Bernoulli’s Principle 

Bernoulli's Principle is a phenomenon in fluid dynamics that describes what happens to a moving fluid—liquid or gas. It was a Swiss mathematician, Daniel Bernoulli, who formulated this theorem in the 18th century. The principle is based on the fluid flow conservation of energy and is usually referred to as Bernoulli's theorem.

1.0Bernoulli’s Equation Explained

State and Prove Bernoulli’s Principle

Before jumping straight into the concept of Bernoulli’s Principle let’s take a look at an important component of Bernoulli’s principle which is Ideal fluid. Ideal fluid is a hypothetical concept that refers to an ideal fluid exhibiting properties like: 

• No Viscosity: An ideal fluid has no internal friction, meaning it flows without any resistance to shear.
• Incompressible: The density of an ideal fluid does not change with pressure.
• Steady Flow: For an ideal fluid, the flow is steady; it does not vary with time so that at any given point, the fluid velocity is unchanging.
• Non-Turbulent Flow: Ideal fluids always move in laminar flow, implying that their movement is smooth and ordered without fluctuations.

We can derive the behavior of an ideal fluid using the two important concepts of fluid mechanics namely Bernoulli’s equation and Continuity equation.

2.0Bernoulli’s Theorem

Statement: The total mechanical energy of a non-viscous fluid in steady, incompressible flow consists of pressure energy, kinetic energy, and potential energy, and it remains constant along a streamline. This is the Bernoulli effect.

In maths, the Bernoulli equation can be written as: 

$P+\frac{1}{2}\rho v^2+\rho gh=constant$

here:

• P = Pressure energy per unit volume (Pa)
• ρ = Density of the fluid (kg/m³)
• v = Velocity of the fluid (m/s)
• g = Acceleration due to gravity (m/s²)
• h = Height above a reference point (m)

This equation shows the relation of pressure, velocity, and elevation of a fluid and how the Bernoulli equation in fluid mechanics helps in building this relation. 

Prove:

According to Bernoulli’s theorem, the work done by fluid can be written as, 

dW = F₁dx₁ - F₂dx₂

dW = p₁A₁dx₁ - p₂A₂dx₂ 

dW = p₁dv - p₂dv = (p₁ - p₂)dv 

As known, the work done on the fluid was due to the conservation of change in gravitational potential energy and change in kinetic energy. The change in the fluid's kinetic energy is given by:

$dK=\frac{1}{2}{m}_2^2{v}_2^2-\frac{1}{2}{m}_1^2{v}_1^2$

$dK=\frac{1}{2}\rho dv(v_2^2-v_1^2)$

Change in the potential energy can be written as: 

$dU=m_{2}g{y}_2-m_{1}g{y}_1=\rho gdv(y_2-y_1)$

Now, according to the theorem, the energy of the system is the same as the summation of the change in the kinetic fluid's energy and the change in the fluid's potential energy. This can be mathematically written as: 

$dW=dK+dU$

Putting values by the above equations: 

$(p1-p2)dv=\frac{1}{2}\rho dv(v_2^2-v_1^2)+\rho gdv(y_2-y_1)$

$p1-p2=\frac{1}{2}\rho (v_2^2-v_1^2)+\rho g(y_2-y_1)$

By rearranging the equation, we will get the Bernoulli's equation: 

$p_1+\frac{1}{2}\rho v_1^2+\rho g{y}_1=p_2+\frac{1}{2}{v}_2^2+\rho g{y}_2$

$p+\frac{1}{2}\rho v+\rho gy=constant$

3.0Components of Bernoulli’s Equation 

• Pressure Head: The pressure head is the height of a fluid column that corresponds to the pressure at a given point in the fluid. It is defined as:

Pressure Head = $\frac{P}{\rho g}$

• Velocity Head: The velocity head is the height equivalent to the kinetic energy per unit weight of the fluid. It is calculated as:

Velocity Head = $\frac{v^2}{2g}$

• Potential Head: The potential head is the height of the fluid column and essentially is due to the gravitational potential energy. It is given as:

Potential head = y

The head in fluid mechanics is defined as the height of a fluid column which represents certain energy in the fluid system. That is, fluid pressure, velocity, or potential can be expressed conveniently in terms of energy and hence can be compared relatively more easily across fluid systems under practical applications, such as pipes, pumps, and turbines.

Bernoulli’s Equation with Head terms: 

Using all the three head terms Bernoulli’s Equation can be re-written as: 

$\frac{P}{\rho g}+\frac{v^2}{2g}+y=constant$

4.0Continuity equation

The continuity equation is based on the principle of mass conservation. For an incompressible, steady-flowing fluid, the rate at which mass enters a pipe must be equal to the rate at which mass exits. The equation is:

A₁v₁ = A₂v₂

Here:

• A₁​ and A₂​ are the cross-sectional areas at two points in the flow.
• v₁ and v₂ are the velocities of the fluid at those points.

