# Thermal Radiation inside a Vacuum (without Gas)

Surface element dA • is located on the spatial surface A] sending radiation to the surface A,-. The radiation power from the surface element dA• to the ele­ment dAi according to Eq. (4.211) is

Dfyjf = Lj(a)j)dAj cos fti doij. (4.214)

The solid angle da>j = dAi cos ft/r2. Equation (4.211) gives

Cos fti ft:

= L(W/)——— — A— = dA, = dEjj, (4.215)

Where Ejj is the proportion of the radiation from dA, relative to the radiation in­tensity of dAt. If Lambert’s cosine law is valid, L is not dependent on direction and

7r Lj = Mj. (4.216)

Thus

Jr cos ft{ COS ftj j A

DEij = M——— ‘-j——- 1 dAi (4.217)

TTY

DEtj = Mjipjj dAj. (4.218)

The visibility factor of the surface element depends on the geometry and gives that part of the radiation intensity of dAj that falls directly on the surface dA, or vice versa.

The radiation intensity of a surface element is the sum of emission and reflection!

M, = ЂiMmt + pjEj, (4.219)

Where p = reflectance and

E, = | dA(4.220)

Giving an integral formula for the radiation intensity function,

Mi = e{Mmi + Pi( dA,. (4.221)

JA,

Equation (4.221) is difficult to solve, and for practical cases approximate methods are used.

Usually the surface is divided into zones, and with sufficient accuracy M is considered constant over this area, giving

Mi = e, M„„ + p,^ Mkfik, (4.222)

Where

Ftk = f hk dAk (4.223)

)a,

Also, by integration,

M, A, = Mf dAt = e, AlMmj + pj’yMkAiFkh (4.224)

A————————————————————————— 4

• Ai where

Fkl = flk = Vlk dAk dAi (4.225)

F hi is the visibility factor between two finite surfaces, and AtFki = A^ is the geometrical radiation surface. Equation (4.225) shows that the geometrical ra­diation surface is symmetric and therefore

Aki = Alk. (4,226)

Integrated over the semispace, the integrals (4.223) are 1. Thus in the hollow,

‘я’ Fki = 1 and ‘я’ Akt — A/{. (4.22/)

This sum includes all the hollow surfaces and also the surface k, if it is concave, in which case the portion Fkk of the received radiation is re­flected back.

The net radiation power falling the surface is the difference between the incom­ing and outgoing radiation and the difference between absorption and emission:

^ = q = E — M = aE — eM„,. (4.228)

For direct net radiation between two blackbodies (from Eq. (4.220)),

Фі2пеі = (Mmi — Mml)A 12 = Aria(Tі — 7 2) • (4.230)

Radiation heat transfer in a hollow can be represented by electrical analogy as

 Фпа = дА = ^АМ-^Мг, (4.231; = G(M-U) current = conductance x potential difference

Where (e/a)Mm = U = radiation potential, which is dependent on the temperature; e is dependent on the radiation properties of the surface and the temperature; and a is dependent on the spectrum of the incoming ra­diation. (a/p)A = G = radiation conductance between the potentials U and M.

For a gray body

\$ = G(M-U)

When the surface is black, G = °° or R = 0, while a — 1 and a = 0; points U and M unite and the potential is Mm.

When the surface is thermally insulated, <п>net = 0. Points U and M unite and the potential is M = (e/a)Mm = UE = M..

For two surfaces, 0;; net = Ј;;v4, — E^Aj = A;1(M; — M,), so by analogy A is the radiation conductance between potentials M,- and M; (see Fig. 4.32).

When there are only two surfaces in the hollow, the net thermal radiation is

 6? , , M„,2 GL-> л
 A] mi
 _Јi_ +J^ + _P2_ O? i A] A i9 ct 2 A 2
 U 1-І/;
 2
 111 G] A G2  (4.232)

If surface 1 is convex or planar, all the incoming radiation is from surface 2, and F]2 = 1, while the visibility factor expresses that part of radiation coming from this surface. If surface 2 is concave, a part of the radiation is also from this surface.

 ^1^*12 ~ ^]2 ^21 = ^2-^21 ~ ^ 1 FII — f21 — 1 -^12 ~— ^2 OL-[ Q-2 ci:2 Јx| 1*2 ^ _
 Pi + i + fe ki^i + i + k^2 J — + J—1 Aj a2 aj a2 a2
 Mml-M
 M2
 125

 HEAT AND MASS TRANSFER

 (4.233)

 A

 + 1 + — J1 Otj -^»2

 When p| = 1 — aj,

 — —M ■ _ aj ml a2 m — 13,1 X + «1 a2 ‘^2

 (4.234)

 Equation (4.234) is valid for two coaxial cylinders and spheres. If A2»A1, a2 ~ e2, as almost all the radiation from surface 2 is reflected back to it. <7i = MmlЂ!-Mm2 (4.236)

 It is difficult to estimate the absorption ratio. Approximately «, = e1(T2) and A2 ~ el(Ti)- When the absorption relations are not dependent on temperature, the fol­lowing approximations can be used (a = e = constant = 1 — p). For two co­axial cylinders and spheres, (4.237)

 L

 El + ^2

 Ђi Ai

 For two parallel planes, A] ~ A2,

 ^ M m i — Mm2 I+i-i ‘ 6i ^2

 (4.238)

 When A2»Al5

 (4.239)

 <7 e](Mmi — Mm2).

 In HVAC technology the following formula is used for small temperature dif­ferences with sufficient accuracy (see Fig. 4.33):

 (4.240)

 Qt = asAT = 6] amuAT.

 Example 4 Radiation heat transfer. The radiation heat transfer between two parallel planes is reduced by placing a parallel aluminum sheet in the middle of the gap. The surface temperatures are 61 = 40 °C and 62 = 5 °C, respectively; the emissivities are e1 = e2 = 0.8.5. The emissivity of both sides of the aluminum FIGURE 4.33 Heat transfer factor representing blackbody radiation for various mean temperatures and temperature differences.

Is ea = 0.05. Calculate by how much the radiation heat transfer is reduced due to the aluminum sheet; surface temperatures remain constant, and the surfaces are assumed to be gray.

Without the aluminum sheet,

_ — Mm2 _ <j(T] — Ti)

<7,2 — — j—— ;————— y———-

TOC o "1-5" h z — + — — 1 = I

Ђ, Ђ2 Ђ

5.27—^————- a(3.134-2.784)(100 K)4

= m (100 K)_____________________________ = 151 9 W

__ i m2

0.85

■’ i.. = ‘V. : ‘ ~i—- —— •

1 + 1-1 Ђ e.:

With the aluminum,

= Au<r(Ti-Tl,) = q*2 = Aa2a(T4-T2) =>

T4 = Tl ^ Tl = 3134 + 27g4 = 7.785 x 109 K4 .

Thu?

 S Qu = — j—————————- 5.67—,—————— 4(3.13 — 77.85)(100 K)4 = ,5.1 ^