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Hi guys,<div><br></div><div><br></div><div>[pre-scriptum: have you tried to compute Nupdate in just one of your simulation (as pointed out in my previous post)? I am really interested to know this number that you can compute in a couple of seconds (please write the used modules, Hydro, Gravity, etc.). </div><div>Nupdate is simply the number of cells updated per second and per processor. Could you be so kind to provide this number?]</div><div><br></div><div><br></div><div>I was doing a benchmark to test FLASH (+Hydro) with the Sedov problem and I found that the computed analytic solution was incorrect.</div><div>There are two corrections to do:</div><div><br></div><div><br></div><div>1) Major. In sedov_vfunc of Simulation_initBlock.F90 it is missing Vpfe in the last line (s<font class="Apple-style-span" size="2">ee Landau & Lifshitz, Fluid Mechanics, p.405)):</font></div><div><br></div><div><br></div><div><div> xi = nu1*alog(5.-(3.*gamma-1.)*Vfpe) + & </div><div> nu2*alog(gamma*Vfpe-1.) - & </div><div> nu1*alog(gam7/gamp1) - nu2*alog(gamm1/gamp1) - & </div><div> 2.*alog(0.5*gamp1 *Vfpe) !<--- Vfpe WAS MISSING (see Landau p. 405)</div></div><div><br></div><div><br></div><div>2) Minor. The solution for xi --> 0. gives estimate for P about two orders of magnitude lower (in set_analytic_sedov). </div><div>Therefore, I have taken the solution given in the book of Sedov (1959), which provides the correct asymptotic value for pressure (constant).</div><div><div style="text-indent: 0px !important; ">This should be used at least when:</div><div style="text-indent: 0px !important; "><br></div><div style="text-indent: 0px !important; "> if (xi .le. 0.075) then ! Use asymptotic r->0 solution. </div><div style="text-indent: 0px !important; "><br></div><div style="text-indent: 0px !important; ">Note that ".le. 1.e-6" is too low.</div></div><div style="text-indent: 0px !important; "><br></div><div><br></div><div>OLD (what is the reference?):</div><div><br></div><div><div> kappa = ( (0.5*gamp1/gamma)**2. * & </div><div> (gamp1/gam7* & </div><div> (5.-(3.*gamma-1.)/gamma))**nu1 )**(1./nu2) * & </div><div> gamm1/gamp1 / gamma</div><div> </div><div><br></div><div> epsilon = k**(nu5+1.) * (k*gamma*kappa)**nu3 * & </div><div> (gamp1/gam7*(3.*gamma-1.))**nu4 * & </div><div> ((2.*gamma+1)/gamma/(3.*gamma-1.)) * & </div><div> (gamm1/gamma)</div><div> zeta = gamm1*gamm1/(2.*gamma*gamma*kappa)</div><div> p(i) = rho_ambient/gamma * 0.16*(R0/t)**2 * epsilon*zeta</div><div><br></div><div><br></div><div>NEW (ok): </div><div><br></div><div> kappa = (0.32/2.**(6./5.)) *&</div><div> ((gamma+1.)**((gamma+1.)/(3.*gamma-1.))/gamma**(4.*gamma/(3.*gamma-1.))) *&</div><div> ((2.*gamma+1.)/(7.-gamma))**((13.*gamma**2-7.*gamma+12.)/&</div><div> (5.*(2.-gamma)*(3.*gamma-1.))) *&</div><div> beta**2 !--> Sedov assumed beta =1, see Eq. 11.11</div><div> p(i) = kappa*rho_ambient*(E/rho_ambient)**(2./5.) *t**(-6./5.)</div></div><div><br></div><div><br></div><div><br></div><div><br></div><div><br></div><div>Now the analytic and simulated profiles are consistent (within errors).</div><div>The plots in the user's guide seem somehow correct (e.g. Fig. 25.9); did you use other routines?</div><div><br></div><div><br></div><div>I would also advise to use higher n_iter_max (>1000) and to add only V = 1./gamma + 1.e-18, instead of</div><div>V = 1./gamma + 1.e-8 (in compute_sedov_v).</div><div><br></div><div><br></div><div><br></div><div>I hope this can help if you are testing Sedov problem.</div><div>(Let me know if I missed something)</div><div><br></div><div><br></div><div>Cheers,</div><div><br></div><div><br></div><div><br></div><div> Max</div><div><br></div> </div></body>
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