9-81 Review Problems 9-99 A cold cylinder is placed horizontally in hot air The rates of heat transfer from the stack with and without wind cases are to be determined Assumptions Steady operating conditions exist Air is an ideal gas with constant properties The local atmospheric pressure is atm Properties The properties of air at atm and the film temperature of (Ts+T∞)/2 = (40+10)/2 = 25°C are (Table A-15) k = 0.02551 W/m.°C Air T∞ = 40°C ν = 1.562 × 10 −5 m /s Pr = 0.7296 Ts = 10°C D = 10 cm 1 β= = = 0.003356 K -1 Tf (25 + 273)K L = 10 m Analysis (a) When the stack is exposed to 10 m/s winds, the heat transfer will be by forced convection We have flow of air over a cylinder and the heat transfer rate is determined as follows: Re = VD ν = (10 m/s)(0.1 m) 1.562 ×10 −5 m /s = 64,020 Nu = 0.027 Re 0.805 Pr / = 0.027(64,020) 0.805 (0.7296)1 / = 179.8 h= (from Table 7-1) k 0.02551 W/m.°C Nu = (179.8) = 45.87 W/m °C D 0.1 m Q& forced conv = hA(T∞ − Ts ) = (45.87 W/m °C)(π × 0.1×10 m )(40 − 10)°C = 4323 W (b) Without wind the heat transfer will be by natural convection The characteristic length in this case is the outer diameter of the cylinder, Lc = D = 0.1 m Then, Ra = gβ (T∞ − Ts ) D ν2 Pr = (9.81 m/s )(0.003356 K -1 )(40 − 10 K )(0.1 m) ⎧ 0.387 Ra / ⎪ Nu = ⎨0.6 + ⎪⎩ + (0.559 / Pr )9 / 16 [ h= (1.562 × 10 −5 m /s) 2 ] / 27 (0.7296) = 2.953 × 10 ⎫ ⎫ ⎧ 0.387(2.953 × 10 )1 / ⎪ ⎪ ⎪ ⎬ = 19.86 ⎬ = ⎨0.6 + / 16 / 27 ⎪⎭ ⎪⎭ ⎪⎩ + (0.559 / 0.7296 ) [ ] k 0.02551 W/m.°C Nu = (19.86) = 5.066 W/m °C D m Q& nat conv = hA(T∞ − Ts ) = (5.066 W/m °C)(π × 0.1×10 m )(40 − 10)°C = 477 W PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 9-82 9-100 A spherical vessel is completely submerged in a large water-filled tank The rates of heat transfer from the vessel by natural convection, conduction, and forced convection are to be determined Assumptions Steady operating conditions exist The surface temperature is constant Properties The properties of water at the film temperature of (Ts+T∞)/2 = (30+20)/2 = 25°C are (Table A-9) ρ = 997 kg/m k = 0.607 W/m.°C μ = 0.891× 10 −3 Water T∞ = 20°C kg/m ⋅ s ν = μ / ρ = 8.937 × 10 −7 m /s Ts = 30°C D = 30 cm Pr = 6.14 β = 0.247 × 10 −3 K -1 Analysis (a) Heat transfer in this case will be by natural convection The characteristic length in this case is Lc = D = 0.3 m Then, Ra = gβ (Ts − T∞ ) D ν Nu = + Pr = (9.81 m/s )(0.247 × 10 −3 K -1 )(30 − 20 K )(0.3 m) 0.589 Ra / [1 + (0.469 / Pr ) ] / 16 / (8.937 × 10 = 2+ −7 m /s) 0.589(5.029 × 10 )1 / [1 + (0.469 / 6.14) ] / 16 / (6.14) = 5.029 × 10 = 144.8 Then k 0.607 W/m.