# Chapter 2 : Water Systems¶

## Example 2.1 Page No : 2-5¶

In :

#initialisation of variables
w = 3000.	#sq ft
w1 = 2000.	#sq ft
w2 = 1000.	#sq ft
r = 15. 	#in
a = 12.	    #in
h = 7.5	    #in

#CALCULATIONS
G = w*(r/a)*h	#gal
g = w1*(r/a)*h	#gal
g1 = w2*(r/a)*h	#gal

#RESULTS
print 'the normally be stored to tide the supply over dry spells = %.0f gal'%(G)
#wrong ans in book

the normally be stored to tide the supply over dry spells = 28125 gal


## Example 2.2 Page No : 2-9¶

In :

#initialisation of variables
m = 17.378   	#mg
h = 20.  	    #in/sq mile
d = 365.	    #in
s = 0.75	    #percent
a = 100.	    #sq miles
p = 750000.	    #gpd per sq mile
t = 180.	    #in
c = 150.	    #gpcd
n = 64600. 	    #gpd per sq mile

#CALCULATIONS
R = h*m	#mg
A = R/d	#mgd
S = s*a*t	#billion gal
Q = a*p/c	#gpd
P = a*n/c	#people against

#RESULTS
print 'the surface water sheds and storage requirements = %.0f mg'%(R)
print 'the well watered sections of north america = %.1f billion gal'%(S/1000)
print 'the average consumption populations = %.0f gpd'%(Q)
print 'the presence of proper storage = %.0f people against'%(round(P,-3))

the surface water sheds and storage requirements = 348 mg
the well watered sections of north america = 13.5 billion gal
the average consumption populations = 500000 gpd
the presence of proper storage = 43000 people against


## Example 2.3 Page No : 2-14¶

In :

#initialisation of variables
w = 20  	#ft
r = 3	    #ft a day
g = 500	    #ft
g1 = 1000	    #ft
h = 7.5/1440	#ft
p = 7.5/1000000	#ft
r1 = 2	        #ft a day

#CALCULATIONS
W1 = w*g1*r*h	#gpm
W2 = w*g1*r1*r*p	#mgd

#RESULTS
#wrong ans in book
print 'the ground water laterally  = %.2f gpm'%(W1)
print 'the water from both sides = %.2f mgd'%(W2)

the ground water laterally  = 312.50 gpm
the water from both sides = 0.90 mgd


## Example 2.4 Page No : 2-18¶

In :
import math

#initialisation of variables
p1 = 10.	    #mgd
p2 = 6940.	#gpm
w = 67000.	#people
d = 2.	    #min
v = d*p2/d	#gal
v1 = 98.2	#cu ft each
q = 30.	    #min
q1 = q*p2/d	#gal
q2 = 13900.	#cu ft
a = 1390.	#sq ft
s = 2.	    #hr
s1 = 120*p2/d	#gal
s2 = 55700.	#cu ft
s3 = s2/8.	#sq ft
r = 3	    #gpm/sq ft
r1 = 6	    #rapid

#CALCULATIONS
D = math.sqrt(v1*4/math.pi)	#ft
S = p2/s3	#gpm/sq ft
A = p2/(r1*r)	#sq ft

#RESULTS
print 'the capacity of the components of a rapid sand filtration plant = %d sq ft'%(A)

the capacity of the components of a rapid sand filtration plant = 385 sq ft


## Example 2.5 Page No : 2-21¶

In :
import math

#initialisation of variables
r = 10000.	#ft
l = 400000	#people
q = 1000000.	#mgd
w = 100	#gpcd
w1 = 150	#gpcd
m = 50	#percent
g = 1.5	#ft
h1 = 2.32	#cfs
h2 = 139	#cfs
d = 12	#ft
c = 100	#ft
l = 10.8	#ft
l2 = 0.85	#ft
l1 = 1000	#ft

#CALCULATIONS
a = r*w/q	#mgd
b = l*w1/q	#mgd
a1 = a*g	#mgd
b1 = b*g	#mgd
D = d*math.sqrt(h1/math.pi)	#in
D1 = d*math.sqrt(h2/math.pi)	#in
L = l/l1	#ft
L1 = l2/l1	#ft

#RESULTS
print 'the higher loss of head in small conduits at equal velocity = %f ft'%(L1)

the higher loss of head in small conduits at equal velocity = 0.000850 ft


## Example 2.6 Page No : 2-25¶

In :

#initialisation of variables
a = 12	#in
b = 24	#in
r = 500	#gpm
d = 200.	#gpcd
d1 = 150.	#gpcd
p1 = 113	#sq in
p2 = 425	#sq in
v1 = 3	    #fps
v2 = 2.35	#cfs
v3 = 9.42	#cfs
h = 646000	#gpd
w = 720000	#gpd

#CALCULATIONS
D1 = v2*h	#gpd
D2 = v3*h	#gpd
W1 = D1-w	#gpd
W2 = D2-w	#gpd
R1 = D1/d	#people
R2 = D2/d	#people
S = W1/d1	#people
S1 = W2/d1	#people

#RESULTS
print 'the absence of fire service for a maximum draft = %d gpd'%(round(D2,-3))
print 'The residential fire flow requirements = %.0f gpd'%(round(W2,-3))

# note: answers are correct.  please check using calculator.

the absence of fire service for a maximum draft = 6085000 gpd
The residential fire flow requirements = 5365000 gpd


## Example 2.7 Page No : 2-28¶

In :

#initialisation of variables
w = 100000.	#ft
c = 250.  	#per capita
p1 = 0.3	#percent
p2 = 0.1	#percent
p3 = 0.60	#percent
w1 = 15.	#mgd

#CALCULATIONS
T = c*w	#$W = p1*T #$
W1 = p2*T	#$W2 = p3*T #$
W3 = ((w1)**(2./3))*(T/100)	#$#RESULTS print 'the replacement cost of the water of a city = %.0f$'%(round(W3,-5))

the replacement cost of the water of a city = 1500000 \$