Bernoulli's Principle

Last Updated : 23 Apr, 2025

Bernoulli's Principle is a very important concept in Fluid Mechanics which is the study of fluids (like air and water) and their interaction with other fluids. Bernoulli's principle is also referred to as Bernoulli's Equation or Bernoulli Theorem.

This principle was first stated by Daniel Bernoulli and then formulated in Bernoulli's Equation by Leonhard Euler in 1752, which provides the relationship between the pressure (P) of the fluid flowing, at a height (h) of the container having kinetic and gravitational potential energy.

The conservation of energy was found to be true for flowing fluids by the statement of Bernoulli's Principle. It may seem contradictory, but Bernoulli's principle describes how a fluid's velocity and pressure are related to each other.

In this article, we have discussed about Bernoulli's principle, Bernoulli's equation, its derivation, examples, and proof in details.

What is Bernoulli’s Principle?

Bernoulli's Principle says that when a fluid is flowing horizontally, the points where the speed is higher exhibit low pressure, while the points where the speed is lower exhibit high pressure. According to Bernoulli's principle, the gravitational potential energy of elevation, the energy related to fluid pressure, and the kinetic energy of the fluid motion combine up to give the total mechanical energy of a flowing fluid and are all constant.

In real life, Bernoulli’s principle can be observed in rivers. In some places, the width of the river was found to be changed. When the width of the river increases the speed of the water flowing through it decreases. While the speed of water increases in the narrower regions.

Bernoulli's Principal Definition

Bernoulli's principle is a fundamental principle in fluid dynamics that relates the pressure within a fluid to the speed of the fluid's motion

Also Check: Application of Bernoulli’s Theorem

Bernoulli’s Principal Formula

Bernoulli's Principle is formulated into an equation called Bernoulli's Equation. Bernoulli's Equation is a relationship between kinetic energy, gravitational potential energy, and the pressure of the fluid inside the container.

Thus, Bernoulli’s Principle Formula is stated as:

\bold{P+\dfrac{1}{2}\rho v^2+\1/2 rho gh=\text{Constant}}
where,
P is the Pressure exerted by the fluid,
ρ is the Density of the fluid,
v is the Velocity of the fluid,
g is the Acceleration due to gravity and
h is the Height of the container.

Bernoulli’s Equation Derivation

Let us consider a container in the shape of a pipe, whose two edges are placed at different heights and varying diameters. The relationship between the areas of cross-sections A, the flow speed v, height from the ground y, and pressure P at two different points 1 and 2 are given in the figure below.

As illustrated in the diagram, the pipe should be at various heights. According to the equation of continuity, its velocity must fluctuate. To achieve this acceleration and force are required, which is produced by the fluid surrounding it, and the pressure in different places must be different.

Before, deriving Bernoulli’s Equation, we assume that the fluid has no viscous forces, so the fluid's energy will be conserved. Also, the density of the fluid flowing at both points of the container must be uniform.

To determine Bernoulli's equation, we must first determine how much work done experienced by the fluid,

