the tube and larger in the narrower parts.

or V_{2}>V_{1} as A_{2}<A_{1}

Note – The product Av is the volume flow rate dV/dt , the rate at which volume crosses a section of the tube.

Hence,

dV/dt = volume flow rate = Av

The mass flow rate is the mass flow per unit time through a cross-section . This is equal to density (ρ) times the volume flow rate dV/dt .

We can generalize the continuity equation for the case in which the fluid is not incompressible. If ρ1 and ρ2 are the densities at sections 1 and 2 then ,

ρ_{1}A_{1}V_{1} = ρ_{2}A_{2}V_{2}

So, this is the continuity equation for a compressible fluid .

**Solved Examples**

**Example 1.**In a horizontal tube of non-uniform cross section area , water is flowing . At a place, the radius of the tube is 5.0 cm and the velocity of water is 9 m/s. What will be the velocity of water where the radius of the pipe is 3.0 cm ?

**Solution ** – Using equation of continuity, A_{1}V_{1} = A_{2}V_{2}

V_{2}= (A_{1}/A_{2})V_{1}

or V_{2} = ( πR_{1}^{2}/πR_{2}^{2} ) V1 = ( R_{1}/R_{2} )^{2} V_{1 }

Substituting the values , we get

V_{2} = ( 5.0×10^{-2}/3.0×10^{-2})^{2}(9)

**V2 = 25 m/s**

**Example 2**. A broad pipe having a radius 10 cm branches into two pipes of radii 5 cm and 3 cm. If the velocity of flowing water in the pipe of radius 3 cm is 5 cm/s, determine the velocities of water in the remaining two pipes. Given that the rate of discharge through the main branch is 600π cm^{3}/s.

**Solution** – Consider any three sections (1), (2) and (3) in the three pipes of different radii as shown in fig. 2

If v_{1} , and v_{2} , be the velocities of water at sections (1) and (2), respectively; then ,

a_{1} = π(10)^{2 }cm^{2} , a_{2} = π(5)^{2} cm^{2} and a_{3} = π(3)^{2} cm^{2}

v_{1 }= ? , v_{2} = ? , v_{3} = 5 cm/s

Discharge rate through the three pipes are

a_{1}v_{1 }= 100πv_{1} cm^{3}/s

a_{2}v_{2} = 25πv_{2} cm^{3}/ s

a_{3}v_{3} = 9π×5 cm^{3} /s

Now , Q = a_{1}v_{1} = a_{2}v_{2} + a_{3}v_{3}

600π= 100πv_{1} = 25πv_{2} + 45π

Solving we get , **v _{1} = 6 cm/s and v_{2} = 22.2 cm/s**

**Example 3 **. Water coming from a tap , as shown in the fig. 3 If it’s speed near the mouth of the tap is Vo= 10m/s , and Area of cross section of tap is 1cm2 . Then calculate the area of the stream water at point M , that is below h= 5 m from tap .

**Solution – **Velocity when it fall 5m

v^{2} = u^{2} + 2as

v^{2} = (10)^{2} + 2(10)(5)

v = 10√2 m/s

Apply equation of continuity, A_{1}V_{1} = A_{2}V_{2 }

1×10 = A_{2} × 10√2**A _{2} = 1/√2 cm^{2}**