RK4 Routine depends on Pr 3 1 B p B 2 1 1 B K y B 2 Y Y Y x p x K y x y p B y y

Rk4 routine depends on pr 3 1 b p b 2 1 1 b k y b 2 y

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RK4 Routine depends on Pr 3 1 B p=- B 2 1 1 B ' K' y - B 2 Y Y Y x p x K y x y p B y y K y K y = φ = φ η = φ = − ρ + ρ ∂ + = − = − = ρ + ρ ∂ ρ ( ) ( ) ( ) 2 2 2 0 2 2 2 0 0 2 2 0 x=0 , y=0 p=p stagnation pressure 2 c=p 2 Bernoulli Eq. 2 p B x y C p p B x y p p u v ρ = − + + ρ = + ρ = +
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N N N ( ) ( ) ( ) ( ) ( ) 2 1/s m m/s 2 2 0 1 u = Bx v=-By =Bxy s y=0 u 0 slips on the wall =xf y u=xf ' y v=-f y 2 . 0 m B p p x F y u v C E x y ψ ψ = −ρ + + = ( ) ( ) ( ) ( ) ( ) 2 2 2 2 f ' y f ' y 0 1 ' " 0 "' ' " "' ODE xf f xf B x xf f ff B f = + − = − −ρ + ν + ρ = + ν x y ( ) ( ) ( ) ( )( ) ( ) 2 2 2 2 2 ' " 0 "' ' " "' 1 ' .0 ' ' 0 " 2 1 ff '= ' " 2 xf f xf xf f ff B f B xf y f f F f B F f + − = +ν + = + ν + − = − −ρ + ν ρ − ν
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( ) ( ) ( ) ( ) [ ] 0 y f ' y=0 u=0, f ' 0 0 v=0, f 0 0 0 p=p F 0 0 0 length scale : B u Bx B x y proper m = → ∞ ∞ = = = = = = ν [ ] ( ) ( ) 2 2 2 2 y = B scale B : / B B ' " =B + "' ' - "=1+ "' velocity m s x f B B B ψ 2 η ν ν = φ η ν = νφ η Β φ νφΒφ ν φ ν ν φ φφ φ ( ) ( ) ( ) 2 2 3 3 ' / ' " " "' B df B B dy y f d B B B y dy f d B B B B B y dy νφ η ∂η = = νφ ν ∂η = φ = φ ν = φ = φ ν ν ν
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( ) ( ) ( ) ( ) ( ) ( ) N ( ) ( ) 0 u=0=xB ' ' 0 0 v=0=- B 0 0 u=xB ' ' 1 Fig s.11 Schlichting u=xf ' y ' v=-f y i U xB xB B η = φ φ = νφ φ = η → ∞ φ η → ∞ = φ η → ∞ = = φ η = − νφ ( ) η inviscid profile
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Unsteady Motions of a Plate : 0 y x u=u(y,t) v=0 Parallel flow, v=w=0 • initially both the plate & fluid are at rest u(y,0)=0 for y>0 • the plate is jerked into motion in its own plane • no-slip at the plate : u(y=0,t)=U(t) for t>0 Two-cases 1- U=constant (Stoke’s First Problem) 2- U(t)=U 0 cos ω t (Stoke’s Second Problem)
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Steady oscillation of the plane at Ucos ω t. zero pressure gradient governing P.D.E. reduces 0 p=const. dp dx = ( ) 2 2 1-D heat conduction u u t y = ν CASE I: U=const. ( ) ( ) 0 0 0, 0 , for t u y t U for t u y t finite = = = > = Unsteady heat conduction equation Carslaw and Jaeger (1959) – Conduction of heat in solids Solution methods -Laplace transforms -Similarity methods 2 2 2 2 u u u t x y = ν + 1 2 2 u y y erf erfc U t t = = ν ν
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( ) 2 0 2 x erf e dx β β π = Similarity Solutions : • applicable for non-linear problems similarity solutions exist for: parabolic P.D.E with two independent variables when there is no geometric length scale in the problem. give ODE y U u t=0 t 1 t 2 increasing time erf x erfc x 1 1 2 x
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1 1 2 2 Eg: At t=t u=0.5U at y=y At t=t u=0.5U at y=y Expect solution exists in the form ( ) ( ) ( ) ( ) ( ) ( ) n n , y where = t , : similarity variable : constant of proportionality: will be found later to make dimensionless. u , =const. u=const. y t u y t f U y t y t Uf when α α = η η η η = η η IF similarity solution exists, it will result in an ODE with f as the dependent variable and η as the independent variable.
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PDE ( ) ( ) ( ) 2 2 n n+1 n n 2 2 n 1 , Let u=Uf y = t y t ' ' t t ' '' t u u t y u u df U n t t d u u Uf U f y y u u u U f y y y y y Un f t α α α α α α = ν η η ∂η = = = ∂η ∂ η ∂η = = = ∂η ∂ ∂η = = =
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