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Heat Transfer to a laminar flow in a microchannel
Posted 23 mai 2016, 07:41 UTC−4 Heat Transfer & Phase Change, Computational Fluid Dynamics (CFD), Microfluidics, Modeling Tools & Definitions, Parameters, Variables, & Functions Version 4.4 3 Replies
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Hello,
I am currently simulating the heat transfer in a bent microchannel using the Conjugate Heat Transfer Physics. The simulation worked satisfactorily. Now I want to calculate heat transfer coefficients by means of the post-processing tools. To calculate the local heat transfer coefficient at a certain length of the cannel, I have to calculate the mean temperature of the fluid at that length in the first place.
A meaningfull definition of the mean temparature would be the surface integral of the temperature times the flux normal to the surface. I defined some cut planes as well as surface integrals (Derived Values) of T*(u*nx+v*ny+w*nz) to be evaluated on the respective cut plane. To get the mean temperature, one only has to devide this integral by the total volume flow of the inlet as all cut planes are defined in a way that the whole volume flow has to pass them.
In my example, the inlet temperature was 293.15 K and all walls were constant at 303.15 K. Unfortunately, the aforementioned evaluation of the mean temperatures gave totally useless values even below the inlet temperature. I attached a file which is a simplified version of the real problem showing the same problems. In this example, the inlet is 0.01 m/s and the cross section is 1mm*1mm so the integral has to be multiplied by 1e8 to receive the temperature. In my "real" problem, the cut planes are rather arbitrarily orientated in space.
I hope someone can help me. Maybe I just overlooked something.
Kind regards,
Alexander Rave from Hamburg, Germany
I am currently simulating the heat transfer in a bent microchannel using the Conjugate Heat Transfer Physics. The simulation worked satisfactorily. Now I want to calculate heat transfer coefficients by means of the post-processing tools. To calculate the local heat transfer coefficient at a certain length of the cannel, I have to calculate the mean temperature of the fluid at that length in the first place.
A meaningfull definition of the mean temparature would be the surface integral of the temperature times the flux normal to the surface. I defined some cut planes as well as surface integrals (Derived Values) of T*(u*nx+v*ny+w*nz) to be evaluated on the respective cut plane. To get the mean temperature, one only has to devide this integral by the total volume flow of the inlet as all cut planes are defined in a way that the whole volume flow has to pass them.
In my example, the inlet temperature was 293.15 K and all walls were constant at 303.15 K. Unfortunately, the aforementioned evaluation of the mean temperatures gave totally useless values even below the inlet temperature. I attached a file which is a simplified version of the real problem showing the same problems. In this example, the inlet is 0.01 m/s and the cross section is 1mm*1mm so the integral has to be multiplied by 1e8 to receive the temperature. In my "real" problem, the cut planes are rather arbitrarily orientated in space.
I hope someone can help me. Maybe I just overlooked something.
Kind regards,
Alexander Rave from Hamburg, Germany
Attachments:
3 Replies Last Post 24 mai 2016, 07:09 UTC−4
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Posted:
9 years ago
Hi
there are a few things that puzzles me:
1) your inlet T is lower than the imposed outlet T, I would have expected the other way around
2) the "solid" sides seem to have no heat exchange and are not really solid, its just a fluid wall, I would have put a certain thickness of solid material and defined a heat exchange value or a fixed T on the outer "solid boundary".
3) absolute and relative pressure, gauge pressure: normally by setting the Conjugated Heat transport node reference pressure to 1[atm] you work only in local reference pressure (with respect to the pA, so the fluid absolute pressure is rather from "nitf" and not user defined (tis might be an V5/4.4 issue).
4) the same the temperatures are normally not user defined by the HT fluid temp is from "nitf" and HT in solid the same (but you have no solid in your model) but this might be a change in my V5 versus your v4.4 model
5) in V5 I would have expected a multiphysics node, but in v4.4 they were probably not there, yet.
Then to calculate your average values its not enough to integrate the T*(u.n) you need to divide by
norm(u.n) too
I would also have started with a simple tube, its easier to check the values to be sure your model is set up correctly. And thereafter tried a bent tube
You are almost there, so its just to continue :)
--
Good luck
Ivar
there are a few things that puzzles me:
1) your inlet T is lower than the imposed outlet T, I would have expected the other way around
2) the "solid" sides seem to have no heat exchange and are not really solid, its just a fluid wall, I would have put a certain thickness of solid material and defined a heat exchange value or a fixed T on the outer "solid boundary".
