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Oops. My mistake.Tom.G said:Uhmm... Something doesn't match here.![]()
This discussion focuses on calculating the temperature at the base of a heat sink with straight fins, given a known power input at the bottom surface and convection conditions on the remaining surfaces. Key formulas for heat dissipation from fins and surfaces between fins are provided, including the heat transfer coefficient and thermal conductivity. The conversation highlights the need for analytical solutions to determine the temperature at the base, as existing formulas typically assume known temperatures rather than power inputs. Participants suggest using FEA simulations and reference tools like COMSOL and online calculators for verification.
PREREQUISITESEngineers, thermal analysts, and researchers involved in electronic cooling solutions, particularly those working with heat sinks and thermal management systems.
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.