

In [1]:

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

Circular area-sink

Circular area-sink with radius 100 m, located at the origin.



In [2]:

    
N = 0.001
R = 100
ml = ModelMaq(kaq=5, z=[10, 0], Saq=2e-4, tmin=1e-3, tmax=1e4)
ca = CircAreaSink(ml, 0, 0, 100, tsandN=[(0, 0.001)])
ml.solve()
ml.xsection(-200, 200, 0, 0, t=[0.1, 1, 10], figsize=(12, 4), sstart=-200)









    



self.neq  0
No unknowns. Solution complete



In [3]:

    
x = np.linspace(-200, 200, 200)
qx = np.zeros_like(x)
for t in [0.1, 1, 10]:
    for i in range(len(x)):
        qx[i], qy = ml.disvec(x[i], 1e-6, t)
    plt.plot(x, qx, label='time is ' + str(t))
qxb = N * np.pi * R ** 2 / (2 * np.pi * R)
plt.axhline(qxb, color='r', ls='--')
plt.axhline(-qxb, color='r', ls='--')
plt.xlabel('x (m)')
plt.ylabel('Qx (m^2/d)')
plt.legend(loc='best');

Circular area-sink and well

Discharge of well is the same as total infiltration rate of the circular area-sink. Well and center of area-sink area located at equal distances from $y$-axis, so that the head remains zero along the $y$-axis. Solution approaches steady-state solution.



In [4]:

    
N = 0.001
R = 100
Q = N * np.pi * R ** 2
ml = ModelMaq(kaq=5, z=[10, 0], Saq=2e-4, tmin=1e-3, tmax=1e4, M=10)
ca = CircAreaSink(ml, -200, 0, 100, tsandN=[(0, 0.001)])
w = Well(ml, 200, 0, rw=0.1, tsandQ=[(0, Q)])
ml.solve()









    



self.neq  1
solution complete



In [5]:

    
ml.xsection(-400, 300, 0, 0, t=[0.1, 1, 10, 100, 1000], figsize=(12, 4), sstart=-400)



In [6]:

    
t = np.logspace(-3, 4, 100)
h = ml.head(-200, 0, t)
plt.semilogx(t, h[0])
plt.xlabel('time')
plt.ylabel('head')
plt.title('head at center of area-sink');



In [7]:

    
N = 0.001
R = 100
Q = N * np.pi * R ** 2
ml = ModelMaq(kaq=5, z=[10, 0], Saq=2e-4, tmin=10, tmax=100, M=10)
ca = CircAreaSink(ml, -200, 0, 100, tsandN=[(0, 0.001)])
w = Well(ml, 200, 0, rw=0.1, tsandQ=[(0, Q)])
ml.solve()
ml.contour([-300, 300, -200, 200], ngr=40, t=20)









    



self.neq  1
solution complete

Two layers

Discharge of well is the same as total infiltration rate of the circular area-sink. Center of area-sink and well are at the origin. Circular area-sink in layer 0, well in layer 1.



In [8]:

    
N = 0.001
R = 100
Q = N * np.pi * R ** 2
ml = ModelMaq(kaq=[5, 20], z=[20, 12, 10, 0], c=[1000], Saq=[2e-4, 1e-4], tmin=1e-3, tmax=1e4, M=10)
ca = CircAreaSink(ml, 0, 0, 100, tsandN=[(0, 0.001)])
w = Well(ml, 0, 0, rw=0.1, tsandQ=[(0, Q)], layers=1)
ml.solve()









    



self.neq  1
solution complete



In [9]:

    
ml.xsection(-200, 200, 0, 0, t=[0.1, 100], layers=[0, 1], sstart=-200)



In [10]:

    
ml.xsection(-500, 500, 0, 0, t=[0.1, 100, 1000], layers=[0, 1], sstart=-500)