**3. Integration process for circular emitters**

#### **3.1. Direct integration for a differential element to a circular disk on a plane parallel to that of the element**

Let us consider the proposed figure. In order to determine the radiant interchange between an emitting circle, which lies in the plane ZX, and a point P situated in another parallel plane XY, the following coordinate system is proposed (Figure 2).

Terms depicted in figure 2 are:

In this chapter, an exact analytical solution derived from complex double integration is presented. The expression obtained significantly soothes the calculation of the configuration factor between a circular emitter and a point that lies in a plane located at any position to the former, not only in an axis perpendicular to its center. Those results were checked against more conventional formulas. Based on such calculus procedures, an entirely new factor for the semicircle to a perpendicular plane that contains the straight edge has been deducted. Likewise, the solution has been converted into an original algorithm and programmed in simulation software developed by the authors so that interactive maps of the radiative field can be visualized in a consistent and accurate way. Thus, computer simulation techniques,

The reciprocity principle enunciated by Lambert in his paramount book Photometria, written

q q 1 2

*dA dA E E cos cos <sup>r</sup>* (1)

p

· · · ·

1 2

*A A*

12 1 2 1 2 2

engineering and image applications will be greatly enhanced and benefitted.

in Latin (Lambert, 1760), yields the following well-known integral equation:

( )

Æ= - - *b b* ò ò

**Figure 1.** The reciprocity principle and quantities' significance for surfaces A1 and A2

Relevant terms in equation (1) are depicted in figure 1.

**2. Outline of the problem & objectives**

158 Solar Radiation Applications

d: vertical distance between the center of the emitting circle and the plane XY.

b: horizontal distance between differential element dA1 and the plane ZX that contains the said circle.

r: Emitting surface radius.

S: Distance between differential elements (in the canonical equation (1) of the configuration factor, it is denoted as r, but in order to differentiate it from radius of the disk (emitting surface, r), we shift to this denomination).

According to figure 2, the differential element dA2 is expressed in terms of r and θ. Thus, to receive a proper integral, the rest of elements inside the integration sign of the canonical equation (1) should be expressed using the said variables. Basic construction for this expression

**Figure 2.** Calculation parameters for the parallel plane.

can be found in numerous manuals of radiative transfer [4]. Furthermore, mathematical support for the integration process is described in references [5]

*dA*1 represents the point source

$$dA\_2 = r dr d\theta \tag{2}$$

$$\cos \theta\_l = \frac{\left(d + r \cos \theta\right)}{S} \tag{3}$$

$$\cos \theta\_2 = \frac{b}{S} \tag{4}$$

$$S = r^2 + d^2 + b^2 - 2dr \cdot \cos\theta \tag{5}$$

Substituting terms from (2) in accordance to figure 2, in the canonical equation of radiative transfer (1), the main integral that we need to solve is,

$$\int\_{A\_1 A\_2} \int\_{\mathbf{r}} \cos \theta\_1 \cos \theta\_2 \frac{dA\_1 dA\_2}{\pi r^2} = \iint\_{\mathbf{0}} \frac{br \cdot (b - r \cos \theta)}{\left(r^2 + d^2 + b^2 - 2dr \cos \theta\right)^2} d\theta dr \tag{6}$$

Operating in the numerator we can decompose this integral in two parts:

$$\int\_{0}^{2\pi} \int\_{0}^{2r} \frac{b^2 r}{\left(r^2 + d^2 + b^2 - 2dr \cdot \cos\theta\right)^2} d\theta dr - \int\_{0}^{2\pi} \int\_{0}^{r} \frac{br^2 \cdot \cos\theta}{\left(r^2 + d^2 + b^2 - 2dr \cdot \cos\theta\right)^2} d\theta dr \tag{7}$$

The limits for inner and outer integral are, respectively: from 0 to 2π, and from 0 to a, that is, the whole extension of the radius of the emitting circle.

In order to solve this double integral, we first integrate with respect to θ. Proceeding with the first integral implies taking out all the constants that are independent of θ, and this yields:

$$b^2r \cdot dr \int\_0^{2\pi} \frac{d\theta}{\left(r^2 + d^2 + b^2 - 2dr \cdot \cos\theta\right)^2} \tag{8}$$

Such expression corresponds to a type, which yields the solution:

$$\int \frac{d\mathbf{x}}{\left(B + C\cdot\cos(A\mathbf{x})\right)^{2}} = \frac{C\cdot\sin(A\mathbf{x})}{A\left(B^{2} - C^{2}\right)\left(B + C\cos(A\mathbf{x})\right)} - \frac{B}{\left(C^{2} - B^{2}\right)}\int \frac{d\mathbf{x}}{B + C\cdot\cos(A\mathbf{x})}\tag{9}$$

The change of variables is defined thus:

can be found in numerous manuals of radiative transfer [4]. Furthermore, mathematical

q

( )

Substituting terms from (2) in accordance to figure 2, in the canonical equation of radiative

cos

q

q

(2)

*<sup>S</sup>* (3)

*<sup>S</sup>* (4)

(5)

