OT: Thermal question

Plywood thermal conductivity is about 0.13 W/m-K

160 sq. in is about 0.103 sq. meter, and 1/2 inch is .0127 meters

That comes to 1.05 K, if I didn't fumble a number.

I would build a custom-made analog computer for this. Start with some

1/2" plywood and a resistor...

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com

I get the inverse, 0.948 K. No? Same thing for an estimate. If the dimensions change it will matter a lot more.

--

Rick

Sno-o-o-ort >:-} ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
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I love to cook with wine.     Sometimes I even put it in the food.

Sez the man with a fully outfitted shop. ;-)

--sp

--
Best regards,  
Spehro Pefhany 
Amazon link for AoE 3rd Edition:            http://tinyurl.com/ntrpwu8 
Microchip link for 2015 Masters in Phoenix: http://tinyurl.com/l7g2k48

I did fumble; area is 192 sq. in == 0.124 m^2 (moving up from a Post-it to an envelope back does improve my accuracy...)

temperature difference = 0.8 K, this time

Temperature rise measured where?

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

Lol

Temp rise = power * thickness / (conductivity * area )

Temp rise = 1 W * 0.0127 m / (0.13 * 0.124) = 0.79 K

Buuuttt, when I looked up the thermal conductivity of plywood I found values ranging from 0.12 and 0.17 W/m-K. The construction of plywood varies a lot, so the temperature rise can vary a lot.

--

Rick

That's the temp. drop across the plywood only. Don't forget thermal resistance from load to plywood, and plywood to ambient (if we're after temp rise of the load).

Cheers, James Arthur

Now the geometry of the load gets involved.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

The right answer involves laminar flow approximation and the Nusselt number.. maybe there is a calculator out there. I think I would just calculate the effect of the four vertical walls and leave it at that for an upper bound.

It won't be much though for something ~5" cube. I'd guess a few degrees C total.

--sp

--
Best regards,  
Spehro Pefhany 
Amazon link for AoE 3rd Edition:            http://tinyurl.com/ntrpwu8 
Microchip link for 2015 Masters in Phoenix: http://tinyurl.com/l7g2k48

I like Tim's analog computer.

--

Rick

So do I, but not enough to build it for Jim.

--sp

--
Best regards,  
Spehro Pefhany 
Amazon link for AoE 3rd Edition:            http://tinyurl.com/ntrpwu8 
Microchip link for 2015 Masters in Phoenix: http://tinyurl.com/l7g2k48

Seems to me that there is no answer until the question is better defined. What temperature rise is to be measured?

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

Where's your sense of adventure ?

We often do thermal breadboards. Cardboard chassis mockups, fans, heat sinks, resistors, duct tape, whatever. Fiddle to optimize. Air flow is especially hard to model. It's fun to build stuff.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

That fine old tradition is the reason old (Cray-1?) supercomputers used three kinds of chips. A gate, a flip-flop, and a dummy chip. All the airflow cooling was designed for each board to be the same size, with a regular grid of component ICs... So if you didn't need M x N chips for the board's function, it was filled in with dummies.

Outer surface area of box with inner dims of 4in x 4in x 10 in:

You rarely get the labelled material thickness in commercial building materials.

but thickness of 1/2in plywood is roughly 12mm (you can measure it).

10.2cm x 10.2cm x 25.4cm internal

becomes 12.6cm x 12.6cm x 27.8cm external

External average surface rise:

external surface area is

12.6 x 12.6 x2 = 318 12.6 x 27.8 x4 = 1400 1720 cm^2

power transfer of 1000mW

produces 1000/1720mW/cm^2 burden or .59mW per cm^2

so external surface rise above ambient is ~ 0.6 degrees.

Difference through the walls:

plywood thermal conductivity = k = 0.13W/mK s= distance (meters) ~ wall thickness A= xsection (meters^2) ~ internal surface area q= power burden (watts) ~ sourced power

q = (T1 - Tn) / (s / k A)

.001W = deltaTC / (.012meters / (0.13W/mC * .0001m^2))

deltaTC = .001W x .012meters / (0.13W/mC * .0001m^2) = 0.92degrees

So the inner wall temperature is 0.9 degrees above the outer surface wall.

0.6 + 0.9 = 1.5 degrees above external ambient.

Internal air is going to be somewhere close. Temperature difference between internal air and the internal wall are roughly the same, as the internal air represents a limited source that is responsible for surface temperature, rather than some infinite sink, but there may be some kind of penalty for the transfer.

Temperature of the source's surface temperature will depend on its size.

RL

Best answer.