# I really need some help on this problem

 Register Blogs Members List Search Today's Posts Mark Forums Read

January 23, 2013, 22:47
I really need some help on this problem
#1
New Member

Zhu
Join Date: Jan 2013
Posts: 18
Rep Power: 4
Hi everyone,
I am trying to simulate the heat transfer performance of supercritical water flow in vertical upward round tube with 3.2 mm diameter. Here is a problem that bothers me for long time.
The distribution patterns of the calculated wall temperature and heat transfer coefficient are not reasonable. The wall temperature increases too fast near the inlet of the tube and then increases almost linearly along the length of the tube. Since the coolant temperature increases almost lnearly along the length of the tube, the heat transfer coefficint, which is defined as
h=q/T, decrease very fast near the inlet of the tube and then almost keeps constant.

I can not find the reason why the wall temperature increases so fast near the inlet of the tube. It should increase step by step, and the difference between the wall temperature and coolant temperature should increase with the increasing of the coolant temperature.

I tried several other k-e models. I got similar unreasonable distribution patterns.
I attached a figure of the distributions of the temperature and heat transfer coefficient.
Would some one please tell me what possible reasons can cause such problem?
Attached Images
 1.jpg (82.8 KB, 37 views)

 January 24, 2013, 03:41 #2 Senior Member     siamak rahimi ardkapan Join Date: Jul 2010 Location: Copenhagen, Denmark Posts: 218 Rep Power: 9 Can you explain what is the coolant here? __________________ Good luck Siamak

 January 24, 2013, 03:44 #3 New Member   Zhu Join Date: Jan 2013 Posts: 18 Rep Power: 4 I am sorry that I didn't describe clearly. The coolant is supercritical water. The temperature of coolant is higher than 390 oC

 January 24, 2013, 12:16 #4 Senior Member   Pauli Join Date: Mar 2009 Posts: 181 Rep Power: 8 Not knowing your geometry, consider this more an interpretation of the graph you shared. The graph indicates the entrance (length < 0.1m) has a much higher heat transfer coefficient than the remainder of the tube. This is driving the computed temperatures. The question is why does the entrance have higher heat transfer coefficient. From a flow physics standpoint, my initial assumption would be entrance effects (separation & mixing) cause additional mixing (varying velocity profile). At length > 0.1 the flow is fully developed. This provides a constant velocity profile & the h=q/dt relationship you expect. From a numerical calculation standpoint, you should be asking questions regarding how the heat transfer coefficient is calculated. Are you using wall functions or resolving the boundary layer? Is grid resolution adequate? Etc, etc. Computing heat transfer coefficients under pipe entrance conditions is not wall functions strength.

 January 24, 2013, 13:26 #5 New Member   Prashanth Avireddi Join Date: Sep 2012 Location: Farmington Hills, MI Posts: 10 Rep Power: 4 Could you put up a schematic of the model you are working with, I have worked with numerically modeling similar flows (multi-phase flows, lot of empiricism). I have got results which are analogous to yours, I was working on Micro-Processor Cooling Techniques, Which is reversal of your application.

 January 24, 2013, 21:06 all informations of my simulation-1 #6 New Member   Zhu Join Date: Jan 2013 Posts: 18 Rep Power: 4 The geometry is very simple. The length of tube is 500 mm. The diameter is 3.2mm. There is no wall thickness, and we only consider the fluid region. The uniform heat flux is 800kW/m2, the mass velocity is 1000kg/m2s , the temperature of inlet is 670 K and the pressure at outlet is 25 MPa. Flow direction is upward.

