# Transient case with no inlet region

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 March 8, 2011, 09:20 Transient case with no inlet region #1 New Member   Join Date: Mar 2011 Posts: 7 Rep Power: 13 I am trying to simulate how the temperature and movement of air inside an oven change in time. More specifically, my project is similar to the one of Tutorial 6 (BuoyantFlow). I have a domain, which is initially full with air at 25oC. A wall is defined as a source of heat flux and as a result the air starts to heat and moves upward since its density changes. In this case, I do not have an inlet region, but I do have an opening region at the top of the domain (like a chimney) through which air is allowed to escape. It is obvious that this is a transient case. My first question is: From literature, I know that in order the temperature to reach a certain value, 8 hours are required. So, I should run a transient case with total time of 8 hours. But with what timestep, if the only thing I want to see is if the temperature reaches the above value after 8 hours? The second question: Since new volume of air does not enter the domain, what happens inside the domain when air leaves it? I mean, is the air that leaves the domain somehow compensated by the solver? Normally, shouldn't, after a certain time, the domain be completely empty? Thank you in advance

March 8, 2011, 12:01
#2
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Attesz
Join Date: Mar 2009
Location: Munich
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Quote:
 Originally Posted by Ancient_Nik But with what timestep, if the only thing I want to see is if the temperature reaches the above value after 8 hours?
use adaptive time step to find the most appropriate timestep for you. But 8 hours is very long, you have to define minutes as timestep, which causes propably large CFL numbers. One of my simulation which is same like this diverges with 30s timestep as well!

Quote:
 Originally Posted by Ancient_Nik Since new volume of air does not enter the domain, what happens inside the domain when air leaves it? I mean, is the air that leaves the domain somehow compensated by the solver? Normally, shouldn't, after a certain time, the domain be completely empty?
If you have only an opening defined, only a slightly mass flow will occur at there. So the air won't exit from the domain, why would it do that? Think at the continuity law. You have to define inlet(s): for exaple gaps at windows, an opened door or something. The domain can not be completely empty, it would be physically and numerically impossible.

 March 8, 2011, 17:52 #3 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,266 Rep Power: 136 Best to run adaptive time stepping homing in on 3-5 coeff loops per iteration. Then the solver will automatically home in the correct time step size.

 March 9, 2011, 03:27 #4 New Member   Join Date: Mar 2011 Posts: 7 Rep Power: 13 Thank you both for your prompt reply! I guess I am covered regarding the timestep selection. Dear Attesz, I agree with you that if the domain starts to empty that would lead to numerical problems. However, I believe this is not physically impossible. The continuity equation states that mass that enters the system equals the one that leaves the system. However, this is true only in a steady state case. It is similar to a tank full with water which starts to empty if you have a hole at its bottom. You cannot have a steady state, since the water volume inside the tank is always time dependent. In my case, I have a domain, let's say a box, which has a hole (chimney at the top). The bottom side of the box is heated (either has a constant high temperature or a heat flux enters the domain). Thus, air (which has 25oC in the beginning) starts to heat and due to that fact it starts to ascend. Consequently, it is logical that since it rises it will eventually leave the box when it reaches the chimney. It is similar to the operation of a fireplace (if you neglect the combustion and the reactions); fumes rise up and leave the fireplace. According to the general mass balance equation: Input + Generation = Output + Accumulation + Consumption. So, in my case, it will always be: -Accumulation = Output. The negative symbol represents the reduction of mass inside the system. Am I somewhere wrong? I hope I haven't confused you. Looking forward for new ideas/feedback. Thanks again

 March 9, 2011, 04:08 #5 Senior Member     Attesz Join Date: Mar 2009 Location: Munich Posts: 368 Rep Power: 15 In a domain, where bouyancy is taken into account the pressure will not constant in direction of gravity according to the hidrostatic laws. In steady state, every outlet or opening boundary conditon has the same pressure value to the air inside the domain except there will be flow because of the pressure differences. If you define heat source, in steady the temperatures will be higher, some gas will exit the domain to be again balance in pressures. In transient case you can study this phenomenon in time. So if you start to heat the domain, the mass flowing in or out will end when the pressure of the air near the boundary condition is the same what you have defined there. All of the mass will never quit the domain and yes, in this case it is physically impossible as well. What you have written is absolutely correct, but note that this is time dependent: -accumulation means that mass in the domain decreases and it is equal to the output, but it is time dependent or it is valid only for an infinite small time intervall!

 March 9, 2011, 05:14 #6 New Member   Join Date: Mar 2011 Posts: 7 Rep Power: 13 Thanks again Attesz. Correct me if I am wrong, but according to your sayings, I understand that a steady state solution can also be achieved in this case. To reach steady state, accumulation must become zero. So, in this case the output should also become zero (since: -accumulation = output). If this is correct and in my case I could also run a steady state, it is of major importance to me to know the time required to reach the steady state. As I wrote in my first post, I want to check the time required for the domain to reach a certain temperature. Is this time the Elapsed Pseudo-Time printed out in the Solver output file? If yes, is it given in seconds? If not, is there a way to calculate this time? Thank you very much Attesz. I really appreciate your help

 March 9, 2011, 08:15 #7 Senior Member     Attesz Join Date: Mar 2009 Location: Munich Posts: 368 Rep Power: 15 Yes You can reach steady state, but if you have only a heat source wall and the others are adiabatic, it could be difficult, because there are also heat source balance. However I think you should run transient because you want to know what happens after an exact time, not in the infinite future... So do transient simulation. If it doesn't matter what happening during time, use as large timestep as you can. Last edited by Attesz; March 9, 2011 at 11:24. Reason: grammar correction :)

 March 9, 2011, 09:01 #8 New Member   Join Date: Mar 2011 Posts: 7 Rep Power: 13 Thanks a lot Attesz. I 'll follow your advice.

 March 18, 2011, 07:29 #9 New Member   Join Date: Mar 2011 Posts: 7 Rep Power: 13 Dear Attesz and Glenn, I tried to do transient run of my case, but in vain...Unfortunately, I cannot get convergence and the solver terminates with error in FINMES. I have tried two ways. I found an earlier POST by Glenn Horrocks and followed his advice. More specifically, I used adaptive time step with minimum time 1e-20 and max 1e20 seconds in order to allow the solver to find the most suitable timestep for my case. However, the solver kept decreasing the timestep, reached 1e-05 seconds and in the 70th step finally terminated with error in FINMES. In a second try, I used fixed timestep of 1s, but in the 100th step I got the same error. Any ideas of how to tackle this problem? Could that be a mesh issue? Perhaps, I should also mention that the Courant number initially is about 20, but gradually increases to the value of 999.99. Unfortunately, I cannot run a steady state case and use it as an initial run, since I have heat transfer and only one wall is heated (as already explained in my first Post) Thank you in advance

 March 19, 2011, 07:23 #10 Senior Member     Attesz Join Date: Mar 2009 Location: Munich Posts: 368 Rep Power: 15 I think FINMES is related to the quality of the mesh, but I'm not sure. Glenn knows it probably The courrant number depends on the mesh sizes and the time step value. The higher the step size the higher the courrant number. Post mesh pictures if you can.

 March 20, 2011, 21:59 #11 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,266 Rep Power: 136 The FINMES error means the linear solver is having problems converging. You need to improve numerical stability - try to fix any boundary or physical problems, smaller timesteps, better initial conditions, better quality mesh.

 March 21, 2011, 02:46 #12 New Member   Join Date: Mar 2011 Posts: 7 Rep Power: 13 Thank you guys. I 'll keep you posted.