Chapter 9 : Groundwater Flow

Example 9.1 Page No : 9-11

In [1]:
	
#initialisation of variables
t = 10	#C
s = 74.2	#days
c = 0.01	#mm
d = 245	#mm
	
#CALCULATIONS
h = s/(d*c)	#cm
	
#RESULTS
print 'the high will water at a temperature  = %.1f cm'%(h)

the high will water at a temperature  = 30.3 cm

Example 9.2 Page No : 9-13

In [2]:
	
#initialisation of variables
p1 = 1000	#ft
p2 = 50	#ft
g = 20	#ft/mile
v = 5280.	#ft
q = 7.5*10**-6	#ft
t = 60	#F
k = 2835	#ft/days
p = 7.5	#ft
	
#CALCULATIONS
S = g/v	#ft
W = k*(g/v)	#ft/day
Q = W*p1*p2*q	#mgd
P = k*p	#ft
P1 = P*p2	#mgd
	
#RESULTS
print 'the velocity of flow  = %.2f mgd'%(Q)
print 'the standard coefficient pf permeability = %.2e mgd'%(P1)

the velocity of flow  = 4.03 mgd
the standard coefficient pf permeability = 1.06e+06 mgd

Example 9.3 Page No : 9-19

In [5]:
import math 
	
#initialisation of variables
p = 40. 	#ft
d = 56. 	#ft
d1 = 140. 	#ft
p1 = 30. 	#ft
w = 3.28*10**-4	#fps
	
#CALCULATIONS
Q = w*(p/d1)*2*d*p	#cfs
q = Q/p	#cfs
K = w*(p/d1)	#fps
x0 = q/(2*math.pi*K)	#ft
Z = 2*math.pi*x0	#ft
	
#RESULTS
print 'the yield of the well if the coefficient of permeability = %.1f ft'%(x0)
print 'the distance of the point of stagnation  = %.0f ft'%(Z)

the yield of the well if the coefficient of permeability = 17.8 ft
the distance of the point of stagnation  = 112 ft

Example 9.4 Page No : 9-23

In [6]:
	
#initialisation of variables
p = 5.*10**6	      #ft
Q = 350.            	#gpm
x = 225.            	#ft
u = 10.**-2	#ft
g = 1.87	#ft
p2 = 7.*10**2	#ft
p3 = 10.9	#ft
w = 4.0	#ft
t = 114.6	    #ft
d = 10.      	#ft
p1 = 5.      	#ft
w1 = 3.2*10**4	#ft
W = 21.75	#ft
	
#CALCULATIONS
T = t*Q*4/p1	#gpd/ft
S = u*(w1)/(g*(p))	#ft
U = g*((S)/(T))*(x**2/d)	#ft
P = t*(p2)*p3/(T)	#ft
U1 = g*((S)/(T))*(1./d)	#ft
P1 = t*(p2)*W/(T)	#ft
	
#RESULTS
print 'the type curve as if a transparency of the observed data had moved into place over the type = %.0f ft'%(P1)

the type curve as if a transparency of the observed data had moved into place over the type = 54 ft

Example 9.5 Page No : 9-24

In [2]:
import math
	
#initialisation of variables
Q = 350. 	#gpm
x = 225. 	#ft
t = 1.  	#min
p = 1.6	    #ft
t2 = 10. 	#min
p2 = 4.5	#ft
p3 = 700. 	#gpm
T = 3.2*10**4	#gpd/ft
t0 = 0.3    	#min
u = 1.15*10**-5
	
#CALCULATIONS
S = t0*(T)*t0/((x)**2*1440)	#ft
P = ((114.6*p3)/(T))*(-0.5772*2.3*math.log(u))	#ft

#RESULTS
print 'A straight line being drawn through the ppints for the higher = %.0f ft'%(P)

# note : book answer is wrong. please check.

A straight line being drawn through the ppints for the higher = 38 ft

Example 9.6 Page No : 9-31

In [8]:
	
#initialisation of variables
h = 4.8	             #ft
m = 13.4	         #ft
k = 10.**-1	         #cm/sec
k1 = 3.28*10**-3	#fps
n = 7.            	#ft
n1 = 11.         	#ft
q = 1.0*10**-2
	
#CALCULATIONS
Q = k1*h*n/n1	#cfs/ft
Q1 = 2*q*10**2	#cfs
	
#RESULTS
print 'A satisfactory orthogonal system the flow of into the collector  = %.0f cfs'%(Q1)

A satisfactory orthogonal system the flow of into the collector  = 2 cfs