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incubator temperature variation/boundary condition
Posted 17 mar 2015, 13:28 GMT-4 Heat Transfer & Phase Change, Computational Fluid Dynamics (CFD) Version 5.0 2 Replies
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Dear ALL,
I want to model a metal tube placed in a PID controlled incubator. I assume the temperature in the incubator varies like a sine wave, i.e. T=T0+0.1*sin(2*pi*t/Period), where t is time and Period is the temperature variation period (2 hours).
I first built a model by using a temperature boundary condition, i.e. the temperature at the outer surface of the metal tube is T=T0+0.1*sin(2*pi*t/Period); then I built another model by using convective cooling boundary condition, i.e. T external=T0+0.1*sin(2*pi*t/Period).
However, the results are very different. Is there anyone who can tell me which method is better?
Thank you very much for your kind help!
I want to model a metal tube placed in a PID controlled incubator. I assume the temperature in the incubator varies like a sine wave, i.e. T=T0+0.1*sin(2*pi*t/Period), where t is time and Period is the temperature variation period (2 hours).
I first built a model by using a temperature boundary condition, i.e. the temperature at the outer surface of the metal tube is T=T0+0.1*sin(2*pi*t/Period); then I built another model by using convective cooling boundary condition, i.e. T external=T0+0.1*sin(2*pi*t/Period).
However, the results are very different. Is there anyone who can tell me which method is better?
Thank you very much for your kind help!
2 Replies Last Post 29 apr 2015, 03:23 GMT-4
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Posted:
10 years ago
Sihe:
What is the Biot number of your system? If your Biot number is very small, you should get similar answer with the two simulations. If your Biot number is close to 1 or greater than 1, you will not get the same answer.
In my opinion the convective cooling approach is the correct one. The temperature at the boundary is driven by convective cooling and is not going to be the same as temperature of the incubator. The main wrinkle in your simulation would be the convective heat transfer coefficient. Heat transfer coefficients are notoriously hard to estimate especially for natural convection if the incubator fluid is at rest.
Regards,
Sri.
What is the Biot number of your system? If your Biot number is very small, you should get similar answer with the two simulations. If your Biot number is close to 1 or greater than 1, you will not get the same answer.
In my opinion the convective cooling approach is the correct one. The temperature at the boundary is driven by convective cooling and is not going to be the same as temperature of the incubator. The main wrinkle in your simulation would be the convective heat transfer coefficient. Heat transfer coefficients are notoriously hard to estimate especially for natural convection if the incubator fluid is at rest.
Regards,
Sri.
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Posted:
10 years ago
Dear Sri,
Thank you very much and I really apologize for my late reply. Now I've got a new problem...
In my model there is a copper tube with an inlet and an outlet. The fluid is air. I use 'Convective Cooling' to simulated the situation in an air incubator. The 'External Temperature' is '278.15+0.1*sin(2*pi*t/7200)', i.e. a full cycle lasts 2 hours.
I want to make the air inlet temperature the same as the temperature of one point on the copper outer surface. However, as the 'External Temperature' is not the outer surface temperature, when I set the inlet temperature '278.15+0.1*sin(2*pi*t/7200)' the results show that there is a 4.1 mK difference between the outer surface point and the inlet temperature during 7200 seconds. I also tried to let the inlet temperature be the same as a temperature probe at the outer surface point, but the results are weird, even bigger difference.
Is there any way to do this? Thanks so much!
Sihe
Thank you very much and I really apologize for my late reply. Now I've got a new problem...
In my model there is a copper tube with an inlet and an outlet. The fluid is air. I use 'Convective Cooling' to simulated the situation in an air incubator. The 'External Temperature' is '278.15+0.1*sin(2*pi*t/7200)', i.e. a full cycle lasts 2 hours.
I want to make the air inlet temperature the same as the temperature of one point on the copper outer surface. However, as the 'External Temperature' is not the outer surface temperature, when I set the inlet temperature '278.15+0.1*sin(2*pi*t/7200)' the results show that there is a 4.1 mK difference between the outer surface point and the inlet temperature during 7200 seconds. I also tried to let the inlet temperature be the same as a temperature probe at the outer surface point, but the results are weird, even bigger difference.
Is there any way to do this? Thanks so much!
Sihe