TimML test line-sink discharge



In [1]:

    
from pylab import *
from timml import *
%matplotlib notebook



In [2]:

    
ml1 = ModelMaq(kaq=20)
rf1 = Constant(ml1, xr=0, yr=20, hr=30)
ls1 = LineSinkBase(ml1, x1=-10, y1=-10, x2=10, y2=10, Qls=1000)
ml1.solve()









    



Number of elements, Number of equations: 2 , 1
..
solution complete



In [3]:

    
print('head at center of line-sink:', ml1.head(ls1.xc, ls1.yc))
print('discharge of line-sink:', ls1.discharge())









    



head at center of line-sink: [ 19.19104524]
discharge of line-sink: [ 1000.]



In [4]:

    
ml2 = ModelMaq(kaq=20)
rf2 = Constant(ml2, xr=0, yr=20, hr=30)
N = 20
d = 20 / N
xw = np.arange(-10 + d/2, 10, d)
yw = np.arange(-10 + d/2, 10, d)
for i in range(N):
    Well(ml2, xw[i], yw[i], Qw=1000 / N)
ml2.solve(silent=True)



In [5]:

    
ml1.contour([-20, 20, -20, 20], 50, [0], np.arange(20, 31, 1), color='b')
ml2.contour([-20, 20, -20, 20], 50, [0], np.arange(20, 31, 1), color='r', newfig=False)



In [6]:

    
ml1 = ModelMaq(kaq=[20, 10], z=[20, 12, 10, 0], c=[100])
rf1 = Constant(ml1, xr=0, yr=20, hr=30)
ls1 = HeadLineSink(ml1, -10, -10, 10, 10, 20, order=7, layers=0)
ml1.solve()









    



Number of elements, Number of equations: 2 , 9
..
solution complete



In [7]:

    
ml2 = ModelMaq(kaq=[20, 10], z=[20, 12, 10, 0], c=[100])
rf2 = Constant(ml2, xr=0, yr=20, hr=30)
N = 50
d = 20 / N
xw = np.arange(-10 + d/2, 10, d)
yw = np.arange(-10 + d/2, 10, d)
for i in range(N):
    HeadWell(ml2, xw[i], yw[i], 20, layers=0)
ml2.solve(silent=True)
Qwell = 0
for i in range(N):
    Qwell += ml2.elementlist[i + 1].discharge()



In [8]:

    
print('discharge of line-sink:', ls1.discharge())
print('discharge of wells:', Qwell)









    



discharge of line-sink: [ 9430.28953368     0.        ]
discharge of wells: [ 9527.67795022     0.        ]



In [9]:

    
ml1.contour([-20, 20, -20, 20], 50, [0], np.arange(20, 31, 1), color='b')
ml2.contour([-20, 20, -20, 20], 50, [0], np.arange(20, 31, 1), color='r', newfig=False)



In [10]:

    
x = linspace(-100, 100, 100)
h1 = ml1.headalongline(x, 0)
h2 = ml2.headalongline(x, 0)
figure()
plot(x, h1.T, 'b')
plot(x, h2.T, 'r')









    














    











    Out[10]:





[<matplotlib.lines.Line2D at 0x107cbc198>,
 <matplotlib.lines.Line2D at 0x107d26be0>]

Resistance line-sink



In [11]:

    
ml = ModelMaq(kaq=3)
ls = HeadLineSink(ml, -10, 0, 10, 0, wh=1, res=2, order=2)
rf = Constant(ml, 0, 20, 2)
ml.solve()









    



Number of elements, Number of equations: 2 , 4
..
solution complete



In [12]:

    
for i in range(3):
    print((ml.head(ls.xc[i], ls.yc[i]) - ls.hc) * ls.wh / ls.res)
    print(np.sum(ls.strengthinf[i] * ls.parameters[:, 0]))









    



[ 0.2856123  0.2856123  0.2856123]
0.285612298215
[ 0.25423528  0.25423528  0.25423528]
0.254235278553
[ 0.2856123  0.2856123  0.2856123]
0.285612298215



In [13]:

    
ml = ModelMaq(kaq=[1, 2], z=[20, 10, 10, 0], c=[1000])
lslayer = 0
order = 2
ls = HeadLineSink(ml, -10, 0, 10, 0, order=order, wh=1, res=2, layers=[lslayer])
rf = Constant(ml, 0, 20, 2)
ml.solve()
for i in range(order + 1):
    print((ml.head(ls.xc[i], ls.yc[i]) - ls.hc[i])[lslayer] * ls.wh / ls.res)
    print(np.sum(ls.strengthinf[i] * ls.parameters[:, 0]))









    



Number of elements, Number of equations: 2 , 4
..
solution complete
[ 0.40780978]
0.407809784166
[ 0.39169168]
0.391691676816
[ 0.40780978]
0.407809784166



