#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
#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))
#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)
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)
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)
#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.
#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))