164 ENVIRONMENTAL ENGINEERING- From the nomograph, u = 2.75 ft/s for Q = 3.2 mgd, up/u = 0.48 and
up = 0.48(2.75 ft/s) = 1.32 ft/s.
Even with a slope of 0.002 (from the nomograph), the resulting velocity at minimum
flow is too low. Try s = 0.005 with D = 18 in. for Steps 1-3. Then Q = 4.7mgd
and u = 4.1 ft/s for Step 4. From the hydraulic elements chart, vp = 1.9 ft/s, which is
close enough. Thus the upstream invert elevation of Manhole 3 is
59.9 - (0.005)(200) = 58.9 ft,
and the downstream is at 58.8 ft, still well above the rock. From Manhole 3 to
Manhole 4, the street obviously has too much slope. Tly using D = 12 in. ands = (58.8 - 40)/100 = 0.188,
since 40 ft is the desired invert elevation of Manhole 4. Read Q = 8.5 mgd and u =
19 ft/s, but only 3.2 mgd is required at maximum flow; henceq/Q = 3.2/8.5 = 0.38, and
up = (0.78)(18ft/~) = 14.8ft/s.This is too high. Use a drop manhole, with the invert at, say, elevation 45.0. In this
case s = 0.138 and Q = 7.3 mgd, u = 14.5 s,q/Q = 3.217.3 = 0.44, and
up = (0.85)(14.5 ft/s) = 12.3 ft/s,which is close enough. The upstream invert elevation of Manhole 4 is thus at 45.0 ft
and the downstream invert elevation can be at, say, 40.0 ft. The minimum velocity for
this last step need not be checked. The final design is shown in Fig. 8-10.It is often convenient to organize sewer calculations according to a table such as
Fig. 8-1 1, a slightly simplified version of a table suggested by the American Society for
Civil Engineers (American Society for Civil Engineers 1969). This reference should
be consulted for details of accepted engineering practice in sewer design.CONCLUSIONThe methods discussed in this chapter are hand calculations, and few modem consulting
engineers who design sewerage systems will use such cumbersome methods. Computer
programs that not only solve for the hydraulic parameters, but evaluate the economic
alternatives (e.g., a deeper cut or a bigger pipe) are generally used. Elegant programs
are only useful, however, if the design process is well understood.