ME 335 Exam II Department of Mechanical Engineering | AT 300, Exams of Health sciences

Download ME 335 Exam II Department of Mechanical Engineering | AT 300 and more Exams Health sciences in PDF only on Docsity! ME 335 (HEAT TRANSFER), FALL 2008, EXAM II OPEN BOOK, 2 SHEETS OF HAND WRITTEN NOTES Department of Mechanical Engineering, University of Michigan Name: Pledge: PROBLEM 1 (32%) An indentation surface 1 shown in Figure Pr.1, is used to dissipate heat by radiation only. The inden- tation has sides a and depth L and has an emissivity of r,1. The surface temperature is T1 and that of the surrounding is T∞. (a) Draw the thermal circuit diagram. (b) Using the specifications given below, calculate the heat flow rate for the case of L = 0. (c) Calculate the heat flow rate for the case of L = a. (d) Calculate the heat flow rate for the case of L = 10a. (e) Calculate the heat flow rate for the case of L = 10a and r,1 = 1. (f) With r,1 < 1, when do we have maximum possible radiation heat flow from this indentation with its base remaining a × a? a = 50 cm, r,1 = 0.5 and T1 = 300 ◦C, T∞ = −10 ◦C. Figure Pr.1 L T1 r,1 T∞ Cavity Surrounding ∋ aa 1' (b) L = 0 (c) L = a (d) L = 10a (e) L = 10a , = 1 r,1 ∋ Ar,1' F1'-1 T1 Eb,1 (Rr, )1 (qr,o)1 (qr,o)∞ = Eb,∞ (Rr,F)1'-1 = Qr,1-∞= Qr,1-1' ∋ 1 Ar,1F1-1' = 1 (qr,o)1'= T∞ (a) (b) = T1′ r,1′ = 1 ∋ r,1′ = a2 A T1′ = T∞ SOLUTION: (a) The thermal circuit diagram for the indentation is shown in Figure Pr.1(b). (b) We note that Qr,1-∞ is the same as Qr,1-1′ , withe T1′ = T∞, r,1′ = 1 (blackbody), and Ar,1′ = a2. The radiation heat transfer is given by Eq. (4.47) as Qr,1-∞ = Qr,1-1′ = Eb,1 − Eb,2 Rr,Σ = Eb,1 − Eb,2 (1−rArr )1 + 1 Ar,1F1-∞ + 0 . When L = 0, we use r,1 = 0.5, Ar,1 = a× a = 0.5× 0.5 = 0.25 m2 and also T1 = 573.15 K and T∞ = 263.15 K, then the radiative heat transfer rate is Qr,1-∞ = σSB(T 41 − T 4∞) (1−rArr )1 + 1 Ar,1′F1′-∞ + 0 = 5.67 × 10−8(W/m2-K4)[573.154(K4) − 263.154(K4)] 1−0.5 0.25(m2)×0.5 + 1 0.25(m2)×1 + 0 = 730.8 W. 1 (c) For L = a, we have, Ar,1 = 5 × a × L = 5 × 0.5 × 0.5 = 1.25 m2, so Qr,1-∞ = σSB(T 41 − T 4∞) (1−rArr )1 + 1 Ar,1′F1′-1 + 0 = 5.67 × 10−8(W/m2-K4)[573.154(K4) − 263.154(K4)] 1−0.5 1.25(m2)×0.5 + 1 0.25(m2)×1 + 0 = 1, 220 W. (d) For L = 10a, we have, Ar,1 = 4 × a × L + a × a = 4 × 0.5 × 5 + 0.5 × 0.5 = 10.25 m2, so Qr,1-∞ = σSB(T 41 − T 4∞) (1−rArr )1 + 1 Ar,1′F1′-1 + 0 = 5.67 × 10−8(W/m2-K4)[573.154(K4) − 263.154(K4)] 1−0.5 10.25(m2)×0.5 + 1 0.25(m2)×1 + 0 = 1, 427 W. (e) For L = 10a and r,1 = 1, we have Qr,1-∞ = σSB(T 41 − T 4∞) 0 + 1Ar,1′F1′-1 + 0 = 5.67 × 10−8(W/m2-K4)[573.154(K4) − 263.154(K4)] 1 0.25(m2)×1 = 1, 462 W. When r,1 = 1, surface 1 becomes