Question

A new development is being planned, which will connect to the city ’s sanitary sewer system. This development will have 80 acres of medium-density residential assume 10 DU / acre and 2.8 persons per DU). Assume a per capita sewage rate of 90 gal/d/person. Additionally, a new community hospital will be constructed with 300 beds (assume 300 gal/d/bed). Assume I/I contribution of 10%.

a. Calculate the mean sewer discharge (accounting for residential, commercial, and I/I flows).

b. Calculate the maximum sewer discharge. Assume a peaking factor of 3.0

c. The new sanitary sewer pipe is to be laid at a slope of 0.003 ft/ft and built with pre -cast concrete pipe (n = 0.013). C ity code requires that the pipe can transmit the peak flo w at 75% full, with a velocity between 1.5 and 10 ft/s. What diameter pipe would you use?

d. How full, in inches, will the pipe be under normal conditions (mean discharge)?

Answer 1

Ans a) Given,

Area = 80 acre

10 DU/acre

=> Number of DU in 80 acres = 10 x 80 = 800 DU

Number of people in DU = 2.8 x 800 = 2240

Per capita sewage rate = 90 gal/d/person

=> Sewage generated in residential flow = 2240 x 90 gal/day = 201600 gal/day or 0.312 ft³/s

Now, number of beds = 300

Sewage generated per bed = 300 gal/day/bed

=> Sewage generated in commercial flow = 300 x 300 galday = 90000 gal/day or 0.14 ft³/s

Total sewage generated in day = 201600 + 90000 = 291600 gal/day

I/I flow = 10 % = 0.1 x 291600= 29160 gal/day or 0.045 ft³/s

Hence, mean sewer discharge = 201600 + 90000 + 29160 = 320760 gal/day or 0.50 ft³/s

Ans b) Maximum sewer discharge = peak factor x mean sewer discharge

=> Maximum sewer discharge = 3 x 0.5 = 1.5 cfs

Ans c) Now, according to Manning equation,

Q = (1.49 A/ N) R^2/3 S^1/2

where, Q = discharge = 1.5 cfs

A = Cross-sectional Area = (/4) D²

N = Manning Roughness coefficient = 0.013

R = Hydraulic depth = Area/ Wetted perimeter

S = Bottom slope = 0.003

  For circular pipe, R = Area /Wetted perimeter = (/4) D² /D = D /4 or 0.25 D

=> 1.5 = [(/4)D² / 0.013] (0.25 D)^2/3 (0.003)^1/2

=> 1.5 = 1.31 D^8/3

=> D = 1.05 ft 12 in  

=> Full flow velocity,V = Q/A = 1.5 / (/4)(1.05)² = 1.73 ft/s

Now, since flow is running 75% full, d /D = 0.75

=> Also, for partially flowing sewer , d/D = 0.5 ( 1 - cos/2) , where is angle subtended by water surface from pipe center

=> 0.75 = 0.5 (1 - cos/2)

=> = 240^o

Proportionate velocity, v / V = [ 1 - 360 Sin / 2]

=> v /V = [ 1 - 360 Sin (240) / 2(240)]

=> v /V = 1.20

=> v = 1.2 V = 1.2 x 1.73 = 2 ft /s

Hence, 12 in pipe 75 % full with velocity 2 ft/s is suitable

Ans d) Now, mean discharge , q = 0.5 cfs

Peak discharge , Q = 1.5 cfs

We know, according to proportionate discharge table for q/Q = 0.5 / 1.5 = 0.333 , d/D = 0.4

=> d = 0.4 D = 0.4 x 12 = 4.8 in

% full = (4.8/12) x 100 = 40 %

Hence, under normal conditions , pipe is flowing 40 % full