The work done at point 1 where the force F₁ is exerted to displace the fluid to dx₁ is
dW₁ = F₁dx₁
Here, the force exerted at point 1 is given by, F₁ = P₁A₁
where P₁and A₁are the pressure exerted and cross-sectional area at point 1.
This implies,
dW₁ = P₁A₁dx₁
⇒ dW₁= P₁dv
Similarly, the work done by the fluid at point 2 is:
dW₂ = P₂A₂dx₂
⇒ dW₂ = P₂dv
Now, the total work done by the fluid flowing through the container is,
dW = P₁dv - P₂dv
⇒ dW = (P₁ - P₂) dv . . .(1)
Now, the change in the kinetic energy of the fluid is given by,
\begin{aligned}\text{d}K&=\dfrac{1}{2}m_2v^2_2-\dfrac{1}{2}m_1v^2_1\end{aligned}
But, the mass m of the fluid can be written as,
m = ρ dv
Therefore,
\begin{aligned}\text{d}K=\dfrac{1}{2}\rho dv (v^2_2-v^2_1)\end{aligned} . . . (2)
Similarly, the change in gravitational potential energy is given by,
\begin{aligned}\text{d}U&=\dfrac{1}{2}m_1gy_2-\dfrac{1}{2}m_2gy_1\\\Rightarrow \text{d}U&=\dfrac{1}{2}\rho dv g(y_2-y_1)\end{aligned} . . . (3)
According to the law of conservation of energy:
dW = dK + dU
\begin{aligned}&(P_1 - P_2) dv=\dfrac{1}{2}\rho dv (v^2_2-v^2_1)+\dfrac{1}{2}\rho dv g(y_2-y_1)\\ \Rightarrow & (P_1-P_2)=\dfrac{1}{2}\rho (v^2_2-v^2_1)+\dfrac{1}{2}\rho g(y_2-y_1)\\ &\Rightarrow P_1+\dfrac{1}{2}\rho v_1^2+\rho gy_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gy_2\end{aligned}
This is Bernoulli’s equation and this can be written as a general expression,
\bold{P+\frac{1}{2}\rho v^2+\rho gh=\text{Constant} }

Bernoulli’s Equation at Constant Depth

As the name suggests, let us suppose any fluid is moving at some constant depth, so h₁ = h₂, under this condition Bernoulli's equation be

P_1+\frac{1}{2}\rho v_1^2+\rho gh_1 = P_2+\frac{1}{2}\rho v_2^2+\rho gh_2
\Rightarrow P_1+\frac{1}{2}\rho v_1^2= P_2+\frac{1}{2}\rho v_2^2 (As h₁ = h₂ ⇒ ρgh₁ = ρgh₂)

Bernoulli’s Equation for Static Fluids

Any fluid is said static if there is no motion in it, for example, oil in a container, a calm lake, the air in a room with no air current, etc. For static fluid there is no velocity gradient throughout the fluid, thus v₁ = v₂. So, under this condition Bernoulli's equation be.

P_1+\frac{1}{2}\rho v_1^2+\rho gh_1 = P_2+\frac{1}{2}\rho v_2^2+\rho gh_2
\Rightarrow P_1+\rho gh_1 = P_2+\rho gh_2
Now, to simplify it much further, h₂ = 0 (any height can be chosen for the reference height to be zero)
\Rightarrow P_2 = P_1+\rho gh_1 ,
This equation tells us that the static pressure increase from point 1 to 2 by ρgh₁.

Principle of Continuity

The primary consideration when representing the flow of fluids or gases is continuity. It shows how mass conservation applies to gases and liquids. The law of conservation of mass in fluid dynamics is demonstrated using the continuity equation.

The principle of continuity is stated as, when a incompressible fluid is flowing in a streamline flow then mass passing through the system at different cross-sectional areas are equal.

The flow and behavior of fluids in a pipe are explained by the continuity equation. This formula is used to explain the flow of liquids and gases through pipes, ducts, rivers, etc.

Consider a pipe as shown above, with different cross-sectional areas A₁ and A₂. Let the rate of mass entering and leaving the given system be M₁ and M₂ respectively, with a speed of flow v, and density ρ of the fluid.
principal of continuity
Then, according to the principle of continuity, the rate of mass of fluid entering must be equal to the rate of mass of fluid leaving the system.
Since, the rate of mass entering, M₁ = ρA₁V₁ Δt and
The rate of mass entering, M₂ = ρA₂V₂ Δt
Therefore,
ρA₁V₁ Δt = ρA₂V₂ Δt
ρA₁V₁= ρA₂V₂
The above equation is now termed as Equation of Continuity.

Applications of Bernoulli’s Principle and Equation

The following are the important applications of Bernoulli's Principle:

Venturi Meter

A venturi meter is a device used to measure the rate of flow in a pipe with a non-uniform cross-section.

To the wide and narrow sections of the pipe, we apply Bernoulli's equation, where h₁=h₂. The pressure P₂ is less than P₁ because A₁ is bigger than A₂, and v₂ is greater than v₁. The fluid is accelerated as it enters the tube, which is narrow, and slowed as it exits by a net force to the left.