a blackbody and the surface grayness resistance is zero. However, com- pared to (d) the increase in heat transfer is small, because Ar,1 for L = 10a is sufficiently large, making the surface grayness resistance contribution in (d) very small. (f) The radiation heat transfer is rewritten as Qr,1-∞ = σSB(T 41 − T 4∞) 1 Ar,1′ [Ar,1′(1−r)Ar,1r,1 + 1 F1′-1 ] . The maximum heat transfer occurs when L → ∞, Ar,1′/Ar,1 → 0. Then radiation heat transfer becomes Qr,1-∞ = σSB(T 41 − T 4∞) 1 Ar,1′F1′-1 , for Ar,1  Ar,1′ . This is equivalent as case (e) Qr,1-∞,max = 1, 462 W. As L becomes very large, the cavity appears as a blackbody emitting maximum radiation through the opening Ar,1′ to the surrounding. COMMENT: By making L of the indentation large, we create a blackbody having a surface area of Ar,1′ = a × a which emits the maximum possible radiative heat transfer, that having the largest Ar,1-infty. 2 Then the thermobuoyant surface-convection resistance is 〈Rku〉L,1-2 = L Aku〈Nu〉Lkf = 3(m) 3 × 3(m2) × 458.8× 0.0267(W/m-K) = 0.0359 K/W. For radiation heat transfer the radiation thermal resistance between two-surface enclosure is given by (4.47) Rr,Σ = ( 1 − r Arr )1 + 1 Ar,1F1-2 + ( 1 − r Arr )2 = ( 1 − 0.9 9 × 0.9) + 1 3 × 3 × 1 + ( 1 − 0.9 5 × 9 × 0.9) = 0.126 m −2. Now, we use the thermal resistances in the energy equation, using T1 = 308.15 K, and T2 = 293.15 K and solve for Tf,∞,3, i.e., 308.15(K)− Tf,∞,3 0.00247(K/W) + 308.15(K)− 293.15(K) 0.0359(K/W) + 5.67 × 10−8[308.154(K4) − 293.154(K4)] 0.126 (m−2) = 0 308.15(K)− Tf,∞,3 0.00247(K/W) + 417.8 + 734.6 = 0 Tf,∞,3 = 311.0 K = 37.85 ◦C. COMMENT: The surface-convection resistance for forced water flow is small compared to the thermobuoyant surface- convection resistance of air. Therefore the temperature difference between water and the metallic wall is 2.85 ◦C, which is small compared to 15 ◦C for air and the wall. 5 PROBLEM 3 (34%) A square plate (a×a) housing an electrical switch producing Joule heating (Ṡe,J)1 is cooled by a parallel flow of air (uf,∞ and Tf,∞), as shown in Figure Pr.3. In order to reduce the plate temperature T1, pin (circular cross section) aluminum fins of diameter D are added to the surface. There are two fin options, with the fins made out of same total length L of aluminum fin material. (i) First option uses sixteen fins of the same length, L/16. (ii) Second option uses sixty four fins of the same length, L/64. Use the cross-flow Nusselt number for both bare and finned areas. For case (i) reduce the cross-flow Nusselt number for a single cylinder by 15% and for the case (ii) by 25%. (a) Draw a thermal circuit diagram. (b) Calculate the resulting T1 for