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Bernoulli’s principle is based on the law of conservation of energy and applies to ideal fluids. Bernoulli’s equation relates the speed of a fluid at a point, the pressure at that point and the height of that point above a reference level. It is just the application of the work-energy theorem in the case of fluid flow.
Key Terms: Bernoulli’s Theorem, Theory, Proof, Potential Energy, Kinetic Energy
Bernoulli’s principle
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Bernoulli's principle states that “the sum of pressure energy, kinetic energy and potential energy per unit volume of an incompressible, non-viscous fluid in a streamlined irrotational flow remains constant along a streamline”.
Mathematically, it can be expressed as
Bernoulli’s Theorem Video Lecture
Bernoulli’s Principle Proof
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Proof: Consider a non-viscous and incompressible fluid flowing steadily between the sections A and B of a pipe of varying cross-section. Let a, be the area of cross-section at A, v, the fluid velocity, P, the fluid pressure, and h, the mean height above the ground level.
Let a, v2?, P? and h? be the values of the corresponding quantities at B.

Bernoulli’s Principle
Let p be the density of the fluid. As the fluid is incompressible, so whatever mass of fluid enters the pipe at section A in time At, an equal mass of fluid flows out at section B in time At. This mass is given by
m= Volume x density
= Area of cross-section x length x density
or, m = a2 v2 Δt ρ = a2 v2 Δt ρ ……(1)
or, a1v1 = a2v2 ……(2)
Therefore, Change in K.E. of the Fluid
= K.E. at B – K.E. at A
= 1/2m (v22 - v12) = 12 a1 v1 Δt ρ (v22 - v12)
[Using (1)]
Change in P.E. of the fluid
= P.E. at B – P.E at A
= mg (h2 – h1) = a1 v1 Δt ρ g (h2 – h1) [Using (1)]
Net work done on the fluid
= Work done on the fluid at A – Work done by fluid at B
= P1 a1 x v1 Δt – P2 a2 x v2 Δt
= P1 a1 v1 Δt - P2 a1 v1 Δt [Using (2)]
= a1 v1 Δt (P1 - P2)
By conservation of energy,
Net work done by the fluid
= Change in K.E. of the fluid + Change in P.E. of the fluid
Therefore, a1 v1 Δt (P1 - P2)
= 1/2 a1 v1 Δt ρ (v22 - v12) + a1 v1 Δt ρ g (h2 – h1)
Dividing both the sides by a1 v1 Δt, we get
P1 - P2 = 1/2 ρ v12 - 1/2 ρ v12 + ρ gh2 - ρ gh1
or, P1 + 1/2 ρ v12 + ρgh1 = P2 + 1/2 ρ v12 + ρgh2
or, P + 12 ρv2 + ρgh = constant ……..(3)
This proves Bernoulli’s principle according to which the total energy per unit volume remains constant.
Equation (3) can also be written as
Pg + 1/2 v2g + h = constant
This is another form of Bernoulli’s principle according to which- “the sum of pressure head, velocity head and gravitational head remains constant in the streamline flow of an ideal fluid.”
Also Read:
| Topics Related Links | ||
|---|---|---|
| Pressure | Pascal’s Law | Hydraulic Machines |
| Venturi-meter | Viscosity | Stokes’ Law |
Limitations of Bernoulli’s equation
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- Bernoulli’s equation ideally applies to fluids with zero viscosity or non-viscous fluids. In case of viscous fluids, we need to take into account the work done against viscous drag.
- Bernoulli’s equation has been derived on the assumption that there is no loss of energy due to friction. But in practice, when fluids flow, some of their kinetic energy gets converted into heat due to work done against the internal forces of friction or viscous forces.
- Bernoulli’s equation is applicable only to incompressible fluids because it does not take into account the elastic energy of fluids.
- Bernoulli’s equation is applicable only to streamline flow of a fluid and not when the flow is turbulent.
- Bernoulli’s equation does not take into consideration the angular momentum of the fluid. So it cannot be applied when the fluid flows along a curved path.
Things to Remember
- Bernoulli’s principle is a fundamental principle of fluid dynamics based on the law of conservation of energy.
- In Bernoulli’s equation : P + 1/2 ρv2 + ρ gh = constant, the term (P + ρ gh) is called static pressure, because it is the pressure of the fluid even if it is at rest, and the term 12 ρv2 is the dynamic pressure of the fluid which is the pressure by virtue of its velocity, v. So Bernoulli’s equation can be written as
Static pressure + Dynamic pressure = constant
- If a liquid is flowing through a horizontal tube, h remains constant and we can write
This shows that if v increases, P decreases and vice versa. Thus for the streamline flow of an ideal liquid flowing horizontally, the pressure decreases when velocity increases and vice versa. This is an important aspect of Bernoulli’s principle which finds many applications.
