Chapter 11 : Surface-Water Collection

Example 11.1 Page No : 11-20

In [2]:
	
#initialisation of variables
s = 20.	#mph
t = 90.	#min
w = 1.31	#ft
h = 7.5	#miles
h1 = 0.22	#ft
t1 = 1100.	#min
t2 = 6.0	#min
p = 32.2	#ft
l = 5.12	#length
l1 = 2.8	#length
p1 = 1400.	#ft
d = 73.	#depth
h3 = 2.06	#ft
e = 173.0	#ft
hi = 0.2	#ft
	
#CALCULATIONS
W = s*w	#mph
hs = h1*((W)**2/p)**0.53*h**0.47	#ft
Ts = t2*(W/p)**0.44*(h/p)**0.28	#sec
Td = t1*h/(p*Ts)	#min
Ls = l1/(l*(Ts)**2)	#ft
D = d/(l*(Ts)**2)	#ft
H = (W)**2*(h*(1/(p1*d)))	#ft
hr = h3*l1	#ft
M = e+hi+hr	#ft
	
#RESULTS
print 'the overwater wind speed = %.0f mph'%(W)
print 'the significant wave height = %.1f ft'%(hs)
print 'the significant wave period = %.1f sec'%(Ts)
print 'the minimum wind duration required to reach the significant wave height = %.0f min'%(Td)
print 'the significant wave lenght adn steepness = %.3f ft'%(Ls)
print 'the reservoir depth ratio = %.1f ft'%(D)
print 'the wind tide or set up = % f ft'%(H)
print 'the run up  = %.1f ft'%(hr)
print 'the maximum elevation reached by the waves = %.1f ft'%(M)

# rounding off error. please check.

the overwater wind speed = 26 mph
the significant wave height = 2.9 ft
the significant wave period = 3.6 sec
the minimum wind duration required to reach the significant wave height = 70 min
the significant wave lenght adn steepness = 0.041 ft
the reservoir depth ratio = 1.1 ft
the wind tide or set up =  0.050375 ft
the run up  = 5.8 ft
the maximum elevation reached by the waves = 179.0 ft

Example 11.2 Page No : 11-23

In [1]:
	
#initialisation of variables
g = 264	#quartz
p = 0.39	#percent
	
#CALCULATIONS
S = (1-p)*(g-1)	#in
	
#RESULTS
print 'the hydraulic gradient and seepage velocity = %.2f in'%(S)
#incorrect answer in textbook

the hydraulic gradient and seepage velocity = 160.43 in

Example 11.3 Page No : 11-26

In [4]:
	
#initialisation of variables
w = 40	#ft
k = 2*10**-3	#cm/sec
p = 3.28*10**-3	#cfs
h = 6.47*10**5	#gpd
p1 = 0.433	#ft
m = 9	#ft
delh = w/(18*9)	#in
k1 = 4.94*10**-4	#cm/sec
	
#CALCULATIONS
Q = k*p*w*(9./18)	#cfs
Q1 = Q*h	#gpd/ft width
P = (1-8./18)*w*p1	#Psig
H = k1/k	#in
	
#RESULTS
print 'the seepage through each foot width of the foundation = %.0f gpd/ft/ width'%(Q1*10)
print 'the excess hydrostatic pressure on the upstream  side of the bottom of the sheet pilling = %.2f Psig'%(P)
print 'the maximum hydraulic gradient and its relations to the coeeficent = %.2f in'%(H)

the seepage through each foot width of the foundation = 849 gpd/ft/ width
the excess hydrostatic pressure on the upstream  side of the bottom of the sheet pilling = 9.62 Psig
the maximum hydraulic gradient and its relations to the coeeficent = 0.25 in

Example 11.4 Page No : 11-13

In [12]:
	
#initialisation of variables
d = 120	#ft
w = 16	#ft
d1 = 120/0.8	#ft
p = 60*0.8	#ft
h = 2	#ft
v = 18.74*0.8	#ft
s = 95.23	#ft
s1 = 0.8	#ft
	
#CALCULATIONS
W = d-h*p	#ft
S = s*s1	#ft
	
#RESULTS
print 'in succession from the intersection of the upstream slop = %.2f ft'%(S)

in succession from the intersection of the upstream slop = 76.18 ft