Bernoulli’s Principle: Law, Derivation, Formula, Limitations

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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

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.”

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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

  1. Bernoulli’s principle is a fundamental principle of fluid dynamics based on the law of conservation of energy.
  2. 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

  1. 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.

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Previous Year Questions

  1. A large drop of oil density … [BITSAT 2012]
  2. A square plate of 0.1 m side moves parallel to a second plate … [AP EAPCET]
  3. A rain drop of radius 0.3mm has a terminal velocity … [VITEEE 2019]
  4. Raindrops are falling from a certain height … [DUET 2009]
  5. Two non-mixing liquids of densities … [NEET 2016]
  6. Two small spherical metal balls, having equal masses, are made … [NEET 2019]
  7. Iceberg floats in water with part of it submerged … [KCET 2020]
  8. The coefficient of viscosity has following dimensions … [KCET 1994]
  9. The ratio of inertial force to viscous force of a fluid … [KEAM]
  10. When the temperature increases, the viscosity of … [JKCET 2006]
  11. If a liquid does not wet glass, its angle of contact is … 
  12. An object of mass 26 kg floats in air and it is in equilibrium state … [UPSEE 2017]
  13. Viscosity decreases with increase in temperature is the reason for … [COMEDK UGET 2009]
  14. The viscous force acting on a rain drop of radius … [COMEDK UGET 2004]
  15. If Y, K and η are the values of Young's modulus … [JEE Mains 2021]
  16. The Young's modulus of steel is twice that of brass … [NEET 2015]
  17. Copper of fixed volume V … [NEET 2014]

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. 

  1. 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.
  2. 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.
  3. Bernoulli’s equation is applicable only to incompressible fluids because it does not take into account the elastic energy of fluids.
  4. Bernoulli’s equation is applicable only to streamline flow of a fluid and not when the flow is turbulent.
  5. 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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                          CBSE CLASS XII Previous Year Papers

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