Question

Bernoulli Theorem explain

class 11 physics​

Answer 1

Bernoulli's theorem, in fluid dynamics, relation among the pressure, velocity, and elevation in a moving fluid (liquid or gas), the compressibility and viscosity (internal friction) of which are negligible and the flow of which is steady, or laminar.

For a streamline fluid flow, the sum of the pressure (P), the kinetic energy per unit volume (ρv2/2) and the potential energy per unit volume (ρgh) remain constant.

Mathematically:- P+ ρv2/2 + ρgh = constant

where P= pressure ,

E./ Volume=1/2mv2/V = 1/2v2(m/V) = 1/2ρv2

E./Volume = mgh/V = (m/V)gh = ρgh

Derive: Bernoulli’s equation

Fluid flow through a pipe of varying width.

Pipe is located at changing heights.

Fluid is incompressible.

Flow is laminar.

No energy is lost due to friction:applicable only to non-viscous fluids.

Mathematically: -

Consider the fluid initially lying between B and D. In an infinitesimal timeinterval Δt, this fluid would have moved.

Suppose v1= speed at B and v2= speedat D, initial distance moved by fluid from to C=v1Δt.

In the same interval Δtfluid distance moved by D to E = v2Δt.

P1= Pressureat A1, P2=Pressure at A2.

Work done on the fluid atleft end (BC) W1 = P1A1(v1Δt).

Work done by the fluid at the other end (DE)W2 = P2A2(v2Δt)

Net work done on the fluid is W1 – W2 = (P1A1v1Δt− P2A2v2Δt)

By the Equation of continuity Av=constant.

P1A1 v1Δt - P2A2v2Δt where A1v1Δt =P1ΔV and A2v2Δt = P2ΔV.

Therefore Work done = (P1− P2) ΔVequation (a)

Part of this work goes in changing Kinetic energy, ΔK = (½)m (v22 – v12) and part in gravitational potential energy,ΔU =mg (h2 − h1).

The total change in energy ΔE= ΔK +ΔU = (½) m (v22 – v12) + mg (h2 − h1). (i)

Density of the fluid ρ =m/V or m=ρV

Therefore in small interval of time Δt, small change in mass Δm

Δm=ρΔV (ii)

Putting the value from equation (ii) to (i)

 

ΔE = 1/2 ρΔV (v22 – v12) + ρgΔV (h2 − h1)  equation(b)

By using work-energy theorem: W = ΔE

From (a) and (b)

(P1-P2) ΔV =(1/2) ρΔV (v22 – v12) + ρgΔV (h2 − h1)

P1-P2 = 1/2ρv22 - 1/2ρv12+ρgh2 -ρgh1(By cancelling ΔV from both the sides).

After rearranging we get,P1 + (1/2) ρ v12 + ρg h1 = (1/2) ρ v22 + ρg h2

This is the Bernoulli’s equation

P+(1/2) ρv2+ρg h = constant.

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Answer 2

Explanation:

Bernoulli theorem that states that as the speed of a moving fluid (liquid or gas) increases, the pressure within the fluid decreases.

An example of Bernoulli's principle is the wing of an airplane.

Bernoulli's equation formula

Pressure + ½ density * square of the velocity + density * gravity. acceleration* height = constant.

The equation is written as

P + ½ ρ v2 +ρ g h = constant.

Where,

p is the pressure exerted by the fluid

v is the velocity of the fluid

ρ is the density of the fluid

h is the height of the container.

One of the most common everyday applications of Bernoulli's principle is in airflight.

• Venturimeter
• Working of an aeroplane:
• When we are standing on a railway station and a train comes we tend to fall towards the train. This can be explained using Bernoulli’s principle.

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