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
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 |
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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 |
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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
5.27—^————- a(3.134-2.784)(100 K)4
= m (100 K)_____________________________ = 151 9 W
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 ^
Radiation heat transfer decreases by
1 — j—u X 100% = 96.6% .
151.9
Posted in INDUSTRIAL VENTILATION DESIGN GUIDEBOOK