°C Nu = (144.8) = 293.0 W/m °C D 0.3 m As = πD = π (0.3 m) = 0.2827 m h= The rate of heat transfer is Q& nat conv = hA(Ts − T∞ ) = (293.0 W/m °C)(0.2827 m )(30 − 20)°C = 828 W (b) When buoyancy force is neglected, there will be no convection currents (since β = 0) and the heat transfer will be by conduction Then Rayleigh number becomes zero (Ra = 0) The Nusselt number in this case is Nu = Then h= k 0.607 W/m.°C Nu = ( 2) = 4.047 W/m °C D m Q& cond = hA(Ts − T∞ ) = (4.047 W/m °C)(0.2827 m )(30 − 20)°C = 11.4 W (c) In this case, the heat transfer from the vessel is by forced convection The properties of water at the free stream temperature of 20°C are (Table A-9) ρ = 998 kg/m k = 0.598 W/m.°C μ ∞ = 1.002 × 10 −3 kg/m.s ν = μ ∞ / ρ = 1.004 × 10 -6 m /s μ s , @ 30°C = 0.798 × 10 −3 kg/m.s Pr = 7.01 PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 9-83 The Reynolds number is Re = VD ν = (0.2 m/s)(0.3 m) 1.004 ×10 −6 m /s = 59,760 The Nusselt number is [ ] ⎛μ hD Nu = = + 0.4 Re 0.5 + 0.06 Re / Pr 0.4 ⎜⎜ ∞ k ⎝ μs [ = + 0.4(59,760) 0.5 + 0.06(59,760) 2/3 ⎞ ⎟ ⎟ ⎠ 1/ ](7.01) 0.4 ⎛ ⎜ 1.002 × 10 −3 ⎜ 0.798 × 10 −3 ⎝ ⎞ ⎟ ⎟ ⎠ 1/ = 439.1 The heat transfer coefficient is h= k 0.598 W/m.°C Nu = ( 439.1) = 875.3 W/m °C D 0.3 m The rate of heat transfer is Q& forced conv = hA(Ts − T∞ ) = (875.3 W/m °C)(0.2827 m )(30 − 20)°C = 2474 W PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 9-84 9-101 A vertical cylindrical vessel looses heat to the surrounding air The rates of heat transfer from the vessel with and without wind cases are to be determined Assumptions Steady operating conditions exist Air is an ideal gas with constant properties The local atmospheric pressure is atm Properties The properties of air at atm and the film temperature of (Ts+T∞)/2 = (60+0)/2 = 30°C are (Table A-15) k = 0.02588 W/m.°C ν = 1.608 × 10 −5 m /s Air T∞ = 0°C Pr = 0.7282 1 β= = = 0.003300 K -1 (30 + 273)K Tf D=1m Ts = 60°C L=3m Analysis (a) When there is no wind heat transfer is by natural convection The characteristic length in this case is the height of the vessel, Lc = L = m Then, gβ (Ts − T∞ ) L3 Ra = ν Pr = (9.81 m/s )(0.003300 K -1 )(60 − K )(3 m) (1.608 × 10 −5 m /s) (0.7282) = 1.477 × 1011 We can treat this vertical cylinder as a vertical plate since 35 L 1/ 35(3) = 11 Gr (1.477 × 10 / 0.7282) The Nusselt number is determined from 1/ = 0.156 < 1.0 and thus D ≥ Nu = 0.1Ra1 / = 0.1(1.477 × 1011 )1 / = 528.6 35 L Gr / (from Table 9-1) Then h= k 0.02588 W/m.°C Nu = (528.6) = 4.560 W/m °C Lc 3m and Q& = hA(Ts − T∞ ) = (4.560 W/m °C)(π ×1× m )(60 − 0)°C = 2579 W (b) When the vessel is exposed to 20 km/h winds, the heat transfer will be by forced convection We have flow of air over a cylinder and the heat transfer rate is determined as follows: Re = VD ν = (20 ×100 / 3600 m/s)(1.0 m) Nu = 027 Re h= 1.608 ×10 805 Pr 1/ −5 m /s = 3.455 × 10 = 027 (3 455 × 10 ) 805 ( 7282 ) / = 698 (from Table 7-1) k 0.02588 W/m.