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How to make UDF Initial temperature for more than one zone ID for Conduction?

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Old   January 29, 2020, 22:01
Default How to make UDF Initial temperature for more than one zone ID for Conduction?
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Dear all,



I simulate ground heat exchanger in Ansys FLUENT 17.2. I divide the main domain into several domain to reduce mesh complexity and different material of the ground (Surface to 15 m depth is sandy and below 15 m is clay). I simulate in transient. Due to temperature distribution in ground, I make UDF for initial temperature. Hoever, I only make for one zone ID. My assumption is no water flow through the ground due to computational limitation. I just want to know the temperature distribution in the ground after several days of heat extraction or injection. For the heat exchanger, I set as a solid body. Hence, I use source term to input heat value based on my previous data. The case is quite complex so In this case, I do only simulation ground. I don't know how to make the other zone id. Do you have any suggestion?

Thank you


Code:
/***********************************************************************

UDF for initializing temperature distribution

************************************************************************/



#include "udf.h"

/* #define SOLID_ID 7 */



DEFINE_INIT(my_init_func,domain)



{

cell_t c;

Thread *t;

real xc[ND_ND];

real y;

Domain *d;

int ID = 7;// ID Ground

 

/* determination of zone ID  */

 

Thread *thread = Lookup_Thread(domain,ID);



/* loop over all cell threads in the domain  */

 

thread_loop_c(t,domain)

 

{

/* loop over all cells  */

begin_c_loop_all(c,t)

{

C_CENTROID(xc,c,t);

y = xc[1]; 



if (-1 < y && y <= 0) /* from 0 to -5m*/

C_T(c,t) = 3 *y*y + 10 * y + 303;

 

else if (-1.5 < y && y <= -1) /* from 0 to -5m*/

C_T(c,t) = 7 * y + 304;



else if (-3 < y && y <= -1.5) /* from -3 to -1.5m*/

C_T(c,t) = 0.2 * y * y + 1.5 * y + 298;

 

else if (-5 < y && y <= -3) /* from -5 to -3m*/

C_T(c,t) = 0.4 * y + 292;



else if (-10 < y && y <= -5) /* from -10 to -5m*/

C_T(c,t) = 0.05 * y + 291;



else            /* from -10 to -20*/

C_T(c,t) = 291;

 

}

end_c_loop_all(c,t)

}

}
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Last edited by teguhtf; January 30, 2020 at 00:08. Reason: adding explaination about heat exchanger
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Old   January 30, 2020, 03:17
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Quote:
I don't know how to make the other zone id
what does it mean?

in code you have
Code:
int ID = 7;// ID Ground
change ID value to any other number
In FLuent GUI you can check ID of each zone in your model:
Setup -> Cell Zone Conditions -> select any zone -> you will see ID value, out this value in UDF

If you want to change apply your code to several zones, for instance for zone 8 -> you can add
Code:
ID = 8;
*thread = Lookup_Thread(domain,ID);
and make additional loop
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Old   January 30, 2020, 04:42
Default ID not being used
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Hi

If you use thread_loop_c, then you do not require cell zone ID. thread_loop_c ensures that the loop is executed over all the cell zones. Check you code and you will find that variable thread is not being used at all.

However, you most likely do not want that. So, remove thread_loop_c and only use the internal loop, begin_c_loop(c, t). The second argument in this loop has to be variable thread and not t, the thread coming from Lookup_Thread command.

If you want the same code to be executed for all the cell zones, then what you are doing is right and you can simply remove the declaration of variable thread as well as command Lookup_Thread that fetches its value.
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Old   February 9, 2020, 22:52
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Dear AlexanderZ,

Thank you for your reply. Yes, I thought like your idea. However, the initialization is only one in FLUENT. I can not make more than one initialization.

After I do simulate with previous code, The code work for all domain.

Thank you for your help
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Old   February 9, 2020, 23:03
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Dear vinerm,

Thank you for your reply. Yes, My code is still working good. But, I am sorry I do not have good solid knowledge about UDF. Here i attach my simulation after 1 day. Leftside temperature contour is initial condition right side is after 24 hours. So far, it is good. Except the low temperature near the coil is too low. and temperature graph below coil is increasing when heat absorbed by coil. The figure 3 should have the similar trends with figures 1 and 2.

by the way, my geometry and meshing i have posted in here Problem in meshing slinky coil burried in the ground

I am still trying to find out the solution.

Do you have any suggestions?

Thank you
Regards
Attached Images
File Type: jpg Temperature contour.jpg (50.5 KB, 8 views)
File Type: jpg Temperature Distribution.jpg (132.9 KB, 12 views)
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Old   February 10, 2020, 00:51
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so do you have problems with UDF or not? explain clearly your issues
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Old   February 10, 2020, 05:13
Default Incomplete Information
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There are a lot of things that we do not know about. Hence, it would be difficult to answer why the temperature trend is as it is and not what you expected. However, it most likely is correct as per your setup. To provide suggestions, we need to know the following

1. Is the energy source term being applied negative or positive? Are these sources within the circular ducts?
2. What are the boundary conditions on the bottom, side, and top surfaces?
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Old   February 10, 2020, 06:02
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Dear vinerm

I am sorry I do not mention it in detail. Thank you for your reply. I do simulation only for the ground side (solid). For circular pipe wall, I set the value is constant heat flux -5kW/m2 (based on my previous data). no fluid flow inside the pipe.

All of the boundary conditions for the wall side are adiabatic. The top wall is constant temperature 302 K. and bottom wall constant heat flux 65 mW/m2.

No water flows through ground. The ground is assumed as purely solid. Actually, my UDF based on this sample I just modify parts in loop side https://www.afs.enea.it/project/nept...udf/node27.htm

Thank you for your help.
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Last edited by teguhtf; February 10, 2020 at 06:10. Reason: addition source of UDF
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Old   February 10, 2020, 06:04
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dear AlexanderZ

Thank you for your reply. I am still checking if there is any mistake in the UDF or any other source errors.

Thank you for your help

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Old   February 10, 2020, 06:27
Default Integral calculation
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By doing a hand calculation, you can find out if the bottom part will heat up or cool down. For -5000 W/m^2 through the pipe, multiply it by the area of the wall where the condition is applied, i.e., area of the pipe wall. Do the same for the bottom wall where 0.065 is applied. If the value at the bottom is lower, then it should eventually cool down or vice-versa.
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