- 23,715
- 5,930
Oops. My mistake.Tom.G said:Uhmm... Something doesn't match here.![]()
The discussion revolves around calculating the temperature at the base of a heat sink with straight fins, given specific boundary conditions such as power at the bottom surface and convection at the remaining surfaces. Participants explore theoretical approaches, numerical simulations, and the potential for analytical solutions.
Participants express uncertainty regarding the availability of analytical solutions for the specific case discussed. Multiple competing views exist regarding the methods of calculation and the assumptions that can be made.
Participants note limitations in existing resources, such as the lack of formulas for specific boundary conditions and the complexity of online calculators that do not align with their needs. There is also mention of the need for accurate input data for thermal resistance calculations.
Oops. My mistake.Tom.G said:Uhmm... Something doesn't match here.![]()
Thank you very much, now the result makes much more sense. I got ##Q_{t}=80 \ W## for ##T_{s}=258^{\circ}C## while the value from simulation (maximum temperature at the base) is ##244^{\circ}C## when ##80 \ W## are applied to the base. Here are my calculations: $$Q_{f}=\frac{\alpha U}{\sqrt{\frac{\alpha U}{\lambda A_{f}}}} \cdot \left( T_{s} - T_{a} \right) \cdot tanh \left( \sqrt{\frac{\alpha U}{\lambda A_{f}}} L \right)$$ $$Q_{f}=\frac{15 \cdot 0.128}{\sqrt{\frac{15 \cdot 0.128}{237 \cdot 0.00024}}} \cdot \left( 258 - 20 \right) \cdot tanh \left( \sqrt{\frac{15 \cdot 0.128}{237 \cdot 0.00024}} 0.04 \right)=17.9563 \ W$$ $$Q_{bf}=\alpha A \cdot \left( T_{s} - T_{a} \right)=15 \cdot 0.000768 \cdot \left( 258 - 20 \right)=2.7418 \ W$$ $$Q_{t}=n_{f} \cdot Q_{f} + n_{bf} \cdot Q_{bf}=4 \cdot 17.9563 + 3 \cdot 2.7418=80.0505 \ W$$ I wonder if there is a way to increse the accuracy of these hand calculations. The approach used here is approximate but maybe there's no better analytical method.Chestermiller said:What these results are telling us is that most of the entering heat leaves through the fins, and most of the heat that enters through the web between the fins flows horizontally to the fins. Only a small fraction is flowing into the air above the web. This means that the web is going to be approximately the same temperature as the base of the fins, and that your original equations in post #1 are going to be the correct ones to use, with Ts=Ts'. So try the problem again using these equations and see what you get. The heat load at the base is going to be equal to the sum of the two heats from the two equations.
-------------------------------------------------------FEAnalyst said:Thanks for the reply. Actually, I’m not doing the calculations for any particular heat sink or cooling system. It’s just an abstract problem that I want to solve with FEA first and then verify using analytical calculations. So it will be used only for educational purposes.
FEAnalyst said:By any chance, do you remember the name of this manufacturer ? Maybe I will be able to get this book.
I checked this site and it's really interesting but unfortunately covers only radiation and natural convection while I need a different set of formulas.