The venturi meter formula is stated as,

\bold{v_1=\sqrt{\dfrac{2gh}{\left(\dfrac{A_1}{A_2}\right)^2-1}}}
where,
v₁ is the rate of fluid flowing,
g is the acceleration due to gravity,
h is the difference in liquid level in the two tubes,
A₁ and A₂ are the cross-sectional areas of the tube.

Principle of Lifting an Aircraft

Bernoulli's theorem governs the operation of aeroplanes. The plane's wings have a certain form. When the plane is moving, the air flows past it at a high rate, despite the plane's low surface wig.

There is a variation in the flow of air above and below the wings due to Bernoulli's principle. As a result of the flow of air on the wing's up surface, this concept causes a change in pressure. If the force is greater than the plane's mass, the plane will ascend as shown in the figure below:

Action of Atomizer or Spray Gun

When the piston is pulled, air flows out of the horizontal tube, causing the pressure to drop which is less than the container's pressure. The consequence is that the liquid in the vertical tube rises. When it runs into the fast air in the tube, it dissipates into a thin mist. The same idea governs filter pumps, Bunsen burners, and sprayers used to apply pesticides or scents.

Relation between Conservation of Energy and Bernoulli’s Equation

Bernoulli’s theorem states that the total amount of these three energies within a constantly moving fluid, i.e., kinetic energy, potential energy, and pressure energy, will remain steady throughout the pipeline. However, several factors must be understood:

The movement of the liquid is assumed to be constant, and there should be no alteration with respect to the pressure, speed, and consistency of the liquid at any point in time. However, if the liquid flow is unstable, there will be a change in speed, and in that case, this equation may not hold true.
The fluid is considered to be incompressible. Since all liquids are incompressible in nature, this principle can be applied to all fluids. However, in the case of gas flow, there should be a constant density, and one can manage with little difference in pressure, velocity, and temperature.
Another consideration is the irrotational feature of liquids, which means that the overall angular momentum throughout the liquid is zero.
The fluid should be ideal for this principle to hold true, meaning there should be no loss of energy due to frictional activity. Hence, there is no excess energy production due to internal particle friction.

Limitations of Bernoulli’s Principle

There are several limitations to Bernoulli’s principle:

Due to friction, the velocity of fluid particles in the middle of a tube gradually decreases in the tube's direction. As a result, the liquid's mean velocity must be used, as the velocity of the particles of the liquid is not constant.
This Bernoulli equation is effective in streamlining liquid supply, but it is ineffective in turbulent or non-steady flow.
The liquid flow will be affected by the liquid's external force.
This theorem is preferably applied to non-viscous fluids, and an incompressible fluid is required.
When a fluid is travelling in a curved path, the energy generated by centrifugal forces must be taken into account.
The liquid flow should remain constant over time.
A small amount of kinetic energy can be converted to heat energy in an unstable flow, and some energy can be lost due to shear stress in a thick flow. As a result, these losses must be overlooked.
The effect of viscosity must be negligible.

Also, Read

Stokes Law
Streamline Flow
Difference between Laminar and Turbulent Flow

Solved Examples on Bernoulli's Principle

Example 1: Water is flowing at a rate of 2m/s through a tube with a diameter of 1 m. If the pressure at this point is 80 kPa, what is the pressure of the water after the tube narrows to a diameter of 0.5m? ρ_water=1.0 kgl^-1

Solution:

