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Bernoulli’s Principle

  • Last Updated : 24 Jun, 2021
Geek Week

Fluids can be found everywhere around us. Earth is surrounded by air, and water covers two-thirds of its surface. Water is not only vital for our survival; it is also essential for the survival of every animal species. Every mammalian body constitutes mostly water. Fluid is different from a solid, it does not have its own distinct shape. A gas covers the whole capacity of its container, whereas solids and liquids have a fixed volume. The volume can be changed due to changes in external pressure is rather small in solids or liquids that is solids and liquids are far less compressible than gases.

Shear stress can alter the geometry of a solid while maintaining its volume. Fluids have a critical property: they have extremely little resistance to shear stress; when a modest amount of shear stress is applied, their shape changes. Fluids have shearing stress that is a million times lower than solids.

Bernoulli’s Principle

Bernoulli’s principle, introduced by Daniel Bernoulli, states that as the speed of a flowing fluid (liquid or gas) rises, so does the pressure within the fluid. Despite the fact that Bernoulli established the rule, it was Leonhard Euler who developed Bernoulli’s equation in its standard form in 1752.

Bernoulli’s principle states that:

The net mechanical energy of the flowing fluid, which includes gravitational potential energy of elevation, fluid pressure energy, and kinetic energy of fluid motion, stays constant.



Fluid flow is a tricky concept however, energy conservation is used to gain some helpful features for steady or streamline flows. The expression of Bernoulli’s principle originated by using the principle of conservation of energy. When this theory is applied to fluids in their ideal state, both density and pressure are inversely proportional. As a result, a fluid moving at a slower rate will exert more force than a fluid moving at a faster rate.

Bernoulli’s Principle Formula

Bernoulli’s equation formula is a relationship between a fluid’s pressure, kinetic energy, and gravitational potential energy in a container.

Bernoulli’s principle is defined as follows:

P+\frac{1}{2}\rho v^2+\rho gh=\text{Constant}

where P is the pressure exerted, v is the velocity, ρ is the density of the fluid and h is the height of the container.

Derivation of Bernoulli’s equation

Consider a fluid running through a pipe with changing cross-sectional area and a fluid that is not compressible running at a constant rate. 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 force is required, which is produced by the fluid surrounding it, and the pressure in different places must be different. 

Bernoulli’s equation is a general formulation that links changes in velocity i.e. kinetic energy changes and height change i.e. potential energy change to pressure differences between two places in a pipe.

The flow of an ideal fluid in a pipe of a varying cross-section.

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 travel 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 traveled a distance of v1∆t to C i.e. 



dx1 = v1Δt

In the same interval ∆t the fluid initially at D moves to E, a distance equal to v2∆t i.e.

dx2 = v2Δt

The areas A1 and A2 have the pressures P1 and P2. The work done on the fluid at BC is,

W1 = P1 ​A1 ​⋅ v1Δt  

      = P1 ​ΔV

where ∆V is the volume that 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 another end DE is,

W2 = P2 ​A2 ​⋅ v2Δt  

      = P2 ​ΔV



So the total work done on the fluid is,

W1 – W2 = P1 ​ΔV – P2 ​ΔV 

              = ΔV (P1 – P2)

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 the 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)\\ \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 be written as a general expression,

P+\frac{1}{2}\rho v^2+\rho gh=\text{Constant}

Application of Bernoulli’s Principle 

  • Venturi meter: A venturi meter is a device that measures the rate of flow of liquid via pipes and is based on Bernoulli’s theorem.
  • Airplane: Bernoulli’s theorem governs the operation of airplanes. 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.
  • Magnus Effect: A revolving ball moves away from its regular course inside the flight once it is thrown. The Magnus effect is the name given to this phenomenon. This impact is important in sports like cricket, soccer, and tennis.

Limitations of Bernoulli’s Principle 

Following are the limitations of Bernoulli’s principle:

  • Because of friction, the fluid particle velocity in the middle of a tube gradually decreases toward the tube’s direction. The velocity of the particles of the liquid is not constant, therefore the liquid’s mean velocity must be used.
  • This Bernoulli equation can be used to streamline liquid supply. It is ineffective in turbulent or non-steady flow.
  • The liquid flow will be affected by the liquid’s external force.
  • This theorem preferably applies to non-viscosity fluids.
  • Incompressible fluid is required.
  • When a fluid is traveling 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 viscous must be negligible.

Sample Problems 

Problem 1: Water is flowing at a rate of 2m3s 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? ρwater=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, 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}\\ A_{1}=\pi\frac{ 1^2}{4}\text{ m}^2\\ A_{1}=\frac{\pi}{4}\text{ m}^2\\

Similarly

 A_{2}=\pi\frac{ d^2}{4}\\ A_{2}=\pi\frac{ 0.5^2}{4}\text{ m}^2\\ A_{2}=\frac{\pi}{16}\text{ m}^2\\

The expression for the velocity for each diameter is 

v_1=\frac{V}{A}\\ v_1=\frac{2}{\frac{\pi}{4}}\frac{\text{m}}{s}\\ v_1=\frac{8}{\pi}\frac{\text{m}}{s}

Similarly,

v_2=\frac{2}{\frac{\pi}{16}}\frac{\text{m}}{s}\\ 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}\\P_{2}=76.2\text{ kPa}

Problem 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.

Problem 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 traveled 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 traveled 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}\\ 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}\\ W_{2}=P_{2}\Delta{V}

So the total work done on the fluid isP+\frac{1}{2}\rho v^2+\rho gv^2=contant

W_{1}-W_{2}=P_{1}\Delta{V}-P_{2}\Delta{V}\\ 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)\\ \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.

This is Bernoulli’s equation and this can written as general expression,

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

Problem 4: What are the limitations of Bernoulli’s principle.

Solution:

Following are the limitations of Bernoulli’s principle:

  • Because of friction, the fluid particle velocity in the middle of a tube  gradually decreases toward the tube’s direction. the velocity of the particles of the liquid are not constant, therefore the liquid’s mean velocity must be used.
  • This Bernoulli’s equation can be used to streamline liquid supply. It is ineffective in turbulent or non-steady flow.
  • The liquid flow will be affected by the liquid’s external force.
  • This theorem preferably applies to non-viscosity fluids.
  • Incompressible fluid is required.
  • When a fluid is traveling 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 viscous must be negligible

Problem 5: Suppose that a huge tank 50m high and filled with water is open to the atmosphere and is hit with 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 \frac{1}{2}ρv^2_1   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}\\ gh_1=\frac{1}{2}v^2_{2}+gh_{2}\\v^2_{2}=2\left(gh_1-gh_2\right)\\v_{2}=\sqrt{2g\left(h_1-h_2\right)}\\

Substitute the values in the above expression,

v_{2}=\sqrt{2\times9.8\left(50-2\right)}\text{ ms}^{-1}\\v_{2}=30.67\text{ ms}^{-1}

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