SOLUTIONS. Question 6-12:

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1 Federal University of Rio Grande do Sul Mechanical Engineering Department MEC Thermal Radiation Professor - Francis Name: Márleson Rôndiner dos Santos Ferreira Question 6-1: SOLUTIONS (a) A circular cylindrical enclosure has black interior surface, each maintained at uniform interior temperature as shown. The outside of the entire cylinder is insulated so that the outside does not radiate to the surroundings. How much Q(W) is supplied to each area as a result of the interior radiative exchange? Note that the areas on the top and on the bottom are given by: and the lateral area is: For the 3 surfaces, we know that A 1 = A = πr = π0.08 = m, (1) A 3 = πrl = π = m. () q o,j = ɛ j e b (T j ) + (1 ɛ j ) n k=1 The interior surfaces are all black, i.e., ɛ j = 1. Thus, F j k q o,k, j {1,, 3}. (3) q o,j = e b (T j ) = σtj 4, j {1,, 3}. (4) and evaluating the equation above, we have q o,1 = σt 4 1 = = W/m, (5) q o, = σt 4 = = W/m, (6) q o,3 = σt 4 3 = = W/m. (7) For the view factors, note that F 1 = F 1, since A 1 = A. Thus, F F 1 + F 1 3 = 1 F 1 = 1 F 1 3 or F 1 3 = 1 F 1, F 1 + F + F 3 = 1 F 3 = 1 F 1 F 3 = 1 F 1 F 3 = F

2 From the Appendix C, we for F 1 : R 1 = 8 30 = 0.667, R = R 1 = 0.667, X = = (8) and follows that: F 1 = Then, we have that We know that, ( ) = = F 1. (9) F 1 3 = 1 F 1 = = = F 3. (10) q R,j = q o,j Thus, we can calculate the values as follows and n F j k q o,k, j {1,, 3}. (11) k=1 q R,1 = q o,1 (F 1 1 q o,1 + F 1 q o, + F 1 3 q o,3 ), (1) q R,1 = q o,1 (F 1 q o, + F 1 3 q o,3 ), (13) ( q R,1 = ), (14) q R,1 = W/m, (15) q R, = q o, (F 1 q o,1 + F q o, + F 3 q o,3 ), (16) q R, = q o, (F 1 q o,1 + F 3 q o,3 ), (17) ( q R, = ), (18) q R, = W/m. (19) Then, we have Q 1 = q R,1 A 1 = = W, (0) Q = q R, A = = W. (1) From the conservation of energy we know that, Therefore, Q 1 + Q + Q 3 = 0. () Q 3 = (Q 1 + Q ) = ( ) (3) Q 3 = W. (4) (b) For the enclosure and the same surface temperatures, divide A 3 into two equal areas A 4 and A 5. What is the Q to each of these two areas, and how do they and their sum compare with Q 3 from part (a)?

3 Note that, The areas are given by: F F 1 6 = 1, and F 5 + F 6 = 1. (5) A 4 = A 5 = π = 0.04π m, (6) A 1 = A = π 0.08 = 0, 0064π m. (7) Since F 1 6 = F 6, from the Appendix C, number 3, we have: Thus, R 1 = = , R = R 1 and X = = F 1 6 = F 6 = ( ) = (8) F 1 4 = 1 F 1 6 = = = F 5 (9) Furthermore, since we have F 4 4 = F 5 5, follows: Then, F F 4 + F 4 5 = 1 and F F 5 + F 5 4 = 1. (30) A F 4 1 = F = π A π From the item (a) we know that F 1 = and noting = (31) F 1 + F 4 + F 5 = 1, F 4 = 1 ( ) = 0.15, (3) by reciprocity relation we have Thus, from the Eq. (30) we have A F 4 = F 4 = π A π = (33) F 4 5 = 1 ( ) = (34) 3

