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264 K= < From Eq. (1), the convection coefficient is then ( )( ) ( )( ) ( ) 4 4 elec,1 1 s sur 2 meas 2 s P A / 2 T T 9.70 W 4.60 W h 8.49 W / m K A / 2 T T 0.01 60 m K \u3b5 \u3c3 \u221e \u2212 \u2212 \u2212 = = = \u22c5 \u2212 × \u22c5 < With RaL = 2.58 × 10 5 , Eq. 9.30 yields ( )1/ 41/ 4 5 2Lk 0.0278 W / m Kh 0.054 Ra 0.54 2.58 10 8.46 W / m KL 0.04m \u22c5= = × = \u22c5 < Again, agreement between the two values of h is well within the experimental uncertainty of the measurements. COMMENTS: Because the semi-circular disks are at the same temperature, the characteristic length corresponds to that of the circular disk, L = D/4. PROBLEM 9.40 KNOWN: Horizontal, circular grill of 0.2m diameter with emissivity 0.9 is maintained at a uniform surface temperature of 130°C when ambient air and surroundings are at 24°C. FIND: Electrical power required to maintain grill at prescribed surface temperature. SCHEMATIC: ASSUMPTIONS: (1) Room air is quiescent, (2) Surroundings are large compared to grill surface. PROPERTIES: Table A-4, Air (Tf = (T¥ + Ts)/2 = (24 + 130)°C/2 = 350K, 1 atm): n = 20.92 ´ 10 - 6 m 2 /s, k = 0.030 W/m×K, a = 29.9 ´ 10-6 m2/s, b = 1/Tf. ANALYSIS: The heat loss from the grill is due to free convection with the ambient air and to radiation exchange with the surroundings. ( ) ( )4 4q A h T T T T .s s s suresé ù= - + -¥ê úë û (1) Calculate RaL from Eq. 9.25, ( ) 3Ra g T T L /L s cb na= - ¥ where for a horizontal disc from Eq. 9.29, Lc = As/P = (pD 2 /4)/pD = D/4. Substituting numerical values, find ( ) ( ) ( )329.8m/s 1/350K 130 24 K 0.25m/4 6Ra 1.158 10 .L 6 2 6 220.92 10 m / s 29.9 10 m / s - = = ´- -´ ´ ´ Since the grill is an upper surface heated, Eq. 9.30 is the appropriate correlation, ( )1 / 41/4 6Nu h L / k 0.54Ra 0.54 1.158 10 17.72L cL L= = = ´ = ( ) 2h Nu k / L 17.72 0.030W/m K / 0.25m/4 8.50W/m K.L cL= = ´ × = × (2) Substituting from Eq. (2) for h into Eq. (1), the heat loss or required electrical power, qelec, is ( ) ( ) ( ) ( )( )W W8 42 4 4q 0.25m 8.50 130 24 K 0.9 5.67 10 130 273 24 273 K2 2 44 m K m K p -= - + ´ ´ + - + × × é ù ê ú ë û q 44.2W 46.0W 90.2W.= + = < COMMENTS: Note that for this situation, free convection and radiation modes are of equal importance. If the grill were highly polished such that e » 0.1, the required power would be reduced by nearly 50%. PROBLEM 9.41 KNOWN: Plate dimensions and maximum allowable temperature. Freestream temperature. FIND: Maximum allowable power dissipation. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible heat loss from sides and bottom, (4) Isothermal plate. PROPERTIES: Table A-4, Air (Tf = 325K, 1 atm): n = 18.4 ´ 10 -6 m2/s, k = 0.028 W/m×K, a = 26.2 ´ 10-6 m2/s. ANALYSIS: The power dissipated by convection is ( )P q hA T T .elec s s= = - ¥ With ( ) ( )2L A / P 1.2m / 4 1.2m 0.3ms= = = ( ) ( ) ( ) ( ) ( ) ( ) 1 33 2g T T L 9.8m/s 325K 50K 0.3msRaL 6 2 6 218.4 10 m / s 26.2 10 m / s b na -- ¥= = - -´ ´ 7Ra 8.44 10 .L = ´ With the upper surface heated, Eq. 9.31 yields hL 1 / 3Nu 0.15Ra 65.8L Lk = = = 0.028W/m K 2h 65.8 6.14W/m K 0.3m ×= = × and the power dissipated is ( ) ( )22q 6.14W/m K 1.2m 50K= × P q 442W.elec = = < COMMENTS: This result corresponds to an average surface heat flux of 442 W/1.44 m 2 = 307 W/m 2 = 0.03 W/cm 2 , which is extremely small. Heat dissipation by free convection in this manner is a poor option compared to the heat flux with forced convection (u¥ = 15 m/s) of 0.15 W/cm 2. PROBLEM 9.42 KNOWN: Material properties, inner surface temperature and dimensions of roof of refrigerated truck compartment. Solar irradiation and ambient temperature. FIND: Outer surface temperature of roof and rate of heat transfer to compartment. SCHEMATIC: ASSUMPTIONS: (1) Negligible irradiation from the sky, (2) Ts,o > T\u221e (hot surface facing upward) and RaL > 10 7 , (3) Constant properties. PROPERTIES: Table A-4, air (p = 1 atm, Tf \u2248 310K): \u3bd = 16.9 × 10-6 m2/s, k = 0.0270 W/m\u22c5K, Pr = 0.706, \u3b1 = \u3bd/Pr = 23.9 × 10-6 m2/s, \u3b2 = 0.00323 K-1. ANALYSIS: From an energy balance for the outer surface, s,o s,i S S conv cond tot T T G q E q R \u3b1 \u2212 \u2032\u2032 \u2032\u2032 \u2212 \u2212 = = \u2032\u2032 ( ) s,o s,i4S S s,o s,o p i T T G h T T T 2R R \u3b1 \u3b5\u3c3 \u221e \u2212 \u2212 \u2212 \u2212 = \u2032\u2032 \u2032\u2032+ where ( ) ( )5 2 2p 1 p i 2 iR t / k 2.78 10 m K / W and R t / k 1.923 m K / W.