In [14]:

    
ml = ModelMaq(kaq=[1, 2], z=[20, 12, 10, 0], c=[1000])
order = 2
ls = HeadLineSink(ml, -10, 0, 10, 0, order=order, hls=1, wh=1, res=2, layers=[0, 1])
rf = Constant(ml, 0, 2000, 2)
ml.solve()
for i in range(order + 1):
    for ilay in range(2):
        print(((ml.head(ls.xc[i], ls.yc[i]) - ls.hc[i]) * ls.wh / ls.res)[ilay])
        print(np.sum(ls.strengthinf[2 * i + ilay] * ls.parameters[:, 0]))









    



Number of elements, Number of equations: 2 , 7
..
solution complete
0.25436055556
0.25436055556
0.324018215661
0.324018215661
0.242013921174
0.242013921174
0.317190051938
0.317190051938
0.25436055556
0.25436055556
0.324018215661
0.324018215661



In [15]:

    
print(ml.headalongline(ls.xc, ls.yc))
print(ls.hc)









    



[[ 1.50872111  1.48402784  1.50872111]
 [ 1.64803643  1.6343801   1.64803643]]
[ 1.  1.  1.  1.  1.  1.]

Specifying heads along line-sinks

Give one value that is applied at all control points



In [16]:

    
ml1 = ModelMaq(kaq=[20, 10], z=[20, 12, 10, 0], c=[100])
rf1 = Constant(ml1, xr=0, yr=20, hr=30)
ls1 = HeadLineSink(ml1, -10, 0, 10, 0, hls=20, order=2, layers=[0])
ml1.solve()
print(ml1.headalongline(ls1.xc, ls1.yc))









    



Number of elements, Number of equations: 2 , 4
..
solution complete
[[ 20.          20.          20.        ]
 [ 39.70195706  39.68141058  39.70195706]]

Give order + 1 values, which is applied at the order + 1 control points. This may not be so useful, as the user needs to know where those control points are.



In [17]:

    
ml1 = ModelMaq(kaq=[20, 10], z=[20, 12, 10, 0], c=[100])
rf1 = Constant(ml1, xr=0, yr=20, hr=30)
ls1 = HeadLineSink(ml1, -10, 0, 10, 0, hls=[20, 19, 18], order=2, layers=[0])
ml1.solve()
print(ml1.headalongline(ls1.xc, ls1.yc))









    



Number of elements, Number of equations: 2 , 4
..
solution complete
[[ 20.          19.          18.        ]
 [ 40.67787514  40.64955164  40.66643038]]



In [18]:

    
ml1 = ModelMaq(kaq=[20, 10], z=[20, 12, 10, 0], c=[100])
rf1 = Constant(ml1, xr=0, yr=20, hr=30)
ls1 = HeadLineSink(ml1, -10, 0, 10, 0, hls=[19, 20], order=2, layers=[0])
ml1.solve()
print(ml1.headalongline(ls1.xc, ls1.yc))









    



Number of elements, Number of equations: 2 , 4
..
solution complete
[[ 19.14644661  19.5         19.85355339]
 [ 40.18503174  40.16548111  40.18907808]]

LineSinkDitch



In [19]:

    
ml1 = ModelMaq(kaq=[20, 10], z=[20, 12, 10, 0], c=[100])
rf1 = Constant(ml1, xr=0, yr=20, hr=30)
ls1 = LineSinkDitch(ml1, -10, -10, 10, 10, Qls=1000, order=2, layers=[0])
ml1.solve()
print(ml1.headalongline(ls1.xc, ls1.yc))
print(ls1.discharge())









    



Number of elements, Number of equations: 2 , 4
..
solution complete
[[ 28.88125025  28.88125025  28.88125025]
 [ 31.48813041  31.48283125  31.48813041]]
[ 1000.     0.]



In [20]:

    
ml1 = ModelMaq(kaq=[20, 10], z=[20, 12, 10, 0], c=[100])
rf1 = Constant(ml1, xr=0, yr=20, hr=30)
ls1 = LineSinkDitch(ml1, -10, -10, 10, 10, Qls=1000, order=2, layers=[0, 1])
ml1.solve()
print(ml1.headalongline(ls1.xc, ls1.yc))
print(ls1.discharge())









    



Number of elements, Number of equations: 2 , 7
..
solution complete
[[ 29.30489961  29.30489961  29.30489961]
 [ 29.30489961  29.30489961  29.30489961]]
[ 615.38461538  384.61538462]

Head line-sink string



In [21]:

    
ml1 = ModelMaq(kaq=[20, 10], z=[20, 12, 10, 0], c=[100])
rf1 = Constant(ml1, xr=0, yr=20, hr=30)
ls1 = HeadLineSinkString(ml1, xy=[(-10, 0), (0, 0), (10, 0), (10, 10)], hls=20, order=5, layers=[0])
ml1.solve()
ml1.contour([-20, 20, -20, 20], 50, [0], 40)









    



Number of elements, Number of equations: 2 , 19
..
solution complete



In [22]:

    
ml1 = ModelMaq(kaq=[20, 10], z=[20, 12, 10, 0], c=[100])
rf1 = Constant(ml1, xr=0, yr=20, hr=30)
ls1 = HeadLineSinkString(ml1, xy=[(-10, 0), (0, 0), (10, 0), (10, 10)], hls=[20, 22], order=5, layers=[0])
ml1.solve()
ml1.contour([-20, 20, -20, 20], 50, [0], 40)









