Re: rho vs. rho +/- 1 (esp. for computePressureField())
Due to recent bot attacks we have changed the sign-up process. If you want to participate in our forum, first register on this website and then send a message via our contact form.
› Forums › OpenLB › General Topics › rho vs. rho +/- 1 (esp. for computePressureField()) › Re: rho vs. rho +/- 1 (esp. for computePressureField())
July 3, 2013 at 3:20 pm
#2107
Randy Roberts
Member
Dear Mathias,rnrnThank you for your reply.rnrnI need to sit down and calculate some values for f^eq and f to get a better feel for question 1).rnrnI’m still a bit uncertain as to question 2). In He/Lou page 942, right after eq. (A13b) it saysrn P = c_s^2 rho / rho_0rnrnSince you are calculating the actual rho (not rho+1) inrn block.get(iX,iY).computeRho(),rnshould you not need thern fields.pressureField.get(iX,iY) -= (T)1rnexpression in order to calculate the pressure?rnrnRandy