\u2212\u2032\u2032 \u2032\u2032= = × \u22c5 = = \u22c5 For a hot surface facing upward and ( ) 3 7L s,oRa g T T L / 10 , h\u3b2 \u3b1\u3bd\u221e= \u2212 > is obtained from Eq. 9.31. Hence, with cancellation of L, ( ) 1/ 32 1 1/ 31/ 3 L s,o12 4 2 k 9.8 m / s 0.00323 K h 0.15 Ra 0.15 0.0270 W / m K T T L 16.9 23.9 10 m / s \u2212 \u221e \u2212 × = = × \u22c5 \u2212 × × \uf8eb \uf8f6\uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8 ( )4 / 32 4 / 3 s,o1.73 W / m K T 305 K= \u22c5 \u2212 Hence, ( ) ( ) ( ) s,o2 2 4 / 3 8 2 4 44 / 3 s,o s,o 5 2 T 263K 0.5 750 W / m K 1.73 W / m K T 305 0.5 5.67 10 W / m K T 5.56 10 1.923 m K / W \u2212 \u2212 \u2212 \u22c5 \u2212 \u22c5 \u2212 \u2212 × × \u22c5 = × + \u22c5 Solving, we obtain s,oT 318.3K 45.3 C= = ° < Hence, the heat load is ( ) ( ) ( )t cond 2 45.3 10 C q W L q 3.5m 10m 1007 W 1.923m K / W + ° \u2032\u2032= \u22c5 = × = \u22c5 < COMMENTS: (1) The thermal resistance of the aluminum panels is negligible compared to that of the insulation. (2) The value of the convection coefficient is ( )1/ 3 2s,oh 1.73 T T 4.10 W / m K.\u221e= \u2212 = \u22c5 PROBLEM 9.43 KNOWN: Inner surface temperature and composition of a furnace roof. Emissivity of outer surface and temperature of surroundings. FIND: (a) Heat loss through roof with no insulation, (b) Heat loss with insulation and inner surface temperature of insulation, and (c) Thickness of fire clay brick which would reduce the insulation temperature, Tins,i, to 1350 K. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through the composite wall, (3) Negligible contact resistance, (4) Constant properties. PROPERTIES: Table A-4, Air (Tf \u2248 400 K, 1 atm): k = 0.0338 W/m\u22c5K, \u3bd = 26.4 × 10-6 m2/s, \u3b1 = 38.3 × 10-6 m2/s, Pr = 0.69, \u3b2 = (400 K)-1 = 0.0025 K-1; Table A-1, Steel 1010 (600 K): k = 48.8 W/m\u22c5K; Table A-3 Alumina-Silica blanket (64 kg/m3, 750 K): k = 0.125 W/m\u22c5K; Table A-3, Fire clay brick (1478 K): k = 1.8 W/m\u22c5K. ANALYSIS: (a) Without the insulation, the thermal circuit is Performing an energy balance at the outer surface, it follows that cond conv radq q q= + ( ) ( )s,i s,o 4 4s,o s,o sur 1 1 3 3 T T hA T T A T T L k A L k A \u3b5\u3c3 \u221e \u2212 = \u2212 + \u2212 + (1,2) where the radiation term is evaluated from Eq. 1.7. The characteristic length associated with free convection from the roof is, from Eq. 9.29 2sL A P 16m 16 m 1m= = = . From Eq. 9.25, with an assumed value for the film temperature, Tf = 400 K, ( ) ( )( )( ) ( )3123 s,os,o 7L s,o6 2 6 29.8m s 0.0025K T T l mg T T LRa 2.42 10 T T26.4 10 m s 38.3 10 m s \u3b2 \u3bd\u3b1 \u2212 \u221e \u221e \u221e \u2212 \u2212 \u2212 \u2212 = = = × \u2212 × × × Hence, from Eq. 9.31 ( ) ( )1/ 3 1/ 31/ 3 7L s,ok 0.0338 W m Kh 0.15Ra 0.15 2.42 10 T TL 1 m \u221e\u22c5= = × \u2212 = \u2212 \u22c5\u221e1 1 3 2.47 , /T T W m Ks o\ufffd \ufffd .(3) Continued... PROBLEM 9.43 (Cont.) The energy balance can now be written ( ) ( ) ( ) 4 /3s,o s,o 1700 T K 1.47 T 298K 0.08m 1.8 W m K 0.005m 48.8W m K \u2212 = \u2212 \u22c5 + \u22c5 ( )4 48 2 4s,o0.3 5.67 10 W m K T 298K\u2212 \uf8ee \uf8f9+ × × \u22c5 \u2212\uf8ef \uf8fa\uf8f0 \uf8fb and from iteration, find Ts,o \u2248 895 K. Hence, ( ) ( ) ( ){ }4 / 3 4 42 2 8 2 4q 16m 1.47 895 298 W m 0.3 5.67 10 W m K 895K 298 K\u2212= \u2212 + × × \u22c5 \u2212\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb { }2 2 5q 16m 7,389 10,780 W m 2.91 10 W= + = × . < (b) With the insulation, an additional conduction resistance is provided and the energy balance at the outer surface becomes ( ) ( )s,i s,o 4 4s,o s,o sur 1 1 2 2 3 3 T T hA T T A T T L k A L k A L k A \u3b5\u3c3