CHEMICAL ENGINEERING

294 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

PROBLEM 11.11

Derive the momentum equation for the flow of a fluid over a plane surface for conditions
where the pressure gradient along the surface is negligible. By assuming a sine function
for the variation of velocity with distance from the surface (within the boundary layer)
for streamline flow, obtain an expression for the boundary layer thickness as a function
of distance from the leading edge of the surface.

Solution

The total mass flowrate through plane 1 – 2 is:

∫l

0 uxdy.

The total momentum flux through plane 1 – 2 is:

∫l

0 u

2

xdy

Change in mass flowrate from 1–2 to 3–4 is:

∂

∂x

{∫

l

0 uxdy

}

dx

Change in momentum flux from 1 – 2 to 3 – 4 is:

∂

∂x

{∫

l

0 u

2

xdy

}

dx.

Change in momentum flux is attributable, in the absence of pressure gradient, to:

(a) Momentum of fluid entering through 2 – 4

Since all this fluid has velocityus, the momentum flux is:

{

∂

∂x

[∫l

0

uxdy

]

dx

}

us

(b) Force due to shear stress at surfaceDR 0 dx.

Thus a momentum balance gives:

∂

∂x

{∫l

0

u^2 xdy

}

dxD

∂

∂x

{∫l

0

uxdy

}

dx.usCR 0 dx.

∴ 

∂

∂x

{∫l

0

ux

usux dy

}

DR 0

Representing velocity within boundary layer by a sine function, then:

ux

us

Dsin

(

2

y

υ

)

Thus:
∫l

0

ux

usux dyDu^2 sυ

∫ 1

0

ux

us

(

1 

ux

us

)

d

(y

υ

)

Du^2 sυ

∫ 1

0

(

sin



2

Ð

y

υ

sin^2



2

y

υ

)

d

y

υ