**2. Radiative transfer between spherical surfaces**

Let us start by discussing radiative exchanges in simple volumes. The sphere is a volume enclosed by only one surface and with some restrictions it is used both in engineering and architecture. If the inner surface of the sphere emits under Lambertian diffusion, the total fraction of energy reaching the same surface will be one hundred percent, that is: the unity.

In mathematical terms this is expressed as:

$$\mathbf{F}\_{11} = \mathbf{1} \tag{1}$$

The fraction of energy leaving surface 1 and arriving at surface 1, is one. The former gives rise to a new algebra defined by:

$$\mathbf{F\_{i1}} + \mathbf{F\_{i2}} + \dots + \mathbf{F\_{in}} = \mathbf{1} \tag{2}$$

In a closed volume, by the principle of energy conservation, and disregarding transmission losses, radiant energy emitted by surface *i* is necessarily distributed in its entirety among the surrounding surfaces. As an example, in a cube, where all faces are equal, the fraction of energy leaving from an internal source to any of the other five is 1/5=0.2.

As the radiant flux is originated in a given surface and bears only nominal relationship with the medium in which the phenomena take place, the distribution of such flux will be a function of the dimensions of the surfaces involved. Thus we could anticipate a second and final property for our algebra (Lambert, 1764).

$$\mathbf{A}\_{i}"\mathbf{F}\_{\overline{\mu}} = \mathbf{A}\_{f}"\mathbf{F}\_{\overline{\mu}} \tag{3}$$

Where A i is the area of surface *i* and Fij is the fraction of energy that leaves *i* and reaches *j*.

Returning to the sphere, it is easy to see that although interior radiation may occur, unless we are able to pierce the surface by means of some kind of section, interaction with the environment remains negligible. Any planar section of the sphere will produce a spherical cap with a circular base that works as an aperture. (See Figure 1).

The size of the aperture is determined by the height of the cap, *h*. Beginning with the case of a hemisphere, the said height coincides with the radius of the sphere *R*=*a*=*h*.

Fig. 1. Surfaces generated by a spherical cap

In this situation, there are two surfaces involved in the radiation problem. Let us apply the former algebra to them.

If for the whole sphere F11 =1, taking into account the dependence on area of radiative transfer, it is assumed that for the hemisphere F11= 1/2= 0.5, and the demonstration will be given later in the text.

By equation (2) the former also implies that F12 =0.5 . Thus, half of the flux of the hemisphere is distributed over itself and the other half is ceded to the circular base that serves as an aperture of the sphere.

The fraction of energy leaving surface 1 and arriving at surface 1, is one. The former gives

 Fi1 + Fi2 +...+Fi*n* =1 (2) In a closed volume, by the principle of energy conservation, and disregarding transmission losses, radiant energy emitted by surface *i* is necessarily distributed in its entirety among the surrounding surfaces. As an example, in a cube, where all faces are equal, the fraction of

As the radiant flux is originated in a given surface and bears only nominal relationship with the medium in which the phenomena take place, the distribution of such flux will be a function of the dimensions of the surfaces involved. Thus we could anticipate a second and

A i\*Fij = A j\*Fji (3)

Where A i is the area of surface *i* and Fij is the fraction of energy that leaves *i* and reaches *j*. Returning to the sphere, it is easy to see that although interior radiation may occur, unless we are able to pierce the surface by means of some kind of section, interaction with the environment remains negligible. Any planar section of the sphere will produce a spherical

The size of the aperture is determined by the height of the cap, *h*. Beginning with the case of

In this situation, there are two surfaces involved in the radiation problem. Let us apply the

If for the whole sphere F11 =1, taking into account the dependence on area of radiative transfer, it is assumed that for the hemisphere F11= 1/2= 0.5, and the demonstration will be

By equation (2) the former also implies that F12 =0.5 . Thus, half of the flux of the hemisphere is distributed over itself and the other half is ceded to the circular base that serves as an

energy leaving from an internal source to any of the other five is 1/5=0.2.

cap with a circular base that works as an aperture. (See Figure 1).

a hemisphere, the said height coincides with the radius of the sphere *R*=*a*=*h*.

rise to a new algebra defined by:

final property for our algebra (Lambert, 1764).

Fig. 1. Surfaces generated by a spherical cap

former algebra to them.

given later in the text.

aperture of the sphere.

