Continuity Equation

The Continuity Equation is a key concept in fluid dynamics that governs the conservation of mass in a fluid flow. It asserts that for an incompressible fluid, the mass flow rate must be consistent across different cross-sections of a pipe or channel. Understanding and designing medical devices like artificial hearts and stents. The continuity equation helps in modeling blood flow through arteries and veins. The continuity equation is a powerful tool for ensuring that mass is conserved in fluid systems, which is crucial for designing and analyzing a wide range of systems across different industries.

1.0Ideal Fluid

An ideal fluid is characterized by its lack of viscosity, incompressibility, and its flow being both steady and irrotational.

2.0Statement of Continuity Equation

It states that for a non-viscous and incompressible fluid flowing smoothly through a pipe with varying cross-sectional areas, the product of the cross-sectional area and the fluid's velocity remains constant along the flow.

3.0Assumption of Continuity Equation

1. Steady flow-The fluid velocity at each point remains constant over time, both in terms of magnitude and direction.
2. Incompressible Flow-The fluid density remains unchanged throughout its flow.
3. Non-viscous Flow-The fluid has no internal friction, so an object moving through it does not encounter any retarding force..
4. Irrotational Flow-The fluid possesses no angular momentum about any point.
5. The tube should have only one entry point and one exit point for the fluid.

4.0Derivation of Continuity Equation

• The continuity equation serves as the mathematical representation of the mass conservation principle in fluid dynamics.
• In the steady flow mass of the fluid entering into a tube of flow in a particular time interval is equal to the mass of fluid leaving the tube.

$\frac{\Delta m_1}{\Delta t}=\frac{\Delta m_2}{\Delta t}$

         Or

${\rho }_1{A}_1{v}_1={\rho }_2{A}_2{v}_2$

For Incompressible Fluid,

${\rho }_1={\rho }_2$

$A_1{v}_1=A_2{v}_2\Rightarrow Av=\text{ Constant }$

Volume flux=Rate of flow=Volume of liquid flowing per second

$\mathrm{Q}=\frac{dV}{dt}=Av$

• Unit of $\text{ Q }-m^3/s\text{ or }\frac{\text{ Litre }}{\text{ Sec }}\text{ or }\frac{ml}{s}$

5.0Application of Continuity Equation

1. Blood Flow in Arteries and Veins: Understanding how the continuity equation applies to blood flow can help in understanding how blood pressure and velocity change as blood moves through arteries and veins of varying diameters. This is crucial for grasping concepts related to blood circulation and cardiovascular health.
2. Airflow through the Lungs: The continuity equation can be used to analyze how air flows through the branching airways of the lungs. This can help in understanding how different airway sizes affect airflow and pressure, which is relevant to respiratory physiology.
3. The continuity equation helps in solving such problems by relating the flow rate to the velocity and cross-sectional area of the pipe.

NOTE:

• Equation of Continuity is a special case of the law of Conservation of Mass.
• Equation of Continuity shows that v ∝1aThe velocity of the liquid at any section of the pipe is inversely related to the cross-sectional area of the pipe at that section.

Example 1. Water flows through a horizontal pipe with an internal diameter of 2 cm at a velocity of 1 m/s. To achieve an exit speed of 4 m/s, Calculate the diameter of the nozzle?

Solution:

$d_1=2\mathrm{cm}=0.02\mathrm{m},v_1=1\mathrm{m}/\mathrm{s},v_2=4\mathrm{m}/\mathrm{s},d_2=?$

Using Equation of Continuity, $a_1{v}_1=a_2{v}_2$

$\frac{\pi }{4}{d}_1^2\times v_1=\frac{\pi }{4}{d}_2^2\times v_2$

$d_2^2=\frac{v_1}{v_2}\times d_1^2=\frac{1}{4}\times (0.02{)}^2=(0.01{)}^2$

$d_2=0.01\mathrm{m}=1\mathrm{cm}$

Example 2.Water from a tap emerges vertically downwards with an initial speed of 3m/s. The cross-sectional area of tap is 2 m2assuming that the pressure remains constant throughout the water flow and the flow is steady. Find the cross-sectional area of stream 80 cm below the tap. $\left(g=10\mathrm{m}/{\mathrm{s}}^2\right)$ 

Solution: By using $3^{rd}$ equation of motion

$v^2=u^2+2as$

$v_2^2=3^2+2(10)(0.8)$

$v_2=5\mathrm{m}/\mathrm{s}$

By using Equation of Continuity,

$A_1{v}_1=A_2{v}_2$

$\text{ (2) (3) }=\left(A_2\right)(5)\Rightarrow A_2=\frac{6}{5}=1.2{\mathrm{m}}^2$

Example 3.In an adult, the average velocity of blood flow through the aorta (which has a radius of 0.9 cm) is 0.33 m/s. As the blood moves from the aorta into the major arteries—30 in total, each with a radius of 0.5 cm—calculate the speed of blood flow through these arteries.

Solution:

$a_1{v}_1=30a_2{v}_2$

$\pi r_1^2{v}_1=30\pi r_2^2{v}_2$

$v_2={\left(\frac{r_1}{r_2}\right)}^2\frac{v_1}{30}={\left(\frac{0.9}{0.5}\right)}^2\times \frac{0.33}{30}=0.036{\mathrm{ms}}^{-1}$

6.0Sample Questions on Continuity Equation

Q-1: Why does the velocity of water increase as it flows out of a PVC pipe when it is suppressed slightly?

Solution: When we press its outlet with our fingers and hence decreases its area of cross-section $v\propto \frac{1}{a}$,so the velocity of water coming out increases.

Q-2. An open tank has two holes in its wall: one is a square hole with side length L  located at a depth  y from the top view, and the other is a circular hole with radius R located at a depth 4y from the top. When the tank is filled to the top with water, the rate of flow from both holes is the same. What is the value of  R ?

Solution: Volume of water flowing out per second=velocity ✕ Area of cross section of the hole=v A

By using Equation of Continuity,

$A_1{v}_1=A_2{v}_2$

$A_1=L^2,v_1=\sqrt{2gy},v_2=\sqrt{2g\times 4y}{A}_2=\pi R^2$

$L^2\times \sqrt{2gy}=\pi R^2\times \sqrt{2g\times 4y}$

$L^2=2\pi R^2\Rightarrow R=\frac{L}{\sqrt{2\pi }}$

Q-3.An incompressible fluid flows steadily through a cylindrical pipe with a radius of 2R at point A and a radius of  R  at point B, which is further along the flow direction. If the velocity of the fluid at point A is v , what is its velocity at point B?

Solution:

By using Equation of Continuity,

$a_1{v}_1=a_2{v}_2$

$\pi (2R{)}^{2}v=\pi R^2{v}_2$

$v_2=4v$

Q-4.The spray pump features a cylindrical tube with a radius of R. and one end features n fine holes, each with a radius r. If the speed of the liquid inside the tube is V, what is the speed of the liquid ejected through the holes?

Solution:

By using Equations of Continuity,

$a_1{v}_1=a_2{v}_2$

$\pi r^{2}V=n\pi r^2{v}_2$

$v_2=\frac{V{R}^2}{n{r}^2}$