§10.3 Radiant heat exchange between two finite black bodies 539 Of course, Q net 1–2 =−Q net 2–1 . It follows that A 1 F 1–2 σ  T 4 1 −T 4 2  =−A 2 F 2–1 σ  T 4 2 −T 4 1  or A 1 F 1–2 = A 2 F 2–1 (10.15) This result, called view factor reciprocity, is very useful in calculations. Example 10.1 A jet of liquid metal at 2000 ◦ C pours from a crucible. It is 3 mm in di- ameter. A long cylindrical radiation shield, 5 cm diameter, surrounds the jet through an angle of 330 ◦ , but there is a 30 ◦ slit in it. The jet and the shield radiate as black bodies. They sit in a room at 30 ◦ C, and the shield has a temperature of 700 ◦ C. Calculate the net heat transfer: from the jet to the room through the slit; from the jet to the shield; and from the inside of the shield to the room. Solution. By inspection, we see that F jet–room = 30/360 = 0.08333 and F jet–shield = 330/360 = 0.9167. Thus, Q net jet–room = A jet F jet–room σ  T 4 jet −T 4 room  =  π(0.003) m 2 m length  (0.08333)(5.67 ×10 −8 )  2273 4 −303 4  = 1, 188 W/m Likewise, Q net jet–shield = A jet F jet–shield σ  T 4 jet −T 4 shield  =  π(0.003) m 2 m length  (0.9167)(5.67 ×10 −8 )  2273 4 −973 4  = 12, 637 W/m The heat absorbed by the shield leaves it by radiation and convection to the room. (A balance of these effects can be used to calculate the shield temperature given here.) To find the radiation from the inside of the shield to the room, we need F shield–room . Since any radiation passing out of the slit goes to the 540 Radiative heat transfer §10.3 room, we can find this view factor equating view factors to the room with view factors to the slit. The slit’s area is A slit = π(0.05)30/360 = 0.01309 m 2 /m length. Hence, using our reciprocity and summation rules, eqns. (10.12) and (10.15), F slit–jet = A jet A slit F jet–room = π(0.003) 0.01309 (0.0833) = 0.0600 F slit–shield = 1 −F slit–jet −F slit–slit    0 = 1 −0.0600 −0 = 0.940 F shield–room = A slit A shield F slit–shield = 0.01309 π(0.05)(330)/(360) (0.940) = 0.08545 Hence, for heat transfer from the inside of the shield only, Q net shield–room = A shield F shield–room σ  T 4 shield −T 4 room  =  π(0.05)330 360  (0.08545)(5.67 ×10 −8 )  973 4 −303 4  = 619 W/m Both the jet and the inside of the shield have relatively small view factors to the room, so that comparatively little heat is lost through the slit. Calculation of the black-body view factor, F 1–2 . Consider two elements, dA 1 and dA 2 , of larger black bodies (1) and (2), as shown in Fig. 10.8. Body (1) and body (2) are each isothermal. Since element dA 2 subtends a solid angle dω 1 , we use eqn. (10.6) to write dQ 1to2 = (i 1 dω 1 )(cos β 1 dA 1 ) But from eqn. (10.7b), i 1 = σT 4 1 π Note that because black bodies radiate diffusely, i 1 does not vary with angle; and because these bodies are isothermal, it does not vary with position. The element of solid angle is given by dω 1 = cos β 2 dA 2 s 2 §10.3 Radiant heat exchange between two finite black bodies 541 Figure 10.8 Radiant exchange between two black elements that are part of the bodies (1) and (2). where s is the distance from (1) to (2) and cos β 2 enters because dA 2 is not necessarily normal to s. Thus, dQ 1to2 = σT 4 1 π  cos β 1 cos β 2 dA 1 dA 2 s 2  By the same token, dQ 2to1 = σT 4 2 π  cos β 2 cos β 1 dA 2 dA 1 s 2  Then Q net 1–2 = σ  T 4 1 −T 