How can I calculate the thermal resistivity of 6061 Aluminum?

[Teece]

Hi,

I normally use commercially available heat sinks, with characteristics that
are easily determined. In order to be able to determine if a specific heat
sink is usable for a given amount of dissipated power, it is necessary to
know the thermal resistivity in Watts/degrees C.

In my current application I am limited to using a rectangular sheet of cast
aluminum alloy 6061 T6. From available tables I have found the thermal
conductivity of this material to be 167 W/m-C. I don't know how to do the
conversion to thermal resistivity because I am unsure of the units used for
thermal conductivity. In the definition, what does W/m-C really mean in that
how does the m (meters) relate to the volume of the aluminum? '"m" is
single dimensional while the heat sink is a three dimentional piece of
Aluminum.

Help is welcome.

Tom
[email protected]

 

[Ian Stirling]

Hi,

I normally use commercially available heat sinks, with characteristics that
are easily determined. In order to be able to determine if a specific heat
sink is usable for a given amount of dissipated power, it is necessary to
know the thermal resistivity in Watts/degrees C.

In my current application I am limited to using a rectangular sheet of cast
aluminum alloy 6061 T6. From available tables I have found the thermal
conductivity of this material to be 167 W/m-C. I don't know how to do the
conversion to thermal resistivity because I am unsure of the units used for
thermal conductivity. In the definition, what does W/m-C really mean in that
how does the m (meters) relate to the volume of the aluminum? '"m" is
single dimensional while the heat sink is a three dimentional piece of
Aluminum.

A 1m cube will have 167W of heat flowing through it when opposite faces
have a 1C temperature difference.
A 1mm cube similarly 167mw.

 

[mike]

A 1m cube will have 167W of heat flowing through it when opposite faces
have a 1C temperature difference.
A 1mm cube similarly 167mw.

Heat sinking is always an interesting topic.
The heat has to go somewhere, so you have to define a point somewhere
in the system that has a defined temperature. That's often not easy.
The conductivity number makes some assumptions about uniformity of
surface temperature, which usually isn't the case at the source end.
Since you said "sheet" and not "block" of aluminum, one might think
it's thin relative to it's other dimensions. There may be a BIG thermal
gradient across the sheet. This can surely be modeled to come up with
a solution; but a single number may well lead you to be too optimistic
about the thermal behavior.

And if there's convection, your characterization problems multiply.
mike

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[Terry Given]

Hi,

I normally use commercially available heat sinks, with characteristics that
are easily determined. In order to be able to determine if a specific heat
sink is usable for a given amount of dissipated power, it is necessary to
know the thermal resistivity in Watts/degrees C.

In my current application I am limited to using a rectangular sheet of cast
aluminum alloy 6061 T6. From available tables I have found the thermal
conductivity of this material to be 167 W/m-C. I don't know how to do the
conversion to thermal resistivity because I am unsure of the units used for
thermal conductivity. In the definition, what does W/m-C really mean in that
how does the m (meters) relate to the volume of the aluminum? '"m" is
single dimensional while the heat sink is a three dimentional piece of
Aluminum.

Help is welcome.

Tom
[email protected]

using the following formula:

sigma = 167 [W/mK] = thermal conductivity, Watts per meter-Kelvin
Rho = 1/sigma = 1/167 [mK/W]

a piece of aluminium of cross sectional area A [m^2] and length L [m] has
end-to-end thermal resistance given by:

R = Rho*L/A = [mK/W]*[m]/[m^2] = [K/W]

That is of course for conduction through the solid. In general your heatsink
thermal resistance comprises two factors:

1) conduction through the heatsink material itself, to get the heat from the
source to the fins - this is what is calculated above. Note that thermal
conduction is actually a diffusion process, so beware of "spreading
resistance" - a small heat source mounted on a large block gradually spreads
out a finite distance as it penetrates the solid.

2) convection from the fins to the surrounding environment (in some cases
radiation may be significant, often it is not, but if so then it can be
considered a parallel heat loss path)

It is usually a lot easier to measure than calculate (unless you happen to
be an expert in fluid dynamics, and even then real problems are still solved
computationally)

Cheers
Terry

 

[Gregory L. Hansen]

Hi,

I normally use commercially available heat sinks, with characteristics that
are easily determined. In order to be able to determine if a specific heat
sink is usable for a given amount of dissipated power, it is necessary to
know the thermal resistivity in Watts/degrees C.

In my current application I am limited to using a rectangular sheet of cast
aluminum alloy 6061 T6. From available tables I have found the thermal
conductivity of this material to be 167 W/m-C. I don't know how to do the
conversion to thermal resistivity because I am unsure of the units used for
thermal conductivity. In the definition, what does W/m-C really mean in that
how does the m (meters) relate to the volume of the aluminum? '"m" is
single dimensional while the heat sink is a three dimentional piece of
Aluminum.

Help is welcome.

Tom
[email protected]

If K is the thermal Konductivity, then the heat that flows through a
length of the metal is

P = KA/L * delta T

A cross-sectional area, L length. You'll conduct more heat with a larger
area, less heat with a longer distance. [A/L] = m, so it's appropriate
that K has one m in the denominator.

Then

delta T = L/AK P

= RL/A P

where R=1/K is the resistivity, with units of m-C/W.

 

[John Larkin]

Hi,

I normally use commercially available heat sinks, with characteristics that
are easily determined. In order to be able to determine if a specific heat
sink is usable for a given amount of dissipated power, it is necessary to
know the thermal resistivity in Watts/degrees C.

