Q = 0.265 mgd dc/D = 0.26 < 0.30

B-3. Design 1000 feet of lateral sewer for a small tributary area on the above installation with the

following flows.

Offices

2 buildings with 100 employees working 8-hour shifts (30 gpcd is a typical allowance)

Theater

300 seats--open 10 hours per day (3 gpd/seat typical allowance)

Shop

30 employees working 8-hour shifts (30 gpcd is a typical allowance)

No industrial wastes

Domestic Flows

Offices, Theater and Shop

100 x 30

= 3000 gpd over 8 hours

300 x 3

= 900 gpd over 10 hours

30 x 30

= 900 gpd over 8 hours

total average daily flow

= 4800 gpd

Use an 8-hour basis since tributary area is small and all occupants are short term.

Average hourly flowrate

= 4800/8 = 600 gph

Extreme peak flowrate

= 22 5/(600)0.167 = 7.73

R

7.73 x 600

= 4637 gph

Peak diurnal flowrate

= 1/2 x 4637 = 2319 gph

Infiltration Allowance

Assume an 8-inch sewer (minimum size).

1000 x 8 x (1000/5280)

= 1515 gpd = 63 gph

Design Flows

Extreme peak

= 4637 + 63 = 4700 gph

Diurnal peak

= 2319 + 63 = 2382 gph

Average hourly

= 600 + 63 = 663 gph

Typical Sewer Design

Try an 8-inch sewer on a 3.0 percent slope (n = 0.013).

Flow depths and velocities

Q = 4700 gph

d/D = 0.20

V = 3.5 fps

Q = 2382 gph

d/D = 0.13

V = 2.7 fps > 2.5 fps

Q = 663 gph

d/D = 0.05

V = 2.0 fps = minimum

Critical depths

Q = 4700 gph

dc/D = 0.29 > 0.20*

Q = 2382 gph

dc/D = 0.20 > 0.13*

* Note that supercritical flow will result here. The critical slope equals 0.65 percent.

However, a slope of 3.0 percent is required to produce the minimum velocity of 2.0 fps.

Considering that the sewer is of minimum size, and that a flatter slope is not feasible if adequate

velocity is to be provided, supercritical flow would be justified in this case.

B-3

Integrated Publishing, Inc. |