In accordance with this reasoning, for a cap whose area equals one third (1/3) of the whole surface, F11 =1/3 and F12 =2/3 and so on.

When the respective areas of surfaces 1 and 2 are introduced, the fraction of energy that leaves 1 and arrives at 2 equates the area-ratio of such potential sources. In this way, equation 3 is proved if we remember that, for the spherical cap, F21 has to be one. The circle is a planar figure and gives one hundred percent of its energy to the surrounding cap.

$$F\_{12} = \frac{a^2}{a^2 + h^2} \tag{4}$$

It is inferred that F11 = 1- F12 , and its value would be,

$$F\_{11} = \frac{h^2}{a^2 + h^2} \tag{5}$$

Substituting into equation 5, the trigonometric relation for the radius of the sphere, R,

$$a^2 + h^2 = 2 \ast R \ast h \tag{6}$$

We obtain an extremely important and beautiful expression,

$$F\_{11} = \frac{h}{2 \ast R} \tag{7}$$

Thus, by simple logic, and with hardly any calculus, the author has solved for the first time in this field one of the most complex integral equations of environmental science (Eq. 8).

$$\otimes\_{1-2} = E\_{b1} \int\_{A\_1} \int\_{A\_2} \cos \theta\_1 \* \cos \theta\_2 \* \frac{dA\_1 \* dA\_2}{\pi r^2} = E\_{b1} \* \pi \* a^2 \tag{8}$$

Where Φ1-2 is the radiative flux exchange between the surfaces considered, and Eb1 is the radiant energy emitted by surface 1. The relative ease of the solution is partly due to the fact that the quantities termed cosθi, which represent the cosines of the angles between the line going from the center of surface 1 to the center of surface 2 and their respective normals, are simpler to find in this case as the said normals always pass through the center of the sphere.

It is convenient to use the former results in an ample variety of ways.

For instance, if the aforementioned division of the sphere is performed by means of two planar sections, like in a quarter of sphere (see Figure 2), the solutions are still valid, in the sense that the quarter of a sphere is giving itself one fourth of its emissive power F33 =1/4.

Fig. 2. Surfaces generated by a quarter of a sphere

As we already found, the two semicircles receive in total 3/4 of the flux, but provided that they are of equal area, each one of them receives 3/8= F32 = F31 .

Thus, equation 2 is fulfilled.

By equation 3, the so-called principle of reciprocity, F13 = F23=2\*π\*R2/ π\*R2\* F31=3/4

And this implies that F12 = F21 =1/4

A second complex integral equation has been solved by the author without calculus.

In a similar manner, adjusting the fragment of sphere which the problem may demand, the radiative exchange between semicircles with a common edge, forming any angle from 0 to 180 degrees, can be found. The above example is valid for 90 degrees.

The author has been the first to propose the following equation, previously unheard of in the literature, to obtain the energy balance between the said semicircles, where x represents the value of the internal angle (Fig. 3),

$$F\_{12} = 1 - \frac{\chi}{90} + \frac{\chi^2}{32400} \tag{9}$$

Fig. 3. Radiative exchanges between two semicircles with a common edge and forming an internal angle *x*

So far so good. The former expression solves a whole set of integral equations and anybody can understand how the radiant flux is transferred from circular or semicircular apertures to the interior of the sphere as a total fraction. However, it is often useful to examine this transfer in more detail, i.e. point by point.

#### **2.1 Radiative transfer between spherical surfaces and points**

Referring again to the sphere in respect with the canonical equation 8; if you look at figure 4, it is easy to find the relationship between r, *cosθ* and the radius of the sphere R.

$$\mathfrak{G}\_{1-2} = \frac{E\_{b1}}{4 \ast \pi \ast R^2} \int\_{A\_1} \int\_{A\_2} \, dA\_1 \ast \, dA\_2 \tag{10}$$

$$A\_2 = \pi \ast \left(a^2 + (\mathcal{Z} \ast R - h)^2\right) \tag{11}$$

$$F\_{21} = \frac{\pi \ast a^2}{\pi \ast (a^2 + (2 \ast R - h)^2)}\tag{12}$$

$$a^2 + h^2 = 2 \ast R \ast h \tag{13}$$

$$F\_{21} = \frac{a^4 \ast h^4}{a^4 + 2 \ast h^4 - 2 \ast h^4 + 3 \ast a^2 \ast h^2 - 2 \ast a^2 \ast h^2} \tag{14}$$

$$F\_{21} = \frac{a^2 \star h^2}{a^2 (a^2 + h^2)} = \frac{h^2}{a^2 + h^2} = \frac{h}{2 \ast R} \tag{15}$$

