Radiation Exchange Between Surfaces: Enclosures with Nonparticipating Media Chapter 13 Sections 13.1 through 13.3 Basic Concepts Basic Concepts Enclosures consist of two or more surfaces that envelop a region of space (typically gas-filled) and between which there is radiation transfer. Virtual, as well as real, surfaces may be introduced to form an enclosure. A nonparticipating medium within the enclosure neither emits, absorbs, nor scatters radiation and hence has no effect on radiation exchange between the surfaces. Each surface of the enclosure is assumed to be isothermal, opaque, diffuse and gray, and to be characterized by uniform radiosity and irradiation. View Factor Integral The View Factor (also Configuration or Shape Factor) The view factor, Fij , is a geometrical quantity corresponding to the fraction of the radiation leaving surface i that is intercepted by surface j. qi j Fij

Ai J i The view factor integral provides a general expression for Fij . Consider exchange between differential areas dAi and dA j : dqi j I i cosi dAi d j i J i Fij 1 A Ai i A j cosi cos j cosi cos j R 2 R2 dAi dA j dAi dAj View Factor Relations View Factor Relations Reciprocity Relation. With cosi cos j

1 F ji A A dAi dA j Aj i j R2 Ai Fij Aj F ji Summation Rule for Enclosures. N Fij 1 j 1 Two-Dimensional Geometries (Table 13.1) For example, An Infinite Plane and a Row of Cylinders 2 D Fij 1 1 s 1/ 2 2

2 1/ 2 D tan s 2D s D 1 View Factor Relations (cont) Three-Dimensional Geometries (Table 13.2). For example, Coaxial Parallel Disks 2 Fij 1 S S 2 4 rj / ri 2 S 1 1/ 2 1 R 2j

Ri ri / L Ri2 R j rj / L Blackbody Enclosure Blackbody Radiation Exchange For a blackbody, J i Ebi . Net radiative exchange between two surfaces that can be approximated as blackbodies net rate at which radiation leaves surface i due to its interaction with j or net rate at which surface j gains radiation due to its interaction with i qij qi j q j i qij Ai Fij Ebi A j F ji Ebj qij Ai Fij Ti 4 T j4

Net radiation transfer from surface i due to exchange with all (N) surfaces of an enclosure: N qi Ai Fij Ti 4 T j4 j 1 General Enclosure Analysis General Radiation Analysis for Exchange between the N Opaque, Diffuse, Gray Surfaces of an Enclosure i i 1 i Alternative expressions for net radiative transfer from surface i: qi Ai J i Gi Fig. (b) (1) qi Ai Ei iGi Fig. (c) (2) qi Ebi J i Fig. (d) 1

/ A i i i (3) Suggests a surface radiative resistance of the form: 1 i / i Ai General Enclosure Analysis (cont) N N j 1 j 1 qi Ai Fij J i J j Ji J j AF 1

i ij (4) Suggests a space or geometrical 1 resistance of the form: Ai Fij Equating Eqs. (3) and (4) corresponds to a radiation balance on surface i: N J J Ebi J i j i 1 i / i Ai j 1 Ai Fij 1 which may be represented by a radiation network of the form (5) General Enclosure Analysis (cont) Methodology of an Enclosure Analysis Apply Eq. (4) to each surface for which the net radiation heat rate qi is known. Apply Eq. (5) to each of the remaining surfaces for which the temperature Ti , and hence Ebi , is known. Evaluate all of the view factors appearing in the resulting equations. Solve the system of N equations for the unknown radiosities, J1 , J 2 ,...., J N . Use Eq. (3) to determine qi for each surface of known Ti and Ti for each surface of known qi .

Treatment of the virtual surface corresponding to an opening (aperture) of area Ai , through which the interior surfaces of an enclosure exchange radiation with large surroundings at Tsur : Approximate the opening as blackbody of area, Ai , temperature, Ti Tsur , and properties, i i 1. Two-Surface Enclosures Two-Surface Enclosures Simplest enclosure for which radiation exchange is exclusively between two surfaces and a single expression for the rate of radiation transfer may be inferred from a network representation of the exchange. T14 T24 q1 q2 q12 1 1 1 2 1 1 A1 A1F12 2 A2 Two-Surface Enclosures (cont) Special cases are presented in Table 13.3. For example, Large (Infinite) Parallel Plates A1 A2 A F12 1 A1 T14 T24

q12 1 1 1 1 2 Note result for Small Convex Object in a Large Cavity. Radiation Shield Radiation Shields High reflectivity (low ) surface(s) inserted between two surfaces for which a reduction in radiation exchange is desired. Consider use of a single shield in a two-surface enclosure, such as that associated with large parallel plates: Note that, although rarely the case, emissivities may differ for opposite surfaces of the shield. Radiation Shield (cont) Radiation Network: T14 T24 q12 q1 q2 1 3,1 1 3,2 1 1 1 2