According to Bernoulli’s expression:
P_{1}+\frac{1}{2}\rho v^2_{1}+\rho gh_{1}=P_{2}+\frac{1}{2}\rho v^2_{2}+\rho gh_{2}
The height is same (h₁=h₂), so the expression can be written as
P_{2}=P_{1}+\frac{1}{2}\rho \left(v^2_{1}-v^2_{2}\right)
The expression for the cross-sectional area is
A_{1}=\pi\frac{ d^2}{4}\\ \Rightarrow A_{1}=\pi\frac{ 1^2}{4}\text{ m}^2\\ \Rightarrow A_{1}=\frac{\pi}{4}\text{ m}^2\\
Similarly
A_{2}=\pi\frac{ d^2}{4}\\ \Rightarrow A_{2}=\pi\frac{ 0.5^2}{4}\text{ m}^2\\ \Rightarrow A_{2}=\frac{\pi}{16}\text{ m}^2\\
The expression for the velocity for each diameter is
v_1=\frac{V}{A}\\ \Rightarrow v_1=\frac{2}{\frac{\pi}{4}}\frac{\text{m}}{s}\\ \Rightarrow v_1=\frac{8}{\pi}\frac{\text{m}}{s}
Similarly,
v_2=\frac{2}{\frac{\pi}{16}}\frac{\text{m}}{s}\\ \Rightarrow v_2=\frac{32}{\pi}\frac{\text{m}}{s}
Substitute the value in Bernoulli’s expression:
P_{2}=80000\text{ Pa}+\frac{1}{2}\times1000\frac{\text{kg}}{\text{m}^3}\times\left(\frac{8}{\pi}-\frac{32}{\pi}\right)\frac{\text{m}}{s}\\ \Rightarrow P_{2}=76.2\text{ kPa}

Example 2: Explain why:

(a) The blood pressure in humans is greater at the feet than at the brain.

(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at sea level, though the height of the atmosphere is more than 100 km.

(c) Hydrostatic pressure is a scalar quantity even though the pressure is force divided by area.

Solution:

(a) The height of the blood column in the foot is greater than that in the brain. As a result, human blood pressure is higher in the feet than in the brain.
(b) The relationship between air density and height is not linear. As a result, pressure does not decrease linearly with height. P = P₀e^–αh gives the air pressure at a height h, where P₀ is the pressure of air at sea level and α is a constant.
(c) When a force is applied to a liquid, the pressure is distributed evenly throughout the liquid. As a result, the pressure due to liquid has no definite direction. As a result, hydrostatic pressure is a scalar value.

Example 3: Drive the expression for Bernoulli’s principle.

Solution:

The flow of an ideal fluid in a pipe of varying cross-section. The fluid in a section of length v₁∆t moves to the section of length v₂∆t in time ∆t
Consider the flow at two regions BC and DE. Take a look at the flow in two different regions: BC and DE. Consider the fluid that present between B and D would travelled in an infinitesimal time interval ∆t. If v₁ is the speed at B and v₂ is the speed at D, the fluid at B has travelled a distance of v₁∆t to C i.e.
dx_{1}=v_{1}\Delta{t}
In the same interval ∆t the fluid initially at D moves to E, a distance equal to v2∆t i.e.
dx_{2}=v_{2}\Delta{t}
The areas A₁ and A₂ has the pressures P₁ and P₂. The work done on the fluid at BC is
W_{1}=P_{1}A_{1}\cdot v_{1}\Delta{t}\\ \Rightarrow W_{1}=P_{1}\Delta{V}
where ∆V is the volume passes through region BC.
Since the same volume ∆V passes through both the regions BC and DE. therefore the expression for the work done by the fluid at the another end DE is
W_{2}=P_{2}A_{2}\cdot v_{2}\Delta{t}\\ \Rightarrow W_{2}=P_{2}\Delta{V}
So the total work done on the fluid is
W_{1}-W_{2}=P_{1}\Delta{V}-P_{2}\Delta{V}\\ \Rightarrow W_{1}-W_{2}=\left(P_{1}-P_{2}\right)\Delta{V}
Let the density of the fluid is ρ therefore the expression ∆m = ρ∆V is the mass passing through the pipe in time ∆t, then change in gravitational potential energy is
\Delta{U} =\rho g\Delta{V}\left(h_{2}-h_{1}\right)
The change in its kinetic energy is given by
\Delta{K}=\frac{1}{2}\rho\Delta{V}\left(v^2_{2}-v^2_{1}\right)
According to he work – energy theorem'
\left(P_{1}-P_{2}\right)\Delta{V}=\frac{1}{2}\rho \Delta{V}\left(v^2_{2}-v^2_{1}\right)+\rho g\Delta{V}\left(h_{2}-h_{1}\right)\\ \Rightarrow \left(P_{1}-P_{2}\right)=\frac{1}{2}\rho\left(v^2_{2}-v^2_{1}\right)+\rho g\left(h_{2}-h_{1}\right)
Rearrange the above expression,
P_{1}+\frac{1}{2}\rho v^2_{1}+\rho gh_{1}=P_{2}+\frac{1}{2}\rho v^2_{2}+\rho gh_{2}
This is Bernoulli’s equation and this can written as general expression.
P + 1/2ρv² + ρgh = constant