°C Nu = (698.1) = 18.07 W/m °C D m Q& = hA(Ts − T∞ ) = (18.07 W/m °C)(π ×1× m )(60 − 0)°C = 10,220 W Discussion There is about four-fold increase in heat transfer due to winds PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 9-85 9-102 A solid sphere is completely submerged in a large pool of oil The rates of heat transfer from the sphere by natural convection, conduction, and forced convection are to be determined Assumptions Steady operating conditions exist The surface temperature is constant Properties The properties of oil are given in problem statement Ts = 60°C Analysis (a) For conduction heat transfer, β = Oil and Ra = Then the Nusselt number is T∞ = 20°C Nu = D = 50 cm Then k 0.22 W/m.°C h = Nu = ( 2) = 0.88 W/m °C D m Q& = hA(T − T ) = (0.88 W/m °C)[π × (0.5 m) ](60 − 20)°C = 27.6 W ∞ s cond (b) In this case, the heat transfer from the vessel is by forced convection Note that the fluid properties in this case are to be evaluated at the free stream temperature T∞ except for μs, which is evaluated at the surface temperature, Ts The Prandtl number and Reynolds number are μc p (0.010 kg/m ⋅ s)(1880 J/kg ⋅ °C) = Pr = = 85.45 k 0.22 W/m ⋅ °C Re = VDρ μ = (1.5 m/s)(0.5 m)(888 kg/m ) = 66,600 0.010 kg/m ⋅ s The Nusselt number is Nu = [ ] ⎛μ hD = + 0.4 Re 0.5 + 0.06 Re / Pr 0.4 ⎜⎜ ∞ k ⎝ μs [ ⎞ ⎟ ⎟ ⎠ 1/ ] 1/ ⎛ 0.010 ⎞ = + 0.4(66,600) 0.5 + 0.06(66,600) / (85.45) 0.4 ⎜ = 1506 ⎟ ⎝ 0.004 ⎠ The heat transfer coefficient is k 0.22 W/m.°C h = Nu = (1506) = 662.4 W/m °C D m The rate of heat transfer is Q& = hA(T − T ) = (662.4 W/m °C)[π × (0.5 m) ](60 − 20)°C = 20,810 W s forced conv ∞ (c) Heat transfer in this case will be by natural convection Note that the fluid properties in this case are to be evaluated at the film temperature of 40ºC The characteristic length in this case is Lc = D = 0.5 m Then, Ra = where Pr = gβ (Ts − T∞ ) D ν μc p k = Pr = (9.81 m/s )(0.00070 K -1 )(60 − 20 K )(0.5 m) (0.007 / 876 m /s) (65.5) = 3.522 × 1010 (0.007 kg/m ⋅ s)(1965 J/kg ⋅ °C) = 65.5 0.21 W/m ⋅ °C Then Nu = + h= 0.589 Ra / [1 + (0.469 / Pr ) ] / 16 / = 2+ 0.589(3.522 × 1010 )1 / [1 + (0.469 / 65.5) ] / 16 / = 250.4 k 0.21 W/m.°C Nu = (250.4) = 105.2 W/m °C D m Q& nat conv = hA(Ts − T∞ ) = (105.2 W/m °C)[π × (0.5 m) ](60 − 20)°C = 3304 W Discussion The heat transfer from the sphere by forced convection is much greater than that by natural convection and the heat transfer by natural convection is much greater than that by conduction PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 9-86 9-103E A small cylindrical resistor mounted on the lower part of a vertical circuit board The approximate surface temperature of the resistor is to be determined Assumptions Steady operating conditions exist Air is an ideal gas with constant properties The local atmospheric pressure is atm Radiation effects are negligible Heat transfer through the connecting wires is negligible Properties The properties of air at atm and the anticipated film temperature of (Ts+T∞)/2 = (220+120)/2 = 170°F are (Table A-15E) Air T∞ = 120°F Resistor 0.1 W D = 0.2 in k = 0.01692 Btu/h.ft.