Chapter8-Seepage

Ex1-pg203

In [7]:
import math
#calculate rate of water flow
##initialisation of variables
H1= 12. ##in
H2= 20. ##in
z= 8. ##in
h1= 24. ##in
h= 20. ##in
k1= 0.026 ##in/sec
D= 3. ##in
##calculations
k2= H2*k1/((z/(1.-h/h1))-H1)
i= h1/(H1+H2)
A= math.pi/4.*D**2
keq= (H1+H2)/((H1/k1)+(H2/k2))
q= keq*A*i*3600.
##results
print'%s %.2f %s'% ('rate of water flow = ',q,' in^3/hr ')
rate of water flow =  330.81  in^3/hr 

Ex2-pg208

In [2]:
#calculate a)How high (above the ground surface) the water will rise if piezometers are placed at points aandb.
#b.The total rate of seepage through the permeable layer per unit length
#c. The approximate average hydraulic gradient at c.
##initialisation of variables
Nd= 6.
H1= 5.6 ##m
H2= 2.2 ##m
k= 5e-5 ##cm/sec
dL= 4.1 ##m
##calculations
H= (H1-H2)/Nd
h1= 5.61-H
h2= 5.61-5.*H
q= 2.38*(H1-H2)*k/Nd
i= H/dL
##results
print'%s %.3f %s'% ('at point a,water will rise to height of = ',h1,' m ')
print'%s %.3f %s'% ('at point b,water will rise to height of =',h2,' m ')
print'%s %.e %s'% ('total rate of seepage per unit lenghth = ',q,' m^3/sec/m ')
print'%s %.3f %s'% ('average hydraulic gradient at c = ',i,' ')
at point a,water will rise to height of =  5.043  m 
at point b,water will rise to height of = 2.777  m 
total rate of seepage per unit lenghth =  7e-05  m^3/sec/m 
average hydraulic gradient at c =  0.138  

Ex3-pg210

In [9]:
import math
#calculate average rate of flow
##initialisation of variables
k1= 5.67 ##ft/day
k2= 11.34 ##ft/day
##from graph
Nd= 8
Nf= 2.5
H= 20
##calculations
q= math.sqrt(k1*k2)*H*Nf/Nd
##results
print'%s %.2f %s'% ('average rate of flow = ',q,' ft^3/day/ft ')
average rate of flow =  50.12  ft^3/day/ft 

Ex4-pg 212

In [4]:
import math
#calculate seepage under the dam 
##initialisation of variables
B= 6. ##m
L= 120. ##m
s= 3. ##m
T= 6. ##m
x= 2.4 ##m
H= 5. ##m
k= 0.008 ##cm/sec
##calculations
b=B/2.
a1= b/T
a2= s/T
a3= x/b
Q= 0.378*k*H*L*36*24
##results
print'%s %.2f %s'% ('seepage under the dam = ',Q,' m^3/day ')
seepage under the dam =  1567.64  m^3/day 

Ex5-pg217

In [6]:
import math
#calculate seepage rate
##initialisation of variables
b= math.pi/4. ##degrees
a= math.pi/6.##degrees
B= 10. ##ft
H= 20. ##ft
h= 25. ##ft
k= 2e-4 ##ft/min
##calculations
r= H/math.tan(b)
d= 0.3*r+(h-H)/math.tan(b)+B+h/math.tan(a)
L= d/math.cos(a)-math.sqrt((d/math.cos(a))**2-(H/math.sin(a))**2)
q= k*L*math.tan(a)*math.sin(a)*24.*60
##results
print'%s %.4f %s'% ('seepage rate = ',q,' ft^3/day/ft ')
seepage rate =  0.9724  ft^3/day/ft