Mecanica De Fluidos

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, c o = (gy), is valid for anyfluid 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 cmhigh 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 δ y terms in Eq. (10.9):
c = gy(1 + δ y/ y)(1 + δ y/2 y)
= 9.81(1.1)(1 + 0.12/1.1)[1 + 0.12/{2(1.1)}] = 3.55 m/s

δV =

(3.55 m/s)(0.12 m)
cδ y
=
= 0.35 m/s
y +δy
1.1 + 0.12 m

Ans. (a)

Ans. (b)

10.3 Narragansett Bayis 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 wavephenomenon, the time delay would be
∆t =

∆L (21 mi)(5280 ft/mi)
=
≈ 3015 s ≈ 50 min
co
32.2(42)

Ans.???

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 water-channel flow in Fig. P10.4has a free surface in three places. Does it
qualify as an open-channel flow? Explain.
What does the dashed line represent?
Solution: No, this is not an open-channel
Fig. P10.4
flow. The open tubes are merely piezometer
or pressure-measuring 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 Waterflows rapidly in a channel
25 cm deep. Piercing the surface with a
pencil point creates a wedge-like wave of
included angle 38°, as shown. Estimate the
velocity V of the water flow.
Fig. P10.5
Solution: This is a “supercritical” flow,
analogous to supersonic gas flow. The Froude number is analogous to Mach number here:

gy
(9.81 m/s2 )(0.25 m)
1
c
sin θ = sin(19°) =
==
=
,
Fr VV
V
m
Solve V = 4.81
Ans. (Fr = 3.1)
s
10.6 Pebbles dropped successively at the
same point, into a water-channel 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.
Fig. P10.6

Solution: The center of each circle moves at stream velocity V. For the small circle,
4V
9V
small circle:X =
; large circle: X + 4 + 6 + 9 =
;
co
co
V
= 3.8 Ans. (a)
co
m
m
Compute c o = gh = 9.81(0.42) = 2.03 , Vcurrent = 3.8c o ≈ 7.7
s
s
Solve Fr =

Ans. (b)

Chapter 10 • Open Channel Flow

735

10.7 Pebbles dropped successively at the
same point, into a water-channel flow of
depth 65 cm, create two circular ripples,
as in Fig. P10.7. From this information,
estimate (a)the Froude number; and (b) the
stream velocity.
Fig. P10.7

Solution: If the pebble-drop-site is at distance X ahead of the small-circle center,

small circle: X =

3V
9V
V2
=
; large circle: X + 4 =
; solve Fr =
co
co
co 3

co = gh = 9.81(0.65) = 2.53

m
2
m
; Vcurrent = co ≈ 1.68
s
3
s

Ans. (a)

Ans. (b)

10.8 An earthquake near the Kenai Peninsula, Alaska,...