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Solution%20to%20HW#6

# Solution%20to%20HW#6 - EGM6812/FluidMechanicsI Homework#6...

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EGM 6812/ Fluid Mechanics I Homework # 6      Fall  2009 1. ( Bernoulli equation in axisymmetric flows )   For steady axisymmetric flows of an  inviscid fluid of constant density  ρ , show that  + +  + F( ψ ) = constant where   is the potential for the body force,  ψ  is the Stokes stream function.   Hint : you need to show first, from the vorticity transport equation, that the vorticity  η  (in  the  φ  direction) is related to  ψ  as η /r = f( ψ ).    Solution: Euler equation at steady state:   u ∇⋅  u  = 1 p f ρ - +   (1) Note: u ∇⋅  u  =  (u u ) - u x ϖ ;   f = -∇Ω 1 p p ρ ρ = ∇ (2) => 2 2 u p u ϖ ρ + + Ω = × (3) For 2-D axisymmetric flows,  u  =( ( , , 0) z r u u ,   ϖ = (0, 0, ) η The mass conservation  ∇⋅  u  =0  gives  u  =  1 e r θ ψ × 1 1 z r e e r r r z ψ ψ - (4) η  =  r z u u z r -  =  2 2 2 2 1 1 r r r r z ψ ψ ψ - - + (5) The vorticity transport equation is  ( / ) 0 D r Dt η =   (see problem HW#5.3) Thus,  η /r = constant along a streamline when the flow is steady. That is η  = r f( ψ ) (6) Thus, LHS of Eq. (3) becomes u ϖ ×  =  1 1 ( ) z r e e rf e r r r z θ ψ ψ ψ - × ( ) f ψ ψ - (7) (3) & (7) => 2 2 u p ρ + + Ω = ( ) f ψ ψ - (8)

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Let  ( ) ( ) F f d ψ ψ ψ = , then   '( ) ( ) F F f ψ ψ ψ ψ = = Hence, 2 2 u p ρ + + Ω = ( ) F ψ -∇ which integrates to    + +  + F( ψ ) = constant. 2. ( Compressible   Bernoulli   equation )     From   the   Bernoulli   equation   for   isentropic  compressible flows, show that  = [1-v 2 /v] γ /( γ -1) For v 2 /v«1, show that   p = p 0  - [1- v 2 /(4a)+ ...]  using Taylor series expansion.   If you use the incompressible Bernoulli equation to  compute the pressure at v=68 (m/s) when a 0 =340 (m/s), how much relative error in p  results in comparison with the dynamic head  ρ 0 v 2 /2? Solution: i) Bernoulli equation for isentropic compressible flows :  2 0 2 V h h + =   or   2 2 max 2 2 V V h + = with  2 0 max 0 0 1 2 p V h γ γ ρ = = - 1 p h γ γ ρ = - => 2 2 2 max max 1 2 V h V V = - 0 0 / 1 / p p ρ ρ - Since    0 0 p p γ γ ρ ρ = , we have  1 1/ 2 2 0 max 1 p V p V γ - = -  =>  /( 1) 2 2 0 max 1 p V p V γ γ - = - (1) ii) Since  /( 1) 2 2 2 2 2 2 2 max max max 1 1 1 1 ...
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Solution%20to%20HW#6 - EGM6812/FluidMechanicsI Homework#6...

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