Chapter 10
•
Open Channel Flow
10.1
The formula for shallowwater wave propagation speed, Eq. (10.9) or (10.10), is
independent of the physical properties of the liquid, i.e., density, viscosity, or surface
tension. Does this mean that waves propagate at the same speed in water, mercury,
gasoline, and glycerin? Explain.
Solution:
The shallowwater wave formula,
o
c(
g
y
)
,
=
is valid for any fluid except
for
viscosity and surface tension effects
. If the wave is very small, or “capillary” in size,
its propagation may be influenced by surface tension and Weber number [Ref. 5
−
7]. If
the fluid is very viscous, its speed may be influenced by Reynolds number. The formula
is accurate for water, mercury, and gasoline but would be inaccurate for
glycerin
.
10.2
A shallowwater wave 12 cm high propagates into still water of depth 1.1 m.
Compute (a) the wave speed; and (b) the induced velocity
δ
V
.
Solution:
The wave is high enough to include the
y
terms in Eq. (10.9):
(1
/ )(1
/2 )
9.81(1.1)(1 0.12/1.1)[1 0.12/{2(1.1)}]
/
. (a)
cg
y
y
y
y
y
Ans
δδ
=+
+
+
=
3.55 m s
==
=
++
(3.55 m/s)(0.12 m)
. (b)
1.1 0.12 m
cy
VA
n
s
yy
0.35 m/s
10.3
Narragansett Bay is approximately 21 (statute) mi long and has an average depth
of 42 ft. Tidal charts for the area indicate a time delay of 30 min between high tide at the
mouth of the bay (Newport, Rhode Island) and its head (Providence, Rhode Island). Is
this delay correlated with the propagation of a shallowwater tidalcrest wave through the
bay? Explain.
Solution:
If it
is
a simple shallowwater wave phenomenon, the time delay would be
o
L
(21 mi)(5280 ft/mi)
t
3015 s
???
c
32.2(42)
Ans.
∆
∆=
=
≈
≈
50 min
This doesn’t agree with the measured
∆
t
≈
30 min. In reality, tidal propagation in estuaries
is a
dynamic
process, dependent on estuary shape, bottom friction, and tidal period.
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Solutions Manual
•
Fluid Mechanics, Fifth Edition
10.4
The waterchannel flow in Fig. P10.4
has a free surface in three places. Does it
qualify as an openchannel flow? Explain.
What does the dashed line represent?
Solution:
No
, this is
not
an openchannel
flow. The open tubes are merely piezometer
Fig. P10.4
or pressuremeasuring devices, there is no flow in them. The dashed line represents the
pressure distribution in the tube, or the “hydraulic grade line” (HGL).
10.5
Water flows rapidly in a channel
25 cm deep. Piercing the surface with a
pencil point creates a wedgelike wave of
included angle 38
°
, as shown. Estimate the
velocity
V
of the water flow.
Solution:
This is a “supercritical” flow,
Fig. P10.5
analogous to supersonic gas flow. The Froude number is analogous to Mach number here:
θ
=°
=
=
=
=
==
2
(9.81 m/s )(0.25 m)
1
sin
sin(19 )
,
Solve
. (Fr
3.1)
gy
c
Fr
V
V
V
VA
n
s
m
4.81
s
10.6
Pebbles dropped successively at the
same point, into a waterchannel flow of
depth 42 cm, create two circular ripples,
as in Fig. P10.6. From this information,
estimate (a) the Froude number; and (b) the
stream velocity.
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 Spring '09
 Lobban
 Fluid Dynamics, Orders of magnitude, Fig, open channel flow

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