Page 503 - Fair, Geyer, and Okun's Water and wastewater engineering : water supply and wastewater removal

JWCL344_ch13_457-499.qxd 8/7/10 8:49 PM Page 461

13.2 Flow in Filled Sewers 461

Minimum grades S and capacities Q of sewers (N 0.013) up to 24 in. (600 mm) in
diameter flowing full at velocities of 2.0, 2.5, 3.0, and 5.0 ft/s (0.60, 0.75, 0.90, and 1.5 m/s)
are listed for convenience in Appendix 8.

EXAMPLE 13.1 DETERMINATION OF FLOW RATE AND VELOCITY IN SEWER USING DESIGN EQUATIONS
1. Given a 12 in. (304.8 mm) sewer, N 0.013, laid on a grade of 4.05% (4.05 ft/1,000 ft or
4.05 m/1,000 m), find its velocity of flow and rate of discharge.
2. Given a velocity of 3 ft/s (0.9144 m/s) for this sewer, find its (minimum) gradient for flow
at full depth.
Solution 1 (U.S. Customary System):

1. n = 0.013
D = 12 in. = 1 ft
s = 0.00405
v = (0.59>n) (D) 0.67 (s) 0.5
= (0.59>0.013) (1) 0.67 (0.00405) 0.5
2.89 ft/s
Q = (0.46>n) (D) 2.67 (s) 0.5
= (0.46>0.013) (1) 2.67 (0.00405) 0.5
3
2.25 ft /s
2. From Eq. 13.5a, the following equation is derived by re-arrangement:
2
2
s = 2.87 v * n * D -1.34
2
2
= 2.87 (3) (0.013) (1) -1.34
= 0.004 or 4%
Solution 2 (SI System):
1. n = 0.013
D = 0.3048 m
s = 0.00405
v = (0.40>n) (D) 0.67 (s) 0.5
= (0.40>0.013) (0.3048) 0.67 (0.00405) 0.5
= (0.40>0.013) (0.4511) (0.0636)
0.88 m/s
Q = (0.31>n) (D) 2.67 (s) 0.5
= (0.31>0.013) (0.3048) 2.67 (0.00405) 0.5
= (0.31>0.013) (0.0419) (0.0636)
3
= 0.064 m /s or 64 L/s
2. From Eq. 13.5b, the following equation is derived by re-arrangement:
2 2 -1.34
s = 6.25 v n D
2
2
= 6.25 (0.9144) (0.013) (0.3048) -1.34
= 6.25 (0.8361) (0.000169) (4.9140)
= 0.004 or 4%