# Chapter 3 : Wastewater Systems¶

## Example 3.1 Page No : 3-12¶

In :
import math

#initialisation of variables
v = 2.5	#fps
c = 0.013	#gpd
p = 300	#gpd
d = 50	#per care
m = 5.20	#ft
a = 1000	#ft

#CALCULATIONS
C = (math.pi*64)/(4*144)*v*646000	#gpd
M = m/a	#ft
P = C/p
A = P/d	#acres

#RESULTS
print 'the acres will it drain if the population density = %.1f acres'%(A)

# rounding off error

the acres will it drain if the population density = 37.6 acres


## Example 3.2 Page No : 3-14¶

In :
import math

#initialisation of variables
a = 37.4	#acres
r = 2.	#in
p = 30.	#min
v = 3	#fps
r1 = 0.6	#in
h = 1.0	#cfs
p1 = 50	#percent
q = 646000	#gpd

#CALCULATIONS
R = r*r1*a	#cfs
A = R/v	#sq ft
D = 12*math.sqrt(4*A/math.pi)	#in
P = r*r1*q/p1	#gpcd

#RESULTS
print 'the per capita storm runoff for a population density = %.0f gpcd'%(round(P,-1))

the per capita storm runoff for a population density = 15500 gpcd


## Example 3.3 Page No : 3-16¶

In :
import math

#initialisation of variables
w = 1.0	#cfs
w1 = 3.0	#cfs
w2 = 45.0	#cfs
v = 3.0	#fps
h = 144	#ft
D = 12*math.sqrt(4*w/(math.pi*w1))	#in
d1 = 1.95	#cfs
D1 = 12*math.sqrt(4*d1)/(math.pi*v)	#in
d2 = 41.6	#cfs
D2 = 12*math.sqrt(4*d2)/(math.pi*w1)	#ins

#CALCULATIONS
C = math.pi*(D)**2*3/(4*h)	#cfs
C1 = math.pi*(1./4)*3	#cfs
V = (d2*4)/(math.pi*4**2)	#fps

#RESULTS
print 'The minimum dry-weather flow  = %.2f cfs'%(C)
print 'The maximum dry-weather flow in excess actual capacity = %.1f cfs'%(C1)
print 'the storm flow in axcess of maximum dry-weather flow = %.1f fps'%(V)

The minimum dry-weather flow  = 1.00 cfs
The maximum dry-weather flow in excess actual capacity = 2.4 cfs
the storm flow in axcess of maximum dry-weather flow = 3.3 fps


## Example 3.4 Page No : 3-26¶

In :

#initialisation of variables
t = 0.8	#mgd
d = 8000.	#people
a = 2.	#hr
v = 800000.	#ft
h = 10.	#ft
a1 = 4.	#in
d1 = 1.	#sq ft per capita

#CALCULATIONS
V = v*(a/24)/a	#gal
S = V/h	#sq ft
S1 = (v/h)/S	#gpd per sq ft
V1 = a*d/h	#cu ft
D = d/S	#ft
E = d1*d/a1	#sq ft
A = t/a3	#acre

#RESULTS
print 'the capacity of the components of a small trickling-filter  = %.2f acre'%(A)

the capacity of the components of a small trickling-filter  = 0.27 acre


## Example 3.5 Page No : 3-28¶

In :

#initialisation of variables
w = 2000.	#sq miles
r = 0.1 	#cfs
d = 80000.	#ft
p = 100.	#gpd
p1 = 80.	#ft
p2 = 340.	#percent
h = 646000.	#ft

#CALCULATIONS
L = r*w	#cfs
R = 6*p1/1.4	#cfs
P = p1*(p2-L)/p2	#percent
D = (d*p)	#gpd
D1 = (L*h)	#gpd

#RESULTS
print "Low-water flow = %d cfs"%(0.1*w)
print "Required flow for disposal of domestic sewage if it i left untreated = %d cfs"%(round(6*80/1.4,-1))
print 'the percent of removal of pollutional load needed = %.0f percent'%(P)
print "Dilution ratio = %d : %d"%((d*p)/(d*p),(200*h)/(d*p))

Low-water flow = 200 cfs
Required flow for disposal of domestic sewage if it i left untreated = 340 cfs
the percent of removal of pollutional load needed = 33 percent
Dilution ratio = 1 : 16


## Example 3.6 Page No : 3-28¶

In :

#initialisation of variables
p = 10000.	#people
p1 = 4. 	#ft
w = 10. 	#in
s = 80. 	#gpcd
h = 43560.	#ft
p2 = 20. 	#ft

#CALCULATIONS
A = p*s/R	#acres
A1 = 1.7	#sq miles
P = p/500	#acres
D = p2*h*4*7.48/(p*s)	#days

#RESULTS
print 'the detention period in ponds  = %.0f days'%(D)

the detention period in ponds  = 33 days


## Example 3.7 Page No : 3-31¶

In :

#initialisation of variables
p = 100000.	#people
a = 75. 	#$a2 = 47. #in b = 10. #in #CALCULATIONS P = a*p #people S = ((a2)*(b**5))/(b)**(1./4) #$

#RESULTS
print 'the money is inversed in the sanitary sewerage system = %.0f $'%(round(S,-5))  the money is inversed in the sanitary sewerage system = 2600000$