I would like to calculate the irradiance incident upon a cca. 2x2 cm surface from 2 different radiating sources.

  1. a 30x32 cm plane surface at 30.5 °C, at a distance of 70 cm,
  2. a 5x10 cm plane surface, 33 °C, at a distance of 16.5 cm. Both surfaces have high emissivity values, cca. 0.98.

As far as I know, I calculate the total power from the surface using the Stefan-Boltzmann law: $F = A \epsilon σT^4$, where $A$ is the surface area, $\epsilon$ the emissivity, $\sigma$ the Stefan-Boltzmann constant, and $T$ the temperature of the surface.

However, I'm not sure how to calculate the radiation intensity falling on my surface of interest (2x2 cm, at a given distance). I know it has to do with the solid angle subtended by the surfaces, but I don't know the exact calculations.


1 Answer 1


I know it has to do with the solid angle subtended by the surfaces, but I don't know the exact calculations.

Not to worry-- most people don't because the math gets hairy fast! Luckily for you, engineers typically rely on charts generated by numerical methods for commonly encountered geometries.

Here is the one for the situation you describe:View factor for small surface away from large rectangle

The "view factor" is a decimal less than 1 which tells you, for example, what fraction of energy emitted by the big rectangle will fall on the small surface. Noting that the chart above gives $F_{dA_1-A_2}$, get your view factor from the graph above as we will need it later.

Let's first define $E_{b}$ as the emissive power of a black (emissivity $\epsilon = 1$) surface:

$E_{b}=\sigma T^4$

$E_{b}$ = emissive power of a black surface in $Watts/m^2$
$\sigma$ = Stefan-Boltzmann constant, units $Watts/m^2 K^4$
$T$ = temperature of black surface in $K$

Between any two black surfaces, the net radiant heat exchange rate from surface 2 to surface 1 is given by:

$$q_{2-1}\ =\ A_1 F_{1-2}(E_{b_2}-E_{b_1})$$

$q_{2-1}$ = net radiant heat exchange rate from surface 2 to surface 1 in $Watts$
$A_1$ = surface area of surface 1 in $m^2$
$F_{1-2}$ = view factor-- how much of what surface 1 "sees" is surface 2 (unitless)
$E_{b_2}$ = emissive power of black surface 2 in $Watts/m^2$
$E_{b_1}$ = emissive power of black surface 1 in $Watts/m^2$

For your case, we will represent the 2x2 cm square by $dA_1$ and the large rectangle by $A_2$ to follow the conventions in the chart. Therefore the surface area of the 2x2 cm square will be represented by $A_{dA_1}$ for example. The equation is updated as follows:

$$q_{A_2-dA_1}\ =\ A_{dA_1} F_{dA_1-A_2}(E_{A_2}-E_{dA_1})$$

This is the net radiant heat exchange rate. Since you only want to know the $Watts/m^2$ from the large rectangle at the small square, we can set $E_{dA_1}=0$ and remove the $A_{dA_1}$ term entirely (this latter part is to get an answer in $Watts/m^2$):

$$E_{A_2\ \text{at}\ dA_1}\ =\ F_{dA_1-A_2}(E_{A_2})$$

$E_{A_2\ \text{at}\ dA_1}$ will be in units of $Watts/m^2$. You are duly warned that this is no longer the net radiant heat exchange. If you are trying to use this to calculate how much the small square will heat up, it will not work unless you consider the net.

Figure and equations are from "Chapter 23: Radiation Heat Transfer" in Fundamentals of Momentum, Heat, and Mass Transfer 5th Edition by James Welty, Charles E. Wicks, Gregory L. Rorrer, and Robert E. Wilson (2008).

  • $\begingroup$ Thank you very much, this graph is indeed really practical! $\endgroup$
    – anna
    Aug 6, 2017 at 21:07
  • $\begingroup$ However, what I don't entirely understand is how the 'receiver' surface area (in this case 2x2 cm) is taken into account? $\endgroup$
    – anna
    Aug 6, 2017 at 21:09
  • $\begingroup$ @anna my previous answer was unclear at best. Please let me know if this version still does not answer your question. $\endgroup$
    – pentane
    Aug 7, 2017 at 5:27
  • $\begingroup$ Thank you very, now it is clear! I would like to use it to compare the radiant energy reaching my small, 2x2 surface in the two cases, not to calculate how much it heats up. So this should work fine. Thank you very much again, huge help! $\endgroup$
    – anna
    Aug 7, 2017 at 8:02
  • $\begingroup$ @anna if I've answered your question, feel free to mark it as answered and toss me an upvote if you're feeling generous :) $\endgroup$
    – pentane
    Aug 7, 2017 at 13:39

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