1 1 1 A1 A1F13 3,1 A3 3,2 A3 A3 F32 2 A2 The foregoing result may be readily extended to account for multiple shields and may be applied to long, concentric cylinders and concentric spheres, as well as large parallel plates. Reradiating Surfaces The Reradiating Surface An idealization for which GR J R . Hence, qR 0 and J R EbR . Approximated by surfaces that are well insulated on one side and for which convection is negligible on the opposite (radiating) side. Three-Surface Enclosure with a Reradiating Surface: q1 q2 T14 T24 1 1 1 2 1 1 A1 A1F12 1 / A1F1R 1 / A2 F2 R 1 2 A2 Reradiating Surfaces (cont)

Temperature of reradiating surface TR may be determined from knowledge of its radiosity J R . With qR 0 , a radiation balance on the surface yields J1 J R J J2 R 1 / A1F1R 1 / A2 F2 R J TR R 1/ 4 Multimode Effects Multimode Effects In an enclosure with conduction and convection heat transfer to or from one or more surfaces, the foregoing treatments of radiation exchange may be combined with surface energy balances to determine thermal conditions. Consider a general surface condition for which there is external heat addition (e.g., electrically), as well as conduction, convection and radiation. qi ,ext qi ,rad qi ,conv qi ,rad qi ,rad Appropriate analysis for N -surface, two-surface, etc., enclosure. Problem: Furnace in Spacecraft Environment Problem 13.88: Power requirement for a cylindrical furnace with two

reradiating surfaces and an opening to large surroundings. KNOWN: Cylindrical furnace of diameter D = 90 mm and overall length L = 180 mm. Heating elements maintain the refractory lining ( = 0.8) of section (1), L1 = 135 mm, at T1 = 800 C. The bottom (2) and upper (3) sections are refractory lined, but are insulated. Furnace operates in a spacecraft vacuum environment. FIND: Power required to maintain the furnace operating conditions with the surroundings at 23 C. Problem: Furnace in Spacecraft Environment (cont) SCHEMATIC: ASSUMPTIONS: (1) All surfaces are diffuse gray, and (2) Uniform radiosity over the sections 1, 2, and 3. ANALYSIS: By defining the furnace opening as the hypothetical area A4, the furnace can be represented as a four-surface enclosure. The power required to maintain A1 at T1 is q1, the net radiation leaving A1. To obtain q1, we must determine the radiosity at each surface by simultaneously solving radiation energy balance equations of the form N J J E bi Ji j j q i

1/ Ai Fij 1 i / i Ai j 1 (1,2) Problem: Furnace in Spacecraft Environment (cont) However, since 4 = 1, J4 = Eb4, and only three energy balances are needed for A1, A2, and A3. A1: A2: A3: E b1 J1 J J J J2 J J4 1 1 3 1 1 1 / 1A1 1/ A1F12 1/ A1F13 1/ A1F14 J J3 J J J J4 0 2 1 2 2 1/ A 2 F21 1/ A 2 F23 1/ A 2 F24 J J

J J2 J J4 0 3 1 3 3 1/ A3 F31 1/ A3 F32 1/ A 3 F34 (3) (4) (5) where q2 = q3 = 0 since the surfaces are insulated (adiabatic) and hence reradiating. From knowledge of J1, q1 can be determined using Eq. (1). Of the N2 = 42 = 16 view factors, N(N 1)/2 = 6 must be independently evaluated, while the remaining can be determined by the summation rule and appropriate reciprocity relations. The six independently determined Fij are: By inspection: (1) F22 = 0 (2) F44 = 0 Problem: Furnace in Spacecraft Environment (cont) Coaxial parallel disks: From Table 13.2, (3)

2 F24 0.5 S S2 4 r4 / r2 1/ 2 0.05573 where 2 S 1 1 R4 2 R2 1 1 0.250 0.250 2

2 18.00 R 2 r2 / L 45 / 180 0.250 R 4 r4 / L 0.250 Enclosure 1-2-2: From the summation rule for A2, (4) F21 = 1 F22 = 1 0.09167 = 0.9083 where F22 can be evaluated from the coaxial parallel disk relation, Table 13.2, with R2 = r2/L1 = 45/135 = 0.333, R2 = r2/L1 = 0.333, and S = 11.00. From the summation rule for A1, (5) F11 = 1 F12 F12 = 1 0.1514 0.1514 = 0.6972 From symmetry F12 = F12 and using reciprocity 2 F12 A 2 F21 / A1 0.090m / 4 0.9083 / 0.090m 0.135m 0.1514 Enclosure 2 -3-4: From the summation rule for A4, (6)

F43 = 1 F42 F44 = 1 0.3820 0 = 0.6180 where F44 = 0 and using the coaxial parallel disk relation from Table 13.2, F42 =0.3820 with R4 = r4/L2 = 45/45 = 1, R2 = r2/L2 = 1, and S = 3. Problem: Furnace in Spacecraft Environment (cont) The View Factors: Using summation rules and appropriate reciprocity relations, the remaining 10 view factors can be evaluated. Written in matrix form, the F ij are 0.6972* 0.1514 0.09704 0.05438 0.9083* 0* 0.03597 0.05573* 0.2911 0.01798 0.3819 0.3090 0.3262 0.05573 0.6180* 0* The Fij shown with an asterisk were independently determined.

