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478 Chapter 13 Hydraulics of Sewer Systems

v distance traveled
w
in unit time

v 2 d 2
v 1
d 1 Figure 13.8 Profile of Surge
Front.

As shown in Fig. 13.8, the momentum principle can be adduced to identify also the
propagation of discontinuous waves in open-channel flow. Waves of this kind may rush
through conduits when a sudden discharge of water from a localized thunderstorm or the
quick release of a large volume of industrial wastewater, for example, enters a drainage
system. In cases such as these, the volume of water undergoing a change in momentum in
unit time and unit channel width is (v w v 1 )d 1 . The celerity of propagation, which is the
wave velocity or speed or propagation of the surge front, relative to the fluid velocity
[equating force to momentum change 1>2gd 2 2 1>2 gd 1 2 (v v )(v v )d in a
2
w
1
1
1
channel of unit width. Continuity of flow, moreover, requires that v d v (d d ) v d ]
1 1
1
2 2
2
w
being c (v v ), it follows that
w
1
2
(c>1gd ) = 1>2(d >d )31 + (d >d )4 (U.S. Customary or SI Units) (13.28)
1
1
1
2
2
where c rate of propagation of a discontinuous surge front, ft/s or m/s; g gravity constant,
2
2
32.2 ft/s or 9.806 m/s ; d flow depth before surge, ft or m; d flow depth raised by the
2
1
surge, ft or m.
EXAMPLE 13.12 HYDRAULIC JUMP AND DISCONTINUOUS SURGE FRONT
Find the rate of propagation of a discontinuous surge front that raises the flow depth from 1 to 2 ft
(0.3048 to 0.6096 m).
Solution 1 (U.S. Customary System):
From Eq. 13.28,
1
2
(c>1gd 1 ) = (d 2 >d 1 )31 + (d 2 >d 1 )4
2
Given, d 1 1 ft (0.3 m) and d 2 2 ft (0.6 m),
2
c gd 1 [1>2 d 2 >d 1 (1 d 2 >d 1 )]
1>2
c = 1g3(1>2) * (2>1)(1 + 2>1)4
c = 13g = 9.8 ft>s (3 m>s)
Solution 2 (SI System):
2
c = gd 1 [0.5 * d 2 >d 1 (1 + d 2 >d 1 )] = 0.5 gd 2 (1 + d 2 >d 1 )
c = [0.5 * 9.806 * 0.6096 (1 + 0.6096>0.3048)] 0.5
3 m/s