# Heat Transfer - Radiation - Net heat transfer between two mirrors

## Homework Statement:

There are two mirrors which act as grey bodies which are a short distance away from one another. There is no transmission at each of the mirrors. What is the net heat transfer between the mirrors?

## Relevant Equations:

$q = \epsilon \sigma T^4$
$\alpha + \tau + \rho = 1$
Hi,

So there is already a written solution which I have, but this is more a question about why we omit reflection that come back to the same mirror?

Method:

Let us consider one of the mirrors, we know it will emit a heat flux given by: $q_{1} = \epsilon_{1} \sigma T_{1}^4$. Given that we are dealing with grey bodies, then we know that $\alpha = \epsilon$ and we are told that $\tau = 0$. If we want to consider the heat that reaches and is absorbed by the other mirror, then we get the following sum (as a result of successive reflections):
$$q_{12} = \epsilon_{2} q_{1} + \rho_{2} \rho_{1} \epsilon_{2} q_{1} + \left( \rho_{2} \rho_{1} \right)^2 \epsilon_{2} q_{1} + \left( \rho_{2} \rho_{1} \right)^3 \epsilon_{2} q_{1} + ..... = \frac{\epsilon_{2} q_{1}}{1 - \rho_{2} \rho_{1}}$$

By symmetry, we can find that $q_{21} = \frac{\epsilon_{1} q_{2}}{1 - \rho_{2} \rho_{1}}$.
Then we can find $Q_{net} = Q_{12} - Q_{21} = \frac{A \epsilon_{1} \epsilon_{2} \sigma (T_{1}^4 - T_{2}^4) } {1 - \rho_{2} \rho_{1}}$ (assuming areas are equal and that T_1 is greater than T_2). This is the answer that I am 'supposed' to get.

However, I was wondering why we don't also consider the rays that are emitted from a mirror and are reflected back towards itself (i.e. the $\epsilon_{1} \rho_{2} q_{1}, etc...$)? I think including these terms would change the q12 term to be: $q_{12} = \frac{\epsilon_{2} ( q_{1} + \rho_{1} q_{2})}{1 - \rho_{2} \rho_{1}}$

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DrClaude
Mentor
For the same reason that you don't include in ##q_{12}## the heat that never left 2 because ##\epsilon_2 < 1##.

You are looking at transfers from 1 to 2. Energy that stays in one object, for whatever reason, is irrelevant.

For the same reason that you don't include in ##q_{12}## the heat that never left 2 because ##\epsilon_2 < 1##.

You are looking at transfers from 1 to 2. Energy that stays in one object, for whatever reason, is irrelevant.
Ah okay that makes sense - we have defined q as the heat that has already left the mirror.

Thank you