4 2   A 1  A 2 cos β 1 cos β 2 πs 2 dA 1 dA 2 (10.16) The view factors F 1–2 and F 2–1 are immediately obtainable from eqn. (10.16). If we compare this result with Q net 1–2 = A 1 F 1–2 σ(T 4 1 − T 4 2 ),we get F 1–2 = 1 A 1  A 1  A 2 cos β 1 cos β 2 πs 2 dA 1 dA 2 (10.17a) 542 Radiative heat transfer §10.3 From the inherent symmetry of the problem, we can also write F 2–1 = 1 A 2  A 2  A 1 cos β 2 cos β 1 πs 2 dA 2 dA 1 (10.17b) You can easily see that eqns. (10.17a) and (10.17b) are consistent with the reciprocity relation, eqn. (10.15). The direct evaluation of F 1–2 from eqn. (10.17a) becomes fairly in- volved, even for the simplest configurations. Siegel and Howell [10.4] provide a comprehensive discussion of such calculations and a large cat- alog of their results. Howell [10.5] gives an even more extensive tabula- tion of view factor equations, which is now available on the World Wide Web. At present, no other reference is as complete. We list some typical expressions for view factors in Tables 10.2 and 10.3. Table 10.2 gives calculated values of F 1–2 for two-dimensional bodies—various configurations of cylinders and strips that approach in- finite length. Table 10.3 gives F 1–2 for some three-dimensional configu- rations. Many view factors have been evaluated numerically and presented in graphical form for easy reference. Figure 10.9, for example, includes graphs for configurations 1, 2, and 3 from Table 10.3. The reader should study these results and be sure that the trends they show make sense. Is it clear, for example, that F 1–2 → constant, which is < 1 in each case, as the abscissa becomes large? Can you locate the configuration on the right-hand side of Fig. 10.6 in Fig. 10.9? And so forth. Figure 10.10 shows view factors for another kind of configuration— one in which one area is very small in comparison with the other one. Many solutions like this exist because they are a bit less difficult to cal- culate, and they can often be very useful in practice. Example 10.2 A heater (h) as shown in Fig. 10.11 radiates to the partially conical shield (s) that surrounds it. If the heater and shield are black, calcu- late the net heat transfer from the heater to the shield. Solution. First imagine a plane (i) laid across the open top of the shield: F h−s +F h−i = 1 But F h−i can be obtained from Fig. 10.9 or case 3 of Table 10.3, Table 10.2 View factors for a variety of two-dimensional con- figurations (infinite in extent normal to the paper) Configuration Equation 1. F 1–2 = F 2–1 =  1 +  h w  2 −  h w  2. F 1–2 = F 2–1 = 1 − sin(α/2) 3. F 1–2 = 1 2   1 + h w −  1 +  h w  2   4. F 1–2 = (A 1 +A 2 −A 3 )  2A 1 5. F 1–2 = r b − a  tan −1 b c −tan −1 a c  6. Let X = 1 +s/D. Then: F 1–2 = F 2–1 = 1 π   X 2 −1 + sin −1 1 X −X  7. F 1–2 = 1,F 2–1 = r 1 r 2 , and F 2–2 = 1 − F 2–1 = 1 − r 1 r 2 543 Table 10.3 View factors for some three-dimensional configurations Configuration Equation 1. Let X = a/c and Y = b/c. Then: F 1–2 = 2 πXY    ln  (1 +X 2 )(1 +Y 2 ) 1 +X 2 +Y 2  1/2 −X tan −1 X −Y tan −1 Y +X  1 +Y 2 tan −1 X √ 1 +Y 2 +Y  1 +X 2 tan −1 Y √ 1 +X 2    2. Let H = h/ and W = w/. Then: F 1–2 = 1 πW      W tan −1 1 W −  H 2 +W 2 tan −1  H 2 +W 2  −1/2 +H tan −1 1 H + 1 4 ln     (1 +W 2 )(1 +H 2 ) 1 +W 2 +H 2  ×  W 2 (1 +W 2 +H 2 ) (1 +W 2 )(W 2 +H 2 )  W 2  H 2 (1 +H 2 +W 2 ) (1 +H 2 )(H 2 +W 2 )  H 2         3. Let R 1 = r 1 /h, R 2 = r 2 /h, and X = 1 +  1 +R 2 2  R 2 1 . Then: F 1–2 = 1 2  X −  X 2 −4(R 2 /R 1 ) 2  4. Concentric spheres: F 1–2 = 1,F 2–1 = (r 1 /r 2 ) 2 ,F 2–2 = 1 − (r 1 /r 2 ) 2 544 Figure 10.9 The view factors for configurations shown in Table 10.3 545 Figure 10.10 The view factor for three very small surfaces “looking at” three large surfaces (A 1  A 2 ). 