# Transient simulation : Static temperature and time averaged static temperature

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 July 2, 2013, 19:39 Transient simulation : Static temperature and time averaged static temperature #1 Senior Member   --------- Join Date: Oct 2010 Posts: 303 Rep Power: 9 I have a transient simulation where I monitored the volume averaged static temperature and the volume averaged mean (time averaged) static temperature i.e.., for instance : Monitor1 - volumeAve(Temperature)@fluiddomain Monitor2 - volumeAve(Temperature.Trnavg)@fluiddomain I observed that the plots of Monitor1 and Monitor2 do not exhibit the same trend. i.e.., the monitor1 which is the temperature at each time step shows a steady trend with the time. so I expected that the Monitor2 which is the time averaged temperature also to show the same trend. Which is however not the case, monitor1 is steady and monitor2 increases in time from a very low value. This is how I actually proceeded - - Carried out 20 iterations in steady state just to initialize the transient simulation - Initialized the transient run with the steady state solution (20 iterations) - Ran the transient simulation with a time step size of 0.1 [s] - Increased the time step size to 0.5 [s] - The simulation was allowed to run continuously ( i.e.., there was never a reinitialization of any statistic) - The quantity volumeAve(Temperature.Trnavg)@fluiddomain - Monitor2 was monitored right from the timestep1 Could some one please explain these trends of the monitor points __________________ Best regards, Santhosh. Last edited by saisanthoshm88; July 3, 2013 at 03:33.

 July 3, 2013, 07:13 #2 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 12,638 Rep Power: 98 Do some simple hand calculations and you will work it out - Take the monitor 1 values for 1, 2, 3, .... , last time steps and calculate the trnavg - You will find it increases from zero and asymptotically approaches the steady value like you are seeing.

 July 3, 2013, 11:05 #3 Senior Member   --------- Join Date: Oct 2010 Posts: 303 Rep Power: 9 Glenn, I understand time weighted averaging as follows - i.e.., for instance if the time step size = 0.1 [s] and if the volumeAve(Temperature)@fluid = 300 [K] after the first time step (=0.1 [s]) And volumeAve(Temperature)@fluid = 310 [K] after the second time step (= 0.2[s]) Then during the third time step, volumeAve(Temperature.Trnavg)@fluid = ((300*0.1) + (310*0.2))/(0.1+0.2) Please correct me if I'm wrong __________________ Best regards, Santhosh.

 July 3, 2013, 19:08 #4 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 12,638 Rep Power: 98 The time step size for the second time step is still 0.1.

 July 4, 2013, 02:18 #5 Senior Member   --------- Join Date: Oct 2010 Posts: 303 Rep Power: 9 Glenn, I got the point. Thank you very much for the clarification. __________________ Best regards, Santhosh.

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