Example 4: Suppose that a huge tank 50m high and filled with water is open to the atmosphere and is hit by a bullet that pierces one side of the tank, allowing water to flow out. The hole is 2m above the ground. If the hole is very small in comparison with the size of the tank, how quickly will the water flow out of the tank?

Solution:

According to Bernoulli’s expression:
P_{1}+\frac{1}{2}\rho v^2_{1}+\rho gh_{1}=P_{2}+\frac{1}{2}\rho v^2_{2}+\rho gh_{2}
According to question it is assumed that the top of the container as point 1, and the hole where water is flowing out as point 2. Both points are open to the atmosphere. Therefore, the pressure on each side of the above equation is equal to 1 atm, and thus it got cancel. The size of the hole on the side of the tank is so small compared to the rest of the tank, the velocity of the water at point 1 is nearly equal to 0. Hence, we can cancel out the 1/2ρ(v₁)² term on the left side of the equation. The expression can rewrite as,
\rho gh_1=\frac{1}{2}\rho v^2_{2}+\rho gh_{2}\\ \Rightarrow gh_1=\frac{1}{2}v^2_{2}+gh_{2}\\ \Rightarrow v^2_{2}=2\left(gh_1-gh_2\right)\\ \Rightarrow v_{2}=\sqrt{2g\left(h_1-h_2\right)}\\
Substitute the values in the above expression,
v₂ = √{2×(9.8)×(50-2)}
v₂ = 30.67 ms^-1

Practice Problems on Bernoulli's Theorem

Problem 1: A fluid is flowing steadily through a pipe. At one point in the pipe, the fluid has a velocity of 10 m/s and a pressure of 200,000 Pa. If the pipe narrows downstream, and the velocity of the fluid increases to 15 m/s, what is the pressure at that point?

Problem 2: An airplane is flying at a constant altitude. The air pressure on the upper surface of the wing is 90,000 Pa, while the air pressure on the lower surface of the wing is 100,000 Pa. If the speed of the air over the top of the wing is 250 m/s, what is the speed of the air under the wing?

Problem 3: Water is flowing through a hose with a diameter of 2 cm at a speed of 4 m/s. If the hose narrows to a diameter of 1 cm, what is the speed of the water at that point? Assume incompressible flow.

Problem 4: A hydraulic lift is used to lift a car weighing 12,000 N. The piston on which the car sits has an area of 0.2 square meters. If a smaller piston connected to the same hydraulic system has an area of 0.02 square meters, how much force must be applied to the smaller piston to lift the car?

Conclusion

Bernoulli's principle states that a fluid's pressure increases with slowing down and decreases with increasing speed. This concept aids in our understanding of many fascinating phenomena, like the functioning of automobile carburetors and water flowing through pipes and airplanes.

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Article Tags :

Bernoulli's Principle

What is Bernoulli’s Principle?

Bernoulli's Principal Definition

Bernoulli’s Principal Formula

Bernoulli’s Equation Derivation

Bernoulli’s Equation at Constant Depth

Bernoulli’s Equation for Static Fluids

Principle of Continuity

Applications of Bernoulli’s Principle and Equation

Venturi Meter

Principle of Lifting an Aircraft

Action of Atomizer or Spray Gun

Relation between Conservation of Energy and Bernoulli’s Equation

Limitations of Bernoulli’s Principle

Solved Examples on Bernoulli's Principle

Practice Problems on Bernoulli's Theorem

Conclusion

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