Chapter 13 : Water Distribution

Example 13.2 Page No : 13-14

In [1]:
import math 
	
#initialisation of variables
p1 = 7.8	#ft
p2 = 6.0	#ft
p3 = 7.4	#ft
p4 = 6.5	#ft
p = 7.6	#ft
h = 1.0	#ft
h1 = 6.7	#ft
p5 = 3.3	#ft
	
#CALCULATIONS
D = p1-p2	#mgd
D1 = p1-p3	#mgd
D2 = p-p4	#mgd
D3 = p4+h	#mgd
D4 = h1-p5	#mgd
D5 = 2.0 - 1.6
	
#RESULTS
print "1 . Deficiency : %.1f mgd"%D
print "2 . Deficiency : %.1f mgd"%D1
print "3 . Deficiency : %.1f mgd"%D2
print "    Net added capacity = %.1f mgd"%(1.2 - 0.2)
print "    Reinforced capacity : %.1f mgd"%(D3)
print "4 . Deficiency : %.1f mgd"%D4
print "    Net added capacity : %.1f mgd"%(p5 - 0.3)
print "    Reinforced capacity : %.1f mgd"%(p5+3)
print "5.  Deficiency : %.1f mgd"%D5 
print "    Net added capacity : %.1f mgd"%(0.6 - 0.2 )
print "    Reinforced capacity : %.1f mgd"%(1.6+0.4)
print 'the demand is taken  = %.1f mgd'%(D3)

1 . Deficiency : 1.8 mgd
2 . Deficiency : 0.4 mgd
3 . Deficiency : 1.1 mgd
    Net added capacity = 1.0 mgd
    Reinforced capacity : 7.5 mgd
4 . Deficiency : 3.4 mgd
    Net added capacity : 3.0 mgd
    Reinforced capacity : 6.3 mgd
5.  Deficiency : 0.4 mgd
    Net added capacity : 0.4 mgd
    Reinforced capacity : 2.0 mgd
the demand is taken  = 7.5 mgd

Example 13.3 Page No : 13-15

In [1]:
	
#initialisation of variables
w = 500. 	#ft
p = 20. 	#psig
h = 40. 	#psig
h1 = 1000. 	#in
q = 1250.	#ft
g = 2.308/0.75	#ft
g1 = 2.308/1.00	#ft
s = 5200.	#gpm
a = 250. 	#gpm
	
#CALCULATIONS
H = (h1-(1./2)*(w))	#ft
H1 = (h-p)*g	#percent
Q = (q-(1./2)*(w))	#ft
Q1 = (h-p)*g1	#percent
S = s/a	#gpm
	
#RESULTS
print 'the number of standard fire streams = %.1f gpm'%(S)

the number of standard fire streams = 20.8 gpm

Example 13.6 Page No : 13-21

In [3]:
	
#initialisation of variables
h1 = 2.1*3	#ft
h2 = 2.1	#ft
h = 8.4	#ft
p = 1000.	#ft
h3 = 5.7	#ft
h4 = 4.2*3	#ft
q = 4.2	#ft
s = 1.68	#ft
q1 = 1.33	#ft
	
#CALCULATIONS
A = p*h/h2	#ft
B = p*(h3+h4)/q	#ft
C = p*(h1+h2)/s	#ft
	
#RESULTS
print 'the equilent pipe for the Hazen willians coefficent = %.0f ft'%(A)
print 'the equilent pipe for the Hazen willians coefficent = %.0f ft'%(round(B,-1))
print 'the equilent pipe for the Hazen willians coefficent = %.0f ft'%(C)

the equilent pipe for the Hazen willians coefficent = 4000 ft
the equilent pipe for the Hazen willians coefficent = 4360 ft
the equilent pipe for the Hazen willians coefficent = 5000 ft

Example 13.8 Page No : 13-27

In [4]:
	
#initialisation of variables
d = 10	#hr
p = 50000	#in
a = 7.5	#mgd
w = 0.75	#mg
s = 5.03	#mg
	
#CALCULATIONS
S = s/w	#mg
P = S-s	#mg
	
#RESULTS
print 'a steady gravity supply equal to maximum daily = %.2f mg'%(P)

a steady gravity supply equal to maximum daily = 1.68 mg