From knowledge of the relevant view factors, the energy balances, Eqs. (3, 4, 5), can be solved simultaneously to obtain the radiosities, J1 73, 084 W / m 2 J 2 67, 723 W / m 2 J 3 36, 609 W / m 2 The net heat rate leaving A1 can be evaluated using Eq. (1) written as q1 E b1 J1 1 1 / 1A1 75,159 73, 084 W / m 2 1 0.8 / 0.8 0.03817 m 2 317 W < where Eb1 = T14 = (800 + 273K)4 = 75,159 W/m2 and A1 = DL1 = 0.090m 0.135m = 0.03817 m2. COMMENTS: Recognize the importance of defining the furnace opening as the hypothetical area A4 which completes the four-surface enclosure representing the furnace. The temperature of A4 is that of the surroundings and its emissivity is unity since it absorbs all radiation incident

on it. Problem 13.93 Problem 13.93: Assessment of ceiling radiative properties for an ice rink in terms of ability to maintain surface temperature above the dew point. KNOWN: Ice rink with prescribed ice, rink air, wall, ceiling and outdoor air conditions. FIND: (a) Temperature of the ceiling, Tc, for an emissivity of 0.05 (highly reflective panels) or 0.94 (painted panels); determine whether condensation will occur for either or both ceiling panel types if the relative humidity of the rink air is 70%, and (b) Calculate and plot the ceiling temperature as a function of ceiling insulation thickness for 0.1 t 1 m; identify conditions for which condensation will occur on the ceiling. Problem 13.93 (cont) SCHEMATIC: In s u la tio n , k = 0 .0 3 5 W /m - K , th ic k n e s s t = 0 .3 m T oo , o = - 5 o C L = 10 m C e i l i n g ( c ) , T c , = 0 . 0 5 o r 0 . 9 4

W a lls ( w ) T w = 15 oC R in k a ir T oo ,i = 1 5 o C h i = 5 W /m 2 -K Ic e ( i) , T i = - 5 o C D = 50 m q O u td o o r s E n e r g y b a la n c e o n c e ilin g o q C e ilin g , k R in k a ir q c o n v ,c q ra d ,c

A c, T c T oo ,o T c o R cond q c o n v,c + q ra d ,c ASSUMPTIONS: (1) Rink comprised of the ice, walls and ceiling approximates a threesurface, diffuse-gray enclosure, (2) Surfaces have uniform radiosities, (3) Ice surface and walls are black, (4) Panels are diffuse-gray, and (5) Thermal resistance for convection on the outdoor side of the ceiling is negligible compared to the conduction resistance of the ceiling insulation.

Problem 13.93 (cont) PROPERTIES: Psychrometric chart (Atmospheric pressure; dry bulb temperature, T db = T,i = 15 C; relative humidity, RH = 70%): Dew point temperature, T dp = 9.4 C. ANALYSIS: Applying an energy balance to the inner surface of the ceiling and treating all heat rates as energy outflows, E in E out 0 q o q conv,c q rad,c 0 (1) where the rate equations for each process are q o Tc T,o / R cond R cond t / kA c q conv,c h i A c Tc T,i q rad,c E b Tc Ac A w Fwc E b Tw Ai Fic E b Ti Since the ceiling panels are diffuse-gray, = .

From Table 13.2 for parallel, coaxial disks Fic 0.672 From the summation rule applied to the ice (i) and the reciprocity rule, Fic Fiw 1 Fiw Fcw (symmetry) Fcw 1 Fic Fwc Ac / A w Fcw Ac / A w 1 Fic 0.410 where Ac = D2/4 and Aw = DL. (2,3) (4) (5) Problem 13.93 (cont) Using the foregoing energy balance, Eq. (1), and the rate equations, Eqs. (2-5), the ceiling temperature is calculated using radiative properties for the two panel types, Ceiling panel Reflective Paint Tc ( C) 0.05

0.94 14.0 8.6 Tc < Tdp . Condensation will occur on the painted panel since T c < Tdp. (b) Applying the foregoing model for 0.1 t 1.0 m, the following result is obtained Ceiling temperature, Tc (C) 15 10 5 0 0.2 0.4 0.6 0.8

Ceiling insulation thickness, t (m) Painted ceiling, epsc = 0.94 Reflective panel, epsc = 0.05 1 < Problem 13.93 (cont) For the reflective panel ( = 0.05), the ceiling surface temperature is considerably above the dew point. Therefore, condensation will not occur for the range of insulation thicknesses. For the painted panel ( = 0.94), the ceiling surface temperature is always below the dew point, and condensation occurs for the range of insulation thicknesses. COMMENTS: From the analysis, recognize that radiative exchange between the ice and the ceiling has the dominant effect on the ceiling temperature. With the reflective panel, the rate is reduced nearly 20-fold relative to that for the painted panel. With the painted panel ceiling, condensation will occur for most of the conditions likely to exist in the rink.