*dA rdrd* <sup>2</sup> =

<sup>+</sup> <sup>=</sup> *d r*

<sup>2</sup> cosq<sup>=</sup> *<sup>b</sup>*

2 22 *S r d b dr* =+ +- 2 ·cos

1

q

cos

transfer (1), the main integral that we need to solve is,

support for the integration process is described in references [5]

*dA*1 represents the point source

160 Solar Radiation Applications

**Figure 2.** Calculation parameters for the parallel plane.

$$d\,d\theta = d\mathbf{x} \quad B = r^2 + d^2 + b^2 \quad C = -2dr \quad A = \mathbf{l} \tag{10}$$

Before operating, and in order to simplify the otherwise tedious calculations, this expression can be put in simpler form by applying logical deductions. Focusing our attention in the first term of (9):

$$\frac{C \cdot \sin(A\mathbf{x})}{A\left(B^2 - C^2\right)\left(B + C\cos\left(A\mathbf{x}\right)\right)}\tag{11}$$

It can be observed that sin(Ax) is in the numerator; if A=1 the former means that we have sin(X); but we need to bear in mind that the limits for our defined integral are 2π and 0, thus, sin(2π), sin(0), equal nil and so does the integral,

$$\left. \frac{C \sin(A\mathbf{x})}{A\left(B^2 - C^2\right)\left(B + C \cos\left(A\mathbf{x}\right)\right)} \right|\_{0}^{2\pi} = 0 - 0 = 0\tag{12}$$

Subsequently, we focus our attention in the second term of equation (9);

$$-\frac{B}{\left(C^2 - B^2\right)} = -\frac{r^2 + d^2 + b^2}{4d^2r^2 - \left(r^2 + d^2 + b^2\right)^2} = +\frac{r^2 + d^2 + b^2}{\left(r^2 + d^2 + b^2\right)^2 - 4d^2r^2} \tag{13}$$

So far, we have solved all terms outside the integration sign of equation (9). What remains inside the integral admits this change:

$$\int \frac{d\mathbf{x}}{B + C\cos(A\mathbf{x})} = \frac{2}{A\sqrt{B^2 + C^2}} \arctan\frac{(B - C)\tan(A\mathbf{x}/2)}{\sqrt{B^2 - C^2}}\tag{14}$$

Therefore, substituting all terms we receive:

$$\frac{r^2 + d^2 + b^2}{\left(r^2 + d^2 + b^2\right) - 4d^2r^2} \left[ \frac{2}{\sqrt{\left(r^2 + d^2 + b^2\right)^2 - 4d^2b^2}} \arctan\frac{\left(\left(r^2 + d^2 + b^2\right) + 2dr\right)\tan(\theta/2)}{\sqrt{\left(r^2 + d^2 + b^2\right)^2 - 4d^2b^2}} \right] \tag{15}$$

Once more, some logics were employed in order to compact the calculations; concentrating on the third term of (15), we find an the *arctangent* and a *tangent* expression. Bearing in mind that the limits of integration are (2π, 0), the result of arctangent is obviously π, and that produces:

$$\frac{r^2 + d^2 + b^2}{\left(r^2 + d^2 + b^2\right) - 4d^2r^2} \cdot \frac{2\pi}{\sqrt{\left(r^2 + d^2 + b^2\right)^2 - 4d^2b^2}}\tag{16}$$

The value of π is taken out of the integration mark and eliminated by means of the canonical equation of the configuration factor. That yields:

$$2b^2 \int\_0^a \frac{r \cdot (r^2 + d^2 + b^2)}{\left(\left(r^2 + d^2 + b^2\right)^2 - 4d^2r^2\right)^{\frac{3}{2}}} dr$$

Again making some arrangements to these elements to produce an expression that enables easy integration, let us multiply the numerator and denominator by 4 and add and subtract a new term, -2rd2 , always bearing in mind to reproduce the original expression in (17); that gives the following equation.

$$2b^2 \int\_0^a \frac{4\left(r\left(r^2 + d^2 + b^2\right) - 2rd^2 + 2rd^2\right)}{4\left(\left(r^2 + d^2 + b^2\right)^2 - 4d^2r^2\right)^{\frac{3d}{2}}} dr\tag{18}$$

Decomposing and operating again:

New Computational Techniques for Solar Radiation in Architecture http://dx.doi.org/10.5772/60017 163

$$2b^2\int\_0^a \frac{4\left(r^2+d^2+b^2\right)-8rd^2+8rd^2}{4\left(\left(r^2+d^2+b^2\right)-4d^2r^2\right)^{\frac{\lambda^2}{2}}}dr =$$

$$=\frac{4\left(r^2+d^2+b^2\right)-8rd^2}{4\left(\left(r^2+d^2+b^2\right)-4d^2r^2\right)^{\frac{\lambda^2}{2}}}+\frac{8rd^2}{4\left(\left(r^2+d^2+b^2\right)-4d^2r^2\right)^{\frac{\lambda^2}{2}}}dr =$$

$$2b^2\int\_0^a \frac{4r\left(r^2+d^2+b^2\right)-8rd^2}{4\left(\left(r^2+d^2+b^2\right)-4d^2r^2\right)^{\frac{\lambda^2}{2}}}dr + \int\_0^a \frac{2rd^2}{\left(\left(r^2+d^2+b^2\right)-4d^2r^2\right)^{\frac{\lambda^2}{2}}}dr\tag{19}$$

