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Prove that Bernoulli's Theorem

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Prove that Bernoulli's Theorem

Posted by Bhaven Kumar 5 years, 7 months ago

CBSE > Class 11 > Physics

1 answers

Yogita Ingle 5 years, 7 months ago

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 A_1, 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.
- P₁A₁v₁Δt - 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
P+(1/2) ρv²+ρg h = constant.
This is the Bernoulli’s equation.

ANSWER

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