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#1 |
New Member
Martijn
Join Date: Jan 2022
Posts: 16
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I am currently validating an heat transfer case of a small aluminium block in a windtunnel. I build the windtunnel myself with a MAF-sensor at the front to PID control the mass flow trough the tube (with a fan at the back). A 30 Watt heat element is used to generate the heat for the forced convection. I used thermal paste and checked if the heat element is generating 30 Watts. A K-type thermocouple is used to measure the heat (calibrated at boiling water 100C).
The simulation itself is set up with 40 g/s at the mass flow inlet, Ideal gas, Al-7075-T6 as material, pressure outlet 0 [Pa], k-w SST, Gravity, Segregated Flow. The solution converges nicely to a steady state but the final result is about 10 degrees C too high in comparison with the wind tunnel data. I ran an unsteady simulation to check the temperature development over time, this can be seen in 'Unsteady Heat Graph'. It behaves very corresponding for the first few minutes, but the last part of the test is just completely off. Mesh is relatively fine, around 5 million cells, y+ is kept below 1. Does anyone know if I need to use additional models or do I need to take another approach? I know that I should not trust my windtunnel data by 100% but I think this error is to large. |
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#2 |
Senior Member
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
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I'm sorry, did you think heat transfer experiments were easy?
Try rerunning your simulation but with 27 W. And then I'll let you think about why it might be 27 W and not 30. |
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#3 |
New Member
Martijn
Join Date: Jan 2022
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Thank you for your reply. Sorry, I am new to heat transfer in CFD. Currently running the 27 W case, is the reason electrical inefficiency or a layer of air between the heat element and the aluminium block? Or did you have another explanation for your correction?
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#4 | |
Senior Member
Join Date: Jun 2011
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Quote:
2. Also, is this the MAX temperature? In my opinion, without knowing much details of your experimental setup, I would say that sources of inaccuracy would be thermal properties values of the Aluminum used, the thermal paste - thermal properties values and thickness of the layer, the airflow - value and uniformity at the inlet |
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#5 | |
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Martijn
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#6 |
Senior Member
Join Date: Jun 2011
Posts: 205
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did you enable the radiation in your thermal model?
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#7 |
Senior Member
Join Date: Jul 2009
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A co-worker of mine has done extensive work looking at heat transfer using CFD, and one of the most important results that he has found is that the y+ value typically needs to be at least an order of magnitude smaller for accurate heat transfer simulation compared to a momentum boundary layer simulation. If you can re-grid relatively easily it would be worth a try to drop the y+ down to ~0.1.
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#8 |
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Martijn
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#9 | |
New Member
Martijn
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#10 | |
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Continuum
Join Date: Aug 2009
Posts: 19
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Quote:
I would say that your experimental setup sounds pretty decent. And your transient solution looks pretty good. It captures the basic capacitive response of the block. Before you dive into the CFD side of things, how are you modeling the block? Are you meshing the Aluminum block or are you treating this as a lumped mass? I noted the T/C is not centered, nor is the heater so this could lead to gradients. Even if this is isothermal, the variation (~8.5%) is well within typical heat transfer uncertainties (15%). Can you post a pic of your setup? Sounds interesting. Regards |
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#11 | |
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Martijn
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#12 |
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Filippo Maria Denaro
Join Date: Jul 2010
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Considering you are solving a case where a physical transient is the goal of your simulation, I suggest to check if you have enough computational power for solving the problem using LES (or DNS if possible...).
On the other hand, your problem requires to prescribe a suitable velocity inflow profile or letting it developint correctly. And the initial condition of the experimental device should be matched. Finally, the y+ should be referred to the thermal boundary layer. For air (Pr=0.71) you should have similar conditions but I suggest to take care of the wall resolution of the grid because the heat flux should be carefully described. Finally, see the sensibility of the solution to a grid refinement. |
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#13 |
Senior Member
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
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So 30 W... have you accounted for the the heat conducted through the 4 dowel pins holding your block in the wind tunnel? Have you accounted for the head conducted through the wire leads? Fin effects are very real because heat conduction (through solids) vastly exceeds typical convection rates.
