Heat Transfer. Homework Solutions

May 9 Homework Solutions 13-8 Determine the view factors F13 and F23 between the rectangular surfaces shown in the figure at the right. We can find the view factor F31 from Figure 13-6 in the text where surface 3 in our problem is surface 1 in the text figure and surface 1 in our problem is surface 2 in the text figure. With this correspondence we have the following terms in to use in Figure 13-6, L1/W = L2/W = 1/3. For these values I read F1→2 = 0.28 from the figure. Alternatively we could calculate the shape factor from the final equation in Table 13-1 which gives a value of 0.257 for the shape factor. I will use the value from the equation for the shape factor F31. We can find F13 from the reciprocity relation A1F13 = A3F31. Since A1 = A2, F13 = F31 = 0.257 . We can find F23 by finding F32 and using the reciprocity relation. In order to find F32 we can use the same figure or equation used before to find F3→1+2 and then use the superposition theorem to find the desired shape factor F32 = F3→1+2 – F31. The problem of finding F3→1+2 has the following dimensions in terms of the labels in Figure 13-6: L1/W = 1/3 and L2/W = 2/3. For these values I read the view factor from Figure 13-6 as 0.34. From the equation for the view factor I find a value of 0.319. I will use this value for F3→1+2. I can then find F32 = F3→1+2 – F31 = 0.319 – 0.257 = 0.0620. Finally we can apply the reciprocity relation A2F23 = A3F32. Since A2 = A3, F23 = F32 = 0.0620 . 13-14 Consider a conical enclosure of height h and base diameter D. Determine the view factor from the conical side surface to a hole of diameter d located at the center of the base. We can determine the required shape factor by considering only the summation and reciprocity relations. Label the small circle as surface (1), the outer ring on the bottom of the cone as surface (2) and the conical side as surface (3). In terms of this numbering scheme we are trying to find F31. We can do this by applying the reciprocity relation A3F31 = A1F13. The areas in this equation are A1 = πd2/4 and A3 = πDh/2. (If you are not familiar with the area for the surface of a cone, you can see that the result is intuitively appealing. It is simply the area of a cylinder with half the height of the cone.) (3) (1) (2) We can deduce the value of F13. From the diagram we see that radiation leaving the small center has no place to go except the conical surface. Therefore, F13 must equal one. With F13 = 1 we can find F31 = A1F13/A3 = (πd2/4)(1)/(πDh/2) or F31 = d2/(2Dh) . 