Question

derivation of Bernoulli theorem

Answer 1

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                                      BERNOULLI'S THEOREM 


When a incompressible non-viscous fluid flows in a stream line motion then at every point the total energy (K.E,P.E, Pressure energy) per unit volume remains constant that is 

                       { P +1/2ev² + egh } = constant 


Consider a tube of flows from X to Y₂

                     A₁  > A₂  → V₂  > V₁

By equation of continuity with AV = constant 
  
                       V α 1/A

Where the fluid flows from x to y, the fluid gains KE as V₂ > V₁

Energy/ unit ve = constant 

Pressure energy 

  KE and PE      [ PE = POTENTIAL ENERGY , KE = KINETIC ENERGY ]

∴  Gain in KE = 1/2 mv₂² - 1/2 mv₁² ----------------(1)


As h₁ > h₂  PEx > PEy

The fluid losses its potential energy when flows from X to Y

∴ Loss in P.E is mgh₁ - mgh₂

At point X, displacement and pressure are in same direction. So, therefore work doe is +ve ( work done by the fluid)

At point Y, displacement and pressure are in opposite direction. So, therefore work done is -ve

When fluid flows from x to y it losses pressure Energy

∴ But by equation of continuity 

          A₁V₁ = A₂V₂ = AV

  Loss of pressure energy 

          ( P₁ - P₂) AV Δt

As AV Δt  = m/e

          ∴ loss of pressure

Energy is ( P₁ - P₂) m/e ----------- (3)


∴ Net gain in energy = loss in energy 

1/2mv²₂ - 1/2mv²₁ = (mgh₁ - mgh₂ ) + (P₁ - P₂) m/ρ ( all the m's will get cancel)

1/2 v₂² - 1/2 v₂² = gh₁ - gh₂  + (P₁ - P₂₂)/ρ        ( multiplying ρ with given equation)

1/2 ρv₂² - 1/2 ρv₁² = egh₁ - egh₂ + P₁ - P₂

1/2 ev₂² + egh₂ + P₂ = 1/2 ev₁² + egh₁ + P₁

 
                     P + 1/2ev + egh = constant 


Hope it helps you !