546 §10.3 Radiant heat exchange between two finite black bodies 547 Figure 10.11 Heat transfer from a disc heater to its radiation shield. for R 1 = r 1 /h = 5/20 = 0.25 and R 2 = r 2 /h = 10/20 = 0.5. The result is F h−i = 0.192. Then F h−s = 1 −0.192 = 0.808 Thus, Q net h−s = A h F h−s σ  T 4 h −T 4 s  = π 4 (0.1) 2 (0.808)(5.67 ×10 −8 )  (1200 +273) 4 −373 4  = 1687 W Example 10.3 Suppose that the shield in Example 10.2 were heating the region where the heater is presently located. What would F s−h be? Solution. From eqn. (10.15) we have A s F s−h = A h F h−s But the frustrum-shaped shield has an area of A s = π(r 1 +r 2 )  h 2 +(r 2 −r 1 ) 2 = π(0.05 +0.1)  0.2 2 +0.05 2 = 0.09715 m 2 548 Radiative heat transfer §10.3 and A h = π 4 (0.1) 2 = 0.007854 m 2 so F s−h = 0.007854 0.09715 (0.808) = 0.0653 Example 10.4 Find F 1–2 for the configuration of two offset squares of area A,as shown in Fig. 10.12. Solution. Consider two fictitious areas 3 and 4 as indicated by the dotted lines. The view factor between the combined areas, (1+3) and (2+4), can be obtained from Fig. 10.9. In addition, we can write that view factor in terms of the unknown F 1–2 and other known view fac- tors: (2A)F (1+3)–(4+2) = AF 1–4 +AF 1–2 +AF 3–4 +AF 3–2 2F (1+3)–(4+2) = 2F 1–4 +2F 1–2 F 1–2 = F (1+3)–(4+2) −F 1–4 And F (1+3)–(4+2) can be read from Fig. 10.9 (at φ = 90, w/ = 1/2, and h/ = 1/2) as 0.245 and F 1–4 as 0.20. Thus, F 1–2 = (0.245 −0.20) = 0.045 Figure 10.12 Radiation between two offset perpendicular squares. [...]... / (A1 F1–2 ) (10 .35 ) §10.4 Heat transfer among gray bodies A specified wall heat flux The heat flux leaving a surface may be known, if, say, it is an electrically powered radiant heater In this case, the lefthand side of one of eqns (10 .34 ) can be replaced with the surface’s known Qnet , via eqn (10 .33 b) For the adiabatic wall case just considered, if surface (1) had a specified heat flux, then eqn (10 .35 )... electrical circuit analogy for radiation among three gray surfaces An insulated wall If a wall is adiabatic, Qnet = 0 at that wall For example, if wall (3) in Fig 10.15 is insulated, then eqn (10 .33 b) shows that eb3 = B3 We can eliminate one leg of the circuit, as shown on the right-hand side of Fig 10.15; likewise, the left-hand side of eqn (10 .34 c) equals zero This means that all radiation absorbed by an... consider heat transfer from one infinite gray plate to another parallel to it Radiant energy flows past an imaginary surface, parallel to the first infinite plate and quite close to it, as shown as a dotted line 550 Radiative heat transfer §10.4 Figure 10. 13 The electrical circuit analogy for radiation between two gray infinite plates in Fig 10. 