In my current application I am limited to using a rectangular sheet of cast
aluminum alloy 6061 T6. From available tables I have found the thermal
conductivity of this material to be 167 W/m-C. I don't know how to do the
conversion to thermal resistivity because I am unsure of the units used for
thermal conductivity. In the definition, what does W/m-C really mean in that
how does the m (meters) relate to the volume of the aluminum? '"m" is
single dimensional while the heat sink is a three dimentional piece of
Aluminum.

Help is welcome.

Tom
[email protected]

Ian's explanation is correct, but that's bulk thermal conductivity,
which doesn't tell you much about how well a sheet will heatsink your
gadget. *That* is a very complex calculation... it depends on the
sheet size, thickness, shape, and environment, and on the size and
shape of the thing you want to sink.

The best way to do this is to build a rough prototype and test it.
Simulation software would cost $5K to $30K, and take longer to learn
than it would take to build a proto.

Oh: a TO-220 transistor bolted to an infinite sheet of 0.062 thick
aluminum has a theta of about 2 k/w in still air, limited by the
thermal spreading resistance of the thin sheet.

John

 

[Tim Wescott]

Hi,

I normally use commercially available heat sinks, with characteristics that
are easily determined. In order to be able to determine if a specific heat
sink is usable for a given amount of dissipated power, it is necessary to
know the thermal resistivity in Watts/degrees C.

In my current application I am limited to using a rectangular sheet of cast
aluminum alloy 6061 T6. From available tables I have found the thermal
conductivity of this material to be 167 W/m-C. I don't know how to do the
conversion to thermal resistivity because I am unsure of the units used for
thermal conductivity. In the definition, what does W/m-C really mean in that
how does the m (meters) relate to the volume of the aluminum? '"m" is
single dimensional while the heat sink is a three dimentional piece of
Aluminum.

Help is welcome.

Tom
[email protected]

So, everyone's post was right; let me see if I can summarize it for you:

1. You want a thermal resistance to something (ambient?), and you have
a thermal resistivity of the aluminum itself, which only tells you how
the heat gets around inside the aluminum.

2. You are not considering the thermal drop from the aluminum to the
surrounding air (will there be air?). This usually dominates a
heat-sink's performance, although small, high-power components sometimes
require a copper heat spreader to help out.

3. You haven't mentioned the thermal drop from your package to the heat
sink -- I hope you know it.

4. Calculating thermal dissipation up front is way hard.

4a. Except that I know several mechanical engineers who can do it with
a spreadsheet and get reliable estimates that are consistently 20% less
than reality, which is exactly what you want.

5. 4a notwithstanding, you should probably do this by experiment. Bolt
your part (or a same-package part) onto your proposed heatsink or a
model thereof. Establish the same airflow that you can expect in the
real thing, and make sure that the surface finish is representative
(black, polished, sand-rough, whatever). Dissipate some heat and
measure some temperatures.

 

[Frithiof Andreas Jensen]

Help is welcome.

*Try it* - there is no simulation better than reality.

Usually the size and orientation of such a panel is defined by the
mechanical construction - which was given on clay tablets from a burning
bush - so the easy way out is to bolt some power resistors in the locations
of whatever you want to cool, load them up with the estiamated power, and
then simply measure the temperature distribution on the plate with
thermistors taped to it.

The big advantage is that you can quicly make up the rest of the
surroundings from f.ex. cardboard and it will not change the result very
much compared to the "real" materials.

You can also calculate - which requires that you get to know some fluid
mechanics; there are three flow "modes" for natural convection where the
equations change and you have to account for radiation too. And you must
program it. Done that!!

- or just shell the USD 20000 (and on up) for a thermal FEM program ;-)

 

[Rene Tschaggelar]

Hi,

I normally use commercially available heat sinks, with characteristics that
are easily determined. In order to be able to determine if a specific heat
sink is usable for a given amount of dissipated power, it is necessary to
know the thermal resistivity in Watts/degrees C.

In my current application I am limited to using a rectangular sheet of cast
aluminum alloy 6061 T6. From available tables I have found the thermal
conductivity of this material to be 167 W/m-C. I don't know how to do the
conversion to thermal resistivity because I am unsure of the units used for
thermal conductivity. In the definition, what does W/m-C really mean in that
how does the m (meters) relate to the volume of the aluminum? '"m" is
single dimensional while the heat sink is a three dimentional piece of
Aluminum.

The thermal conductivity of a material tells something about the
temperature gradient, once the heat flows.
It is less the conductivity of the aluminum, that matters but the
interface between aluminum and the air.
Usual finned heatsinks are specified under the seldom written
assumption that the fins are oriented to the vertical, along the
convection, with 1 foot or so above and below unrestricted.
When mounted horizontal, onto a pcb, the convection is restricted
and the achievable figures are lower.

Rene

 

[John Larkin]

*Try it* - there is no simulation better than reality.

Usually the size and orientation of such a panel is defined by the
mechanical construction - which was given on clay tablets from a burning
bush - so the easy way out is to bolt some power resistors in the locations
of whatever you want to cool, load them up with the estiamated power, and
then simply measure the temperature distribution on the plate with
thermistors taped to it.

One caution: thin-metal systems are dominated by spreading resistance,
and then the footprint of the heat source becomes important. If a few
power resistors are used as the test heat source, and they don't match
the size and location of the semiconductors to be used later, results
can be way off. We like to use Caddock TO-220 resistors for testing,
as they have a proper TO-220 footprint. Of course, the best heat
source is the actual, operating circuit.

John