Radiation Heat Transfer

Solutions to Exercise Twelve –Radiation Heat Transfer 1. A furnace is of cylindrical shape with R = H = 2 m. The base, top, and side surfaces are all black and are maintained at uniform temperatures of 500 K, 700 K, and 1400 K, respectively. Determine the net rate of radiation heat transfer to or from the top surface during steady operation. (Problem and figure P13-30 from Çengel, Heat and Mass Transfer.) Since all the surfaces are black, the heat transfer from the top surface (surface 1) is given by the following equation. ( ) ( Q&1 = Q&12 + Q&13 = A1 F12σ T14 − T24 + A1 F13σ T14 − T34 ) We have all the data required except for the view factors. We can find the view factor from surface 1 to surface 2 from Figure 13.7 with L/r1 = H/R = (2 m)/(2 m) = 1and r2/L = (2 m)/(2 m) = 1. I read a value of F12 = 0.38 from Figure 13-7. Alternatively we can find the view factor from the following equation for two coaxial, circular surfaces with radii ri and rj separated by a distance L from Table 131. 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 = 2 m and L = 2 m, so r1/L = r2/L = (2 m)/(2 m) =1. This gives 1 + (r2 L ) 1 + (1) = 1+ =3 2 (1)2 (r1 L ) 2 2 ⎛ ⎛ r2 ⎞ ⎞⎟ 1 ⎛⎜ ⎛ 2 m ⎞ ⎞⎟ 1⎜ 2 2 ⎟ = 0.3820 F12 = ⎜ S − S − 4⎜⎜ ⎟⎟ ⎟ = ⎜ 3 − 3 − 4⎜⎜ r1 ⎠ ⎟ 2 ⎜ 2⎜ 2 m ⎟⎠ ⎟⎟ ⎝ ⎝ ⎝ ⎠ ⎝ ⎠ 2 2 S = 1+ We can find F13 from the summation rule: F11 + F12 +F13 = 1. Since surface 1 is flat, F11 = 0 and we have F13 = 1 – 0 – 0.3820 = 0.6280. We can now solve for the heat transfer from surface 1 whose area is πR2 = π(2 m)2 = 12.566 m2. [ ( ) ( )] Q&1 = A1σ F12 T14 −T24 + F13 T14 −T34 = −8 2 5.670x10 W (0.3820)(700K)4 −(500K)4 −(0.6280)(700K)4 −(1400K)4 12.566m 2 4 m ⋅K ( ) { [ Q&1 = –1.538x106 W ] [ ]} Page 2 2. Consider a 20-cm diameter hemispherical enclosure. The dome is maintained at 600 K and heat is supplied from the base at a rate of 50 W while the base surface with an emissivity of 0.55 is maintained at 400 K. Determine the emissivity of the dome. This is a two surface problem for which we can find the heat transfer from the following equation. Q&12 = ( ) Eb1 − Eb 2 A1σ T14 − T24 = 1 − ε1 1 − ε 2 1 − ε1 A 1− ε 2 1 1 + + + + 1 A1ε 1 A1 F12 A2ε 2 F12 A2 ε 2 ε1 Define surface 1 as the flat surface (the base) and surface 2 as the hemispherical dome. Since the bottom surface is flat, all its radiation reaches the hemispherical dome so F12 = 1. The area of the base, A1 = πD2/4 = π(0.2 m)2/4 = 0.03142 m2. The area of the hemisphere is half the area of a sphere: A2 = πD2/2 = π(0.2 m)2/2 = 0.06283 m2 From the formulas we see that the area ratio A1/A2 = 1/2. Substituting F12 = 1 and A1/A2 = 1/2 into our heat transfer equation and solving for e2 gives. Q& 12 = 1 ε1 + ( ( ) ( ) ( ) A1σ T14 − T24 A1σ T14 − T24 A1σ T14 − T24 = = A 1− ε 2 1 − ε1 1 ⎞ 1 + 1 −1 1 1 1⎛ 1 + + 1 − 1 + + ⎜⎜ − 1⎟⎟ F12 A2 ε 2 ε1 ε1 1 2 ⎝ε2 ⎠ ε 1 2ε 2 2 1 1 Aσ T −T − = 1 2ε 2 2 Q&12 4 1 4 2 ) ( Aσ T −T 1 = 1 2ε 2 Q&12 ⇒ 4 1 4 2 )− 1 + 1 ε1 2 1 2 ⇒ ε2 = 4 A1σ T1 − T24 1 1 − + & ε1 2 Q12 ( ) Substituting the given data for all the terms on the right side of this equation gives the desired emissivity. 