    



Number of elements, Number of equations: 2 , 19
..
solution complete



In [23]:

    
xls1 = np.linspace(-10, 10, 50)
yls1 = np.linspace(0, 0, 50)
hls1 = ml1.headalongline(xls1, yls1)
figure()
plot(xls1, hls1[0])
xls2 = np.linspace(10, 10, 50)
yls2 = np.linspace(0, 10, 50)
hls2 = ml1.headalongline(xls2, yls2)
plot(10 + yls2, hls2[0])









    














    











    Out[23]:





[<matplotlib.lines.Line2D at 0x107f42828>]



In [24]:

    
for ls in ls1.lslist:
    print(ml1.headalongline(ls.xc, ls.yc)[0])
    print(ls.hc)









    



[ 20.03301038  20.1255034   20.25915969  20.40750698  20.54116327
  20.63365629]
[ 20.03301038  20.1255034   20.25915969  20.40750698  20.54116327
  20.63365629]
[ 20.69967704  20.79217007  20.92582636  21.07417364  21.20782993
  21.30032296]
[ 20.69967704  20.79217007  20.92582636  21.07417364  21.20782993
  21.30032296]
[ 21.36634371  21.45883673  21.59249302  21.74084031  21.8744966
  21.96698962]
[ 21.36634371  21.45883673  21.59249302  21.74084031  21.8744966
  21.96698962]

add resistance



In [25]:

    
ml1 = ModelMaq(kaq=[20, 10], z=[20, 12, 10, 0], c=[100])
rf1 = Constant(ml1, xr=0, yr=200, hr=2)
ls1 = HeadLineSinkString(ml1, xy=[(-10, 0), (0, 0), (10, 0), (10, 10)], hls=[0, 1], 
                          res=2, wh=5, order=5, layers=[0])
ml1.solve()









    



Number of elements, Number of equations: 2 , 19
..
solution complete



In [26]:

    
xls1 = np.linspace(-10, 10, 50)
yls1 = np.linspace(0, 0, 50)
hls1 = ml1.headalongline(xls1, yls1)
figure()
plot(xls1, hls1[0])
xls2 = np.linspace(10, 10, 50)
yls2 = np.linspace(0, 10, 50)
hls2 = ml1.headalongline(xls2, yls2)
plot(10 + yls2, hls2[0])









    














    











    Out[26]:





[<matplotlib.lines.Line2D at 0x107f81160>]



In [27]:

    
for ls in ls1.lslist:
    print(ml1.headalongline(ls.xc, ls.yc)[0])
    print(ls.hc)









    



[ 1.73602412  1.72073513  1.70756748  1.69921681  1.69538532  1.6943048 ]
[ 0.01650519  0.0627517   0.12957984  0.20375349  0.27058163  0.31682814]
[ 1.69420359  1.69490577  1.69752156  1.70261389  1.70968187  1.71733186]
[ 0.34983852  0.39608503  0.46291318  0.53708682  0.60391497  0.65016148]
[ 1.71953228  1.71919368  1.72456317  1.73498697  1.74775754  1.75930814]
[ 0.68317186  0.72941837  0.79624651  0.87042016  0.9372483   0.98349481]



In [28]:

    
ls.res









    Out[28]:





2

Ditch string



In [29]:

    
ml1 = ModelMaq(kaq=[20, 10], z=[20, 12, 10, 0], c=[100])
rf1 = Constant(ml1, xr=0, yr=20, hr=1)
ls1 = LineSinkDitchString(ml1, xy=[(-10, 0), (0, 0), (10, 0)], Qls=100, wh=2, res=5, order=2, layers=[0])
ml1.solve()
print('discharge:', ls1.discharge())









    



Number of elements, Number of equations: 2 , 7
..
solution complete
discharge: [ 100.    0.]



In [30]:

    
ml1.contour([-20, 20, -20, 20], 50, [0], 20)

Ditch in different layers



In [31]:

    
ml1 = ModelMaq(kaq=[20, 10], z=[20, 12, 10, 0], c=[100])
rf1 = Constant(ml1, xr=0, yr=20, hr=1)
ls1 = LineSinkDitchString(ml1, xy=[(-10, 0), (0, 0), (10, 0), (10, 20)], Qls=100, wh=2, res=5, order=2, layers=[0,1,0])
ml1.solve()









    



Number of elements, Number of equations: 2 , 10
..
solution complete

Angle well



In [32]:

    
ml = Model3D(kaq=1, z=np.arange(10, -0.1, -0.2), kzoverkh=0.1, topboundary='semi', topres=0, topthick=2, hstar=7)
xy = list(zip(np.linspace(-10, 10, 21), np.zeros(21)))
ls = LineSinkDitchString(ml, xy=xy, Qls=100, wh=2, res=5, order=2, layers=np.arange(10, 30, 1))
ml.solve()









    



Number of elements, Number of equations: 2 , 60
..
solution complete



In [33]:

    
ml.vcontour([-20, 20, 0, 0], 100, 20)