Chapter 5 : Water and Wastewater Volumes

Example 5.1 Page No : 5-6

In [1]:
	
#initialisation of variables
t1 = 5.25	#yr
t2 = 10.00	#yr
yi = 171000	#in
ye = 111000	#in
yt = 5.23300	#in
yl = 5.04532	#in
yn = 31500	#in
ym = 0.09853	#in
tm = 9.25	#yr
tn = 10.00	#yr
	
#CALCULATIONS
T = t1/t2	#yr
T1 = tm/tn	#yr
Y = yi-ye	#in
Yt = yt-yl	#in
	
#RESULTS
print 'the fifth intercensal year  = %.3f yr'%(T)
print 'the ninth postcensal year  = %.3f yr'%(T1)

the fifth intercensal year  = 0.525 yr
the ninth postcensal year  = 0.925 yr

Example 5.2 Page No : 5-8

In [3]:
from sympy import Symbol
import math

#initialisation of variables
y0 = 30000	#in
y1 = 172000	#in
y2 = 292000	#in
a = 172	     #ft
p = 30.	    #ft
y = 292	    #ft
q = 322000	#ft
g = 313	    #ft
n = 0.05	#ft
d = -2.442	#ft
t = Symbol('t')

#CALCULATIONS
L = (2*p*a*y2-(a)**2*q)/(p*y-(a)**2)	#moreover
m = (g-p)/p	#ft
N = n*d	#in
Y = g/(1+9.43*math.exp(N)*t)	#in
	
#RESULTS
print 'the saturation populations = %.2f moreover'%(round(L,-3))
print 'the coefficients = %.3f in'%(N)
print 'the equation of a logistic curve = ',(Y)

the saturation populations = 313000.00 moreover
the coefficients = -0.122 in
the equation of a logistic curve =  313/(8.34611446180163*t + 1)

Example 5.4 Page No : 5-19

In [8]:
import math 
	
#initialisation of variables
p = 100000	#in
d = 150.	#in
h = 1000000.	#in
a1 = 2.0	#draft
a2 = 3.0	#draft
a3 = 1.6	#draft
m = 1.5	    #in
q = 2.5	    #in
v = 1020.	#in
w = 100. 	#in
t = 0.01	#in
v1 = 13.2	#mgd
	
#CALCULATIONS
A = d*p/h	#mgd
M = m*A	#mgd
M1 = q*A	#mgd
V = v*math.sqrt(w)*(1-t*math.sqrt(w))	#gpm
D = M+v1	#mgd
L = a1*A	#mgd
L1 = (4./3)*M	#max
H = a2*A	#mgd
H1 = (4./3)*M1	#max
F = a3*A	#mgd
F1 = (1.6)*15	#max
	
#RESULTS
print 'the resulting capacities of the four system  = %.f max'%(F1)

the resulting capacities of the four system  = 24 max

Example 5.6 Page No : 5-22

In [2]:
	
#initialisation of variables
r = 48	#in
A = 450	#gpd/acre
B = 8000.	#gpd/mile
S = round(5280./350)	#manholes/mile

#CALCULATIONS
C = (B-S*100)/12	#gpd/mile

#RESULTS
print 'the ground a quarter of it eventually  = %.2f gpd/mile'%(C)

# note : answer in book is wrong. please check.

the ground a quarter of it eventually  = 541.67 gpd/mile

Example 5.7 Page No : 5-23

In [15]:
	
#initialisation of variables
p1 = 20.	#ft
p2 = 30.	#ft
w = 5.  	#person
s = 17800.	#in
h = 1200.	#in
q = 100. 	#in
i = 1.	#in
	
#CALCULATIONS
S = p1*p2*i*s/(h*w)	#gpcd
P = (q*w*10/S)	#percent


#RESULTS
print 'the degree of separation of stormwater = %.0f percent'%(P)

the degree of separation of stormwater = 3 percent

Example 5.8 Page No : 5-24

In [16]:
	
#initialisation of variables
s = 105	#gpcd
m = 315	#gpcd
m1 = 35	#gpcd
Q1 = 360	#gpcd
Q2 = 30	#gpcd
p1 = 20	#pecent
p2 = 15	#persons/acer
D = 21	#persons/acre
I = 2000	#gpd/acre
	
#CALCULATIONS
A = D*(s+Q2)+I	#gpd/acre
R = D*(m+Q2)+I	#gpd/acre
L = D*(m1+Q2)+I	#gpd/acre
	
#RESULTS
print 'the average peak and low rates of flow  = %.0f gpd/acre'%(round(L,-1))

the average peak and low rates of flow  = 3370 gpd/acre