5.0Applications of Bernoulli’s Principle 

The application of Bernoulli’s equation is many across different fields. Here are some important applications:

Application	Diagram
Aeroplane Wings (Lift): The Bernoulli effect explains how an aeroplane wing produces lift. It's due to the curvature of the wing: air moving above it has to move faster than air below it, and so, by Bernoulli, it has to have lower pressure above and higher pressure below. This is what produces the upward force.
Venturi Effect: Where the fluid velocity increases when a pipe narrows, in varying cross-section pipes, the pressure reduces. This is an application of Bernoulli's theorem and is applied in carburettors, flow meters, and atomizers.
Blood flow in Arteries: Bernoulli's principle explains the reason behind how pressure varies with blood vessels with respect to changes in blood flow velocity as a result of constriction or obstruction. More speed at constriction results in reduced pressure.
Sports (Football, Cricket): The Bernoulli theorem can be used in the trajectory of balls. For instance, a soccer ball is kicked with spin curves because of the difference in pressure created by the air's velocity.
Water Flow in Pipes: The Bernoulli equation in fluid mechanics is applied to analyze the flow of water in pipes of different diameters. The flow rate is constant, but pressure and velocity vary with a change in diameter.

6.0Solved Examples 

Problem 1: Water is flowing from a tank through a pipe. The pressure at the bottom of the tank (point 1) is 150,000 Pa, and the height is 10 m. Find the pressure at the top of the tank (point 2), where the height is 0 m. The velocity at the bottom of the tank is 1 m/s, and at the top of the tank, the velocity is negligible.

Solution: Given that: P₁ = 150000Pa, h₁=10m, v₁ = 1m/s, h₂ = 0m, v₂ = 0m/s 

Using Bernoulli's equation between points 1 and 2 

$p_1+\frac{1}{2}\rho v_1^2+\rho g{y}_1=p_2+\frac{1}{2}v22+\rho g{y}_2$

$150000+\frac{1}{2}(1000)(1{)}^2+(1000)(9.8)(10)=P_2+\frac{1}{2}(1000)(0{)}^2+(1000)(9.8)(0)$

$P_2=150000+500+98000=2,48,500$

Problem 2: An aeroplane wing is designed such that the air above the wing moves faster than the air below the wing. The velocity of the air above the wing is 200 m/s, and the velocity of the air below the wing is 150 m/s. Given that the air density is 1.225 kg/m³, calculate the difference in pressure between the upper and lower surfaces of the wing that produces lift.

Solution: Let the pressure below the wing be P₁ and the pressure above the wing be P₂.

It is given that the velocity above the wing is v₁ = 200m/s and the velocity below the wing v₂ = 150m/s, so here we apply Bernoulli’s equation between the two points: 

$p_1+\frac{1}{2}\rho v_2^2=p_2+\frac{1}{2}{v}_1^2$

$P_1-P_2=\frac{1}{2}\rho (v_1^2-v_2^2)$

$P_1-P_2=\frac{1}{2}\times 1.225\times (200^2-150^2)$

$P_1-P_2=\frac{1}{2}\times 1.225\times (40000-22500)$

$P_1-P_2=\frac{1}{2}\times 1.225\times 17500$

$P_1-P_2=10718.75Pa$

Hence, the pressure difference between the upper and lower wing is 10718.75Pa.

Problem 3: Water flows vertically through a pipe. The water velocity at a point 4 m above the ground is 2 m/s, and the pressure at this point is 250 kPa. Calculate the velocity of water 6 m below this point, where the pressure is 300 kPa.

Given: 

Height h₁ = 4m on point 1. 

Height h₂ = –6m on point 2.

v₁ = 2m/s, =1000kg/m3, g = 9.8m/s², 

P₁ = 250kPa = 250000Pa

P₂ = 300kPa = 300000Pa

Solution: By applying Bernaulli’s equation: 

$p_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=p_2+\frac{1}{2}{v}_2^2+\rho g{h}^2$

$\frac{1}{2}\rho v_2^2+=p_2-p_1+\frac{1}{2}{v}_1^2+\rho g(h_2-h_1)$

$v_2+=\sqrt{\frac{2}{\rho }(p_2-p_1+\frac{1}{2}{v}_1^2+\rho g(h_2-h_1))}$

$v_2=\sqrt{\frac{2}{1000}(250000-300000+\frac{1}{2}1000\times 2^2+10009.8(4-(-6)))}$

$v_2=\sqrt{\frac{2}{1000}\times 50000}$

$v_2=\sqrt{100}$

$v_2=10m/s$