Calculate Thermal Resitance from Heat Sink Geometry

AI Thread Summary
Calculating the thermal resistance of a heat sink involves understanding its geometry, material properties, and airflow conditions. Free convection complicates predictions, making direct measurement with a thermocouple and power resistors a more practical approach. While theoretical calculations are possible, they often require advanced knowledge and resources, such as textbooks on heat transfer. Comparing the custom heat sink to existing off-the-shelf profiles can also provide useful insights. Ultimately, practical measurement is recommended due to the complexities involved in thermal resistance calculations.
mcouch
Messages
5
Reaction score
0
Hi all,

I have a heat sink for which I know the exact geometry (have the CAD model). This heat sink was custom designed and machined from a hunk of aluminum, though it is very basic and I have a hunch I can find an extrusion or OTS part that will match its performance. Rather than trying to just match the surface area, I'd like to know what the thermal resistance of it is so that I can try to match it.

How do I calculate the thermal resistance of the heat sink given its geometry and material?

Here's a rough description of the heat sink:

18 square fins spaced evenly in a single row at a pitch of 0.36". Each fin is 2" tall by 2" wide and is 0.08" thick.

Thanks
 
Last edited:
Engineering news on Phys.org
mcouch said:
Hi all,

I have a heat sink for which I know the exact geometry (have the CAD model). This heat sink was custom designed and machined from a hunk of aluminum, though it is very basic and I have a hunch I can find an extrusion or OTS part that will match its performance. Rather than trying to just match the surface area, I'd like to know what the thermal resistance of it is so that I can try to match it.

How do I calculate the thermal resistance of the heat sink given its geometry and material?

Here's a rough description of the heat sink:

18 square fins spaced evenly in a single row at a pitch of 0.36". Each fin is 2" tall by 2" wide and is 0.08" thick.

Thanks

The thermal resistance will depend on the orientation (horizontal or vertical), and whether there is forced airflow over it.

Often it is easier to just measure the thermal resistance θ with a thermocouple and power resistors...
 
I'm sorry, I forgot to mention: this is free convection. So I can make whatever assumptions I need to about ambient temperature, air density, air viscosity, etc.
 
It's about impossible to calculate, and certainly unreasonable, sorry. Measure it.

The heat exchange depends fundamentally on how much air circulates and how, which is already inaccessible to hand calculation and which computer programmes are bad at.

Add to it that the exchange is limited by the fine layer of air right against the metal that circulates less good... A few formulas are known for very simple shapes and that's all.

Some people specialize in that and, after months and years, are capable of giving an estimate. If you read French: Sacadura, "Initiation aux transferts thermiques". Hard, lengthy, and it won't help so much in a practical case.

Besides measuring, you could compare with existing similar heatsinks. Take a catalogue with data, say like Fischer, find the most resembling profile. By the way, I'm surprised someone machines a heat sink instead of cutting it from a bought profile.

Free convection is the most difficult to predict.
 
Enthalpy said:
It's about impossible to calculate, and certainly unreasonable, sorry. Measure it.

The heat exchange depends fundamentally on how much air circulates and how, which is already inaccessible to hand calculation and which computer programmes are bad at.

Engineering calculations w.r.t. fins and heatsinks (arrays of fins) are very possible and used often for things like electronics cooling and heat exchanger design. Unfortunately an internet forum isn't really the right place to try an teach someone how to do such calculations; you really need to take a college senior-level heat transfer class to do it.

Short of that, a textbook on the subject is a good reference on how such calculations are done, the textbook I have, Introduction to Heat Transfer, covers everything you need to do such calculations.

Suffice to say the general calculation procedure would be:
  • Conceptualize a thermal-equivalent circuit for the component being cooled, with thermal resistance variables for the heatsink, component, and convection.
  • Estimate ambient temperature and required power dissipation.
  • Using fin geometic properties calculate fin efficiency and thermal-equivalent resistance as a function of temperature. This would be accomplished using the "fin efficiency" for your particular geometry, out of a table in the book.
  • Plug the equation into the thermal-equivalent circuit and solve for the resulting temperature of the component and heatsink.
  • Iterate.
 
Posted June 2024 - 15 years after starting this class. I have learned a whole lot. To get to the short course on making your stock car, late model, hobby stock E-mod handle, look at the index below. Read all posts on Roll Center, Jacking effect and Why does car drive straight to the wall when I gas it? Also read You really have two race cars. This will cover 90% of problems you have. Simply put, the car pushes going in and is loose coming out. You do not have enuff downforce on the right...
Thread 'Physics of Stretch: What pressure does a band apply on a cylinder?'
Scenario 1 (figure 1) A continuous loop of elastic material is stretched around two metal bars. The top bar is attached to a load cell that reads force. The lower bar can be moved downwards to stretch the elastic material. The lower bar is moved downwards until the two bars are 1190mm apart, stretching the elastic material. The bars are 5mm thick, so the total internal loop length is 1200mm (1190mm + 5mm + 5mm). At this level of stretch, the load cell reads 45N tensile force. Key numbers...
I'm trying to decide what size and type of galvanized steel I need for 2 cantilever extensions. The cantilever is 5 ft. The space between the two cantilever arms is a 17 ft Gap the center 7 ft of the 17 ft Gap we'll need to Bear approximately 17,000 lb spread evenly from the front of the cantilever to the back of the cantilever over 5 ft. I will put support beams across these cantilever arms to support the load evenly
Back
Top