 January 24, 2013, 21:07 all informations of my simulation-2 #7 New Member   Zhu Join Date: Jan 2013 Posts: 18 Rep Power: 4 1. The needed thermal properties of water are as follow: density Number of interval 3 Interval ranges [663,700, 800, 1073] Numbers of coefficients [5,5,4] Coefficient [20888437.32897, -121466.279457929, 264.892650225018, -0.2567597311521, 0.000093333082540744, 79195.9972248006, -404.619712014594, 0.7784131025619, -0.00066713291581415, 2.1477224361164E-07, 902.76678358608, -2.2681931455009, 0.002078145302, -6.5474813334491E-07] Exponents [0.0,1.0,2.0,3.0,4.0, 0.0,1.0,2.0,3.0,4.0, 0.0,1.0,2.0,3.0] -------------------------------------------------------------------------------- Specific heat Number of interval 3 Interval ranges [663,683,783, 1073] Numbers of coefficients [5,5,5] Coefficient [72098255169.205, -426901983.90771, 947913.913893974, -935.475678541713, 0.3462040914552, 35740652.493319, -189447.825833259, 376.86407580743, -0.3333750897413, 0.00011063602503477, 223117.120300903, -870.188643358111, 1.2944485418665, -0.00085908519886809, 2.1442187362133E-07] Exponents [0.0, 1.0, 2.0, 3.0, 4.0, 0.0, 1.0, 2.0, 3.0, 4.0, 0.0, 1.0, 2.0, 3.0, 4.0] ---------------------------------------------------------------------------------------------------------------------- Dynamic viscosity (\$Temperature<663.0)? 3.15716e-5: ((\$Temperature<683.0)? 6.9224264624736-0.0408832952645*\$Temperature +0.000090549079244015* pow(\$Temperature,2)-8.913669797436E-08* pow(\$Temperature,3)+3.2906108603975E-11* pow(\$Temperature,4): ((\$Temperature<783.0)? 0.0108989512696-0.000057839685831513*\$Temperature +1.1532390544032E-07* pow(\$Temperature,2)-1.0216997356426E-10* pow(\$Temperature,3)+3.3955858119227E-14* pow(\$Temperature,4): ((\$Temperature<1073.0)? -0.0000032427270621927+4.741942361337E-08*\$Temperature +-4.78555217991E-12* pow(\$Temperature,2): 4.21E-5))) ------------------------------------------------------------------------------------------------------------------------- Thermal conductivity (\$Temperature<663.0)? 0.2264: ((\$Temperature<695.0)? 47661.033038555-278.416789654012*\$Temperature +0.6099315425136* pow(\$Temperature,2)-0.00059388587645495* pow(\$Temperature,3)+2.1685650279812E-07* pow(\$Temperature,4) : ((\$Temperature<775.0)? 185.834410219319-0.9793944425447*\$Temperature +0.0019389248748* pow(\$Temperature,2)-0.0000017078378939546* pow(\$Temperature,3)+5.6465757723963E-10* pow(\$Temperature,4): ((\$Temperature<1073.0)? 2.4199512766181-0.0094779285294*\$Temperature +0.000014350317833471* pow(\$Temperature,2)-9.5997271574745E-09* pow(\$Temperature,3)+2.421812079001E-12* pow(\$Temperature,4): 0.123))) ------------------------------------------------------------------------------------------------------------------------ Prantl Number (\$Temperature<663.0)? 3.9489: ((\$Temperature<700.0)? 712655.370282141-4137.78650804068*\$Temperature +9.0095886619281* pow(\$Temperature,2)-0.0087191263008* pow(\$Temperature,3)+0.0000031643427014962* pow(\$Temperature,4): ((\$Temperature<800.0)? 1789.0113166512-9.1712038684109*\$Temperature +0.017667884118* pow(\$Temperature,2) -0.00001514546110914* pow(\$Temperature,3)+4.8732160696751E-09* pow(\$Temperature,4): ((\$Temperature<1073.0)? 35.1933611129059-0.132228857307*\$Temperature +0.00019253019932984* pow(\$Temperature,2)-1.2511500694886E-07* pow(\$Temperature,3)+3.0554723673796E-11* pow(\$Temperature,4): 0.91)))

January 24, 2013, 21:10
all informations of my simulation-3
#8
New Member

Zhu
Join Date: Jan 2013
Posts: 18
Rep Power: 4
2. The generated mesh is shown in Fig.1.
Attached Images
 1.jpg (35.3 KB, 14 views)

January 24, 2013, 21:12
all informations of my simulation-4
#9
New Member

Zhu
Join Date: Jan 2013
Posts: 18
Rep Power: 4
3. The selected physical model
Attached Images
 2.jpg (31.0 KB, 17 views)

January 24, 2013, 21:16
all informations of my simulation-5
#10
New Member

Zhu
Join Date: Jan 2013
Posts: 18
Rep Power: 4
4. the wall y+ is from 16 to 33
Attached Images
 3.jpg (14.7 KB, 15 views)

January 24, 2013, 21:21
all informations of my simulation-6
#11
New Member

Zhu
Join Date: Jan 2013
Posts: 18
Rep Power: 4
There are totally three boundaries: Inlet Outlet and wall
Attached Images
 4.jpg (21.6 KB, 8 views)

January 24, 2013, 21:24
all informations of my simulation-7
#12
New Member

Zhu
Join Date: Jan 2013
Posts: 18
Rep Power: 4
A line-probe is generated on the wall boundary to be used to get the wall temperature.
Nine plane sections are generated along Z direction to get coolant temperature by using Surface Average of the Reports
Attached Images
 5.jpg (15.6 KB, 6 views)

January 24, 2013, 21:37
#13
New Member

Zhu
Join Date: Jan 2013
Posts: 18
Rep Power: 4
Quote:
 Originally Posted by Prashanth.A Could you put up a schematic of the model you are working with, I have worked with numerically modeling similar flows (multi-phase flows, lot of empiricism). I have got results which are analogous to yours, I was working on Micro-Processor Cooling Techniques, Which is reversal of your application.
All informations have been uploaded. As my understanding, you use coolant to cool the micro-processor, right?

