CHEMICAL ENGINEERING

296 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

The mass flux across 1 – 2 is:

∫L

0 uxdy.

The change from 1–2 to 3–4 is:

∂

∂x

{∫

L

0 uxdy

}

dx.

The rate of momentum entering through 2 – 4 is:



∂

∂x

{∫L

0

uxusdy

}

dx,

Assumingus 6 Df
x, then:
∂P
∂x

D 0.

A momentum balance gives:



∂

∂x

{∫L

0

u^2 xdy

}

dxD

∂

∂x

{∫L

0

uxusdy

}

dxCR 0 dx

or: 

∂

∂x

{∫L

0

ux

usux dy

}

DR 0 D

(

∂ux

∂y

)

yD 0

for a Newtonian fluid

The sine function is:

uxDKsinky,so

∂ux

∂y

DKkcoskyand

∂^2 ux

∂y^2

Kk^2 sinky

is satisfied for all finite values ofKandk.
The boundary conditions are:
yD 0 uxD 0

yD 0

∂^2 ux

∂y^2

D 0

yDυuxDus KDus

yDυ

∂ux

∂y

D 0

Thus: kD



2

andkD



2 υ

.

Hence: uxDussin



2

y

υ

,

∂ux

∂y

D

us

2 υ

cos



2

y

υ

and:

{∫L

y

}

D 0

Thus: u^2 s

∂

∂x

{∫υ

0

sin



2

y

υ

(

1 sin



2

y

υ

)

dyD

    us

2 υ

∂

∂x

υ

{∫ 1

0

[

sin



2

y

υ

sin^2



2

y

υ

]

d

y

υ

}

D



2 υus