°F ν = 0.222 ×10 −3 ft /s Pr = 0.7161 1 β= = = 0.001587 R -1 (170 + 460)R Tf Q& Analysis The solution of this problem requires a trial-and-error approach since the determination of the Rayleigh number and thus the Nusselt number depends on the surface temperature which is unknown We start the solution process by “guessing” the surface temperature to be 220°F for the evaluation of the properties and h We will check the accuracy of this guess later and repeat the calculations if necessary The characteristic length in this case is the diameter of resistor, Lc = D = 0.2 in Then, Ra = gβ (Ts − T∞ ) D ν2 Pr = (32.2 ft/s )(0.001587 R -1 )(220 − 120 R )(0.2 / 12 ft ) ⎧ 0.387 Ra / ⎪ Nu = ⎨0.6 + ⎪⎩ + (0.559 / Pr )9 / 16 [ h= (0.222 × 10 −3 ft /s) 2 ⎫ ⎧ 0.387(343.8)1 / ⎪ ⎪ = + ⎬ ⎨ / 27 ⎪⎭ ⎪⎩ + (0.559 / 0.7161)9 / 16 ] [ (0.7161) = 343.8 ⎫ ⎪ = 2.105 / 27 ⎬ ⎪⎭ ] k 0.01692 Btu/h.ft.°F Nu = (2.105) = 2.138 Btu/h.ft °F D 0.2 / 12 ft As = πDL + 2πD / = π (0.2 / 12 ft )(0.3 / 12 ft ) + 2π (0.2 / 12 ft) / = 0.00175 ft and (0.1× 3.412) Btu/h Q& ⎯→ Ts = T∞ + = 120°F + = 211°F Q& = hAs (Ts − T∞ ) ⎯ hAs (2.138 Btu/h.ft °F)(0.00175 ft ) which is sufficiently close to the assumed temperature for the evaluation of properties Therefore, there is no need to repeat calculations PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 9-87 9-104 An ice chest filled with ice at 0°C is exposed to ambient air The time it will take for the ice in the chest to melt completely is to be determined for natural and forced convection cases Assumptions Steady operating conditions exist Air is an ideal gas with constant properties Heat transfer from the base of the ice chest is disregarded Radiation effects are negligible Heat transfer coefficient is the same for all surfaces considered The local atmospheric pressure is atm Properties The properties of air at atm and the anticipated film temperature of (Ts+T∞)/2 = (15+20)/2 = 17.5°C are (Table A-15) Air, T∞ = 20°C Ice chest, 0°C k = 0.02495 W/m.°C ν = 1.493 ×10 −5 m /s Pr = 0.7316 1 β= = = 0.003442 K -1 (17.5 + 273)K Tf 30 cm Q& cm Analysis The solution of this problem requires a trial-and-error approach since the determination of the Rayleigh number and thus the Nusselt number depends on the surface temperature which is unknown We start the solution process by “guessing” the surface temperature to be 15°C for the evaluation of the properties and h We will check the accuracy of this guess later and repeat the calculations if