option (i) and option (ii). (c) Comment on the difference and design selection. (Ṡe,J)1 = 25 W, a = 15 cm, D = 0.2 cm, L = 60 cm, uf,∞ = 1 m/s and Tf,∞ = 20 ◦C. Determine the air (Table C.22) properties at 300 K. Qb Qf T1 Rku, b Rku,f Tf,∞ Qku Figure Pr.3. D qk u qu Base T1 uf,∞ Tf,∞ D qu D qu D qu D qu Pin FinsPin Fins Base T1 (i) (ii) L 16 L 64 a a a a D uf,∞ Tf,∞ (Se,J)1 . (Se,J)1 . D (a) (b) SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.3(b). (b) The total surface-convection heat transfer rate for fins is given by (6.149) Qku = (Ab + Aku,fηf )〈NuD〉kf D (T1 − Tf,∞), Ab = Abase − NfAk, Aku,f = NfPku,fLc. First, we find the Nusselt number for cross-flow using the fluid properties, for air at T = 300 K given in Table C.22, νf,∞,3 = 15.66 × 10−6(m2/s) Table C.22 kf = 0.0267(W/m-K) Table C.22 Pr = 0.69 Table C.22. 6 The Reynolds number for a single fin is ReD = uf,∞D νf = 1(m/s) × 0.002(m) 15.66× 10−6(m2/s) = 127.7. The Nusselt number for cross-flow over cylinders is given in Table 6.4 as 〈Nu〉D = a1Rea2D Pr1/3. For ReD = 127.7, we have a1 = 0.683 and a2 = 0.466. Then the Nusselt number is 〈Nu〉D = 0.683Re0.466D Pr1/3 = 0.683× 127.70.466 × 0.691/3 = 5.784. The fin efficiency for a single fin is given by (6.147), i.e., ηf = tanh(mLc) mLc where m = ( Pku,f 〈Nu〉Dkf AkksD )1/2, Lc = L + D 4 . For case (i) using the thermal conductivity of aluminum from Table C.14 ks = 237 W/m-K, Lc = L + D 4 = 0.6(m) 16 + 0.002(m) 4 = 0.038 m, 〈Nu〉D,eff = 0.85〈Nu〉D = 0.85 × 5.784 = 4.916, m = ( πD〈Nu〉D,effkf AkksD )1/2 = [ π × 0.002(m)× 4.916 × 0.0267(W/m-K) π × 0.0022(m)/4 × 237(W/m-K) × 0.002(m)] 1/2 = 23.53 m−1, ηf = tanh(mLc) mLc = tanh[23.53(m−1) × 0.038(m)] [23.53(m−1) × 0.038(m)] = tanh(0.894) 0.894 = 0.7132 0.894 = 0.7978. The surface areas are Ak = πD2/4 = π × 0.0022/4 = 3.14 × 10−6 m2, Ab = Abase − NfAk = a2 − 16Ak = 0.15 × 0.15(m2) − 3.14 × 10−6(m2) = 0.02245 m2, Aku,f = NfPku,fLc = 16πDLc = 16 × π × 0.002(m)× 0.038(m) = 3.82 × 10−3 m2. Then the T1 is found using the energy equation (2.9) Qku = (Ab + Aku,fηf )〈Nu〉D,eff kf D (T1 − Tf,∞) = Ṡe,J, {0.02245(m2) + 3.82 × 10−3(m2) × 0.7978} × 4.916 × 0.0267(W/m-K) 0.002(m) [T1 − 293.15(K)] = 25 W 1.674 × [T1 − 293.15(K)] = 25 W T1 = 308.1 K = 34.94 ◦C. For case (ii)using the thermal conductivity of aluminum from Table C.14 ks = 237 W/m-K, Lc = L + D 4 = 0.6(m) 64 + 0.002(m) 4 = 0.00988 m, 〈Nu〉D,eff = 0.75〈Nu〉D = 0.75 × 5.784 = 4.338, m = ( πD〈Nu〉D,effkf AkksD )1/2 = [ π × 0.002(m) × 4.338× 0.0267(W/m-K) π × 0.0022(m)/4 × 237(W/m-K) × 0.002(m)] 1/2 = 22.11 m−1, ηf = tanh(mLc) mLc = tanh[22.11(m−1) × 0.00988(m)] [22.11(m−1) × 0.00988(m)] = tanh(0.218) 0.218 = 0.215 0.218 = 0.9844. 7