The integral of a sum is given, which can be treated as the sum of integrals. Dealing with the first term of (19) the following expression is received:

$$2\delta^2 \int\_0^a \frac{4r\left(r^2 + d^2 + b^2\right) - 8rd^2}{4\left(\left(r^2 + d^2 + b^2\right) - 4d^2r^2\right)^{\frac{3}{2}}} dr = 2\delta^2 \int\_0^a \frac{4r\left(r^2 + b^2 - d^2\right)}{4\left(\left(r^2 + d^2 + b^2\right) - 4d^2r^2\right)^{\frac{3}{2}}} dr\tag{20}$$

That offers the solution:

Subsequently, we focus our attention in the second term of equation (9);

inside the integral admits this change:

( )

162 Solar Radiation Applications

new term, -2rd2

the following equation.

Decomposing and operating again:

Therefore, substituting all terms we receive:

( )

equation of the configuration factor. That yields:

2 22

+ +

*rdb*

2

2

ò

ò

( ) ( ) ( )

+ + + + - =- = + - - ++ ++ - *B rdb rdb*

2 2 2 2 22 2 2 2 2 2 2 22 4 4

So far, we have solved all terms outside the integration sign of equation (9). What remains

<sup>2</sup> ·tan( / 2) · ·cos( )

 + + ê ú ++ - ++ - ++ - ë û

Once more, some logics were employed in order to compact the calculations; concentrating on the third term of (15), we find an the *arctangent* and a *tangent* expression. Bearing in mind that the limits of integration are (2π, 0), the result of arctangent is obviously π, and that produces:

The value of π is taken out of the integration mark and eliminated by means of the canonical

2 22

+ +

4

, always bearing in mind to reproduce the original expression in (17); that gives

Again making some arrangements to these elements to produce an expression that enables easy integration, let us multiply the numerator and denominator by 4 and add and subtract a

( ( ) )

++ - +

2 22 2 2

(( ) )

<sup>3</sup> <sup>2</sup> <sup>2</sup> <sup>0</sup> 2 2 2 22 4 · 2 2 2 · 4 4

*<sup>a</sup> r r d b rd rd b dr r d b dr*

++ -

2 2 2 22 2 2 22 2 2 <sup>2</sup> · <sup>4</sup> ( )4

(( ) )

++ -

<sup>3</sup> <sup>2</sup> <sup>2</sup> <sup>0</sup> 2 2 2 22 ·( ) 2 ·

*<sup>a</sup> rr d b <sup>b</sup> dr*

++ - ++ -

*arctan*

2 ·tan( / 2) <sup>2</sup> · <sup>4</sup> ( )4 ( )4 é ù ++ +

*arctan r d b dr r d b db r d b db* (15)

p

*r d b dr r d b db* (16)

*r d b dr* (17)


(( ) ) 2 22 2 22 2 2 2 22 2 2 22 2 2 2 2 22 2 2

*r d b dr rdb*

2 22 2 22

*C B dr r d b r d b dr* (13)

( ) 2 2 2 2

*B C Ax AB C B C* (14)

q

(18)

$$-b^2 \left[ \frac{1}{\left( \left( r^2 + d^2 + b^2 \right)^2 - 4d^2 r^2 \right) \right]\_0^{\frac{1}{2}}} = \frac{b^2}{d^2 + b^2} - \frac{b^2}{\left( \left( a^2 + d^2 + b^2 \right)^2 - 4d^2 a^2 \right)^{\frac{1}{2}}} \tag{21}$$

Now it is time to proceed to the remaining term of (19)

$$\int\_{0}^{a} \frac{2rd^2}{\left(\left(r^2 + d^2 + b^2\right) - 4d^2r^2\right)^{\frac{3}{2}}} dr\tag{22}$$

Expanding the denominator and rearranging its terms with reference to the variable r:

$$2b^2d^2\int\_0^d \frac{2rdr}{\left(r^4 + 2r^2(b^2 - d^2) + (b^2 + d^2)^2\right)^{\frac{3}{2}}}\tag{23}$$

Introducing the change of variable r2 =t, that yields:

$$2b^2d^2\int\_{0}^{l} \frac{dt}{\left(t^2 + 2t\cdot(b^2 - d^2) + (b^2 + d^2)^2\right)^{\frac{3'}{2}}}\tag{24}$$

This integral responds to the following model with the solution:

$$\int \frac{d\mathbf{x}}{X\sqrt{X}} = \frac{2(2a\mathbf{x} + b)}{\Delta\sqrt{X}} \text{ for } X = a\mathbf{x}^2 + b\mathbf{x} + c \quad \text{and} \quad \Lambda = 4ac - b^2 \tag{25}$$