4 The interior surfaces are all black, i.e., ɛ j = 1. Thus, and evaluating the equation above, we have Thus, Then, we have q o,j = e b (T j ) = σt 4 j. (35) q o,1 = σt 4 1 = = W/m, (36) q o, = σt 4 = = W/m, (37) q o,4 = σt 4 4 = = W/m, (38) q o,5 = q o,4. (39) q R,4 = q o,4 (F 4 1 q o,1 + F 4 q o, + F 4 5 q o,5 ), (40) q R,4 = W/m. (41) Q 4 = q R,4 A 4 = π = W. (4) From the conservation of energy we know that, Therefore, and Q 1 + Q + Q 4 + Q 5 = 0. (43) Q 5 = (Q 1 + Q + Q 4 ) = ( ) (44) Q 5 = W. (45) Q 3 = Q 4 + Q 5 = , (46) Q 3 = W. (47) (c) What are Q 6 /A 6 and Q 7 /A 7 for the same enclosure and surfaces temperatures? How do they compare with Q 3 /A 3 from part (a) Q 4 /A 4 and Q 5 /A 5 from part (b)? Note that, F 1 6 = 1 F 1 1. (48) 4

5 From the Appendix C, we have and follows that: R 1 = = 4, R = R 1, X = =.065 (49) F 1 1 = From the Eq. (48) we have by the reciprocity relation, ( ) 4 = , (50) 4 F = F 1 1. (51) F 1 6 = = 0.07 = F 7, (5) A F 6 1 = F = π = A π (53) F 7 = F 6 1. (54) and for F 1, we have from Appendix C, R 1 = = , R = R 1, X = = 14.5 and F 1 = 1 [14.5 ] = F 1 = F 1 = F 1 = F 1 Thus, F 1 + F 7 + F 3 = 1, F 3 = = = F 1 3 F 1 + F F F 1 3 = 1, F 1 7 = = , by symmetry F 6 = F 1 7 = and by relation reciprocity as we know, F 6 = F 6 A A 6 = π 0.003π = = F 7 1. (55) Q k = σa k n j=1(t 4 k T4 j )F k j. (56) Then, Q 6 = π Q 6 = and Q 7 = π Q 7 = [ ] ( ) ( ) , [ ] ( ) ( ) , 5

Therefore, Q 6 = 45567.58 W/m, A 6 (57) Q 7 = 694.96 W/m. A 7 (58) Since we have the lateral area more refined, the regions closer to A (more cold) have more heat transfer due to higher view factor.

6 Therefore, Q 6 = W/m, A 6 (57) Q 7 = W/m. A 7 (58) Since we have the lateral area more refined, the regions closer to A (more cold) have more heat transfer due to higher view factor. Question 6-5: A very long cylinder at temperature T 1 is coaxial with a long square enclosure shown in cross section below. The conditions on surfaces 1 5 are show in the table. Find Q 1 and T. Surface T [K] Q [W/m] ε We know from the problem 5-, F 1 = F 3 1 = F 4 1 = F 5 1 = π 8, F 3 = F 5 = F 4 5 = F 4 3 = F 3 = F 3 4 = F 5 = F 5 4 = 0.48, F 4 = F 4 = F 5 3 = F 3 5 = , A = A 3 = A 4 = A 5 = l m, A 1 = π 1 l = πl m. by reciprocity relation, we have A F 1 = F 1 = π l A 1 8 πl = 0.5. (59) Due to symmetry we can note that F 1 3 = F 1 4 = F 1 5 = F 1 = 0.5, e b,1 = σt 4 1 = = W/m q o,1 = ɛ 1 e b,1 + (1 ɛ 1 ) [q o, F 1 + q o,3 F q o,4 F q o,5 F 1 5 ] q o,1 = [q o, + q o,3 + q o,4 + q o,5 ] since we have q o,3 = q o,4 = q o,5 = 0 (because T 3 = T 4 = T 5 = 0 and ε 3 = ε 4 = ε 5 = 1), results q o,1 = q o,, and noting that Q = 0 q R, = 0, we have q R, = q o, + (q o,1 F 1 + q o,3 F 3 + q o,4 F 4 + q o,5 F 5 ) 0 = q o, F 1 q o,1 q o, = F 1 q o,1. (60) 6