13-26E Consider a 10-ft by 10-ft by 10-ft cubical furnace whose top and side surfaces closely approximate black surfaces and whose base surface has an emissivity ε = 0.7. The base, top, and side surfaces of the furnace are maintained at uniform temperatures of 800 R, 1600 R, and 2400 R, respectively. Determine the net rate of radiation heat transfer between (a) the base and the side surfaces and (b) the base and the top surfaces. Also, determine the net rate of radiation heat transfer to the base surface. In this problem all the side surfaces have identical radiation properties (ε = 1) and the same temperature (2400 R) so we can consider the four side surfaces as a single surface for radiation exchange. In the diagram shown at the left, we call the bottom surface, surface 1, the top surface, surface 2, and the sides of the cube, “surface” 3. The view factors for the heat exchange between the top and bottom surfaces, F12 and F21, are the same because the areas of both surfaces are the same; F12 = F21. We can find this common view factor from Figure 13-5 with L1/D = L2/D = 1 since all sides of the cube are the same. With these parameters we find F12 = F21 = 0.2 from the figure. Using the corresponding equation in Table 13-1 gives F12 = F21 = 0.1998. Use the rounded value of 0.2 in further calculations. We can use the summation rule: F11 + F12 + F13 = 1 and the fact that surface 1 is a flat surface so that F11 = 0 to obtain the view factor F13 = 1 – F12 = 1 – 0.2 = 0.8. By the symmetry of the cube F23 = F13. We can find F31 (which will also be equal to F32 by symmetry) by the reciprocal rule: A1F13 = A3F31. The area A1 is the area on one side of the cube: A1 = (10 ft)(10 ft) = 100 ft2. The area A3 is the area of the four sides counted as a single surface. This area is four times the area of one side or A3 = 400 ft2. Thus, F31 =A1F13/A3 = (100 ft2)(0.2)/(400 ft2) = 0.05. By symmetry, F32 = 0.05. Finally, we can apply the summation rule to the side surfaces: F31 + F32 + F33 = 1 so F33 = 1 – F31 – F32 = 1 – 0.05 – 0.05 = 0.9. Since we know the temperature of all the surfaces, we can compute the black body emissive power of each surface, Eb = σT4. Eb1 = σT14 = Eb 2 = σT24 = Eb3 = σT34 = 0.1714 x10 −8 Btu h ⋅ ft 2 ⋅ R 4 0.1714 x10 −8 Btu h ⋅ ft 2 ⋅ R 4 0.1714 x10 −8 Btu h ⋅ ft ⋅ R 2 4 (800 R )4 = 702 Btu h ⋅ ft 2 (1600 R )4 = 11,233 Btu 2 h ⋅ ft (2400 R )4 = 56,866 2Btu h ⋅ ft Normally we would have to solve three simultaneous equations for three values of radiosity in an enclosure with three surfaces. However, for a black surface, the radiosity is the same as the emissive power. (To convince yourself of this recall the definition of J = ρG + εEb. For a black surface ρ = 0 and ε = 1.) For the two black surfaces in this problem, we can write J2 = Eb2 and J3 = Eb3. We can use the general radiosity equation for surface one in a three surface enclosure from slide 29 of the May 2 lecture presentation. ⎡ 1 − ε1 ⎤ (F12 + F13 )⎥ J1 − 1 − ε1 F12 J 2 − 1 − ε1 F13 J 3 = σT14 = Eb1 ⎢1 + ε1 ε1 ε1 ⎣ ⎦ Setting J2 = Eb2 and J3 = Eb3 in this equation and inserting ε1 = 0.7, F12 = 0.2 and F13 = 0.8 gives. 1 − 0.7 1 − 0.7 ⎡ 1 − 0.7 ⎤ 4 ⎢⎣1 + 0.7 (0.2 + 0.8)⎥⎦ J 1 − 0.7 (0.2)Eb 2 − 0.7 (0.8)Eb3 = σT1 = Eb1 1.429 J 1 − 0.08571Eb 2 − 0.3429 Eb3 = Eb1 ⇒ J1 = Eb1 + 0.08571Eb 2 + 0.3429 Eb3 1.429 Substituting the values for the black body emissive powers found above gives us our solution for J1. 