Page 510 - 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 468

468 Chapter 13 Hydraulics of Sewer Systems

EXAMPLE 13.5 DETERMINATION OF DIAMETER OF PARTIALLY FILLED SEWERS USING
DESIGN EQUATIONS
3
An average sewer pipe with n = 0.015 is laid on a slope of 0.0002 and is to carry 83.6 ft /s
3
(2.3675 m /s) of wastewater when the sewer pipe flows 0.9 full (d/D 0.9). Determine the di-
ameter of the sewer pipe.
Solution 1 (U.S. Customary System):
r = A>P w = [Circle - (Sector AOCE - Triangle AOCD)]>Arc ABC
-1
-1
Angle u = cos (0.4D>0.5D) = cos 0.800
u = 36°52¿
2
Area of Sector AOCE = [2(36°52¿)>360°](pD >4) = 0.1612D 2
Length of Arc ABC = pD - [2(36°52¿)>360°](pD) = 2.498D
Area of Triangle AOCD = 2(0.5 * 0.4D * 0.4D tan 36°52¿) = 0.12 D 2
2
2
2
r = A>P w = [0.25pD - (0.1612D - 0.12D )]>(2.498D)
2
= (0.7442D )>(2.498D)
= 0.298D
2
A = 0.7442D and P w = 2.498D, then r = 0.298D
s
Q = Av = A[(1.49>n) r 2>3 1>2 ]
2
83.6 = (0.7442 D )[(1.49>0.015)(0.298D) 2>3 (0.0002) 1>2 ]
D 8>3 = 180, D 7 ft
E
D C
A

O 0.90 D

B

Solution 2 (SI System):
The relationships of d versus r, A and P w have been established in Solution 1.
A = 0.7442 D 2
P w = 2.498 D
2
r = A>P w = (0.7442 D )>(2.498 D) = 0.298 D
s
Q = Av = A[(1>n) r 2>3 1>2 ]
2
2.3675 = (0.7442 D )[(1>0.015)(0.298 D) 2>3 (0.0002) 1>2 ]
D = 2.13 m = 2,130 mm

EXAMPLE 13.6 DETERMINATION OF DIAMETER OF PARTIALLY FILLED SEWERS USING
DESIGN DIAGRAM
Determine the diameter of the sewer pipe described in Example 13.5 assuming n 0.015, s
3
3
0.002, Q 83.6 ft /s (2.3675 m /s), and d/D 0.9.