7 and q o,1 = (F 1 q o,1 ), q o,1 = π 8 q o,1, q o,1 = W/m, replacing in the Eq. (60), q o, = π q o, = W/m. Furthermore, and Therefore, e b, = 1 q o, 1 ε (F 1 q o,1 ), since q 0,3 = q 0,4 = q 0,5 = 0 ε ε σt 4 = [ π ] T = K q R,1 = q o,1 F 1 q o,, since q 0,3 = q 0,4 = q 0,5 = 0 q R,1 = q R,1 = W/m. Q 1 = q R,1 A 1, (61) Q 1 = πl = l W, (6) Q 1 = kw/m. l (63) Question 6-6: A frustum of a cone gas its base heated as shown. The top is held as 600 K while the side is perfectly insulated. All surfaces are diffuse-gray. What is the temperature attained by surface 1 as a result of radiative exchange within the enclosure? 7

8 The temperature T 1 can be evaluated as and we have e b,1 = 1 ε 1 q o,1 + 1 ε 1 ε 1 3 q o,1 = q 1 + F 1 k q o,k k=1 3 q o, = 0 + F k q o,k k=1 q o,3 = ε 3 σt (1 ε 3) 3 F 1 k q o,k, k=1 3 k=1 F 3 k q o,k. From Appendix C, we have the view factor F 1 3 R 1 = 15/15 = 1, R = 10/15 = , X = F 1 3 = 1 ( ) = =.4444, Thus, and by reciprocity relation, F 1 = 1 F 1 3 = = , (64) and F 1 = A 1 A F 1 F 1 = 15 π = π F 3 1 = A 1 A 3 F 1 3, F 3 1 = 15 π = , π by reciprocity relation we have F 3 = 1 F 3 1 = = , F 3 = A 3 A F 3, F 3 = 10 π = π Then, we have a follow system q o,1 F 1 q o, F 1 3 q o,3 = q 1 F 1 q o,1 + q o, F 3 q o,3 = 0 (1 ε 3 )F 3 1 q o,1 (1 ε 3 )F 3 q o, + q o,3 = ε 3 σt3 4 8

9 replacing the values we have q o, q o, q o,3 = 10000, 0.19q o,1 + q o, q o,3 = 0, q o, q o, + q o,3 = Solving the system above, we have Then, we have q o,1 = 1453 W/m (65) q o, = W/m (66) q o,3 = W/m (67) e b,1 = [ ] 0.4 (68) = 4309 (69) σt 4 1 T 1 = 933 K. (70) We believe that there were some rounding error when we calculate the view factors, that implies in the difference of the results. Question 6-8: An enclosure has four sides that are all equilateral triangles of the same size (i.e., it is an equilateral tetrahedron). The sides are of length L = 4 m and have conditions imposed as a follows: Side 1 is black and is at uniform temperature T 1 = 800 K. Side is diffuse-gray and is perfectly insulated on the outside. Side 3 is black and has a uniform heat flux of 8 kw/m supplied to it. Side 1 is black and is at T 4 = 0 K. Find q 1, T and T 3. For simplicity, do not subdivide the surfaces areas. We note that all the view factors are F j k = 0.33 and all areas are equals. q o,1 = = W/m q o,4 = = 0 W/m 9

from the equation above we have, q o, = 0 + (0.33q o,3 + 0.33q o,1 + 0.33q o,4 ) q o, = 0.33q o,3 + 7664.58, and (71) q o,3 = 8000 + (0.33q o,1 + 0.33q o, + 0.33q o,4 ), q o,3 = 8000 + (0.33q o,1 + 0.33q o, ), q o,3 = 8000 + 0.

