THE BOUNDARY LAYER 295
Du^2 sυ
{[
2
cos
2
y
υ
]l
0
1
2
∫ 1
0
(
1 cos
y
υ
)
d
(y
υ
)
}
Du^2 sυ
{
2
1
2
C
[
1
sin
y
υ
]l
0
}
Du^2 sυ
{
2
1
2
}
D 0. 1366 u^2 sυ.
Differentiating:
dux
dy
DusÐ
(
2 υ
cos
2
y
υ
)
(
dux
dy
)
yD 0
D
(
2
)
Ð
(u
s
υ
)
and: R 0 D
(
dux
dy
)
yD 0
D
2
us
υ
.
Thus:
∂
∂x
0. 1366 u^2 sυD
2
us
υ
∴ υdυD
(
us
)(
0. 2732
)
dx.
∴
υ^2
2
D
(
(^) x
us
)(
0. 2732
)
∴
υ^2
x^2
D
(
usx
)(
0. 1366
)
and:
υ
x
D 4. 796 Rex^1 /^2
PROBLEM 11.12
Derive the momentum equation for the flow of a viscous fluid over a small plane surface.
Show that the velocity profile in the neighbourhood of the surface may be expressed as a
sine function which satisfies the boundary conditions at the surface and at the outer edge
of the boundary layer.
Obtain the boundary layer thickness and its displacement thickness as a function of the
distance from the leading edge of the surface, when the velocity profile is expressed as a
sine function.
Solution
The momentum flux across 1 – 2 is:
∫L
0 u
2
xdy
The change from 1–2 to 3–4 is:
∂
∂x
{∫
L
0 u
2
xdy
}
dx