Chapter7- Permeability

Ex1-pg168

In [1]:
#Calculate the hydraulic conductivity in cm/sec.
import math
##initialisation of variables
L= 30. ##cm
A= 177. ##cm^2
h= 50. ##cm
Q= 350. ##cm^3
t= 300. ##sec
##claculations
k=Q*L/(A*h*t)
##results
print'%s %.4f %s'% ('hydraulic conductivity = ',k,' cm/sec ')
hydraulic conductivity =  0.0040  cm/sec 

Ex2-pg169

In [2]:
#Determine the hydraulic conductivity of the soil in in./sec.
import math
##initialisation of variables
L= 203. ##mm
A= 10.3 ##cm^2
a= 0.39 ##cm^2
h0= 508. ##mm
h180= 305. ##mm
t= 180. ##sec
##calculations
k= 2.303*a*L*math.log10(h0/h180)/(A*t)
##results
print'%s %.2f %s'% ('hydraulic conductivity of sand = ',k,' in/sec ')
hydraulic conductivity of sand =  0.02  in/sec 

Ex3-pg169

In [3]:
#The hydraulic conductivity of a clayey soil is 3  107 cm/sec. The viscosity of water at 25°C is 0.0911  104 g # sec/cm2 
#Calculate the absolute permeability of the soil.
import math
##initialisation of varilables
k= 3e-7 ##cm/sec
n= 0.0911e-4 ##g*sec/cm^2
dw= 1. ##g/cc
##calculations
K= k*n/dw
##results
print'%s %.2e %s'% ('absolute premeability = ',K,' cm^2 ')
absolute premeability =  2.73e-12  cm^2 

Ex4-pg170

In [5]:
#With k  5.3  105 m/sec for the permeable layer, calculate the rate of seepage through it in m3 /hr/m width if H  3 m and a  8°.

import math
##initialisation of variables
k= 5.3e-5 ##m/sec
H= 3 ##m
a= 0.139 ##radians
##calculations
A= H*math.cos(a)
i= math.sin(a)
q= k*i*A*3600
##results
print'%s %.4f %s'% ('rate of seepage = ',q,' m^3/hr/m ')
rate of seepage =  0.0785  m^3/hr/m 

Ex5-pg171

In [8]:
import math
#calculate flow rate
##initialisation of variables
L= 50. ##m
k= 0.08e-2##m/sec
h= 4. ##m
H1= 3. ##m
H= 8. ##m
a= 0.139 ##radians
##calculations
i= h*math.cos(a)/L
A= H1*math.cos(a)
q= k*i*A
##results
print'%s %.5f %s'% ('flow rate = ',q,' m^3/sec/m ')
flow rate =  0.00019  m^3/sec/m 

Ex6-pg174

In [18]:
#calculate hydraulic conductivity at void ratio of 0.65
##initialisation of variables
k1= 0.02 ##cm/sec
e1= 0.5 
e2= 0.65
##calculations
k2= k1*(e2**3/(1.+e2))/(e1**3/(1.+e1))
##results
print'%s %.2f %s'% ('hydraulic conductivity at void ratio of 0.65 =',k2,'cm/sec ')
hydraulic conductivity at void ratio of 0.65 = 0.04 cm/sec 

Ex7-pg176

In [2]:
#calculate the value of grain size and plot the graph
import math
%matplotlib inline
import warnings
warnings.filterwarnings('ignore')
from math import log
import numpy
from math import tan
import matplotlib
from matplotlib import pyplot
#given
e=numpy.array([100,96,84,50,0])
p=numpy.array([0.06,0.0425,0.02,0.015,0.0075])

#calculations


#results

pyplot.plot(p,e)
pyplot.xlabel('Percent passing')
pyplot.ylabel('grain size,mm')
pyplot.title('Graph of percent passinge vs grain size')
pyplot.show()
print('look at the axis reverse in text book')
look at the axis reverse in text book

Ex8-pg177

In [17]:
#calculate hydraulic conductivity
##initialisation of variables
e= 0.6
D10= 0.09 ##mm
##calculations
k= 2.4622*(D10**2*(e**3/(1+e)))**0.7825
##results
print'%s %.4f %s'% ('hydraulic conductivity = ',k,' cm/sec ')
hydraulic conductivity =  0.0119  cm/sec 

Ex9-pg177

In [11]:
#calculate hydraulic conductivity
##initialisation of variables
e= 0.6
D10= 0.09 ##mm
D60= 0.16 ##mm
##calculations
Cu=D60/D10
k= 35*(e**3/(1+e))*(Cu**0.6)*(D10**2.32)
##results
print'%s %.3f %s'% ('hydraulic conductivity =',k,'cm/sec ')
hydraulic conductivity = 0.025 cm/sec 

Ex10-pg179

In [16]:
import math
#calculate hydraulic conductivity
##initialisation of variables
k1= 0.302e-7 ##cm/sec
k2= 0.12e-7 ##cm/sec
e1= 1.1
e2= 0.9
e= 0.75
##calcualtions
n= (math.log10((k1/k2)*((1+e1)/(1+e2))))/math.log10(e1/e2)
C= k1/(e1**n/(1+e1))
k= C*(e**n/(1+e))
##results
print'%s %.e %s'% ('hydraulic conductivity =',k,'cm/sec')
hydraulic conductivity = 5e-09 cm/sec

Ex11-pg185

In [13]:
#calculate ration of equivalent hydraulic conductivity
##initialisation of variables
H1= 2. ##m
H2= 3. ##m
H3= 4. ##m
k1= 1e-4 ##cm/sec
k2= 3.2e-2 ##cm/sec
k3= 4.1e-5 ##cm/sec
##calculations
H= H1+H2+H3
Kh= (1./H)*((k1*H1)+(k2*H2)+(k3*H3))
Kv= H/((H1/k1)+(H2/k2)+(H3/k3))
P= Kh/Kv
##results
print'%s %.2f %s'% ('ration of equivalent hydraulic conductivity =',P,' ')
ration of equivalent hydraulic conductivity = 139.97  

Ex12-pg186

In [15]:
import math
#calculate rate of water supply
##initialisation of variables
H= 450. ##mm
h= 150. ##mm
k1= 1e-2 ##cm/sec
k2= 3e-3 ##cm/sec
k3= 4.9e-4 ##cm/sec
h1= 300. ##mm
##calculations
Kv= H/(h*(1./k1+1./k2+1./k3))
i= h1/H
q= Kv*i*100.*3600.
##results
print'%s %.2f %s'% ('rate of water supply =',q,' cm/hr ')
rate of water supply = 291.01  cm/hr