3) absolute and relative pressure, gauge pressure: normally by setting the Conjugated Heat transport node reference pressure to 1[atm] you work only in local reference pressure (with respect to the pA, so the fluid absolute pressure is rather from "nitf" and not user defined (tis might be an V5/4.4 issue).
4) the same the temperatures are normally not user defined by the HT fluid temp is from "nitf" and HT in solid the same (but you have no solid in your model) but this might be a change in my V5 versus your v4.4 model
5) in V5 I would have expected a multiphysics node, but in v4.4 they were probably not there, yet.
Then to calculate your average values its not enough to integrate the T*(u.n) you need to divide by
norm(u.n) too
I would also have started with a simple tube, its easier to check the values to be sure your model is set up correctly. And thereafter tried a bent tube
You are almost there, so its just to continue :)
--
Good luck
Ivar
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Posted:
9 years ago
Hi Ivar, hello all,
thank you for your reply.
1.) I am not sure if I got your point. In this example, the fluid is heated up by the walls. It could of course also be cooled down...
2.) I defined no thickness of the wall because I want to calculate the pure heat exchange between the wall surface and the fluid under the "constant wall temperature" boundary condition. I am aware that heat conduction within the wall can play a major roll in micro heat exchangers, but I do not want to consider it here.
3.) The calculated pressure distribution is OK. You can of course set the outlet pressure to 1 [atm] rather than 0 [Pa]. I think this caueses only a pressure shift.
My main point is the surface integration to get the mean temperature of a cut plane. U.n = u*nx+v*ny+w*nz should be the flux through an surface element. So the integration of T*(u*nx+v*ny+w*nz) over the cut plane devided by the total volume flow should give the thermodynamic mean temperature. When you for example take the cut plane at the inlet (with area A = 1e-6 m^2), T and u is constant (T=293.15 K, u = 0.01 m/s, v=0, w=0). Then the integration should yield T*u*A = 293.15e-8 Km^3/s but it yields 285.69e-8 Km^3/s.
Strange enough, the integration of "1" over the surface yields the right value of 1e-6 m^2 and also u*nx+v*ny+w*nz gives the right volume flow rate. But with the temperature, it somehow does not work.
I would be grateful for further hints about this.
Greetings,
Alex
thank you for your reply.
1.) I am not sure if I got your point. In this example, the fluid is heated up by the walls. It could of course also be cooled down...
2.) I defined no thickness of the wall because I want to calculate the pure heat exchange between the wall surface and the fluid under the "constant wall temperature" boundary condition. I am aware that heat conduction within the wall can play a major roll in micro heat exchangers, but I do not want to consider it here.
3.) The calculated pressure distribution is OK. You can of course set the outlet pressure to 1 [atm] rather than 0 [Pa]. I think this caueses only a pressure shift.
My main point is the surface integration to get the mean temperature of a cut plane. U.n = u*nx+v*ny+w*nz should be the flux through an surface element. So the integration of T*(u*nx+v*ny+w*nz) over the cut plane devided by the total volume flow should give the thermodynamic mean temperature. When you for example take the cut plane at the inlet (with area A = 1e-6 m^2), T and u is constant (T=293.15 K, u = 0.01 m/s, v=0, w=0). Then the integration should yield T*u*A = 293.15e-8 Km^3/s but it yields 285.69e-8 Km^3/s.
Strange enough, the integration of "1" over the surface yields the right value of 1e-6 m^2 and also u*nx+v*ny+w*nz gives the right volume flow rate. But with the temperature, it somehow does not work.
I would be grateful for further hints about this.
Greetings,
Alex
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Posted:
9 years ago
I think I solved my problem... Even though the density of water does not change that much between 293 K and 303 K, the assumption of a constant density and accordingly a constant volume flow rate was not a good one. Now I calculate the integral of nitf.rho*T*(u*nx+v*ny+w*nz) and devide it by the constant mass flow. That yields more realistic values between the inlet temperatur and the wall temperature. So, indeed, I "overlooked" something.
Kind regards,
Alex
Kind regards,
Alex