Making the substitution:

$$a = \text{l}; \quad b = 2\left(b^2 - d^2\right); \quad c = (b^2 + d^2)^2; \text{ } \Lambda = 4(b^2 + d^2)^2 - \left(b^2 - d^2\right)^2 = 16b^2d^2 \tag{26}$$

The ensuing outcome is given:

$$2b^2d^2\left[\frac{2(2t+2b^2-2d^2)}{16b^2d^2\sqrt{\left(t^2+2t\cdot(b^2-d^2)+(b^2+d^2)^2\right)}}\right]\_0^{d^2} = \frac{a^2+b^2-d^2}{2\sqrt{\left(a^2+d^2+b^2\right)^2-4d^2a^2}} - \frac{b^2-d^2}{2\left(d^2+b^2\right)}\tag{27}$$

Finally, adding both terms, from (21) and (27), in order to obtain the final result:

$$\begin{aligned} \frac{b^2}{a^2+b^2} - \frac{b^2}{\sqrt{\left(\left(a^2+d^2+b^2\right)^2 - 4d^2a^2\right)}} + \frac{a^2+b^2-d^2}{2\sqrt{\left(\left(a^2+d^2+b^2\right)^2 - 4d^2a^2\right)}} - \frac{b^2-d^2}{2\left(d^2+b^2\right)} = \\ \frac{-2b^2+a^2+b^2-d^2}{2\sqrt{\left(\left(a^2+d^2+b^2\right)^2 - 4d^2a^2\right)}} + \frac{2b^2-b^2+d^2}{2\left(d^2+b^2\right)} = \\ \frac{a^2-b^2-d^2}{2\sqrt{\left(\left(a^2+d^2+b^2\right)^2 - 4d^2a^2\right)}} + \frac{b^2+d^2}{2\left(d^2+b^2\right)} = \\ \frac{1}{2} - \frac{b^2+d^2-a^2}{\sqrt{\left(\left(a^2+d^2+b^2\right)^2 - 4d^2a^2\right)}} \end{aligned} \tag{28}$$

#### **3.2. Direct integration for a differential element to a circular disk on a plane perpendicular to that of element**

In the perpendicular plane, that is, the ZX plane, according to the defined coordinate reference system, the main equation to be solved is:

New Computational Techniques for Solar Radiation in Architecture http://dx.doi.org/10.5772/60017 165

$$\int\_{0}^{a2x} \int\_{0}^{b} \frac{br\left(d - r\cos\theta\right)}{\left(r^2 + d^2 + b^2 - 2dr\cos\theta\right)^2} r d\theta dr\tag{29}$$

This can be decomposed into two terms

( )

+ -++ 2 ·( ) ( )

2(2 ) 2 2 4 <sup>+</sup> <sup>=</sup> = + + D= -

( ) (( ) )

Finally, adding both terms, from (21) and (27), in order to obtain the final result:

(( ) ) (( ) )

++ - ++ -

+ = +

+ = +

2 2 22 2 2 2 2 2 2 2 2 2 2 2 2 22 2 2 2 22

*b b abd bd d b d b a d b da a d b da*

+ - - - <sup>+</sup> - = + +

**3.2. Direct integration for a differential element to a circular disk on a plane perpendicular**

In the perpendicular plane, that is, the ZX plane, according to the defined coordinate reference

+ - + - - ê ú <sup>=</sup> - <sup>+</sup> + -++ ++ - ë û

2 2 2 2 2 2 2 2 22 <sup>2</sup> 2 2 2 22 0

2( ) 16 2 ·( ) ( ) 2 4

*d b bd t t b d b d a d b da*

2

*a tb d abd bd b d*

<sup>D</sup> <sup>ò</sup> *dx ax b for X ax bx c and ac b XX X* (25)

( ) ( )<sup>2</sup> 2 2 2 22 2 22 2 2 2 2 *a b b d c b d b d b d bd* = = - = + D= + - - = 1; 2 ; ( ) 4(; ) <sup>16</sup> (26)

2 2 22 2 2 2

2( ) 42 4

(27)

(28)

*t tb d b d* (24)

<sup>3</sup> 2 2 2 2 22 <sup>2</sup> <sup>0</sup>

2 2 2

*<sup>a</sup> dt b d*

ò

This integral responds to the following model with the solution:

2

Making the substitution:

164 Solar Radiation Applications

The ensuing outcome is given:

2(2 2 2 ) <sup>2</sup>

(( ) )

++ -

(( ) )

++ -

<sup>2</sup> 2 4

+ - -

(( ) )

*a d b da*

++ -

system, the main equation to be solved is:

2 22 <sup>2</sup> 2 2 2 22

*bda*

2( ) 2 4

2 2 2( ) 2 4

é ù

222 2 22 2 2 2 <sup>2</sup> 2 2 2 22

*babd bbd d b a d b da*


22 2 2 2 2 2 <sup>2</sup> 2 2 2 22


*abd bd d b a d b da*

2 2

1

**to that of element**

$$\int\_{0}^{2\pi} \frac{brd}{\left(r^2 + d^2 + b^2 - 2dr\cos\theta\right)^2} d\theta dr - \iint\_{0}^{2\pi} \frac{br^2 \cos\theta}{\left(r^2 + d^2 + b^2 - 2dr\cos\theta\right)^2} d\theta dr\tag{30}$$

