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Very Simple Heat Transfer
Posted Mar 22, 2012, 5:39 p.m. EDT Version 4.2a 2 Replies
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Hello,
As always, I really appreciate for your help.
I have a 3D simulation using "Heat Transfer in Solids". It's a extremely simple simulation. I attach mph file.
What I did is applying power to the1-microliter of water to see how much the temperature rises. (10ncal energy (which is 42n Joule) to water for a certain period of time)
1. In Global Definitions, I made "Rectangle" function. (From 0.1 to 0.2 second, 0.1 second period)
2. And then I define a "pulse" in variable under model. Then I simply use "Heat Source" to apply TOTAL POWER in "Heat Transfer in Solids".
3. And the result shows 7.15 micro Kelvin of water temperature, which is quite different from the simple calculation. (Q=m*c*delta T)
4. To confirm the result, I increase time period into 0.2 second and half the power to apply the same energy to the water. (Since Energy is Power times Time.)
5. But the result shows about 9.16 micro kelvin which is very similar with the calculation.
6. So I changed the time period into 0.3, then the result goes back to 7.15 uK.
7. When I changed the time into 0.4 second, then the result shows 9.16 micro Kelvin.
I cannot figure it out. Please let me know what you guys think.
As always, I really appreciate for your help.
I have a 3D simulation using "Heat Transfer in Solids". It's a extremely simple simulation. I attach mph file.
What I did is applying power to the1-microliter of water to see how much the temperature rises. (10ncal energy (which is 42n Joule) to water for a certain period of time)
1. In Global Definitions, I made "Rectangle" function. (From 0.1 to 0.2 second, 0.1 second period)
2. And then I define a "pulse" in variable under model. Then I simply use "Heat Source" to apply TOTAL POWER in "Heat Transfer in Solids".
3. And the result shows 7.15 micro Kelvin of water temperature, which is quite different from the simple calculation. (Q=m*c*delta T)
4. To confirm the result, I increase time period into 0.2 second and half the power to apply the same energy to the water. (Since Energy is Power times Time.)
5. But the result shows about 9.16 micro kelvin which is very similar with the calculation.
6. So I changed the time period into 0.3, then the result goes back to 7.15 uK.
7. When I changed the time into 0.4 second, then the result shows 9.16 micro Kelvin.
I cannot figure it out. Please let me know what you guys think.
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2 Replies Last Post Mar 23, 2012, 1:23 p.m. EDT
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Posted:
1 decade ago
Hi
I'm not behind my COMSOL WS so I cannot open your model, but the most common error with time dependent HT models is that the default time series solver uses an "automatic" time speing, assuming an assymptotical decay behaviour, which is not the case for rapid on off pulses. Therefore, I would advice that you check your time stepping, first to definea fine time step with 3-10 steps over the period, and then to set the lower time solver node to time step tab "intermediate" or "stric"t and not to "automatic".
This forces the solver to take more steps and to better resolve the transient behaviour, you will see taht on your solver log, the "out" are the time values, if several arrive without any calculation steps, then the solver is interpolating its resut to get the demanded time steps stored, this is ok for assymptotical solutions, often error prone for periodic or pulsed solutions
--
Good luck
Ivar
I'm not behind my COMSOL WS so I cannot open your model, but the most common error with time dependent HT models is that the default time series solver uses an "automatic" time speing, assuming an assymptotical decay behaviour, which is not the case for rapid on off pulses. Therefore, I would advice that you check your time stepping, first to definea fine time step with 3-10 steps over the period, and then to set the lower time solver node to time step tab "intermediate" or "stric"t and not to "automatic".
This forces the solver to take more steps and to better resolve the transient behaviour, you will see taht on your solver log, the "out" are the time values, if several arrive without any calculation steps, then the solver is interpolating its resut to get the demanded time steps stored, this is ok for assymptotical solutions, often error prone for periodic or pulsed solutions
--
Good luck
Ivar
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Posted:
1 decade ago
Thanks, Ivar.
You mean
In Study>>Solver Configurations>>Solver>>Time-Dependent Solver>>Time Stepping>> Steps taken by solver
Do you suggest that I should change from "Free" into "Intermediate" or "Strict"?
You mean
In Study>>Solver Configurations>>Solver>>Time-Dependent Solver>>Time Stepping>> Steps taken by solver
Do you suggest that I should change from "Free" into "Intermediate" or "Strict"?