702 Btu J1 = Eb1 + 0.08571Eb 2 + 0.3429 Eb3 h ⋅ ft = 1.429 2 + 0.08571 11,233 Btu h ⋅ ft 1.429 2 + 0.3429 56,866 Btu h ⋅ ft 2 = 14,813 Btu h ⋅ ft 2 The basic equation for the net heat transfer from surface i to surface k is given by the equation involving the radiosities of the two surfaces. Q& ik = Ai Fij ( J i − J k ) If one of the surfaces is black we can replace the radiosity by the black body emissive power. We use this equation to find the net heat transfer from surface one to surface three, with J3 = Eb3 is Q&1↔3 6 3.364 x10 Btu ⎛ 14,813 Btu 56,866 Btu ⎞ ⎟= − = A1 F13 (J 1 − J 3 ) = A1 F13 (J 1 − Eb3 ) = 100 ft (0.2 )⎜ − h ⎜ h ⋅ ft 2 h ⋅ ft 2 ⎟⎠ ⎝ ( 2 ) The negative sign indicates that the net heat transfer is from surface three to surface one. In a similar manner we find the net heat transfer from surface one to surface two, with J2 = Eb2. Q&1↔ 2 4 7.161x10 Btu ⎛ 14,813 Btu 11,233 Btu ⎞ ⎟= = A1 F12 ( J 1 − J 2 ) = A1 F12 (J 1 − Eb 2 ) = 100 ft 2 (0.2 )⎜ − h ⎜ h ⋅ ft 2 h ⋅ ft 2 ⎟⎠ ⎝ ( ) The net heat transfer from surface one is the sum of the two heat interchanges computed above. (This is true because surface one is flat had has no heat interchange with itself. Q&1 = Q&1↔ 2 + Q&1↔3 3.364 x10 6 Btu 7.161x10 4 Btu =− +− = − h h 3.293 x106 Btu h We could also find this heat transfer from the following equation. ( ) Aε 100 ft 2 (0.7 ) ⎛⎜ 702 Btu 14,813 Btu ⎞⎟ 3.293 x10 6 Btu − Q&1 = 1 1 (Eb1 − J 1 ) = = − ⎜ h ⋅ ft 2 1 − ε1 1 − 0.7 h ⋅ ft 2 ⎟⎠ h ⋅ ft 2 ⎝ This result provides a check on the previous calculations. 13-36 A furnace is shaped like a long equilateraltriangular duct where the width of each side is 2 m. Heat is supplied from the base surface, whose emissivity is ε1 = 0.8, at a rate of 800 W/m2 while the side surfaces, whose emissivities are 0.5, are maintained at 500 K. Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure? Since the side surfaces both have the same emissivity and the same temperature we may treat these two sides as a single radiation surface. Call this surface 2 and the base surface 1. We can use the equation for radiative heat transfer in a two-surface enclosure. Q&12 = ( ) Eb1 − Eb 2 A1σ T14 − T24 = 1 − ε1 1 − ε 2 1 − ε1 A 1 − ε2 1 1 + + 1 + + ε1 A1ε1 A1 F12 A2 ε 2 F12 A2 ε 2 ⇒ q&12 = ( ) σ T14 − T24 1 − ε1 A 1− ε2 1 + + 1 ε1 F12 A2 ε 2 Since the base is flat, F11 = 0 and the summation rule gives F12 = 1 = F11 = 1 – 0 = 1. For a given length, L, of the furnace, the area of the bottom is A1 = (2 m)L and the area of the sides is A2 = 2(2 m)L; so, A2 = 2A1 or A1/A2 = 0.5. Solving the heat flux equation for the unknown T1 and substituting these numerical results and the given data into the resulting equation for T1 gives. q&12 σ ⎛ 1 − ε1 A 1 − ε2 ⎞ 1 ⎜⎜ ⎟⎟ + T24 = T14 + + 1 F12 A2 ε 2 ⎠ ⎝ ε1 ⇒ T1 = 4 q&12 σ ⎛ 1 − ε1 A 1 − ε2 ⎞ 1 ⎜⎜ ⎟⎟ + T24 + + 1 F12 A2 ε 2 ⎠ ⎝ ε1 800 W T1 = 4 m2 5.670 x10 −8 W m2 ⋅ K 4 1 − 0 .5 ⎞ ⎛ 1 − .8 1 4 + + 0 .5 ⎜ ⎟ + (500 K ) = 543 K 1 0 .5 ⎠ ⎝ .8 13-43 Consider a circular grill whose diameter is 0.3 m. The bottom of the grill is covered with hot coal bricks at 950 K, while the wire mesh on top of the grill is covered with steaks initially at 5oC. The distance between the coal bricks and the steaks is 0.20 m. Treating both the steaks and the coal bricks as blackbodies, determine the initial rate of radiation heat transfer from the coal bricks to the steaks. Also, determine the initial rate of radiation heat transfer to the steaks if the side opening of the grill is covered by aluminum foil, which can be approximated as a reradiating surface. Assuming that both surfaces are black bodies, the heat transfer from the coal (surface 1) to the steaks (surface2) is ( Q&12 = A1 F12 σ T14 − T24 ) The view factor for two coaxial, circular surfaces with radii ri and rj separated by a distance L is given by the following equation in Table 13-1. (The view factor can also be found from Figure 13-7.) 2⎞ ⎛ ⎛ r2 ⎞ ⎟ 1 + (r2 L )2 1⎜ 2 F12 = ⎜ S − S − 4⎜⎜ ⎟⎟ ⎟ where S = 1 + 2⎜ (r1 L )2 ⎝ r1 ⎠ ⎟ ⎠ ⎝ For this problem r1 = r2 = 0.15 m and L = 0.2 m, so r1/L = r2/L = (0.15 m)/(0.2 m) = 0.75. This gives S = 1+ 1 + (r2 L )2 (r1 L )2 = 1+ 1 + (0.75)2 (0.75)2 = 3.7778 2⎞ 2⎞ ⎛ ⎛ ⎛ r2 ⎞ ⎟ 1 ⎜ ⎛ 0.15 m ⎞ ⎟ 1⎜ 2 2 ⎟⎟ ⎟ = 0.2864 F12 = ⎜ S − S − 4⎜⎜ ⎟⎟ ⎟ = ⎜ 3.7778 − 3.7778 − 4⎜⎜ 2⎜ ⎝ 0.15 m ⎠ ⎟ ⎝ r1 ⎠ ⎟ 2 ⎜ ⎠ ⎝ ⎠ ⎝ The area of the coals is πD2/4 = π(0.3 m)2/4 = 0.07069 m2. The initial heat transfer when the temperature of the steaks is 298 K can now be found. ( ) ( ) [ ] 5.670 x10 −8 W (950 K )4 − (298 K )4 926 W Q&12 = A1 F12 σ T14 − T24 = 0.07069 m 2 (0.2864) 2 4 m ⋅K In the second part of the problem the sides are aluminum foil, which is assumed to be a reradiating surface. A reradiating surface is one in which there is no heat transfer except by radiation and the net radiation heat transfer is zero. According to the basic equation for a gray- & = A ε (E − J ) (1 − ε ) ; so if Q& = 0 then Ji = Ebi. That is a reradiating diffuse surface, Q i i i bi i i i surface has the same radiosity as a black surface, σT4. For this problem, where the other two surfaces are black, we have a heat transfer between three black bodies, one of which has no net heat transfer. Considering this new surface, surface 3,we have the following expression for the zero net heat transfer to this surface. Q& 3 = 0 = Q&13 + Q& 23 = A1 F13 (Eb1 − Eb3 ) + A2 F23 (Eb 2 − Eb3 ) = 0 We can solve this equation for the emissive power of the reradiating surface, Eb3. Eb 3 = A1 F13 Eb1 + A2 F23 Eb 2 =0 A1 F13 + A2 F23 The two new view factors, F13 and F23 introduced here must be the same because of symmetry. From the summation rule, F12 + F13 = 1 so F13= 1 – F12 = 1 – 0.2864 = 0.7136. The black-body emissive powers of surfaces 1 and 2 are Eb1 = σT14 = 5.670 x10 −8 W m ⋅K 2 4 (950 K )4 = 461832 W m Eb 2 = σT24 = 5.670 x10 −8 W m ⋅K 2 4 W (298 K )4 = 444.14 2 m 2 We use these emissive powers with