The first part of this expression has already been solved, but with b2 instead of b d as a constant. The first term was solved in two parts, which were expressed in equations (21) and (27). From equation (21):

$$-b \cdot d \left[ \frac{1}{\left( \left( r^2 + d^2 + b^2 \right)^2 - 4d^2 r^2 \right) \Big|\_{0}^{\frac{\delta}{2}}} \right]\_0^d = \frac{bd}{d^2 + b^2} - \frac{bd}{\sqrt{\left( a^2 + d^2 + b^2 \right)^2 - 4d^2 a^2}} \tag{31}$$

**Figure 3.** Calculation parameters for the perpendicular plane

Now, equation (27) is rearranged as follows:

$$
\begin{aligned} &2bd^3 \left[ \frac{2(2t+2b^2-2d^2)}{16b^2d^2\sqrt{\left(t^2+2d^2(b^2-d^2)+\left(b^2+d^2\right)^2\right)}} \right]\_0^{b^2} = \\ &b = 2bd^3 \left[ \frac{(a^2+b^2-d^2)}{4b^2d^2\sqrt{\left(\left(a^2+d^2+b^2\right)^2\right)^2-4d^2a^2}} - \frac{(b^2-d^2)}{4b^2d^2(d^2+b^2)} \right] \\\\ &\frac{d\cdot(a^2+b^2-d^2)}{2b\sqrt{\left(\left(a^2+d^2+b^2\right)^2\right)^2-4d^2a^2}} - \frac{d(b^2-d^2)}{2b(d^2+b^2)} \end{aligned} \\ \tag{32}
$$

Next, assemble equations (31) and (32), group terms by common denominator and operate:

(( ) ) (( ) ) 22 2 2 2 2 2 2 2 2 2 2 2 2 22 2 2 2 22 ·( ) ( ) 2( ) 42 4 + - - -+ - + + ++ - ++ *bd bd d a b d db d d b bd b a d b da b a d b da* ( ) (( ) ) ( ) ( ) (( ) ) ( ) 22 2 2 2 2 2 22 3 2 3 2 2 2 2 <sup>2</sup> <sup>2</sup> 2 2 2 22 2 2 2 22 · 2 2· 2 2 24 24 +- - - - - - + + = + = + + ++ - ++ *d a b d bd bd d b d ad bd d bd d bd b bd b b a d b da b a d b da* ( ) (( ) ) ( ) ( ) ( ) (( ) ) 22 2 2 2 2 22 2 2 2 2 2 2 2 22 <sup>2</sup> 2 2 2 22 <sup>2</sup> 24 24 --+ + + - +- = + = + <sup>=</sup> <sup>+</sup> ++ - ++ *d a b d db d db d a d bd b <sup>b</sup> b a d b da b a d b da*

$$\frac{1}{2} \frac{d}{b} \left| 1 - \frac{\left(b^2 + d^2 - a^2\right)}{\sqrt{\left(\left(a^2 + d^2 + b^2\right)^2 - 4d^2 a^2\right)}} \right| \tag{33}$$

Also, the second term from equations (30):

#### New Computational Techniques for Solar Radiation in Architecture http://dx.doi.org/10.5772/60017 167

$$\int\_{0}^{a2\pi} \frac{br^2 \cos \theta}{\left(r^2 + d^2 + b^2 - 2dr \cos \theta\right)^2} d\theta dr \tag{34}$$

Again, integrating with respect to θ we receive:

Now, equation (27) is rearranged as follows:

166 Solar Radiation Applications

3

3

( )

(( ) )

· 2 2·

( )

(( ) )

1 · 1

Also, the second term from equations (30):

( )

22 2 2 <sup>2</sup> 2 2 2 2 2 22

4( ) 4 4

*bd d b bd a d b da*

++ - ë û

é ù

22 2 2 2

*bd b b a d b da* (32)

2( ) 42 4

+ = + =

( ) (( ) ) ( )

( )

(( ) )

(33)

2 2 2 2 2 2 22

é ù + - ê ú <sup>=</sup> + -++ ë û

16 2 ·( ) ( )

(( ) )

*d a b d db d*

Next, assemble equations (31) and (32), group terms by common denominator and operate:

(( ) ) (( ) )

( )

( ) ( )


<sup>2</sup> <sup>4</sup>

*<sup>b</sup> a d b da*

*d bda*

22 2 2 2 2 22

2 2 2 2 2 2 2 22 <sup>2</sup> 2 2 2 22 <sup>2</sup> 24 24

( )

é ù ê ú + - -

++ - ë û

2 22

(( ) )

