Heat sink - calculating temperature at the base

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The discussion focuses on calculating the temperature at the base of a heat sink with straight fins, given a known power input and convection conditions. Participants explore existing formulas for heat dissipation from fins and surfaces between fins, but express difficulty in adapting these to scenarios with known power rather than temperature. There is a consensus that while analytical solutions may be limited, numerical methods like FEA can provide insights, and thermal resistance data from heat sink manufacturers can aid in validation. The conversation also touches on the complexity of heat transfer dynamics, emphasizing the need for accurate modeling to compare numerical and analytical results effectively. Overall, the thread highlights the challenges and methodologies in thermal analysis of heat sinks.
  • #31
Tom.G said:
Uhmm... Something doesn't match here. :oldconfused:
Oops. My mistake.
 
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  • #32
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.
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.
 
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  • #33
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.
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Sorry, I just dropped into the thread to see how it's going, and I see that Chet is giving you great help. But I do need to take issue with those 244-258°C numbers. That is way too high for any real semiconductor device or transformer device. I know that this is just a numerical exercise, but a good extension would be to figure out how big the heatsink needs to be for a practical semiconductor product. Also what difference it makes to turn the heatsink to the vertical orientation, and what difference it makes to use forced air flowing across the heatsink.
https://www.analog.com/media/en/training-seminars/tutorials/mt-093.pdf
 
  • #34
If I may interject here, this thread brings to mind a quote from President James A. Garfield: “The ideal college is Mark Hopkins on one end of a log and a student on the other.”

You can look up those people if you want but the point is, a dedicated talented teacher interacting one on one with a dedicated willing student. Awesome to see in practice. PF can be a joy.

Kudos to @Chestermiller and @FEAnalyst
 
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  • #35
@FEAnalyst
Since you seem to have found your solution this is a bit dated, but for completeness, here goes anyhow.

Way back in post #6 I mentioned a book about heat sinks, and you responded:
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.

I stumbled across this thread today, re-searched for the book, and found it!

Title: HEAT SINK APPLICATION HANDBOOK
Publisher: AHAM
968 W. Foothill Blvd., Azusa California, 91702, ph (213) 334-5135​
2 Gill St. Bldg. 5, Woburn, Massachusetts, 01801, ph (617) 935-4350​

Pages: 180
Publication Date: 1974
5¼ x 7¾ inches, perfect-bound paper back, cover price USD $9.50

A Google search (https://www.google.com/search?&q=aham+heatsink) shows it is now "AHAM Tor Inc."

A semiconductor AP Note from 1993 shows:
AHAM-TOR Heatsinks, 27901 Front St, Rancho, California, 92390, (714) 676-4151

A quick and cursory look thru the book indicates that although it addresses Fin Shape, Conductivity, etc., it mostly concerns the interface calculations rather than the heat flow details within the heatsink itself.

Cheers,
Tom
 
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