necessary The characteristic length for the side surfaces is the height of the chest, Lc = L = 0.3 m Then, Ra = gβ (T∞ − Ts ) L3 ν2 ⎧ ⎪ ⎪ ⎪ Nu = ⎨0.825 + ⎪ ⎪ ⎪⎩ h= Pr = (9.81 m/s )(0.003442 K -1 )(20 − 15 K )(0.3 m) (1.493 × 10 −5 m /s) 2 ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ 1/ 0.387 Ra ⎪ ⎪ = ⎬ ⎨0.825 + / 27 ⎡ ⎛ 0.492 ⎞ / 16 ⎤ ⎪ ⎪ ⎢1 + ⎜ ⎥ ⎪ ⎪ ⎟ ⎢⎣ ⎝ Pr ⎠ ⎥⎦ ⎪⎭ ⎪⎩ (0.7316) = 1.495 × 10 ⎫ ⎪ 1/ ⎪ 0.387(1.495 × 10 ) ⎪ = 35.15 / 27 ⎬ ⎡ ⎛ 0.492 ⎞ / 16 ⎤ ⎪ ⎢1 + ⎜ ⎥ ⎪ ⎟ ⎢⎣ ⎝ 0.7316 ⎠ ⎥⎦ ⎪⎭ k 0.02495 W/m.°C Nu = (35.15) = 2.923 W/m °C L m The heat transfer coefficient at the top surface can be determined similarly However, the top surface constitutes only about one-fourth of the heat transfer area, and thus we can use the heat transfer coefficient for the side surfaces for the top surface also for simplicity The heat transfer surface area is As = 4(0.3 m)(0.4 m) + (0.4 m)(0.4 m) = 0.64 m Then the rate of heat transfer becomes Q& = T∞ − T s , i R wall + Rconv ,o = T∞ − T s , i L + kAs hAs = (20 − 0)°C 0.03 m (0.033 W/m.°C)(0.64 m ) + = 10.23 W (2.923 W/m °C)(0.64 m ) The outer surface temperature of the ice chest is determined from Newton’s law of cooling to be Q& 10.23 W = 20°C − = 14.53°C Q& = hAs (T∞ − Ts ) → Ts = T∞ − hAs (2.923 W/m C)(0.64 m ) which is almost identical to the assumed value of 15°C used in the evaluation of properties and h Therefore, there is no need to repeat the calculations PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 9-88 The rate at which the ice will melt is Q& 10.23 × 10 −3 kJ/s = = 3.066 × 10 −5 kg/s Q& = m& hif → m& = 333.7 kJ/kg hif Therefore, the melting of the ice in the chest completely will take m = m& Δt ⎯ ⎯→ Δt = 30 kg m = = 9.786 × 10 s = 271.8 h = 11.3 days m& 3.066 × 10 −5 kg/s (b) The temperature drop across the styrofoam will be much greater in this case than that across thermal boundary layer on the surface Thus we assume outer surface temperature of the styrofoam to be 19 ° C Radiation heat transfer will be neglected The properties of air at atm and the film temperature of (Ts+T∞)/2 = (19+20)/2 = 19.5°C are (Table A-15) k = 0.0251 W/m.°C ν = 1.511× 10 −5 m /s Pr = 0.7310 β= 1 = = 0.00342 K -1 (19.5 + 273)K Tf The characteristic length in this case is the width of the chest, Lc = W =0.4 m Then, Re = VW ν = (50 ×1000 / 3600 m/s)(0.4 m) 1.511×10 −5 m /s = 367,700 which is less than critical Reynolds number ( 5× 10 ) Therefore the flow is laminar, and the Nusselt number is determined from Nu = h= hW = 0.664 Re 0.5 Pr / = 0.664(367,700) 0.5 (0.7310) / = 362.7 k k 0.0251 W/m.°C Nu = (362.7) = 22.76 W/m °C W m Then the rate of heat transfer becomes Q& = T∞ − T s , i R wall + Rconv ,o = T∞ − T s , i L + kAs hAs = (20 − 0)°C 0.03 m (0.033 W/m.