<sup>2</sup> 2 2 2 22

= + = + <sup>=</sup> <sup>+</sup> ++ - ++ *d a b d db d db d a d*

*bd b <sup>b</sup> b a d b da b a d b da*

2 2 24 24

2 2 2 2 2 2 2 2 2 22 2 2 2 22

*bd bd d a b d db d d b bd b a d b da b a d b da*

+ - - -+ - + + ++ - ++ -

22 2 2 2 2 2 22 3 2 3 2 2 2 2 <sup>2</sup> <sup>2</sup> 2 2 2 22 2 2 2 22

+ + ++ - ++ *d a b d bd bd d b d ad bd d bd d bd b bd b b a d b da b a d b da*

+- - - - - - +

( ) () <sup>2</sup>

*abd b d bd*

ê ú + - - <sup>=</sup> - <sup>+</sup>

22 2 2 2 2 2 <sup>2</sup> 2 2 2 22 ·( ) ( )

+ - - - <sup>+</sup>

*bd t t b d b d*

2(2 2 2 ) <sup>2</sup>

*tb d bd*

(( ) )

++ -

2( ) 2 4

2 2

2

*a*

0

22 2 2 2

·( ) ( )

$$\int\_{0}^{a} \frac{4bdr^2}{\left(\left(r^2 + d^2 + b^2\right)^2 - 4d^2r^2\right)^{\frac{3}{2}}} dr\tag{35}$$

Employing the change r2 =t in equation (35),

$$2bd \int\_0^d \frac{t}{\left(\left(t^2 + 2t\left(b^2 - d^2\right)\right) + \left(b^2 + d^2\right)^2\right)^{\frac{3}{2}}} \cdot dt \tag{36}$$

An integral that admits the immediate solution:

$$\begin{aligned} \int \frac{\mathbf{x} \, d\mathbf{x}}{X \sqrt{X}} &= -\frac{2\left(2b\mathbf{x} + 2c\right)}{\Delta \sqrt{X}} \text{ for} \\ X &= a\mathbf{x}^2 + b\mathbf{x} + c \quad \text{and} \ \Delta = 4ac - b^2 \end{aligned} \tag{37}$$

Making the following substitutions:

$$\begin{aligned} a &= 1; \ b = 2\left(b^2 - d^2\right); \ c = \left(b^2 + d^2\right)^2\\ \Delta &= 4\left(b^2 + d^2\right)^2 - \left(b^2 - d^2\right)^2 = 16b^2d^2 \end{aligned} \tag{38}$$

Equation (36) becomes:

$$2bd \left[ \frac{(b^2 - d^2)t + (b^2 + d^2)^2}{4b^2 d^2 \sqrt{t^2 + 2t(b^2 - d^2) + \left(b^2 + d^2\right)^2}} \right]\_0^{d^2} = \frac{a^2 \left(b^2 - d^2\right) + (b^2 + d^2)^2}{2bd \sqrt{(a^2 + b^2 + d^2)^2 - 4d^2a^2}} - \frac{b^2 + d^2}{2bd} \tag{39}$$

Finally, in order to produce the final solution for the perpendicular plane, it is required to assemble equations (33) and (39). Grouping and rearranging by common denominators it yields:

$$\begin{aligned} \frac{d}{2b} - \frac{d\left(b^2 + d^2 - a^2\right)}{2b\sqrt{\left(\left(a^2 + d^2 + b^2\right)^2 - 4d^2a^2\right)}} + \frac{a^2\left(b^2 - d^2\right) + \left(b^2 + d^2\right)^2}{2bd\sqrt{\left(a^2 + b^2 + d^2\right)^2 - 4d^2a^2}} - \frac{b^2 + d^2}{2bd} \\\\ \frac{a^2b^2 - a^2d^2 + b^4 + d^4 - 2b^2d^2 - b^2d^2 - d^4 + a^2d^2}{2bd\sqrt{\left(a^2 + b^2 + d^2\right)^2 - 4d^2a^2}} + \frac{d^2 - b^2 - d^2}{2bd} \\\\ \frac{b^2(a^2 + b^2 + d^2)}{2bd\sqrt{\left(a^2 + b^2 + d^2\right)^2 - 4d^2a^2}} - \frac{b^2}{2bd} \\\\ \frac{b}{2d} \left[\frac{a^2 + b^2 + d^2}{\sqrt{\left(a^2 + b^2 + d^2\right)^2 - 4d^2a^2}} - 1\right] \end{aligned}$$

Again, this result can be checked against usual formulas that appear in numerous configura‐ tion factor catalogues, although those do not completely solve the problem. Only they work when the element is in a plane that passes through the center of the circle. A more general solution of a vector nature had been presented by the authors in other texts [6],[7]. In this chapter a sound relationship between the two fundamentals expressions has been found.

