Example Problem 4.17 y y3 y 4 u≈U (2 −2 3 + 4 ) δ δ δ δ =C x 1 ( ) 2 for y≤δ a) Find an expression for the velocity component in the y direction v(x,y). ● ● We have an expression for u, we should be able to relate that to v through mass conservation Our problem is 2-D, incompressible and steady ∂ρ ∂ + ( ρ u)+ ∂ ( ρ v)+ ∂ ( ρ w)=0 ∂t ∂ x ∂y ∂z ∂ρ ∂ + ( ρ u)+ ∂ ( ρ v)=0 ∂t ∂ x ∂y ∂ ( ρ u)+ ∂ ( ρ v )=0 ∂x ∂y ∂u ∂v ρ +ρ =0 ∂x ∂y ∂v ∂u =−( ) ∂y ∂x 3 ∂v ∂u =−( ) ∂y ∂x 4 y y y u≈U (2 −2 3 + 4 ) δ δ δ ∂v y y3 y 4 ∂ =−( (U ( 2 δ −2 3 + 4 ))) ∂y ∂x δ δ 3 4 ∂v y y y =−U ∂ (2 δ −2 3 + 4 ) ∂y ∂x δ δ 3 4 ∂v y ∂δ y ∂δ ∂ y ∂δ ∂ ∂ =−U (2 ) − (2 3 ) + ( 4) ∂ δ δ ∂ δ ∂ δ ∂y ∂x δ ∂x δ ∂x ∂v y y3 y4 ∂ δ ∂ ∂ ∂ =−U (2 )− ∂ δ (2 3 )+ ∂ δ ( 4 ) ∂y ∂ x ∂δ δ δ δ 3 4 ∂v 2y 6y 4y =−U ∂ δ −( 2 )+( 4 )−( 5 ) ∂y ∂x δ δ δ [ [ [ ] ] Sub in ] Chain rule [ [ 3 4 ∂v 2 y 6 y 4 y =−U ∂ δ −( 2 )+( 4 )−( 5 ) ∂y ∂x δ δ δ 3 4 ∂v y 3 y 2 y =2 U ∂ δ −( 2 )+( 4 )−( 5 ) ∂y ∂x δ δ δ ∂δ = C = δ ∂ x (2 x 0.5 ) 2 x [ ∫[ ∂v y 3 y3 2 y4 δ =U ( 2 )−( 4 )+( 5 ) ∂y x δ δ δ v =U δ x a) [ ] ] y 3 y3 2 y4 ( 2 )−( 4 )+( 5 ) dy δ δ δ ] 2 4 5 y 3y 2y δ v =U ( 2 )−( 4 )+( 5 ) x 2δ 4δ 5δ ] ] 1 ( ) 2 and δ =C x C= δ(0.5) x b) Find the maximum v when x=1m for a flow where: U =3 m/ s δ ( x=1 m)=1.1 cm [ ] 2 4 5 y 3y 2y δ v =U ( 2 )−( 4 )+( 5 ) x 2δ 4δ 5δ The value of v is increasing with increasing y up to a maximum of v max =v ( y= δ ) [ ] 1 3 2 v max =3( 0.011) ( )−( )+( ) =0.00495 m/ s 2 4 5 Solution is really quite pretty when done this way. However, there was a number of places we could have gone an “uglier” route and still arrived at the same answer.