constant) and is initially at rest. At
t
= 0
+
, the fluid is impulsively driven by
a constant shear stress
τ
h
exerted on the horizon (
y
= 0).
(a) Assume that a solution independent of horizontal location exists. Derive
the governing equation for velocity field and state the initial and boundary
conditions.
(b) Solve the partial differential equation in part (a).
(c) The rate of work done on the fluid per unit width is
˙
W
(
t
) =
τ
h
u
(0
, t
)
.
Show that the work done on the fluid per unit width up to time
t
is
W
(
t
) =
ˆ
t
0
˙
W
(
t
0
)
dt
0
=
4
τ
2
h
3
μ
r
νt
3
π
.
(d) The rate of energy dissipation of the fluid per unit width is
˙
D
(
t
) =
ˆ
0
∞
μ
∂u
∂y
2
dy .
Show that the energy per unit width that has been dissipated as heat up
to time
t
is
D
(
t
) =
ˆ
t
0
˙
D
(
t
0
)
dt
0
=
4
τ
2
h
3
μ
r
νt
3
π
2

√
2
.
(e) Show that the kinetic energy per unit width of the flow at time
t
is
K
(
t
) =
ˆ
0
∞
ρ
2
u
2
dy
=
4
τ
2
h
3
μ
r
νt
3
π
√
2

1
.
3
Hint:
(1) erfc(x) is an odd function about (
x, y
) = (0
,
1), i.e., erfc(

x
)

1 =

[erfc(
x
)

1]
(2) erfc(
∞
) = 2, erfc(0) = 1, erfc(
∞
) = 0
(3)
´
erfc(
x
)
dx
=
x
erfc(
x
)

e

x
2
√
π
+
C
(4)
ˆ
∞
0
[erfc(x)]
2
dx
=
2

√
2
√
π
(5) Show that
dK
dt
=
˙
W

˙
D
mathematically.
4
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 Fall '13
 Fluid Mechanics, Work, Surface tension