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By Woo Chang Chung Bernoulli’s Principle  and Simple Fluid Dynamics
Pressure ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Pascal’s Law ,[object Object],[object Object],[object Object],[object Object],PA -  ρ  Ahg – P 0 A = 0 P = P 0  +  ρ  gh
Hydraulics ,[object Object],[object Object],[object Object],[object Object],You have to push down the piston on the left far down to achieve some change in the height of the piston on the right.
Continuity Equation ,[object Object],[object Object],[object Object],A = area v = velocity
Bernoulli’s Equation Where p is the pressure,  ρ  is the density, v is the velocity,  h is elevation, and g is gravitational acceleration
Derivation of Bernoulli’s Equation ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Derivation of Bernoulli’s Equation Consider the change in total energy of the fluid as it moves from the inlet to the outlet. Δ E total  = W done on fluid  - W done by fluid Δ E total  = ( 1 / 2 mv 2 2  + mgh 1 ) – ( 1 / 2 mv 1 2  + mgh 2 ) W done on fluid   - W done by fluid  = ( 1 / 2 mv 2 2  + mgh 1 ) – ( 1 / 2 mv 1 2  + mgh 2 ) P 2 V 2  - P 1 V 1  = ( 1 / 2 mv 2 2  + mgh 1 ) – ( 1 / 2 mv 1 2  + mgh 2 )  P 2  – P 1  = ( 1 / 2 ρ  v 1 2  +  ρ  gh 1 ) – ( 1 / 2 ρ  v 1 2  +  ρ  gh 1 ) E total  =  1 / 2 mv 2  + mgh  W =  F / A *A*d = PV P 2  +  1 / 2 ρ  v 1 2  +  ρ  gh 1  = P 1  +  1 / 2 ρ  v 1 2  +  ρ  gh 1 ∴
Venturi Tube ,[object Object],[object Object],[object Object],P 2  +  1 / 2 ρ  v 1 2  = P 1  +  1 / 2 ρ  v 1 2   ;  ΔP =   ρ / 2 *(v 2 2  – v 1 2 )
[object Object],[object Object],Atomizer (Demonstration)
Torricelli and his Orifice ,[object Object],[object Object],Depending on the contour and shape of the opening, different discharge  coefficients  can be applied to the equation  (of course we assume simpler situation here).
Derivation of Torricelli’s Equation ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],P 2  +  1 / 2 ρ  v 1 2  +  ρ  gh 1  = P 1  +  1 / 2 ρ  v 1 2  +  ρ  gh 1
Pitot – Static Tube ,[object Object],[object Object],[object Object]
 
Pitot-Static Tube ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Misinterpretation of Bernoulli ,[object Object],[object Object],[object Object],[object Object],[object Object]
Streamlines ,[object Object],[object Object],[object Object],[object Object]
Aerodynamic Lift ,[object Object],[object Object],[object Object],[object Object],[object Object]
Misconceptions of Lift ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Coanda Effect ,[object Object],[object Object],Due to viscosity, adjacent air molecules are swept and result in lower pressure. Then the steam follows the boundary This floating ping pong ball owes its levitation to the Coanda Effect. (DEMO)
Sources ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

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Bernoulli's Principle

  • 1. By Woo Chang Chung Bernoulli’s Principle and Simple Fluid Dynamics
  • 2.
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  • 6. Bernoulli’s Equation Where p is the pressure, ρ is the density, v is the velocity, h is elevation, and g is gravitational acceleration
  • 7.
  • 8. Derivation of Bernoulli’s Equation Consider the change in total energy of the fluid as it moves from the inlet to the outlet. Δ E total = W done on fluid - W done by fluid Δ E total = ( 1 / 2 mv 2 2 + mgh 1 ) – ( 1 / 2 mv 1 2 + mgh 2 ) W done on fluid - W done by fluid = ( 1 / 2 mv 2 2 + mgh 1 ) – ( 1 / 2 mv 1 2 + mgh 2 ) P 2 V 2 - P 1 V 1 = ( 1 / 2 mv 2 2 + mgh 1 ) – ( 1 / 2 mv 1 2 + mgh 2 ) P 2 – P 1 = ( 1 / 2 ρ v 1 2 + ρ gh 1 ) – ( 1 / 2 ρ v 1 2 + ρ gh 1 ) E total = 1 / 2 mv 2 + mgh W = F / A *A*d = PV P 2 + 1 / 2 ρ v 1 2 + ρ gh 1 = P 1 + 1 / 2 ρ v 1 2 + ρ gh 1 ∴
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