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:

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

b: horizontal distance between differential element dAj 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 0. 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

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

dA1 represents the point source

Подпись: (2)

image283 Подпись: (3)

dA2 = rdrdO

Подпись:Подпись:cosq

S = r2 + d2 + b2 – 2dr cosq

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,

image287 image288

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

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

image289 Подпись: (8)

In order to solve this double integral, we first integrate with respect to 0. Proceeding with the first integral implies taking out all the constants that are independent of 0, and this yields:

image291 Подпись: C • sin( Ax) A (B2 - C2) (B + C cos (Ax)) image293 Подпись: (9)

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

The change of variables is defined thus:

de = dx B = r2 + d2 + b2 C = -2dr A = 1 (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):

Csin( Ax)

A(B2 – C2) (B + Ccos(Ax)) (11)

Подпись: A(B2 image295 Подпись: (12)

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 2n and 0, thus, sin(2n), sin(0), equal nil and so does the integral,

Подпись:

image298

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

image299 Подпись: 2 (B - C)tan(Ax /2) ■arctan A B2 + C2 B2 - C2 Подпись: (14)

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

Подпись:Подпись: (15)

Подпись: r2 + d2 + b2 (r2 + d2 + b2)-4d 2r2
Подпись: 2 yj(r2 + d2 + b2 f - 4d2b2

Therefore, substituting all terms we receive:

image306 Подпись: (16)

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 (2n, 0), the result of arctangent is obviously n, and that produces:

image308 Подпись: (17)

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

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.

a 4 (r(r2 + d2 + b2) – 2rd2 + 2rd2)

Подпись:2b2 J 3 ■■ dr

0 4 ((r2 + d2 + b2 )2 – 4d V J2

Decomposing and operating again:

image311

a 4(r2 + d2 + b2)-8rd2 + 8rd2 2b1 M———————– dr–

o 4((r2 + d2 + b)-4d2r2p

 

image312
image313

(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:

 

image314

(20)

 

That offers the solution:

 

image315

(21)

 

(22)

 

image316

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

 

a-

2b 2d2 f————-

0 (t1 + lp(b2

dt

d2) + (b2 + d2)2 f2

image317 Подпись: (25)

This integral responds to the following model with the solution:

image319 image320

Making the substitution:

image321 Подпись: (28)

Finally, adding both terms, from (21) and (27), in order to obtain the final result: b2 b2 a2 + b2 – d2 b2 – d2

2^1 ((a2 + d2 + b2)2 – 4d2a2)

2.2. 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:

image323 image324

This can be decomposed into two terms

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):

image325 Подпись: (31)

a

Now, equation (27) is rearranged as follows:

image327

2bd3

 

(b2 – d2)
4b2d 2(d2 + b2)

 

________ (a2 + b2 – d2)_________

4b2dJ((a2 + d2 + b2)2 – 4d2a2)

 

= 2bd3

 

d (a2 + b2 – d2)
2b^((a2 + d2 + b2) – 4d2a2)

 

d(b2 – d2) 2b(d2 +b2)

 

(32)

 

image328

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

d(b2 – d2) 2b(d2+b2)

 

bd

d2 + b2

 

image329

image204

d (a2 + b2 – d2)- 2b2d 2b2d – d {b2 – d2)

2b J((a2 + d2 + b2)2 – 4d2a2) 2b(d2 + b’)

a2d – b2d – d3 b2d + d3

2bJ((a2 + d2 + b2)2 – 4d2a2) 2b(d2 + b’)

-d [-a2 + b2 + d2) d (b2 + d2)

 

-d (tb2 + d2 – a2)

 

image330

(b2 + d2 – a2)

 

1.d

2 b

 

1-

 

: + d2 + b2′) – 4d2a2)

 

image331

Also, the second term from equations (30):

image332
Подпись: (34)
image334

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:

image335

b2 + d2
2bd

 

d_

2b

 

image336

image337

b2(a2 + b2 + d2)________ b_

Подпись: b_ 2d image339 Подпись: (40)

2bdyj(a2 + b2 + d2)2 -4d2a2 2bd

Again, this result can be checked against usual formulas that appear in numerous configura­tion 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.

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