January 24, 2013, 22:14
#14
New Member

Zhu
Join Date: Jan 2013
Posts: 18
Rep Power: 4
Quote:
 Originally Posted by Pauli Not knowing your geometry, consider this more an interpretation of the graph you shared. The graph indicates the entrance (length < 0.1m) has a much higher heat transfer coefficient than the remainder of the tube. This is driving the computed temperatures. The question is why does the entrance have higher heat transfer coefficient. From a flow physics standpoint, my initial assumption would be entrance effects (separation & mixing) cause additional mixing (varying velocity profile). At length > 0.1 the flow is fully developed. This provides a constant velocity profile & the h=q/dt relationship you expect. From a numerical calculation standpoint, you should be asking questions regarding how the heat transfer coefficient is calculated. Are you using wall functions or resolving the boundary layer? Is grid resolution adequate? Etc, etc. Computing heat transfer coefficients under pipe entrance conditions is not wall functions strength.
Thank you very much for your kind, patient analysis.
I can understand the entrance effect at length < 0.1 m. But the wall temperature distribution pattern is more like a simulation result with constant thermal properities, not with changed thermal properities.
In fact, in the high enthalpy region of supercritical water, all thermal properties of water are functions of coolant temperature. I defined all these thermal properties using field function of STAR CCM, or polynomial density and specific heat.

I attached all informations used in my simulation. Are there any important steps that I missed during the simulation process?

Thank you very much anyway!

 January 25, 2013, 01:03 #15 Senior Member   Pauli Join Date: Mar 2009 Posts: 181 Rep Power: 8 Did you use a uniform inlet velocity profile? Or did you apply a parabolic profile representing fully developed flow? From the information you provided, it appears you specified a uniform profile. I believe that is the primary source of your high h at the inlet. Your setup panel shows you chose a 2-layer all y+ wall treatment. The y+ plot shows your y-plus is in the range where wall functions are used. With wall functions, the heat transfer function coefficient is strongly driven by the near wall velocity. Your mesh & y+ values are nice. I'm guessing the red spot in the y+ plot is the inlet. That observation coupled with your setup panel led me to believe you have a uniform inlet velocity profile.

January 25, 2013, 01:31
#16
New Member

Zhu
Join Date: Jan 2013
Posts: 18
Rep Power: 4
Quote:
 Originally Posted by Pauli Did you use a uniform inlet velocity profile? Or did you apply a parabolic profile representing fully developed flow? From the information you provided, it appears you specified a uniform profile. I believe that is the primary source of your high h at the inlet. Your setup panel shows you chose a 2-layer all y+ wall treatment. The y+ plot shows your y-plus is in the range where wall functions are used. With wall functions, the heat transfer function coefficient is strongly driven by the near wall velocity. Your mesh & y+ values are nice. I'm guessing the red spot in the y+ plot is the inlet. That observation coupled with your setup panel led me to believe you have a uniform inlet velocity profile.
Thank you very much for your help.
I did use uniform inlet velocity profile in my simulation.
The problem of my simulation is that the simulation results are more like we use constant thermal properties, rather than changed thermal properties.
I have defined all thermal properities as the functions of the coolant temperature. But it seems all these functions are not rightly used.
I don't know what causes this problem

 January 27, 2013, 10:54 #17 New Member   Prashanth Avireddi Join Date: Sep 2012 Location: Farmington Hills, MI Posts: 10 Rep Power: 4 In the Physics model, change the fluid setting from Liquid (H2o) to Multiphase Mixture with inlet being only Liquid H2o. That should solve the problem if there was any.

January 30, 2013, 04:12
#18
New Member

Zhu
Join Date: Jan 2013
Posts: 18
Rep Power: 4
Quote:
 Originally Posted by Prashanth.A In the Physics model, change the fluid setting from Liquid (H2o) to Multiphase Mixture with inlet being only Liquid H2o. That should solve the problem if there was any.
Thank you very much for your advice. I want to try a case using the physics model of Multiphase mixture.
I met a problem here, as shown in attached figure. If I want to use only Liquid H2O, the vomume fraction of velocity inlet should be defined as [1.0, 0.0]. But I can not define it in STAR CCM. I was informed that I gave an Invalid array input.
How to solve this problem?
Would you please give me a bit more detailed information about how to set multiphase mixture model?
I am looking forward to your help.
Thank you very much!!
Attached Images
 problem.jpg (41.7 KB, 8 views)

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post Wouter Fluent UDF and Scheme Programming 6 June 6, 2012 04:43 JFDC FLUENT 1 July 11, 2011 05:59 Se-Hee CFX 2 June 10, 2007 06:29 ParodDav CFX 5 April 29, 2007 19:13 Thomas P. Abraham Main CFD Forum 5 September 8, 1999 14:52

All times are GMT -4. The time now is 13:32.