13 If the gray plate is diffuse, its radiation has the same geometrical... is black (ε2 = 1) A large enclosure does not reflect much radiation back to the small object, and therefore becomes like a perfect absorber of the small object’s radiation — a black body Heat transfer among gray bodies §10.4 5 53 Additional two-body exchange problems Radiation shields A radiation shield is a surface, usually of high reflectance, that is placed between a high-temperature source and its... three equations Qnet1 , at node B1 : eb1 − B1 B1 − B2 B1 − B3 + = 1 1 1 − ε1 A1 F1 3 ε1 A 1 A1 F1–2 Qnet2 , at node B2 : eb2 − B2 B2 − B1 B2 − B3 = + 1 1 1 − ε2 A1 F1–2 A2 F2 3 ε2 A 2 Qnet3 , at node B3 : eb3 − B3 B3 − B1 B3 − B2 = + 1 − 3 1 1 3 A 3 A1 F1 3 A2 F2 3 (10 .3 4a) (10 .34 b) (10 .34 c) If the temperatures T1 , T2 , and T3 are known (so that eb1 , eb2 , eb3 are known), these equations can be solved.. .Heat transfer among gray bodies §10.4 10.4 549 Heat transfer among gray bodies Electrical analogy for gray body heat exchange An electric circuit analogy for heat exchange among diffuse gray bodies was developed by Oppenheim [10.6] in 1956 It begins with the definition of two new quantities: flux of energy that irradiates the H (W/m2 ) ≡ irradiance = surface and total flux of radiative energy... geometrical distribution as that from a black body, and it will travel to other objects in the same way that black body radiation would Therefore, we can treat the radiation leaving the imaginary surface — the radiosity, that is — as though it were black body radiation travelling to an imaginary surface above the other plate Thus, by analogy to eqn (10. 13) , Qnet1–2 = A1 F1–2 (B1 − B2 ) = B1 − B2 (10.22) 1 A1 ... B3 = 5.67 × 10−8 T3 − 822.6 so T3 = 34 7 K Algebraic solution of multisurface enclosure problems An enclosure can consist of any number of surfaces that exchange radiation with one another The evaluation of radiant heat transfer among these surfaces proceeds in essentially the same way as for three surfaces For multisurface problems, however, the electrical circuit approach is less convenient than a. .. 0.15) and has a uniform temperature of 100◦ C Calculate the rate of heat transfer to the copper base per meter of length of the duct Solution Assume the duct walls to be gray and diffuse and that convection is negligible The view factors can be calculated from configuration 4 of Table 10.2: F1–2 = A1 + A2 − A3 0.5 + 0 .3 − 0.4 = 0.4 = 2A1 1.0 Similarly, F2–1 = 0.67, F1 3 = 0.6, F3–1 = 0.75, F2 3 = 0 .33 , and... 230 4 − Ts Solving, we find Ts = 2 13 K Heat transfer among gray bodies §10.4 The electrical circuit analogy when more than two gray bodies are involved in heat exchange Let us first consider a three-body transaction, as pictured in at the bottom and left-hand sides of Fig 10.15 The triangular circuit for three bodies is not so easy to analyze as the in-line circuits obtained in twobody problems The basic . in practice. Example 10.2 A heater (h) as shown in Fig. 10.11 radiates to the partially conical shield (s) that surrounds it. If the heater and shield are black, calcu- late the net heat transfer. small object’s radiation — a black body. §10.4 Heat transfer among gray bodies 5 53 Additional two-body exchange problems Radiation shields. A radiation shield is a surface, usually of high re- flectance,. gray plate to another parallel to it. Radiant energy flows past an imaginary surface, parallel to the first infinite plate and quite close to it, as shown as a dotted line 550 Radiative heat transfer