Hello Yvan,
Thank you for your reply. Following your suggestions, in addition to the changes I mentioned in clptur, distyp and tridim, I also altered the test in dvvpst.f90 to the following:
if (ineedy.eq.1 .and. abs(icdpar).eq.1) then
ineeyp = 0
do iphas = 1, nphas
if( (itytur(iphas).eq.4.and.idries(iphas).eq.1) &
.or. (iturb(iphas).eq.41.and.idries(iphas).eq.0) ) then
ineeyp = 1
endif
enddo
if (ineeyp.eq.1) then
NAMEVR = 'Yplus'
idimt = 1
ientla = 0
ivarpr = 1
call psteva(nummai, namevr, idimt, ientla, ivarpr, &
!==========
ntcabs, ttcabs, ra(iyppar), rbid, rbid)
endif
endif
However, I still only obtain values for the quad4 cells in the chr.ensight folder. So I also tried the following code in usvpst to output the values of yplus on the volume
namevr = 'yplus_vol'
idimt = 1
ivarpr = 0
do iloc = 1, ncelps
iel = lstcel(iloc)
tracel(iloc) = ra(iyppar+iel-1)
enddo
call psteva(ipart, namevr, idimt, ientla, ivarpr, &
ntcabs, ttcabs, tracel, trafac, trafbr)
and on the boundary
do ii = 1, 32
namevr (ii:ii) = ' '
enddo
namevr = "yplus_sur"
idimt = 1
ivarpr = 0
do iloc = 1, nfbrps
ifac = lstfbr(iloc)
trafbr(iloc) = ra(iyplbr+ifac-1)
enddo
! Output values
call psteva(ipart, namevr, idimt, ientla, ivarpr, &
!==========
ntcabs, ttcabs, tracel, trafac, trafbr)
I now obtain an output for yplus on both the volume and the boundary cells, but I would like to confirm that the above code is correct before I invest days for the actual simulation. On this subject, I would also like to ask a rather basic question about restarts: if I have a total number of time steps of 50000 for example, is there a way to interrupt the calculation before the end and then restart from that point? In other words, if I choose to post-process every ‘n’ time steps and interrupt a calculation before the end, is there a way to restart from the last saved data? So far I have only been able to restart once the previous calculation had completed the total number of time steps, so I’ve had to break the calculation down into ‘intervals’ and then restart each time until I reached the desired total number of time steps.I did this to avoid losing large amounts of computation times if something went wrong before the end of the simulation, but perhaps I’m missing something obvious and I would be very grateful for a clarification.
Thanks again,
Kristin