Environmental Engineering FOURTH EDITION

Collection of Wastewater 163

Figure! 8-9. Sewer layout for Example 8.3.

EXAMPLE 8.3. The system shown in Fig. 8-9 is to be designed given the following

flows: maximum flow = 3.2mgd, minimum flow = 0.2mgd, minimum allowable

velocity = 2 ft /s, and maximum allowable velocity = 12 ft /s.

All manholes should be about 10 ft deep, and there is no additional flow between

Manhole 1 and Manhole 4. Design acceptable invert elevations for this system. (The

nomograph of Fig. 8-6 is used in this solution.)

From Manhole 1 to Manhole 2:

1. The street slopes at Y100, so choose the slope of the sewer s = 0.02.


2. Assume TI = 0.013 and try D = 12 in. From the nomograph of Fig. 8-6, connect


3. Connect the point on the turning line with D = 12 in.


4. Read u = 6.1 ft /s for Q = 3.2 mgd from intersection of the line drawn in Step 3.


5. To check for minimum velocity,

n = 0.013 and s = 0.02, and extend that straight line to the turning line.

This is acceptable for the sewer flowing full.

q/Q = 0.2/3.2 = 0.063,

and from the hydraulic elements chart, Fig. 8-8, q/Q = 0.063 intersects the discharge

curve at d/D = 0.2, which intersects the velocity curve at up/v = 0.48, and

up = 0.48(6.1 ft/~) = 2.9ft/~.

6. The downstream invert elevation of Manhole^1 is ground elevation minus^10 ft,
   or 62.0 ft. The upstream invert elevation of Manhole 2 is thus 62.0 - 2.0 = 60.0ft.
   Allowing 0.1 ft for head loss in the manhole, the downstream invert elevation is
   59.9 ft.
   From Manhole 2 to Manhole 3, the slope will be a problem because of rock. Try
   a larger pipe, D = 18 in. Repeating steps 1 and 2,


7. Connect the point on the turning line with D = 18 in.