Fig. 5. Cap and sphere where the radiative exchange takes place

In figure 5, the cap *S* goes from A to B and its area is π\*2\*R\*h, as mentioned above (Eq. 6). Dividing the former by the total area of the sphere according to equation 10, the result is as shown in Equations 7 and 15. That is, the energy received at any point of the interior sphere wall outside the spherical cap *S* is constant, h/2\*R. This is often expressed as 1/2\*(1-cosα) or 1/2\*(1-cos2β).

With the former property we can replace the cap source by its enclosed circle AB.

Fig. 6. The radiation vector in a sphere

Radiant energy due to the spherical cap or the said circle, in a direction normal to the interior of the sphere, is constant but it is mandatory not to forget that radiation is in truth a vector, meaning that its projection on different planes may present diverse values.

In figure 5, the cap *S* goes from A to B and its area is π\*2\*R\*h, as mentioned above (Eq. 6). Dividing the former by the total area of the sphere according to equation 10, the result is as shown in Equations 7 and 15. That is, the energy received at any point of the interior sphere wall outside the spherical cap *S* is constant, h/2\*R. This is often expressed as 1/2\*(1-cosα) or

Radiant energy due to the spherical cap or the said circle, in a direction normal to the interior of the sphere, is constant but it is mandatory not to forget that radiation is in truth a

vector, meaning that its projection on different planes may present diverse values.

With the former property we can replace the cap source by its enclosed circle AB.

Fig. 5. Cap and sphere where the radiative exchange takes place

1/2\*(1-cos2β).

Fig. 6. The radiation vector in a sphere

Finding the radiation vector that originates in the cap source poses no particular problem (See Figure 6). For reasons of symmetry, its centre has to be at point Q, following the wellknown principle of the circumference by virtue of which, equal arcs subtend equal angles1.

The extreme of the vector lies at the point under study. If the direction of the vector is known, it only remains to calculate its modulus. Using trigonometric properties, as the normal EN is constant, this implies that the vertical component EZ equates the normal for the value of direction angle θ is a half of the angle subtended by the arc OP from the centre of the sphere, being P the point under study and O the horizontal projection of the centre.

The last step to determine the modulus of the vector is to project its vertical component h/(2\*R) onto the horizontal plan, multiplying by the tangent of θ.

The fact that the vertical component is constant has led to the construction of useful graphs in which horizontal radiation EZ at a given point is obtained as a function of the radius of the cap's base and the distance from the circle's centre to the point considered (See figures 7 and 8).

Fig. 7. Perpendicular component of the radiation vector under a disk

If the said quantities are known, the coordinate component EH, which is constant only for the same height over the horizontal in the sphere, in other words, by parallels, can be found employing the circumference's properties.

$$R^2 = a^2 + \left(\frac{b^2 + r^2 - a^2}{2b}\right)^2\tag{16}$$

$$h = R - \sqrt{R^2 - a^2} \tag{17}$$

And the vertical distance from the origin of the radiation vector to the point considered is, evidently,

$$d = b + h \tag{18}$$

<sup>1</sup> Both MacAllister and Sumpner failed to see this point and located the origin of the vector at the centre of the enclosed circle, though this error may be relevant for sizeable sources.

Fig. 8. The MacAllister graph for finding the vertical component of radiation, using only the magnitudes on figure 7 and without need for calculation

To obtain the vectorial field of radiation around a spherical cap source or its equivalent disk (the final aim of radiative transfer and simulation for such common geometry) is tantamount to finding a set of virtual spheres which contains both the source and each one of the points considered in the reticule that represents the field. From this set we extract the components and consequently the radiation vector at each point of the domain analysed.

The author has created original software that simulates the aforementioned fields for planes, spheres and cylinders at every possible inclination angle in spherical coordinates, θ or φ; some results are presented in figures 9 and 10.

