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Question:
State and prove Bernoulli theorem
Answer:

Bernoulli Theorem

  • 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 ,
    • / Volume=1/2mv2/V = 1/2v2(m/V) = 1/2ρv2
    • /Volume = mgh/V = (m/V)gh = ρgh

Bernoulli Theorem Derivation

Derive: Bernoulli’s equation

Assumptions:

  1. Fluid flow through a pipe of varying width.
  2. Pipe is located at changing heights.
  3. Fluid is incompressible.
  4. Flow is laminar.
  5. 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 time interval Δt, this fluid would have moved.
    • Suppose v1= speed at B and v2= speed at 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– W= (P1A1v1Δt− P2A2v2Δt)
  • By the Equation of continuity Av=constant.
  • P1Av1Δt - 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.
  • The flow of an ideal fluid in a pipe ofvarying cross section. The fluid in asection of length v1Δt moves to the section of length v2Δt in time Δt.

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