Also Read:
| Chapter Related links | ||
|---|---|---|
| Surface tension | Surface Energy | Viscosity |
| Mechanical Properties of Fluids | Surface Tension | Barometer |
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Sample Questions
Ques. A beaker of circular cross section of radius 4 cm is filled with mercury up to a height of 10 cm. Find the force exerted by the mercury on the bottom of the beaker. The atmospheric pressure = 105 Nm-2. Density of mercury = 13600 kg m-3. Take g=10 ms-2. (3 marks)
Ans. The pressure at the surface = atmospheric pressure
= 105 N m-2 .
The pressure at the bottom
= 105 N m-2 + hρg
= 105 N m-2 + (0.1 m) (13600 kg m-3) (10 m s-2)
= 105 N m-2 + 13600 N m-2
= 1.136 x 105Nm.
The force exerted by the mercury on the bottom
=(1.136 x 105 N m-2) x (3.14 x 0.04m x 0.04m) = 571 N.
Ques. The density of air near earth's surface is 1.3 kg /m and the atmospheric pressure is 1.0 x 105 Nm If the atmosphere had uniform density, same as that observed at the surface of the earth, what would be the height of the atmosphere to exert the same pressure? (3 marks)
Ans. Let the uniform density be ρ and atmospheric height be h. The pressure at the surface of the earth would be
P = ρgh
or, 1.0 x 105 N m-2 = (1.3 kg m-3) (9.8 m s-2) h
or, h= 1.0 x 105 N m-2 (1.3 kg m-3) (9.8 m s-2) = 7850 m.
Even Mount Everest (8848 m) would have been outside the atmosphere.
Ques. State Bernoulli’s principle. (2 marks)
Ans. Bernoulli's principle states that the sum of pressure energy, kinetic energy and potential energy per unit volume of an incompressible, non-viscous fluid in a streamlined irrotational flow remains constant along a streamline.
Mathematically, it can be expressed as
P+ 1/2ρv² + ρgh = constant
Ques. State the limitations of Bernoulli’s equation. (5 marks)
Ans.
- Ans.Bernoulli’s equation ideally applies to fluids with zero viscosity or non-viscous fluids. In case of viscous fluids, we need to take into account the work done against viscous drag.
- Bernoulli’s equation has been derived on the assumption that there is no loss of energy due to friction. But in practice, when fluids flow, some of their kinetic energy gets converted into heat due to work done against the internal forces of friction or viscous forces.
- Bernoulli’s equation is applicable only to incompressible fluids because it does not take into account the elastic energy of fluids.
- Bernoulli’s equation is applicable only to streamline flow of a fluid and not when the flow is turbulent.
- Bernoulli’s equation does not take into consideration the angular momentum of the fluid. So it cannot be applied when the fluid flows along a curved path.
Ques. Consider a liquid of density 1200 kg m-3 flowing steadily in a tube of varying cross section. The cross section at a point A is 1.0 cm2 and that at B is 20 mm2, the points A and B are in the same horizontal plane. The speed of liquid at A is 10 cm s-1. Calculate the difference in pressure at A and B. (3 marks)
Ans. From equation of continuity, the speed v2 at B is given by,
A1v1 = A2v2
or, (0.1 cm2) (10cm s-1 ) = (20 mm2)v2
or, v2= 1.0 cm220 mm2 x 10 cm s-1.
By Bernoulli equation
P1 + 1/2 ρ vâ2 + ρ gh1 = P2 + 1/2 ρ vâ2 + ρ gh2
Here h1 = h2 . Thus,
P1 – P2 = 1/2 ρ vâ2 - 1/2 ρ vâ2
= 144 Pa.
Ques. What is head loss in Bernoulli’s equation? (2 marks)
Ans. The head loss in Bernoulli’s equation is the representation of the reduction in the total pressure which is the sum of the velocity head, pressure head, and the elevation head of the fluid flowing through the hydraulic system.
Ques. What is the maximum suction head of a pump? (1 marks)
Ans. The maximum suction head of a pump is approximately 15 feet.
Ques. What is the head loss equation? (2 marks)
Ans. The following equation is the mathematical representation of the head loss:
hL = f L/D v2/2g
Where,
- hL is the head loss
- f is the Darcy friction factor
- L is the pipe length
- D is inside pipe diameter
- v is the fluid velocity
- g is the gravitational constant
- d is the inside pipe diameter
- Q is the volumetric flow rate
Ques. In the case of an emergency, a vacuum brake is used to stop the train. How does this brake work? (2 marks)
Ans. Steam at high pressure is made to enter the cylinder of the vacuum brake. Due to high velocity, pressure decreases in accordance with Bernoulli’s principle. Due to this decrease in pressure, the piston gets lifted. Hence the brake gets lifted.
Ques. Why are the wings of an aeroplane rounded outwards (i.e. more curved) while flattened inwards? What is this shape called? (2 marks)
Ans.
The special design of the wings which is slightly convex upward and concave downward increases velocity at the upper surface and decreases it at the lower surface. So according to Bernoulli’s Theorem, the pressure on the upper side is less than the pressure on the lower side.
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