10 from the equation above we have, q o, = 0 + (0.33q o, q o, q o,4 ) q o, = 0.33q o, , and (71) q o,3 = (0.33q o, q o, q o,4 ), q o,3 = (0.33q o, q o, ), q o,3 = (0.33q o, ), q o,3 = , (7) replacing Eq. (7) in Eq. (71), we have q o, = (73) Thus, q R,1 = q o,1 (0.33q o, q o,3 ), q R,1 = 36 ( ), q R,1 = kw/m. Therefore, q o,3 = e b,3, = σt3 4, T 3 = K, and q o, = e b,, = σt 4, T = K, Question 6-31: (...). It is within a thin-walled concentric cylinder of the same length having a diameter 8 cm. The emissivity on the inside of the cylinder is ε = 0.45, and on the outside is ε o = All surfaces are diffuse-gray. The entire assembly is suspended in a large vacuum chamber at T e = 300 K. What is the temperature T of the cylindrical shell? For simplicity, do not subdivide the surfaces areas. (Hint: F 1 = 0.5, F = ) 10

11 As we know, Thus, by reciprocity relation we have: F 1 = 0.5, (74) F = (75) A F 1 = π F 1 = 0.5 A 1 π F 1 = Lets define a imaginary surface (3) on the top and the bottom sections. This way we have, Note that we have: F F 1 + F 1 3 = 1, F 1 3 = 1 F 1, F 1 3 = = , F 1 + F + F 3 = 1, F 3 = 1 F 1 F, F 3 = = also we have, q o,3 = ε 3 σt 4 3 q o,3 = q o,3 = W/m. Then we can write the following equations: q o,1 = ε 1 e b,1 + (1 ε 1 )(F 1 q o,i + F 1 3 q o,3 ), [ q o,1 = ] + (1 0.3)( q o,i q o,3 ), q o,1 = q o,i, (76) q o,i = ε i e b, + (1 ε i )(F i 1q o,1 + F i i q o,i + F 3 q o,3 ) q o,i = T q o, (77) q 0,e = ε o σt 4 + (1 ε o)(f e 3q o,3 ), (external) q 0,e = T , besides that (78) q,e = q 0,e F e 3q o,3, q,e = q 0,e , and by the energy balance, (79) q i + q e = 0, furthermore, (80) q i = q o,i (F i 1q o,1 + F i i q o,i + F i 3q o,3 ) q i = q o,i 0.497q o, (81) Working with Equations (76)-(81), we obtain ( T ) T 4 = = T

12 The last one equation results, T 4 = T = K Question 6-45: Two plate are joined at 90 degrees and are very long normal to the cross section shown. The vertical plate (plate 1) is heated uniformly with a heat flux of 300 W/m. The horizontal plate has a uniform temperature of 400 K. Both plates have an emissivity of 0.6. The environment is at T e = 300 K. The dimensions are shown on the figure. Derive the equation necessary for find the temperature distribution on surface 1, and net radiative heat flux on surface. From the Appendix C (number 15) we have, h 1 = 0.10; W = 0.15; H = 0.10 = Then, 0.15 F 1 = 1 [ ] , F 1 = By reciprocity relation, we have Besides that we have, Furthermore, A F 1 = F 1 = A F 1 = F F 1 + F 1 3 = 1 F 1 3 = 1 F 1 = , F 1 + F + F 3 = 1 F 3 = 1 F 1 = q o,3 = εe b,3 = = W/m q o,1 = q R,1 + (q o, F 1 + q o,3 F 1 3 ) q o,1 = q o,, and (8) q o, = ε e b, + (1 ε )(q o,1 F 1 + q o,3 F 3 ) q o, = (1 0.6)(0.3415q o, ). (83) 1

13 Replacing Eq. (8) in (83), we have and replacing (84) in (8), we have Follows that, q o, = ( q o, ) q o, = W/m, (84) q o,1 = W/m. (85) q R, = ( ) q R, = W/m. For temperature, we have e b,1 = 1 q o,1 1 ε 1 [q o, F 1 + q o,3 F 1 3 ] ε 1 ε 1 σt1 4 = [ ] T 1 = K Porto Alegre - November 11,

SOLUTIONS. F 0 λ1 T = (1) F 0 λ2 T = (2) ε = (6) F 0 λt = (7) F 0 λt = (11)

SOLUTIONS. F 0 λ1 T = (1) F 0 λ2 T = (2) ε = (6) F 0 λt = (7) F 0 λt = (11) Federal University of Rio Grande do Sul Mechanical Engineering Department MEC0065 - Thermal Radiation Professor - Francis Name: Márleson Rôndiner dos Santos Ferreira SOLUTIONS Question 2-5: The total hemispherical