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Commit 8f24cc3f authored by Matthieu Schaller's avatar Matthieu Schaller
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Updated documentation of plotting scripts

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examples/SedovBlast_2D/makeIC.py
+ 3
− 6
View file @ 8f24cc3f
......@@ -31,8 +31,6 @@ E0= 1. # Energy of the explosion
N_inject = 21 # Number of particles in which to inject energy
fileName = "sedov.hdf5"
#L = 101
#---------------------------------------------------
glass = h5py.File("glassPlane_128.hdf5", "r")
......@@ -50,10 +48,9 @@ m = zeros(numPart)
u = zeros(numPart)
r = zeros(numPart)
for i in range(numPart):
r[i] = sqrt((pos[i,0] - 0.5)**2 + (pos[i,1] - 0.5)**2)
m[i] = rho0 * vol / numPart
u[i] = P0 / (rho0 * (gamma - 1))
r = sqrt((pos[:,0] - 0.5)**2 + (pos[:,1] - 0.5)**2)
m[:] = rho0 * vol / numPart
u[:] = P0 / (rho0 * (gamma - 1))
# Make the central particle detonate
index = argsort(r)
......
examples/SedovBlast_2D/plotSolution.py
+ 7
− 1
View file @ 8f24cc3f
......@@ -18,7 +18,7 @@
#
##############################################################################
# Computes the analytical solution of the 3D Sedov blast wave.
# Computes the analytical solution of the 2D Sedov blast wave.
# The script works for a given initial box and dumped energy and computes the solution at a later time t.
# Parameters
......@@ -85,6 +85,8 @@ P = sim["/PartType0/Pressure"][:]
rho = sim["/PartType0/Density"][:]
# Now, work our the solution....
from scipy.special import gamma as Gamma
from numpy import *
......@@ -190,6 +192,8 @@ def sedov(t, E0, rho0, g, n=1000, nu=3):
rho *= rho0
return r, p, rho, u, r_s, p_s, rho_s, u_s, shock_speed
# The main properties of the solution
r_s, P_s, rho_s, v_s, r_shock, _, _, _, _ = sedov(time, E_0, rho_0, gas_gamma, 1000, 2)
# Append points for after the shock
......@@ -202,6 +206,8 @@ v_s = np.insert(v_s, np.size(v_s), [0, 0])
u_s = P_s / (rho_s * (gas_gamma - 1.)) #internal energy
s_s = P_s / rho_s**gas_gamma # entropic function
# Plot the interesting quantities
figure()
......
examples/SedovBlast_2D/sedov.py deleted 100755 → 0
+ 0
− 212
View file @ 938d7745
###############################################################################
# This file is part of SWIFT.
# Copyright (c) 2015 Bert Vandenbroucke (bert.vandenbroucke@ugent.be)
#
# This program is free software: you can redistribute it and/or modify
# it under the terms of the GNU Lesser General Public License as published
# by the Free Software Foundation, either version 3 of the License, or
# (at your option) any later version.
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
# GNU General Public License for more details.
#
# You should have received a copy of the GNU Lesser General Public License
# along with this program. If not, see <http://www.gnu.org/licenses/>.
#
##############################################################################
import numpy as np
import scipy.integrate as integrate
import scipy.optimize as optimize
import os
# Calculate the analytical solution of the Sedov-Taylor shock wave for a given
# number of dimensions, gamma and time. We assume dimensionless units and a
# setup with constant density 1, pressure and velocity 0. An energy 1 is
# inserted in the center at t 0.
#
# The solution is a self-similar shock wave, which was described in detail by
# Sedov (1959). We follow his notations and solution method.
#
# The position of the shock at time t is given by
# r2 = (E/rho1)^(1/(2+nu)) * t^(2/(2+nu))
# the energy E is related to the inserted energy E0 by E0 = alpha*E, with alpha
# a constant which has to be calculated by imposing energy conservation.
#
# The density for a given radius at a certain time is determined by introducing
# the dimensionless position coordinate lambda = r/r2. The density profile as
# a function of lambda is constant in time and given by
# rho = rho1 * R(V, gamma, nu)
# and
# V = V(lambda, gamma, nu)
#
# The function V(lambda, gamma, nu) is found by solving a differential equation
# described in detail in Sedov (1959) chapter 4, section 5. Alpha is calculated
# from the integrals in section 11 of the same chapter.
#
# Numerically, the complete solution requires the use of 3 quadratures and 1
# root solver, which are implemented using the GNU Scientific Library (GSL).
# Since some quadratures call functions that themselves contain quadratures,
# the problem is highly non-linear and complex and takes a while to converge.
# Therefore, we tabulate the alpha values and profile values the first time
# a given set of gamma and nu values is requested and reuse these tabulated
# values.
#
# Reference:
# Sedov (1959): Sedov, L., Similitude et dimensions en mecanique (7th ed.;
# Moscou: Editions Mir) - french translation of the original
# book from 1959.
# dimensionless variable z = gamma*P/R as a function of V (and gamma and nu)
# R is a dimensionless density, while P is a dimensionless pressure
# z is hence a sort of dimensionless sound speed
# The expression below corresponds to eq. 11.9 in Sedov (1959), chapter 4
def _z(V, gamma, nu):
if V == 2./(nu+2.)/gamma:
return 0.
else:
return (gamma-1.)*V*V*(V-2./(2.+nu))/2./(2./(2.+nu)/gamma-V)
# differential equation that needs to be solved to obtain lambda for a given V
# corresponds to eq. 5.11 in Sedov (1959), chapter 4 (omega = 0)
def _dlnlambda_dV(V, gamma, nu):
nom = _z(V, gamma, nu) - (V-2./(nu+2.))*(V-2./(nu+2.))