Fig. 8. The MacAllister graph for finding the vertical component of radiation, using only the

To obtain the vectorial field of radiation around a spherical cap source or its equivalent disk (the final aim of radiative transfer and simulation for such common geometry) is tantamount to finding a set of virtual spheres which contains both the source and each one of the points considered in the reticule that represents the field. From this set we extract the components and consequently the radiation vector at each point of the domain

The author has created original software that simulates the aforementioned fields for planes, spheres and cylinders at every possible inclination angle in spherical coordinates, θ or φ;

magnitudes on figure 7 and without need for calculation

some results are presented in figures 9 and 10.

analysed.

Fig. 9. Distribution of radiation on a sphere of radius 20 m. and reflectance 0.3, under three circular sources of 9 m. diameter and 10000 lumen/m2 (lux) intensity, rotated 120º

Fig. 10. The same three sources as in fig. 9 but radiation values are found instead on a circular disk 7.5 metres under the centre of the sphere

On the understanding that radiation fields are additive due to their vectorial nature, the author has solved with ease the fundamental problem of radiative transfer. To be sure, not all the openings2 in buildings are circular but a significant amount of them can be approximated to one or several emitting disks with sufficient precision, taking into account

<sup>2</sup> The word "open" comes from ancient Greek οπή (opeh), meaning eye. The main aperture or vent in classic buildings is often termed "opaion" or "oculus", its latin equivalent.

$$
\mathfrak{gl}\_{1-2} = E\_{b1} \ast \pi \ast h\_1 \ast h\_2 \tag{19}
$$

$$h = \sqrt{a^2 + b^2} - b \tag{20}$$

$$F\_{12} = F\_{21} = \frac{a^2 + 2 \ast b^2 - 2 \ast b \ast \sqrt{a^2 + b^2}}{a^2} \tag{21}$$

the inevitable lack of accuracy in the construction industry. In any case, the author has adapted his software to triangular and rectangular apertures (see section 4) but in the latter

Once this matter is settled, retrieving the fundamental equation expounded at 8 to be partly solved in 10 and extending it to a second spherical cap as in figure 11, the radiant flux

Fig. 11. Two spherical caps inside a sphere employed to find the radiative transfer

In the special situation that h1 = h2 = h, which often coincides with parallel disks of equal radius *a*, the flux would be Eb1\*π\*h2 and the fraction of energy from disk 1 to disk 2 (or their

If only the perpendicular distance between the disks, called 2b, is known (see figure 12), the

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<sup>3</sup> As a matter of fact only J. H. Lambert was capable of finding a solution for perpendicular rectangles with a common edge without the help of integration, but he often complained in his book of the "insurmountable difficulties" that the

∅��� � ��� ����� � �� (19)

� � √�� � �� � � (20)

�� (21)

situation, solutions are not simple and require complex integration3.

**2.2 Radiative transfer between complex surfaces** 

For the area of the second cap is nothing but π\*2\*R\*h2 .

between the two caps of heights h1 and h2 is,

surrounding caps), equates h2/a2.

Thus, the fraction is obtained as,

height of the cap would be,

process entailed.

Fig. 12. Surfaces defined by a cylindrical volume used to find the radiative transfer

With the former expression, solving the radiative transfer inside cylinders is easy as there are only three surfaces involved and we can form a non-trivial system of two equations with two unknowns. Changing the circular base of the cylinder by a spherical cap (fig.13) will alter some values but not the general problem because all the possibilities of transfer between caps and disks have already been explored. For instance, if the cap is a hemisphere, the values of the factors to the disk and the cylinder need to be affected by a2/(a2 +h2)=0.5 (Eq. 4), and progressively until reaching one which is again the planar disk. 4

The resulting volume in figure 13 has been used by humankind for centuries; mainly in churches but also in libraries, concert halls, banks, markets or pavilions of any sort. With another cap on the base, the form is more recently used for silos, fluid reservoirs and containment vessels at power stations.

Having solved the primary transference problems, the following step is the important subject of interreflections.

