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August 20, 2011, 20:29 |
@ Glenn
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#21 |
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Hareesh R Iyer
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Hi, pretty long time since I posted something here! Lateral sloshing can be simulated easily in CFX, but only managed to do so by lateral external excitation. How can the lateral slosh condition due to pitching be simulated?
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August 22, 2011, 21:38 |
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#22 |
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Glenn Horrocks
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With a moving mesh approach you can move the tank in any direction you like.
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August 22, 2011, 21:54 |
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#23 |
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Hareesh R Iyer
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@ Glenn..
Yes, am trying to crack it that way. If you find something sometime pls add it up here. |
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December 20, 2012, 00:33 |
Help
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#24 |
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Hi im new to CFD and i am trying to simulate tank sloshing for 10s but apparently my time does not run at all. It always get stuck at 0.0s even though the system is running. Can anyone help me with this? Help is greatly appreciated!
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December 20, 2012, 00:34 |
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#25 |
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Glenn Horrocks
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Please post your output file.
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December 20, 2012, 00:43 |
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#26 |
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Hi as seen in the attached pictures, the water does not flow at all. The tank is moving at weird angles thou. Thanks for the help.
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December 20, 2012, 00:52 |
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#27 |
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Glenn Horrocks
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Your simulation is in Fluent. Try the fluent forum.
But two general comments - have you set the gravity vector? And have you run the simulation long enough that the fluid would move? |
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December 20, 2012, 11:56 |
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#28 |
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Hareesh R Iyer
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10s is fair enough for simulation
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December 20, 2012, 16:44 |
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#29 |
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Glenn Horrocks
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Not necessarily, it depends on the length and viscosity scales (and a few other parameters, not sure what non-dimensional number that is). If the fluid is highly viscous it will take a while for the fluid to move enough that you will see it.
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December 20, 2012, 21:00 |
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#30 |
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Hareesh R Iyer
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glenn is right..if viscosity, impulse time etc are high, 10s is insufficient
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December 26, 2012, 03:58 |
How to input 'sin' eqn as part of the operating conditon
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#31 |
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Happy boxing day everyone. With reference to the sloshing tutorial, instead of moving the tank in -9.81m/s in the x direction, may i know how do i input the speed if my speed is in the form Asinwt where 'A= ampitute', 't = seconds' and 'w = 2*pi*f'. For example, how to input 2.5sin(80*pi)t where the tank is undergoing a vibratory motion in the x direction. Thanks for the help!
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December 26, 2012, 05:04 |
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#32 |
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Glenn Horrocks
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Replace the constant value with a CEL expression. Several CFX tutorials show how to use CEL expressions.
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December 26, 2012, 09:52 |
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#33 | |
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Hareesh R Iyer
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Quote:
Use CEL help to write functions.. 1st assign a func name say Gx.. and try as follows: Gx = 2.5 [mm] * sine((80 [rad/s] * pi()) * t [sec]) it depends on your omega, amp & timestep size. @ Glenn, Pl verify on the feasibility of this expression in CFX (actually I posted it without referring back)..so not sure if my memory is correct |
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December 27, 2012, 05:51 |
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#34 |
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Glenn Horrocks
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There are a few minor problems with the CEL, but they would stop it from working and end the simulation with an error:
1) The sine function is "sin", not "sine". 2) The pi variable is pi, not pi(). 3) The variable t is already in units of time. There is no need to define units for t as they are already defined. Note I have not checked these through CFX either, but hopefully they are correct. |
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December 27, 2012, 06:58 |
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#35 |
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Thanks for the help, appreciate it! I will give it a try
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December 27, 2012, 10:56 |
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#36 |
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Hareesh R Iyer
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Glenn, the CEL expressions have to be given such that each parameter is followed strictly by its unit in "[ ]"..otherwise you are correct.
Nathan, actually for acceleration as a variable expression, the 1st term in the sample excitation you asked would be steady state acceleration & not displacement. so the unit need to be "[mm/sec^2]" or so.. |
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December 28, 2012, 04:28 |
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#37 | |
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Glenn Horrocks
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Quote:
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December 29, 2012, 20:41 |
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#38 |
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Hareesh R Iyer
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here is a sample working expression
Gx = 2.5*sin(20*t[s]/1[s]) if after a particular time, this excitation has to stop, use step function for the value of Gx by giving it as a new expression. Then call it into the acceleration definition of the problem. |
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December 30, 2012, 06:25 |
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#39 |
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Glenn Horrocks
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I see, I was not aware an equation like that would even work.
Then I simply say it is bad practise. If the variable t is already a variable of time then the [s] is unnecessary, if the variable is of units other than time then I have no idea what it will do, but it sounds undefined. If you remove the [s] does it still return exactly the same result? |
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December 30, 2012, 20:52 |
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#40 |
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Hareesh R Iyer
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if [s] is removed from t, then 1 [s] can also be removed, but there is inconsistency.
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