Chapter 10
•
Open Channel Flow
10.1
The formula for shallow-water 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 shallow-water 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 shallow-water 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 shallow-water tidal-crest wave through the
bay? Explain.
Solution:
If it
is
a simple shallow-water 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.