Fig. 13. Volume composed of a cylinder and a spherical cap used to find the radiative transfer among those surfaces

<sup>4</sup> Note that values under 0.5 can also be found for this relationship in a sort of globular cap with an area bigger than the hemisphere.

$$E\_{tot} = E\_{dtr} + E\_{ref} \tag{2}$$

$$F\_d = \begin{pmatrix} F\_{11} \ast \rho\_1 & F\_{12} \ast \rho\_2 & F\_{13} \ast \rho\_3 \\ F\_{21} \ast \rho\_1 & F\_{22} \ast \rho\_2 & F\_{23} \ast \rho\_3 \\ F\_{31} \ast \rho\_1 & F\_{32} \ast \rho\_2 & 0 \end{pmatrix} \tag{23}$$

$$F\_r = \begin{pmatrix} 1 & -F\_{12} \ \* \rho\_2 & -F\_{13} \ \* \rho\_3 \\ -F\_{21} \ \* \rho\_1 & 1 & -F\_{23} \ \* \rho\_3 \\ -F\_{31} \ \* \rho\_1 & -F\_{32} \ \* \rho\_2 & 1 \end{pmatrix} \tag{24}$$

$$F\_r \* E\_{ref} = F\_d \* E\_{dlr} \tag{25}$$

$$F\_{rd} = F\_r^{-1} \* F\_d \tag{26}$$

$$E\_{ref} = F\_{rd} \* E\_{dlr} \tag{27}$$

$$E\_{ref} = E \* \frac{W}{A} \* (\rho + \rho^2 + \cdots \rho^n) \tag{28}$$

$$\lim\_{n \to \infty} \left( \frac{\rho^{n+1} - 1}{\rho - 1} - 1 \right) = \frac{\rho}{1 - \rho} \tag{29}$$

$$E\_{ref} \; = E \; \* \frac{W}{A} \* \left(\frac{\rho}{1-\rho}\right) \tag{30}$$

$$f(\mathbf{x}, \mathbf{y}) = \frac{E}{2} \left| 1 - \frac{\mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2 - a^2}{\sqrt{(\mathbf{x}^2 + \mathbf{y}^2)^2 + 2\mathbf{z}(\mathbf{x}^2 + \mathbf{y}^2) \cdot (\mathbf{z}^2 - a^2) + (\mathbf{z}^2 + a^2)^2}}\right| \tag{31}$$

$$f(\mathbf{x}, \mathbf{y}) = \frac{\mathbf{z}}{2} \left[ \frac{\mathbf{y}}{\sqrt{\mathbf{z}\_1^2 + \mathbf{y}}} \left( \arctan \frac{\mathbf{x} + \frac{\mathbf{a}}{\mathbf{z}}}{\sqrt{\mathbf{z}\_2^2 + \mathbf{y}}} - \arctan \frac{\mathbf{x} - \frac{\mathbf{a}}{\mathbf{z}}}{\sqrt{\mathbf{z}\_2^2 + \mathbf{y}}} \right) - \frac{\mathbf{y}}{\sqrt{\mathbf{z}\_1^2 + \mathbf{y}}} \left( \arctan \frac{\mathbf{x} + \frac{\mathbf{a}}{\mathbf{z}}}{\sqrt{\mathbf{z}\_1^2 + \mathbf{y}}} - \arctan \frac{\mathbf{x} - \frac{\mathbf{a}}{\mathbf{z}}}{\sqrt{\mathbf{z}\_1^2 + \mathbf{y}}} \right) \right] \tag{32}$$

$$\propto = \arctan(b/a) \tag{34}$$

$$
\beta\_- = \arctan(b/c) \tag{35}
$$

$$\beta\_1 = \arctan\left(\frac{b}{\sqrt{a^2 + c^2}}\right) \tag{36}$$

$$F\_{12} = \frac{2}{a\*\pi} \left[ a\*\phi\_0(a) + c\*\phi\_0(\beta) - \sqrt{a^2 + c^2}\*\phi\_0(\beta\_1) \right] \tag{37}$$

This situation is convenient for the treatment of several building features that perform as radiation filters such as canopies, awnings, louvers and even courtyards or reflective ponds.

For instance, in a system of louvers it is possible to isolate the volume seen in figure 18 and treat it as a single space with three virtual faces. Once the radiation that reaches the surface of the glass is obtained, the procedure is the same as for a room without louvers but the emissive power used for the window is the previous value and not the one applicable for the unobstructed orientation.

Fig. 18. Subdivision of louvers in a protected window

In this way, most of the problems derived from the geometry of the design are solved and radiation filters can be properly evaluated. Previously they were only considered obstructions without any potential to add for the energy balance.
