Equation Of Continuity

Equation of Continuity

The Equation of Continuity is based on the principle of conservation of mass. It asserts that for an incompressible fluid (one whose density remains constant), the mass flow rate must be the same at any two points along a streamline.

The equation of continuity is derived from the principle of conservation of mass.

Have a look at the flow of ideal fluid through the tube AB.

For the above figure

Let the Mass of the liquid entering per second at $\mathrm{A}=\dot{M}_A$

The mass of the liquid leaving per second at $\mathrm{B}=\dot{M}_B$

From Mass conservation law we can write

$\dot{M}_A=\dot{M}_B$

If the cross-sectional area of the pipe at points A and B is a₁ and a₂ respectively.

Let the liquid enter with normal velocity v₁ at A and leave with velocity v₂ at B.

And $\rho_1$ and $\rho_2$ are the densities of the liquid at points A and B respectively.

Then $\dot{M}_A=\rho_1 a_1 v_1$ and $\dot{M}_B=\rho_2 a_2 v_2$
But $\dot{M}_A=\dot{M}_B$

Since the flow is incompressible so $\rho_1=\rho_2$

So Equation of Continuity for the liquid flow in tube AB is given by

$a_1 v_1=a_2 v_2 o r \quad a v=$ constant

The Equation of Continuity states that for the liquid flow in the tube, the product of cross-section area and velocity remains the same at all points in the tube.

From the Equation of Continuity, we can say that

The velocity of flow will increase if the cross-section decreases and vice-versa.

Recommended Topic Video

Solved Examples Based On Equation Of Continuity

Example 1: From shows how the stream of water emerging from the faucet neeks down as it falls. The area changes from $A_0$ to A through a fall of h. At what rate does the water flow to the tap?

1) $A o \sqrt{\frac{2 g h A^2}{A_0^2-A^2}}$
2) $2 A_o \sqrt{\frac{2 g h A^2}{A_0^2-A^2}}$
3) $\sqrt{\frac{2 g h A^2}{A_0^2+A^2}}$
4) $\sqrt{\frac{g h A^2}{A_0^2-A^2}}$

Solution:

Use,$\quad A_0 V_0=A V$ Also $v^2=v_0^2+2 g h v_0^2=v^2-2 g h v_0=\sqrt{\frac{2 g h A^2}{A_0^2-A^2}}$ or $R=A_0 V_0=A_0 \sqrt{\frac{2 g h A^2}{A_0^2-A^2}}$

Hence, the answer is the option (1).

Example 2: The figure below shows a liquid being pushed out of the tube by a piston having an area of cross-section. The area of the cross-section at the outlet is $10 \mathrm{~mm}^2$. If the piston is pushed at a speed of $4 \mathrm{~cm} \mathrm{~s}^{-1}$. the speed of outgoing fluid is _________$\mathrm{cm} \mathrm{s}^{-1}$

1)80

2)40

3)50

4)60

Solution:

By equation of continuity

$\begin{aligned} & \mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2 \\ & \left(2 \mathrm{~cm}^2\right)(4 \mathrm{~cm} / \mathrm{s})=\left(10 \times 10^{-2} \mathrm{~cm}^2\right)(\mathrm{v}) \\ & \frac{8 \mathrm{~cm}^3}{\mathrm{~s}}=10^{-1} \mathrm{~cm}^2(\mathrm{v}) \\ & \mathrm{V}=80 \mathrm{~cm} / \mathrm{s}\end{aligned}$

Hence, the answer is the option (1).

Example 3: Water flows in a horizontal tube (see figure). The pressure of water changes by 700Nm² between A and B where the area of cross-section is 40cm² and 20 cm², respectively. Find the rate of flow of water through the tube.

(density of water = 1000kgm^-3)

1)3020 cm³/s

2)2420 cm³/s

3)2720 cm³/s

4)1810 cm³/s

Solution:

Using the equation of continuity

$\begin{aligned} & 40 V_A=20 V_B \\ & \Rightarrow 2 V_A=V_B\end{aligned}$

Using Bernoulli's Equation

$\begin{aligned} & P_A+\frac{1}{2} \rho V_A^2=P_B+\frac{1}{2} \rho V_B^2 \\ & P_A-P_B=\frac{1}{2} \rho\left(V_B^2-V_A^2\right) \\ & \Delta P=\frac{1}{2} \times 1000\left(V_B^2-\frac{V_B^2}{4}\right) \\ & \Rightarrow \Delta P=500 \times \frac{3 V_B^2}{4} \\ & \Rightarrow V_B=\sqrt{\frac{\Delta P \times 4}{1500}}=\sqrt{\frac{700 \times 4}{1500}} \mathrm{~m} / \mathrm{s} \\ & \text { Volume flow rate }=20 \times 100 \times V_B=2732 \mathrm{~cm}^3 / \mathrm{s}\end{aligned}$ Hence, the answer is the option (3).

Example 4: Water is flowing at a speed of 1.5 ms^-1 through a horizontal tube of cross-sectional area 10^-2 m² and you are trying to stop the flow by your palm. Assuming that the water stops immediately after hitting the palm, the minimum force (in N) that you must exert should be (density of water=10³ kgm^-3).

1)15

2)22.5

3)33.7

4)45

Solution:

Equation of Continuity

Mass of the liquid entering per second at A = mass of the liquid leaving per second at B.

a₁ v₁ = a₂ v₂

wherein

a₁ and a₂ are the area of the cross-section.

Let us say the speed of water is v and the area of cross-section is A

In one second mass of water that flows is $m=\rho A v$

F = momentum change per second

$m v=\rho A v^2=10^3 \times 10^{-2} \times(1.5)^2=22.5 \mathrm{~N}$

Hence, the answer is the option (2).

Example 5:

Consider a water jar of radius R that has water filled up to height H and is kept on a stand of height h (see figure). Through a hole of radius r (r << R) at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of the water stream when it hits the ground is x. Then :

1) $x=r\left(\frac{H}{H+h}\right)$
2) $x=r\left(\frac{H}{H+h}\right)^{\frac{1}{2}}$
3) $x=r\left(\frac{H}{H+h}\right)^{\frac{1}{4}}$
4) $x=r\left(\frac{H}{H+h}\right)^2$

Solution:

$\begin{aligned} & \mathrm{A}_1 \mathrm{v}_1=\mathrm{A}_2 \mathrm{v}_2 \\ & \pi r^2 \sqrt{2 g H}=\pi x^2 \sqrt{2 g(H+h)} \\ & \therefore \quad x=r\left(\frac{H}{H+h}\right)^{1 / 4}\end{aligned}$

Hence, the answer is the option (2).

Summary

The equation of continuity states that, under conditions in which there are no leaks, the mass flow rate of fluid must be constant from one pipe cross-section to another.It simply means that as the fluid picks up speed, it has to travel through a narrower section, and as the pipes of the fluid slow down, it passes through a wider section. In this very principle lies the efficient design of fluid systems for different industries: water management, chemical processing, or aerodynamics.