1 1 2 2 = ε2 = −8 4 4 A1σ T1 − T2 5.670 x10 W 1 1 (400 K )4 − (600 K )4 0.03142 m 2 − + 2 4 & 1 1 m ⋅K ε1 2 Q12 − + − 50 W 0.55 2 ( ) ( ) ε2 = 0.209 [ ] Page 3 3. Consider a triangular enclosure that has the two rectangular surfaces shown at the right. These two surfaces are perpendicular to each other with a common edge which is 1.6 m long. The horizontal surface is 0.8 m wide and the vertical surface is 1.2 m high. The horizontal surface has an emissivity of 0.75 and is maintained at 400 K. The vertical surface is black and is maintained at 550 K. The back sides of the surfaces are insulated. The remaining three sides of the triangular enclosure (one flat surface that joins the outer edges of the horizontal and vertical surfaces shown and two triangular surfaces at the ends) are at 290 K and can be considered to have an emissivity of 0.85. Determine the net rate of radiation heat transfers between the two surfaces, and between the horizontal surface and the surroundings. (Problem and figure P13-45 from Çengel, Heat and Mass Transfer.) The equations for a three surface enclosure are taken from slides 29 and 30 of the May 2 lecture presentation. ⎛ 1 − ε1 ⎞ 1 − ε1 1 − ε1 1 − ε1 ⎜⎜1 + F12 + F13 ⎟⎟ J1 − F12 J 2 − F13 J 3 = σT14 ε1 ε1 ε1 ε1 ⎝ ⎠ − ⎛ 1 − ε2 ⎞ 1 − ε2 1− ε2 1 − ε2 F21 J1 + ⎜⎜1 + F21 + F23 ⎟⎟ J 2 − F23 J 3 = σT24 ε2 ε ε ε 2 2 2 ⎝ ⎠ − ⎛ 1 − ε3 ⎞ 1 − ε3 1 − ε3 1 − ε3 F31 J1 − F32 J 2 + ⎜⎜1 + F31 + F32 ⎟⎟ J 3 = σT34 ε3 ε3 ε3 ε3 ⎝ ⎠ Because surface 2 is a black body, e2 = 1, and the second equation reduces to J2 = σT24, and, because we know T2, our system of three equations in three unknowns becomes a system of two equations in two unknowns. ⎛ 1 − ε1 ⎞ 1 − ε1 1 − ε1 1 − ε1 ⎜⎜1 + F12 + F13 ⎟⎟ J1 − F13 J 3 = σT14 + F12 σT24 ε1 ε1 ε1 ε1 ⎝ ⎠ − ⎛ 1 − ε3 ⎞ 1 − ε3 1 − ε3 1 − ε3 F31 J1 + ⎜⎜1 + F31 + F32 ⎟⎟ J 3 = σT34 + F32 σT24 ε3 ε3 ε3 ε3 ⎝ ⎠ The areas of the two flat surfaces in the enclosure are A1 = L1W = (0.8 m) (1.6 m) = 1.28 m2 and A2 = L2W = (1.2 m) (1.6 m) = 1.92 m2; the area of the third radiation surface is the area of a single diagonal surface, W(L12 + L22)1/2 plus the area two triangles enclosing the ends: 2[(1/2)L1L2]. This third surface area is thus (1.6 m)[(0.8 m)2 +(1.2 m)2]1/2 + (0.8 m)(1.2 m) = 3.268 m2. We have to find four view factors F12, F13, F31, and F32. We can find F12 from the view factor graph in Figure 13-6. For this problem we have L1/W = (0.8 m) / (1.6 m) = 0.5 and L2/W = (1.2 m) / (1.6 