Are you sure 30 W is 30 W? Do you even actually measure the voltage drop across the heater element? Did you measure the resistance of the heating element at the heated temperature? Note that this requires a four-terminal sensing setup. If you use a nickel based heating element the resistance of the heater would be less sensitive to temperature changes, but in a typical experiment we characterize this because every % counts. I used to do this day in and day out, measure the voltage and resistances of a (at one point a hundred) live heaters to get the most accurate heat rate. We would also spend a ton of time doing heat leakage experiments to characterize the heat losses and easily >75% of the actual experiment was doing these heat loss tests. If I spent 1 day doing the heat transfer test, I'd spend 3 days characterizing heat losses. |
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#14 |
Senior Member
Join Date: Jun 2011
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After seeing the pictures I have the same questions as LuckyTran. The four pins holding the block are quite massive and being in the airflow path will definitely reduce its temperature (especially if they are made of Aluminum). Why don't you include these in your model too.
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#15 | |
New Member
Martijn
Join Date: Jan 2022
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Quote:
You make some very good points about the support pins, it was stupid of me to assume that they where neglectible. Thanks for this tip, I will model it again with this in mind. Heating element is AC and uses the electricity from the wall outlet, I measured the RMS voltage and current of this circuit and came to the conclusion that the heating element gets 30 W of electrical energy. I did a small experiment to roughly check the 30 W again, I isolated the block and let it warm up to 30 deg C. The time it took to get to this temperature was 112 seconds. Initial temperature was 19.5 deg C. The weight of the block is 320 grams. With this information and the specific heat of AL-7075-T6 (960 J/kg*C), it can be calculated that the amound of Joules is: 960*0.32*(30-19.5) = 3302.4 J Which gives a power of: 3302.4/112 = 29.486 W I understand your point about doing the heat leakage experiments, however it is a project with relatively small time left so unfortunatly I have no time to put more research into this. But I will take this information with me for future experiments. |
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#16 | |
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Gerry Kan
Join Date: May 2016
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Quote:
If this is really a heat transfer experiment. I would like to weigh in on it. Many moons ago, the lab I was working in had lots of these setups similar to yours, except we were interested in conductive heat transfer so everything was performed in a near vacuum environment. To minimize radiative heat transfer, we would put reflective heat shields around the test rigs while we were making the runs. On that token, I imagine this could very well be where some (or much) of the missing 0.6 W is going in your case. You might not have time to find out empirically, but a simple radiation heat transfer calculation could help you verify if to what degree this assertion is valid. Gerry. P.S. - On first glance I think your results look pretty good, especially at the start. The CFD results, being idealized, should overpredict your observations. Instead of trying to tune everything and make them match, you could think about what accounted for the final temperature difference and evaluate their plausibility. This is the more rewarding path since it requires you to think about your data. One more thing, you should also look at error bars on the observation, which gives you extra room to play with. |
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#17 | |
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Martijn
Join Date: Jan 2022
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#18 | |
Senior Member
Gerry Kan
Join Date: May 2016
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Quote:
Even at low temperatures it could be more than you think. The Stefan-Boltzmann relation: ![]() Let's make some assumptions to put the figures in your neighborhood. ![]() ![]() ![]() The amount of radiation heat transfer is ... ![]() which is close to the 0.6 W that was previously discussed. I imagine substituting these with values to your situation, you are going to get something not far from these estimates. Gerry. |
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#19 |
New Member
Martijn
Join Date: Jan 2022
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Quote:
Tambient = 19 C Tfinal = 50 C A = 0.0166796 m^2 When calculated, the radiation heat transfer is about 3.7 W... its not neglectible. With the Stefan-Boltzmann relation, the material is not taken into account, does this matter? |
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#20 | |
Senior Member
Gerry Kan
Join Date: May 2016
Posts: 360
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Quote:
![]() For most non-black surfaces ![]() Gerry. |
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Tags |
heat transfer cooling, validation |
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