Calculation For Thermal Conductivity [closed]

Total newbie here way over my head on a project I'm working on. I need to protect some electronics while they are inside an injection mould of liquid latex that measures 180 degrees celsius. The electronics are 'baked' for lack of a better word for 16 minutes, then dropped into cold water out of the mould to cool. I need to make sure I don't melt/destroy the electronics by keeping them below 60 degrees celsius during this step.

My plan is to use a silicone potting compound which measures at 0.06 watt/meter/K. I'm trying to calculate how thick the layer of silicon needs to be between the electronics and the heated latex and what the temperature will reach per minute the ball is exposed to this heat so I don't need to destroy prototype electronics doing trial and error.

Any help would be amazing! In summary:

• 180 degrees Celsius for 16 minutes
• electronics measure 12mm in diameter (sphere)
• Silicone thermal conductivity of 0.06 watt/meter/K

What thickness is required to make sure the electronics don't get above 60 degrees celsius?

Update - I found the specific heat value on the spec sheet of 1200 J/kg.K

closed as off-topic by Kyle Kanos, Jon Custer, user259412, Michael Seifert, user36790 Dec 2 '16 at 3:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question appears to be about engineering, which is the application of scientific knowledge to construct a solution to solve a specific problem. As such, it is off topic for this site, which deals with the science, whether theoretical or experimental, of how the natural world works. For more information, see this meta post." – Jon Custer, user259412, Michael Seifert, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

• You will also need the thermal capacity of the the silicone to know how its temperature will rise. – Gremlin Dec 1 '16 at 10:48
• Without values for the specific heat capacity of the sphere (and its content) there's no way of estimating temperature rise of the sphere. – Gert Dec 1 '16 at 10:58
• Welcome on Physics SE :) You might want to consider whether the engineering SE might not be a better place for this question ... – Sanya Dec 1 '16 at 11:30

Provided some material constants are known, a reasonable estimate could be obtained by lumped thermal analysis. The blue shell is the protective silicone. The surrounding temperature is $T_{\infty}$ ($180\:\mathrm{C}$) and we're looking for the function $T(t)$.

Lumped analysis assumes the temperature of the sphere is uniform (no radial temperature gradients).

Using Newton's law of cooling/heating we can describe the heat flux entering the sphere as:

$$\frac{\mathbf{d}Q}{\mathbf{d}t}=uA[T_{\infty}-T(t)],\tag{1}$$

where $u$ is the overall heat transfer coefficient and $A$ the surface area of the sphere (assuming the silicone layer isn't too thick).

An infinitesimal heat flow $\mathbf{d}Q$ causes the sphere to heat up by $\mathbf{d}T$, acc.:

$$\mathbf{d}Q=mc_p\mathbf{d}T(t),\tag{2}$$ where $m$ is the mass of the sphere and $c_p$ its spcific heat capacity. Substituting $(2)$ into $(1)$ we get:

$$\frac{mc_p\mathbf{d}T(t)}{\mathbf{d}t}=uA[T_{\infty}-T(t)]\tag{3}$$

$(3)$ is a first order linear differential equation, solvable by separation of variables and it yields:

$$\ln\Bigg[\frac{T_{\infty}-T(t)}{T_{\infty}-T_0}\Bigg]=-\frac{uA}{mc_p}t,\tag{4}$$

where $T_0$ is the initial temperature of the sphere and $T(t)$ the temperature after time $t$. So $(4)$ describes the approx. temperature evolution of the sphere.

The overall heat transfer coefficient $u$ can be estimated for not too large thicknesses $\theta$ of the silicone shell by:

$$\frac1u\approx\frac{1}{k_1}+\frac{\theta}{\kappa}+\frac{1}{k_2},\tag{5}$$ where:

1. $k_1$ is the heat transfer coefficient from heating medium to silicon and $k_2$ is the heat transfer coefficient from silicone to sphere.
2. $\kappa$ is the heat conductivity of the silicone.

By setting $T(t)$ to the desired 'safe' value and using $(4)$, $u$ can then be estimated. And by using $(5)$, the minimum $\theta$ can be calculated so $T(t)$ doesn't exceed the safe temperature.

This is a tricky question, and any of the mathematical answers are only going to give you an estimate. If you want an answer that you can rely on, you need to experiment or use something like finite element modelling.

I can't give a complete answer, but maybe point you towards a starting framework.

The problem is governed by the heat equation. It's a partial differential equation that you'll have to solve. We can try thinking it of just one dimension (i.e. the thickness of the insulation).

$\frac{\partial u}{\partial t} (x, t) = k \frac{\partial^2 u}{\partial x^2}(x,t)$

Here the temperature is $u(x,t)$ and $k$ is the thermal conductivity. We'll have to set up some boundary and initial conditions. At the outer surface of the insulation $x=R$, we can set $u = T_{bath}$ for all $t$.

For the inner surface of the insulation, I think a simplification would be to think about a solid sphere of insulation rather than a shell. It makes the maths easier and is probably more conservative. Then we could set an initial condition $u = T_{room}$, and enforce that the first derivative $\frac{\partial u}{\partial x} \rvert_{x=0}$ goes to zero for all $t$.

You also need to choose an initial temperature profile $u(x,0)$. In reality it's going to be discontinuous, but if you choose a smooth functional form it's going to make the maths easier.

Then what you want to find is the thickness $R$ so that the temperature $u$ at $x=0$ is less than your target after the given amount of time.

You can probably tell that this isn't the easiest question, and I wouldn't trust its result in a real-work application without a substantial safety factor.

• Using Fourier's heat equation for a shelled sphere is very hard to do. I say this from personal experience. Using much simpler lumped analysis will allow to at least play around with some numbers without either numerical or analyticl number crunching. – Gert Dec 1 '16 at 11:46

I would try this experimentally. Don't bake your prototype. Bake a thermocouple and measure the temperature.

Off hand, I would guess that anything less than several inches isn't enough.