TTim Exercise 1



In [1]:

    
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
from ttim import *

Consider a three-aquifer system. Aquifer properties are given in Table 1. All aquifers have elastic storage. A well is located at $(x,y)=(0,0)$ and is screened in layer 1. The well starts pumping at time $t=0$ with a discharge $Q=1000$ m$^3$/d. The radius of the well is 0.2 m.

Table 1 - Aquifer properties for exercise 1

Layer	$k$ (m/d)	$c$ (d)	$S_s$ (m$^{-1}$)	$z_t$ (m)	$z_b$ (m)
Aquifer 0	1	-	0.0001	25	20
Leaky layer 1	-	1000	0	20	18
Aquifer 1	20	-	0.0001	18	10
Leaky layer 2	-	2000	0	10	8
Aquifer 2	2	-	0.0001	8	0

Exercise 1a

Compute the head as a function of time at $(x,y)=(50,0)$. Make a plot of the head vs. time from $t=0.1$ till $t=1000$ days using a linear scaling on both axis. Do the same using a logarithmic time axis.



In [2]:

    
ml = ModelMaq(kaq=[1, 20, 2],
              z=[25, 20, 18, 10, 8, 0],
              c=[1000, 2000],
              Saq=[1e-4, 1e-4, 1e-4],
              Sll=[0, 0],
              phreatictop=False,
              tmin=0.1,
              tmax=1000)
w = Well(ml, xw=0, yw=0, rw=0.2, 
         tsandQ=[(0,1000)], 
         layers=1)
ml.solve()

t = np.logspace(-1, 3, 100)
h = ml.head(50, 0, t)
plt.figure(figsize=(12, 6))
plt.subplot(121)
plt.plot(t, h[0], label='layer 0')
plt.plot(t, h[1], label='layer 1')
plt.plot(t, h[2], label='layer 2')
plt.legend(loc='best')
plt.ylabel('head [m]')
plt.xlabel('time [days]')
plt.subplot(122)
plt.semilogx(t, h[0], label='layer 0')
plt.semilogx(t, h[1], label='layer 1')
plt.semilogx(t, h[2], label='layer 2')
plt.legend(loc='best')
plt.xlabel('time [days]');









    



self.neq  1
solution complete

Exercise 1b

Create a plot of the head vs. distance from the well after 10 days of pumping. Plot the head in all three layers up to a distance of 1000 m from the well. Make the same plot after 1000 days of pumping. Is there much difference between 100 and 1000 days of pumping?



In [3]:

    
ml.xsection(x1=-1000, x2=1000, y1=0, y2=0, npoints=100, t=10, layers=[0, 1, 2]) 
plt.xlabel('distance (m)')
plt.ylabel('head (m)')
ml.xsection(x1=-1000, x2=1000, y1=0, y2=0, npoints=100, t=1000, layers=[0, 1, 2]) 
plt.xlabel('distance (m)')
plt.ylabel('head (m)');

Exercise 1c

The well is turned off after 100 days. Compute the head as a function of time at $(x,y)=(50,0)$. Make a plot of the head vs. time from $t=0.1$ till $t=1000$ days using a logarithmic time axis.



In [4]:

    
ml = ModelMaq(kaq=[1, 20, 2],
              z=[25, 20, 18, 10, 8, 0],
              c=[1000, 2000],
              Saq=[1e-4, 1e-4, 1e-4],
              Sll=[0, 0],
              phreatictop=False,
              tmin=0.1,
              tmax=1000)
w = Well(ml, xw=0, yw=0, rw=0.2, tsandQ=[(0,1000), (100,0)], layers=1)
ml.solve()
t = np.logspace(-1, 3, 100)
h = ml.head(50, 0, t)
plt.semilogx(t, h[0], label='layer 0')
plt.semilogx(t, h[1], label='layer 1')
plt.semilogx(t, h[2], label ='layer 2')
plt.legend(loc='best')
plt.ylabel('head [m]')
plt.xlabel('time [days]');









    



self.neq  1
solution complete

Exercise 1d

Compute the head inside the well as a function of time from $t=0.1$ till $t=1000$ days using a logarithmic time axis. On the same graph, plot the head inside the well vs. time when the entry resistance of the well is 0.1 days.



In [5]:

    
h = w.headinside(t)
plt.semilogx(t, h[0], label='res=0')  # head from previous solution
w.res = 0.1
ml.solve()
h = w.headinside(t)
plt.semilogx(t, h[0], label='res=0.1')
plt.legend(loc='best')
plt.ylabel('head [m]')
plt.xlabel('time [days]');









    



self.neq  1
solution complete

Exercise 1e

Conside again the case of a well without skin effect. Compute the head inside the well as a function of time from $t=1$ min till $t=1$ day using a logarithmic time axis. On the same graph, plot the head inside the well vs. time when the wellbore storage is taken into account.



In [6]:

    
tmin = 1.0 / 24 / 60  # 1 minute
ml = ModelMaq(kaq=[1, 20, 2],
              z=[25, 20, 18, 10, 8, 0],
              c=[1000, 2000],
              Saq=[1e-4, 1e-4, 1e-4],
              Sll=[0, 0],
              phreatictop=False,
              tmin=1e-4,
              tmax=1)
w = Well(ml, xw=0, yw=0, rw=0.2, tsandQ=[(0,1000)], layers=1)
ml.solve()
t = np.logspace(np.log10(tmin), 0, 100)
h = w.headinside(t)
plt.semilogx(t, h[0], label='no wellbore storage')  # head from previous solution
w.rc = 0.2
ml.solve()
h = w.headinside(t)
plt.semilogx(t, h[0], label='wellbore storage')
plt.legend(loc='best')
plt.ylabel('head [m]')
plt.xlabel('time [days]');
plt.xticks([tmin, 10 * tmin, 60 * tmin, 1], ['1 min','10 min','1 hr','1 day']);









    



self.neq  1
solution complete
self.neq  1
solution complete