Last change
on this file since 6748 was
4959,
checked in by steve, 17 years ago
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Copied John Jakeman's and Sudi Mungkasi's 1d shallow water code
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File size:
3.8 KB
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1 | |
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2 | |
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3 | class AnalyticDam: |
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4 | |
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5 | def __init__(self, h0 = 5.0, h1 = 10.0, L = 2000.0): |
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6 | from math import sqrt |
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7 | |
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8 | self.h0 = h0 # depth upstream (m) |
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9 | self.h1 = h1 # depth downstream (m) |
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10 | self.L = L # length of domain |
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11 | |
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12 | g = 9.81 # gravity (m/s^2) |
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13 | |
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14 | c0 = sqrt(g*h0) #left celerity |
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15 | c1 = sqrt(g*h1) #right celerity |
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16 | |
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17 | zmin=-100.0 |
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18 | zmax=101.0 |
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19 | for i in range(100): |
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20 | z=(zmin+zmax)/2.0 |
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21 | u2=z-c0*c0/4.0/z*(1.0+sqrt(1.0+8.0*z*z/c0/c0)) |
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22 | c2=c0*sqrt(0.5*(sqrt(1.0+8.0*z*z/c0/c0)-1.0)) |
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23 | func=2.0*c1/c0-u2/c0-2.0*c2/c0 |
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24 | if (func > 0.0): |
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25 | zmin=z |
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26 | else: |
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27 | zmax=z |
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28 | |
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29 | if( abs(z) > 99.0): |
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30 | print 'no convergence' |
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31 | |
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32 | self.u2 = u2 |
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33 | self.c0 = c0 |
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34 | self.c1 = c1 |
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35 | self.c2 = c2 |
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36 | self.g = g |
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37 | self.z = z |
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38 | |
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39 | |
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40 | |
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41 | def __call__(self, C,t): |
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42 | |
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43 | from Numeric import zeros,Float |
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44 | from math import sqrt |
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45 | |
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46 | #t = 0.0 # time (s) |
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47 | h0 = self.h0 |
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48 | h1 = self.h1 |
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49 | L = self.L |
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50 | n = len(C) # number of cells |
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51 | |
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52 | u2 = self.u2 |
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53 | c0 = self.c0 |
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54 | c1 = self.c1 |
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55 | c2 = self.c2 |
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56 | |
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57 | g = self.g |
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58 | z = self.z |
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59 | |
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60 | u = zeros(n,Float) |
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61 | h = zeros(n,Float) |
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62 | uh = zeros(n,Float) |
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63 | x = C-3*L/4.0 |
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64 | #x = zeros(n,Float) |
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65 | #for i in range(n): |
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66 | # x[i] = C[i]-1000.0 |
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67 | |
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68 | # Upstream and downstream boundary conditions are set to the intial water |
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69 | # depth for all time. |
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70 | |
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71 | # Calculate Shock Speed |
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72 | #h2 = 7.2692044 |
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73 | |
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74 | #S2 = 2*h2/(h2-h0)*(sqrt(g*h1)-sqrt(g*h2)) |
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75 | #u2 = S2 - g*h0/(4*S2)*(1+sqrt(1+8*S2*S2/(g*h0))) |
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76 | |
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77 | h2=h0/(1.0-u2/z) |
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78 | x3=(u2-c2)*t |
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79 | x2=z*t |
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80 | x1=-c1*t |
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81 | x1_ = -1*L/2.0-x1 |
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82 | x2_ = -1*L/2.0+2*x1 |
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83 | #x3_ = -1*L/2.0-x3 |
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84 | #t=50 |
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85 | #x = (-L/2:L/2) |
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86 | for i in range(n): |
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87 | # Calculate Analytical Solution at time t > 0 |
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88 | u3 = 2.0/3.0*(sqrt(g*h1)+x[i]/t) |
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89 | h3 = 4.0/(9.0*g)*(sqrt(g*h1)-x[i]/(2.0*t))*(sqrt(g*h1)-x[i]/(2.0*t)) |
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90 | u3_ = 2.0/3.0*((x[i]+L/2.0)/t-sqrt(g*h1)) |
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91 | h3_ = 1.0/(9.0*g)*((x[i]+L/2.0)/t+2*sqrt(g*h1))*((x[i]+L/2.0)/t+2*sqrt(g*h1)) |
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92 | #if t == 30: |
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93 | # x[i] = 500 |
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94 | # print 'x2',x2 |
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95 | # print 'x3',x3 |
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96 | # print 'x1',x1 |
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97 | if ( x[i] <= x2_ ): |
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98 | #print 'here x2_=', x2_ |
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99 | u[i] = 0.0 |
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100 | h[i] = 0.0 |
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101 | uh [i] = u[i]*h[i] |
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102 | #elif ( x[i] <= x3_ ): |
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103 | # print 'here x3_=', x3_ |
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104 | # u[i] = -1*u2 |
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105 | # h[i] = h2 |
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106 | # uh[i] = u[i]*h[i] |
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107 | elif ( x[i] <= x1_ ): |
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108 | #print 'here x1_=', x1_ |
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109 | u[i] = u3_ |
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110 | h[i] = h3_ |
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111 | uh[i] = u[i]*h[i] |
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112 | #else: |
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113 | # u[i] = 0.0 |
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114 | # h[i] = h0 |
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115 | # uh[i] = u[i]*h[i] |
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116 | |
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117 | #elif ( x[i] <= x1/2.0 ): |
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118 | # u[i] = 0.0 |
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119 | # h[i] = h1 |
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120 | # uh[i] = u[i]*h[i] |
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121 | elif ( x[i] <= x1 ): |
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122 | #print 'here x1=', x1 |
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123 | u[i] = 0.0 |
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124 | h[i] = h1 |
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125 | uh[i] = u[i]*h[i] |
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126 | elif ( x[i] <= x3 ): |
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127 | #print 'here x3=', x3 |
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128 | u[i] = u3 |
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129 | h[i] = h3 |
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130 | uh[i] = u[i]*h[i] |
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131 | elif ( x[i] < x2 ): |
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132 | #print 'here x2=', x2 |
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133 | u[i] = u2 |
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134 | h[i] = h2 |
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135 | uh[i] = u[i]*h[i] |
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136 | else: |
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137 | #print 'here the last section' |
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138 | u[i] = 0.0 |
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139 | h[i] = h0 |
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140 | uh[i] = u[i]*h[i] |
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141 | |
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142 | return h , uh |
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