#### **3.3. Resolution of the integral for the third coordinate plane**

Being radiation a vector, the resolution for a third coordinate plane that obviously cuts the emitting circle in two halves is required; the outline of the integral in this case yields:

$$\int\_{0}^{a} \iint \frac{br^2 \sin \theta}{\left(r^2 + d^2 + b^2 - 2dr \cos \theta\right)^2} \cdot d\theta \cdot dr \tag{41}$$

In this particular case, the limits of the integral cannot be extended to 2π, as the value would be nil. If (41) is integrated with respect to θ, in the numerator the derivatives of cosθ, -sinθ could be found. Therefore by making this change:

$$\mathbf{t} \cdot \mathbf{t} = \cos \theta \quad \text{d}\mathbf{t} = -\sin \theta \, d\theta \tag{42}$$

Integral (41) can therefore be expressed as:

$$\int\_{0}^{a} \frac{-br^2}{\left(r^2 + d^2 + b^2 - 2drt\right)^2} dt \, dr \, \tag{43}$$

**Figure 4.** Calculation parameters for the semicircle

( )

168 Solar Radiation Applications

( ) 2 22 2 2 2 2 22 2 2

+ - - ++ +

( )

+

*<sup>d</sup> a b d da* (40)

<sup>2</sup> 2 2 22 2 2 2 2 2 22

22 2 2 4 4 2 2 2 2 4 2 2 2 2 2

*ab ad b d bd bd d ad d b d*

2 ( )4 2 - ++ - - - + --

*bd a b d d a bd*

22 2 2 2 2 2 22 2 2 ( ) 2 ( )4 2

*ba b d b bd a b d d a bd*

+ + -

2 2 2 ( )4 2 4

2 2 22 2 2 2

++ -

22 2 2 2 22 2 2 <sup>1</sup>

é ù + + ê ú - ++ - ë û

Again, this result can be checked against usual formulas that appear in numerous configura‐ tion factor catalogues, although those do not completely solve the problem. Only they work when the element is in a plane that passes through the center of the circle. A more general solution of a vector nature had been presented by the authors in other texts [6],[7]. In this chapter a sound relationship between the two fundamentals expressions has been found.

Being radiation a vector, the resolution for a third coordinate plane that obviously cuts the

· ·

q

 qq

· ·

q

*r d b drcos* (41)

(42)

*r d b drt* (43)

emitting circle in two halves is required; the outline of the integral in this case yields:

( )

*t cos dt sin d* = = q

( )

*<sup>a</sup> br dt dr*

<sup>2</sup> 2 22 0 1


1 2

òò

+ +-

òò

2

In this particular case, the limits of the integral cannot be extended to 2π, as the value would be nil. If (41) is integrated with respect to θ, in the numerator the derivatives of cosθ, -sinθ

q

*<sup>a</sup> br sin d dr*

2 <sup>2</sup> 2 22 0 0

2 ( )4

*b abd*

**3.3. Resolution of the integral for the third coordinate plane**

p

could be found. Therefore by making this change:

Integral (41) can therefore be expressed as:

++ -


(( ) )

Taking out all the constants, and integrating with respect to r, the primitive is just the quotient of the numerator:

$$\frac{-br^2}{2dr} \left[ \frac{1}{r^2 + d^2 + b^2 - 2drr} \right]\_0^{-1} = \frac{br}{2d} \left( \frac{1}{r^2 + d^2 + b^2 - 2dr} - \frac{1}{r^2 + d^2 + b^2 + 2dr} \right) \tag{44}$$

Integrating (44) with respect to r; the last integral to solve is:

$$\int\_{0}^{a} \frac{r}{r^2 + d^2 + b^2 - 2dr} r dr + \int\_{0}^{a} \frac{r}{r^2 + d^2 + b^2 + 2dr} r dr \tag{45}$$

That responds to the form:

$$\int \frac{\mathbf{x} \cdot d\mathbf{x}}{X} = \frac{1}{2a} \cdot \ln X - \frac{b}{2a} \int \frac{d\mathbf{x}}{X} \quad X = a\mathbf{x}^2 + b\mathbf{x} + c \quad \Delta = 4ac - b^2$$

$$\int \frac{d\mathbf{x}}{X} = \frac{2}{\sqrt{\Delta}} \cdot \arctan \frac{2a\mathbf{x} + b}{\sqrt{\Delta}} \tag{46}$$

Substituting in (46) yields for both terms:

$$\begin{aligned} a &= 1; \; b = \pm 2d; \; c = \left(d^2 + b^2\right) \\\\ X &= r^2 \pm 2dr + \left(d^2 + b^2\right) \quad \Delta = 4b^2 \end{aligned} \tag{47}$$

Now, substituting in the first term of (45):

$$\frac{1}{2} \left[ \ln \left( r^2 + d^2 + b^2 - 2dr \right) \right]\_0^a + \frac{d}{b} \left[ \arctan \left( \frac{r - d}{b} \right) \right]\_0^a \tag{48}$$

And bringing equation (48) to the limits of the integral [a,0]:

$$\frac{1}{2} \left( \ln \left( a^2 + d^2 + b^2 - 2da \right) \right) - \ln \left( d^2 + b^2 \right) + \frac{d}{b} \left( \arctan \frac{a - d}{b} + \arctan \frac{d}{b} \right) \tag{49}$$

In a similar way, for the second term of equation (45):

$$-\frac{1}{2}\left(\ln\left(a^2+d^2+b^2+2da\right)\right)+\ln\left(d^2+b^2\right)+\frac{d}{b}\left(\arctan\frac{a+d}{b}-\arctan\frac{d}{b}\right)\tag{50}$$

Finally, multiplying by *<sup>b</sup>* 2*πd* and grouping, the final result can be expressed as:

$$\frac{1}{2\pi} \left( \arctan \frac{a+d}{b} + \arctan \frac{a-d}{b} \right) + \frac{b}{4\pi d} \ln \frac{a^2 + b^2 + d^2 - 2ad}{a^2 + b^2 + d^2 + 2ad} \tag{51}$$

It can be demonstrated that the former equates the area subtended by a circular sector that encompasses the diameter of the emitting disk and the corresponding sector of a hyperbola defined by the intersection of the unit sphere and the cone [7].