m) = 0.75; for these values I read a view factor of 0.27 from Figure 13-6. Using the equation in Table 13-1, the view factor is found to be F12 = 0.2749, which will be used in subsequent calculations. We can find F13 from the summation rule: F11 + F12 + F13 = 1, where F11 = 0 because surface 1 is a flat surface. This gives F13 = 1 – 0 – 0.2749 = 0.7251. Page 4 We can use the reciprocity rule to find F31: A3F31 = A1F13; so, F31 = A1F13/A3 = (1.28 m2)(0.7251)/ (3.268 m2) = 0.2840. We still need F32, but we cannot use the summation rule to find it. Surface 3 is not a flat surface so we cannot say that F33 = 0. If we knew F23 we could use the reciprocity rule to find F32. We can find F23 from the summation rule for surface 2, where F22 = 0: F23 = 1 – F21 – F22 = 1 – F21. And we can use the reciprocity rule to find F21= A1F12/A2 = (1.28 m2)(0.2749)/ (1.92 m2) = 0.1833. So we have F23 = 1 – F21 = 1 – 0.1833 = 0.8167, and F32 = A2F23/A3 = (1.92 m2)(0.8167)/ (3.268 m2) = 0.4799. The emissivity factors in the equations that we have to solve are (1 – ε1)/ ε1 = (1 – 0.75)/0.75 = 1/3 and (1 – ε3)/ ε3 = (1 – 0.85)/0.85 = 0.1765. We can now compute the coefficients and righthand-side terms in our system of two equations. We start by computing the black body emissive power for the three surfaces. Eb1 = σT14 = 5.670 x10 −8 W m2 ⋅ K 4 (400 K )4 = 14522W m E b3 = σT34 = Eb 2 = σT24 = 5.670 x10 −8 W m2 ⋅ K 4 5.670 x10 −8 W m2 ⋅ K 4 (550 K )4 = 51882W m (290 K )4 = 401.02 W m Using these emissive powers we can compute the right-hand side terms of our equations. σT14 + 1 − ε1 1451 W 1 5188 W 1927 W σT24 = + = ε1 3 m2 m2 m2 σT34 + 1 − ε3 401.0 W 5188 W 840.4 W σT24 = + (0.1765) = 2 ε3 m m2 m2 We next compute the left-hand side terms of the two equations. 1 − ε1 1 − ε1 1 (F12 + F13 ) = 1 + 1 (0.2749) + 0.2417 = 1.333 F13 = (0.7251) = −0.2417 1 + ε1 3 ε1 3 − 1 − ε3 1 − ε3 (F31 + F32 ) = 1 + 0.05013 + (0.1765)(0.4799) = 1.135 F31 = −(0.1765)(0.2840) = −0.05013 1 + ε3 ε3 We have the following system of equations to solve. 1.333J 1 − 0.2417 J 3 = 1927 W − 0.005013 J 1 − 1.135 J 3 = m2 840.4 W m2 We see that the left-hand side coefficients are dimensionless and the right-hand side terms have units of W/m2 so our results for radiosity will have units of W/m2. Solving these equations gives J1 = 1592 W/m2 and J3 = 810.9 W/m2. (Recall that we previously had J2 = Eb2 = 5188 W/m2.) We can now find the desired heat transfers. The heat transfer from the vertical surface (2) to the horizontal surface (1) is ( ) ⎛ 5188 W 1592 W ⎞ Q& 21 = A2 F21 ( J 2 − J 1 ) = 1.92 m 2 (0.1833)⎜ − ⎟ = 1265 W/m2 2 2 m ⎝ m ⎠ The heat transfer from the horizontal surface (1) to the surroundings (3) is ( ) ⎛ 1592 W 810.9 W ⎞ − Q&13 = A1 F13 ( J1 − J 3 ) = 1.28 m 2 (0.1833)⎜ ⎟ = 725.1 W/m2 2 2 m ⎝ m ⎠