F13 = F23 = 0.7136 and A1 = A2 = 0.07069 m to find Eb3. W 447.14 W (0.07069 m )(0.7136) 46183 + (0.07069 m )(0.7136) 23315 W m m = = (0.07069 m )(0.7136) + (0.07069 m )(0.7136) m 2 2 2 Eb 3 2 2 2 2 We can now compute the net heat transfer to the steaks (surface 2). We use the reciprocity relation A3F32 = A2F23 in getting the final expression below. Q& 2 = Q&12 + Q& 32 = A1 F12 (Eb1 − Eb 2 ) + A3 F32 (Eb3 − Eb 2 ) = A1 F12 (Eb1 − Eb 2 ) + A2 F23 (Eb3 − Eb 2 ) ( ) ( ) ⎛ 46183 W 447.14 W ⎞ ⎛ 23315 W 447.14 W ⎞ − − Q& 2 = 0.07069 m 2 (0.2864 )⎜ ⎟ + 0.07069 m 2 (0.2864 )⎜ ⎟ 2 2 m m2 ⎝ m ⎠ ⎝ m2 ⎠ Q& 2 = 2079 W 13-59 A radiation shield that has the same emissivity ε3 on both sides is placed between two large parallel plates, which are maintained at uniform temperatures of T1 = 650 K and T2 = 400 K and have emissivities of ε1 = 0.6 and ε2 = 0.9, respectively. Determine the emissivity of the radiation shield if the radiation heat transfer between the plates is to be reduced to 15 percent of that without the radiation shield The heat flux between two large parallel plates is a special case of the general equation for heat transfer in a two surface enclosure shown below. Q&12 = ( ) Eb1 − Eb 2 A1σ T14 − T24 = A 1 − ε2 1 − ε1 1 − ε 2 1 − ε1 1 1 + + + + 1 A1ε1 A1 F12 A2 ε 2 F12 A2 ε 2 ε1 ⇒ q&12 = For infinite parallel plates F12 = 1 and A1 = A2, this equation becomes (q&12 )∞ c = ( ) ( ) ( ) σ T14 − T24 1 − ε1 A 1 − ε2 1 + + 1 ε1 F12 A2 ε 2 ( ) σ T14 − T24 σ T14 − T24 σ T14 − T24 = = 1 − ε1 1 1 − ε 2 1 1 1 1 −1 +1 + −1 + −1 + +1 ε1 ε2 ε1 ε 2 ε1 ε2 1 Substituting the data for the problem gives the heat transfer for no shields. 5.670 x10 −8 W (q&12 )no shield m ⋅K 2 = 4 [(650 K ) 4 − (400 K )4 1 1 + −1 0 .6 0 .9 ] = 4877 W m2 We want a radiation shield that will reduce the heat transfer to (15%)(4877 W/m2) = 731.6 W/m2. We use equation 13-43 of the text for heat transfer with one shield. (q&12 )1 shield = ( ) σ T14 − T24 ⎞ ⎞ ⎛ 1 ⎛1 1 1 ⎜⎜ + + − 1⎟ − 1⎟⎟ + ⎜ ⎜ ⎟ ⎠ ⎝ ε 3,1 ε 3, 2 ⎝ ε1 ε 2 ⎠ In this equation ε3,1 is the emissivity of the shield facing side 1 and ε3,2 is the emissivity of the shield facing side 2. We want to determine the emissivity of the shield that produces a given heat transfer and we will assume that the emissivity of both sides of the shield is the same. Making this assumption (ε3,1 = ε3,2) and solving for the unknown emissivity of the shield gives. (q&12 )1 shield ⎛ (q&12 )1 shield ⎜⎜ 1 ⎝ ε1 + ( ( 1 2⎞ − 2 + ⎟⎟ = σ T14 − T24 ε2 ε3 ⎠ ε3 = ) ( ) σ T14 − T24 σ T14 − T24 = 1 1 1 1 1 1 2 + −1+ + −1 + −2+ ε1 ε 2 ε 3,1 ε 3, 2 ε1 ε 2 ε3 = ) (q&12 )1 shield ⇒ ( ) ⎛1 1 ⎞ 2 = σ T14 − T24 − (q&12 )1 shield ⎜⎜ + − 2 ⎟⎟ ε3 ⎝ ε1 ε 2 ⎠ 2(q&12 )1 shield ( ) ⎛1 ⎞ 1 σ T14 − T24 − (q&12 )1 shield ⎜⎜ + − 2 ⎟⎟ ⎝ ε1 ε 2 ⎠ Entering the known values into this equation gives the answer for the emissivity of the shield. 2 ε3 = 5.670 x10 −8 W m2 ⋅ K 4 [(650 K ) 4 731.6 W m2 − (400 K ) 4 ] 731.6 W ⎛ 1 1 ⎞ − + − 2⎟ ⎜ 2 ⎝ 0 .6 0 .9 ⎠ m = 0.181