If a=d, the factor is,

$$\frac{1}{2\pi}\arctan\frac{2a}{b} - \frac{b}{4\pi d} \ln\frac{4a^2 + b^2}{b^2} \tag{52}$$

Where the former assert is more easily visualized.

2 2 · <sup>+</sup> <sup>=</sup> D D <sup>ò</sup> *dx ax b arctan*

( ) 2 2 *a b dc d b* = =± = + 1; 2 ;

Substituting in (46) yields for both terms:

170 Solar Radiation Applications

Now, substituting in the first term of (45):

2

2

2

Finally, multiplying by *<sup>b</sup>*

If a=d, the factor is,

2p ( ) 2 22

And bringing equation (48) to the limits of the integral [a,0]:

In a similar way, for the second term of equation (45):

2*πd*

( ( )) ( ) <sup>1</sup> 2 22 2 2 ln 2 ln

+ + ç ÷

defined by the intersection of the unit sphere and the cone [7].

p

*arctan arctan ln*

<sup>1</sup> 2 arctan

é ù æ ö - é ù + +- + ê ú ç ÷ ë û ë û è ø *<sup>a</sup> <sup>a</sup> d rd ln r d b dr*

( ( )) ( ) <sup>1</sup> 2 22 2 2 ln 2 ln arctan arctan


1 2

æ ö + - ++ -

è ø ++ +

12 4

 p

<sup>+</sup> - *a b ab arctan ln*

2 4

*d ad d a d b da d b arctan arctan*

and grouping, the final result can be expressed as:

 p

It can be demonstrated that the former equates the area subtended by a circular sector that encompasses the diameter of the emitting disk and the corresponding sector of a hyperbola

*a d a d b a b d ad*

æ ö - + +- - + + <sup>+</sup> ç ÷

*d ad d a d b da d b*

*<sup>X</sup>* (46)

2 22 2 *X r dr d b b* = ± + + D= 2 ( ) 4 (47)

è ø

æ ö +

è ø

22 2 22 2

*<sup>b</sup> b d a b d ad* (51)

*b db* (52)

4 2

2 2 2

*b b* (48)

*bb b* (49)

*bb b* (50)

<sup>0</sup> <sup>0</sup>

If d=0 the expression is undetermined and the limit is passed with l'Hôpital's rule, obtaining the familiar result.

$$\frac{1}{\pi} \left( \arctan \frac{a}{b} - \frac{ab}{a^2 + b^2} \right) \tag{53}$$

Equation (51) is entirely new and has never been mentioned in literatures; equation (52) constitutes a particular case of the former, that is, when a equals d, meaning that the receiving point lies on the edge of the emitting semicircle; equation (53) is the particular case in which the receiving point is aligned with the center of the half-disk.

#### **3.4. Extension to three dimensional emitters – Configuration factor between a sphere and a differential element placed at a random position**

After analyzing the previous form factors for the circle, a new question can be deducted. A sphere can be considered, in terms of radiative transfer as a circle [7], as the viewed area of the said sphere from a distant point equals always a circle, because only half of the emitter is visible. Let us consider as emitting source a sphere of radius r, and a differential element, placed randomly in space at a distance (x,y,z), referenced to the three coordinate directions as shown in figure 5:

**Figure 5.** Calculations parameters for the sphere and a differential element at a random position

The differential element, as in former cases, is defined by its normal, and it is necessary to find the radiation vector *Fr* <sup>→</sup> impinging on it. Obtaining the modulus (configuration factor) is a direct operation, the angles formed by the unit element are already known:

$$\left| \overline{F\_r} \right| = F\_{d1-x} \cdot \cos \alpha + F\_{d1-y} \cdot \cos \beta + F\_{d1-z} \cdot \cos \gamma \tag{54}$$

And expanding each of them,

$$\begin{aligned} F\_{d1-x} &= \frac{r^2 \cdot x}{\sqrt{(x^2 + y^2 + z^2)^3}} \\\\ F\_{d1-y} &= \frac{r^2 \cdot y}{\sqrt{(x^2 + y^2 + z^2)^3}} \\\\ F\_{d1-z} &= \frac{r^2 \cdot z}{\sqrt{(x^2 + y^2 + z^2)^3}} \\ \end{aligned}$$

#### **3.5. Configuration factor between a sphere and a plane.**

Extending the previous deduction to a finite rectangle located at a certain distance to the sphere in a random position (figure 6), a new unknown factor has been deducted:

**Figure 6.** Configuration factor between a sphere and a parallel plane