denom = V*(V-1.)*(V-2./(nu+2.))+nu*(2./(nu+2.)/gamma-V)*_z(V, gamma, nu)
return nom/denom
# dimensionless variable lambda = r/r2 as a function of V (and gamma and nu)
# found by solving differential equation 5.11 in Sedov (1959), chapter 4
# (omega = 0)
def _lambda(V, gamma, nu):
if V == 2./(nu+2.)/gamma:
return 0.
else:
V0 = 4./(nu+2.)/(gamma+1.)
integral, err = integrate.quad(_dlnlambda_dV, V, V0, (gamma, nu),
limit = 8000)
return np.exp(-integral)
# dimensionless variable R = rho/rho1 as a function of V (and gamma and nu)
# found by inverting eq. 5.12 in Sedov (1959), chapter 4 (omega = 0)
# the integration constant C is found by inserting the R, V and z values
# at the shock wave, where lambda is 1. These correspond to eq. 11.8 in Sedov
# (1959), chapter 4.
def _R(V, gamma, nu):
if V == 2./(nu+2.)/gamma:
return 0.
else:
C = 8.*gamma*(gamma-1.)/(nu+2.)/(nu+2.)/(gamma+1.)/(gamma+1.) \
*((gamma-1.)/(gamma+1.))**(gamma-2.) \
*(4./(nu+2.)/(gamma+1.)-2./(nu+2.))
lambda1 = _lambda(V, gamma, nu)
lambda5 = lambda1**(nu+2)
return (_z(V, gamma, nu)*(V-2./(nu+2.))*lambda5/C)**(1./(gamma-2.))
# function of which we need to find the zero point to invert lambda(V)
def _lambda_min_lambda(V, lambdax, gamma, nu):
return _lambda(V, gamma, nu) - lambdax
# dimensionless variable V = v*t/r as a function of lambda (and gamma and nu)
# found by inverting the function lamdba(V) which is found by solving
# differential equation 5.11 in Sedov (1959), chapter 4 (omega = 0)
# the invertion is done by searching the zero point of the function
# lambda_min_lambda defined above
def _V_inv(lambdax, gamma, nu):
if lambdax == 0.:
return 2./(2.+nu)/gamma;
else:
return optimize.brentq(_lambda_min_lambda, 2./(nu+2.)/gamma,
4./(nu+2.)/(gamma+1.), (lambdax, gamma, nu))
# integrand of the first integral in eq. 11.24 in Sedov (1959), chapter 4
def _integrandum1(lambdax, gamma, nu):
V = _V_inv(lambdax, gamma, nu)
if nu == 2:
return _R(V, gamma, nu)*V**2*lambdax**3
else:
return _R(V, gamma, nu)*V**2*lambdax**4
# integrand of the second integral in eq. 11.24 in Sedov (1959), chapter 4
def _integrandum2(lambdax, gamma, nu):
V = _V_inv(lambdax, gamma, nu)
if V == 2./(nu+2.)/gamma:
P = 0.
else:
P = _z(V, gamma, nu)*_R(V, gamma, nu)/gamma
if nu == 2:
return P*lambdax**3
else:
return P*lambdax**4
# calculate alpha = E0/E
# this corresponds to eq. 11.24 in Sedov (1959), chapter 4
def get_alpha(gamma, nu):
integral1, err1 = integrate.quad(_integrandum1, 0., 1., (gamma, nu))
integral2, err2 = integrate.quad(_integrandum2, 0., 1., (gamma, nu))
if nu == 2:
return np.pi*integral1+2.*np.pi/(gamma-1.)*integral2
else:
return 2.*np.pi*integral1+4.*np.pi/(gamma-1.)*integral2
# get the analytical solution for the Sedov-Taylor blastwave given an input
# energy E, adiabatic index gamma, and number of dimensions nu, at time t and
# with a maximal outer region radius maxr
def get_analytical_solution(E, gamma, nu, t, maxr = 1.):
# we check for the existance of a datafile with precomputed alpha and
# profile values
# if it does not exist, we calculate it here and write it out
# calculation of alpha and the profile takes a while...
lvec = np.zeros(1000)
Rvec = np.zeros(1000)
fname = "sedov_profile_gamma_{gamma}_nu_{nu}.dat".format(gamma = gamma,
nu = nu)
if os.path.exists(fname):
file = open(fname, "r")
lines = file.readlines()
alpha = float(lines[0])
for i in range(len(lines)-1):
data = lines[i+1].split()
lvec[i] = float(data[0])
Rvec[i] = float(data[1])
else:
alpha = get_alpha(gamma, nu)
for i in range(1000):
lvec[i] = (i+1)*0.001
V = _V_inv(lvec[i], gamma, nu)
Rvec[i] = _R(V, gamma, nu)
file = open(fname, "w")
file.write("#{alpha}\n".format(alpha = alpha))
for i in range(1000):
file.write("{l}\t{R}\n".format(l = lvec[i], R = Rvec[i]))
xvec = np.zeros(1002)
rhovec = np.zeros(1002)
if nu == 2:
r2 = (E/alpha)**0.25*np.sqrt(t)
else:
r2 = (E/alpha)**0.2*t**0.4
for i in range(1000):
xvec[i] = lvec[i]*r2
rhovec[i] = Rvec[i]
xvec[1000] = 1.001*r2
xvec[1001] = maxr
rhovec[1000] = 1.
rhovec[1001] = 1.
return xvec, rhovec
def main():
E = 1.
gamma = 1.66667
nu = 2
t = 0.001
x, rho = get_analytical_solution(E, gamma, nu, t)
for i in range(len(x)):
print x[i], rho[i]
if __name__ == "__main__":
main()
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