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# Example 4.17

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```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.
```