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particle time step size, number of time steps in DPMPM |
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April 22, 2014, 09:21 |
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#21 |
New Member
Join Date: Jan 2014
Posts: 11
Rep Power: 12 |
Hi Amir sir
In my heat sink cooling problem actually I want relative humidity of 15% at the exit of the sink but I am getting it 7.42% , I tried with different combinitions of particle time step size and number of time steps. In this particular process I have given mass flow rate of 4.22e-7 and stop time of injection as 100s for every combinition, But still my rate of evaporation and hence RH is not increasing. (In my problem the relative reynolds number is 0, air velocity is 1 m/s). Can you suggest me a way to resolve the problem? Regards Ruturaj |
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July 3, 2014, 14:34 |
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#22 |
New Member
Join Date: Jan 2014
Posts: 11
Rep Power: 12 |
Hi amir sir
Actually I am able to validate results of one of paper by using DPM in FLUENT. I drawn geometry with symmetry boundary condition, but I was giving total mass flow rate through duct. But that mass flow rate must be half. So with excess mass my sink was getting overcooled. Because of your previous quick replies, I was able to understand various terms in fluent physically. Thanks a lot. |
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July 29, 2016, 14:03 |
Particle status for DPM iteration
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#23 |
New Member
Sagar
Join Date: Apr 2016
Posts: 23
Rep Power: 10 |
Hello
I have read the Fluent User guide and also have read the posts regarding this but I am not able to comprehend the unsteady particle tracking of Fluent. I am running unsteady simulations for continuous phase and also want to have unsteady particle tracking. In the particle treatment I choose unsteady and I also choose to inject particles with fluid flow time step. Till here its good but now it starts to get messy. Firstly, if I want to inject particles at fluid flow time step why does it still asks about particle time step size and what is this parameter exactly? Secondly, the time step size of my continuous phase is 1e-5 and I keep my particle time step size as 0.0001. When I run the simulation, Fluent recognizes the number of parcels but the status for the parcels it prints on the console is wierd. Updating solution at time level N... done. Injecting 31360 particle parcels with mass 5.375e-09 at t = 0 DPM Iteration .... number tracked = 31360, escaped = 0, aborted = 0, trapped = 0, evaporated = 0, incomplete = 0, incomplete_parallel = 0 My question why the status of the particle is zero? When I try to reduce the time step size of the continuous phase to 1e-4, the same 31360 particle parcels are tracked and gives proper status (in my case they are droplets and should evaporate). The mass flow rate of the DPM phase is 0.5375 g/s and I inject 31360 particle parcels from injection file. |
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November 2, 2016, 03:12 |
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#24 | |
Senior Member
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Quote:
Dear Amir, I want to inject a fluid in my flow inside a nozzle that has gas in it, I used plain atomizer model. I am not sure about the diameter. What diameter should I choose? |
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August 25, 2017, 09:16 |
particle time step size
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#25 |
New Member
ahmed elbanna
Join Date: Oct 2016
Posts: 7
Rep Power: 10 |
hi every body, I have benefited from these fruitful discussions, and I reviewed the common resources those are explaining the DPM, and my convictions as follow
1- either you used unsteady or steady particle tracking, there will be a time advancing for the particle to find the trajectory. don't forget that the particle ODE's independent variable is the time(t).therefore, the 'steady' term does not mean the cancelling out of the time derivative. 2- the main difference between the unsteady or steady treatment of the particles is the timing of DPM source update to the continuous phase governing equations. 3- I think that the particle time step size and the number of time steps below the particle treatment are related to when should the DPM sources be updated to the continuous phase equations. |
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August 25, 2017, 09:36 |
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#26 | |
New Member
ahmed elbanna
Join Date: Oct 2016
Posts: 7
Rep Power: 10 |
Quote:
1- either you used unsteady or steady particle tracking, there will be a time advancing for the particle to find the trajectory. don't forget that the particle ODE's independent variable is the time(t).therefore, the 'steady' term does not mean the cancelling out of the time derivative. 2- the main difference between the unsteady or steady treatment of the particles is the timing of DPM source update to the continuous phase governing equations. 3- I think that the particle time step size and the number of time steps below the particle treatment are related to when should the DPM sources be updated to the continuous phase equations. |
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August 24, 2024, 15:27 |
use of number of time steps
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#27 |
New Member
indrajit
Join Date: May 2024
Posts: 1
Rep Power: 0 |
for an injection is this holds true (stedy state continuous phase - unsteady particle tracking)?
Iteration upto which injection will run = Total injection period specified in sec / (number of time steps * Particle time step size ) |
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