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Transient simulation : Static temperature and time averaged static temperature

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Old   July 2, 2013, 19:39
Default Transient simulation : Static temperature and time averaged static temperature
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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
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Last edited by saisanthoshm88; July 3, 2013 at 03:33.
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Old   July 3, 2013, 07:13
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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.
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Old   July 3, 2013, 11:05
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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
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Old   July 3, 2013, 19:08
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The time step size for the second time step is still 0.1.
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Old   July 4, 2013, 02:18
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Glenn,

I got the point. Thank you very much for the clarification.
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