°C)(0.64 m ) + = 13.43 W (22.76 W/m °C)(0.64 m ) The outer surface temperature of the ice chest is determined from Newton’s law of cooling to be Q& 13.43 W = 20°C − = 19.1°C Q& = hAs (T∞ − Ts ) → Ts = T∞ − hAs (22.76 W/m C)(0.64 m ) which is almost identical to the assumed value of 19°C used in the evaluation of properties and h Therefore, there is no need to repeat the calculations Then the rate at which the ice will melt becomes Q& 13.43 × 10 −3 kJ/s Q& = m& hif → m& = = = 4.025 × 10 −5 kg/s 333 kJ/kg hif Therefore, the melting of the ice in the chest completely will take ⎯→ Δt = m = m& Δt ⎯ m 30 = = 7.454 × 10 s = 207.05 h = 8.6 days m& 4.025 × 10 −5 PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 9-89 9-105 An electronic box is cooled internally by a fan blowing air into the enclosure The fraction of the heat lost from the outer surfaces of the electronic box is to be determined Assumptions Steady operating conditions exist Air is an ideal gas with constant properties Heat transfer from the base surface is disregarded The pressure of air inside the enclosure is atm Properties The properties of air at atm and the film temperature of (Ts+T∞)/2 = (32+15)/2 = 28.5°C are (Table A-15) k = 0.02577 W/m.°C Air −5 T =25°C ∞ ν = 1.594 × 10 m /s 200 W Pr = 0.7286 ε = 0.75 15 cm 1 Ts = 32°C -1 50 cm = = 0.003317 K β= (28.5 + 273)K Tf 50 cm Analysis Heat loss from the horizontal top surface: The characteristic length in this case is δ = Ra = gβ (Ts − T∞ ) L3c ν2 Pr = (0.5 m) A = = 0.125 m Then, P 2[(0.5 m) + (0.5 m)] (9.81 m/s )(0.003317 K -1 )(32 − 25 K )(0.125 m) (1.594 × 10 −5 m /s) (0.7286) = 1.275 × 10 Nu = 0.54Ra1 / = 0.54(1.275 × 10 )1 / = 18.15 h= k 0.02577 W/m.°C Nu = (18.15) = 3.741 W/m °C Lc 0.125 m Atop = (0.5 m) = 0.25 m and Q& top = hAtop (T s − T∞ ) = (3.741 W/m °C)(0.25 m )(32 − 25)°C = 6.55 W Heat loss from vertical side surfaces: The characteristic length in this case is the height of the box Lc = L =0.15 m Then, Ra = gβ (Ts − T∞ ) L3 ν2 Pr = (9.81 m/s )(0.003317 K -1 )(32 − 25 K )(0.15 m) (1.594 × 10 −5 m /s) 2 (0.7286) = 2.204 × 10 ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1/ ⎪ 1/ 0.387(2.204 × 10 ) 0.387 Ra ⎪ ⎪ ⎪ ⎪ Nu = ⎨0.825 + = ⎨0.825 + = 20.55 / 27 ⎬ / 27 ⎬ ⎡ ⎛ 0.492 ⎞ / 16 ⎤ ⎡ ⎛ 0.492 ⎞ / 16 ⎤ ⎪ ⎪ ⎪ ⎪ ⎢1 + ⎜ ⎢1 + ⎜ ⎪ ⎥ ⎪ ⎪ ⎥ ⎪ ⎟ ⎟ ⎢⎣ ⎝ Pr ⎠ ⎢⎣ ⎝ 0.7286 ⎠ ⎥⎦ ⎥⎦ ⎪⎩ ⎪⎭ ⎪⎩ ⎪⎭ k 0.02577 W/m.°C h = Nu = (20.55) = 3.530 W/m °C L 0.15 m Aside = 4(0.15 m)(0.5 m) = 0.3 m and Q& side = hAside (Ts − T∞ ) = (3.530 W/m °C)(0.3 m )(32 − 25)°C = 7.41 W The radiation heat loss is Q& rad = εAs σ (Ts − Tsurr ) = (0.75)(0.25 + 0.3 m )(5.67 × 10 −8 W/m K )[(32 + 273 K ) − (25 + 273 K ) ] = 17.95 W Then the fraction of the heat loss from the outer surfaces of the box is determined to be (6.55 + 7.41 + 17.95) W f = = 0.160 = 16.0% 200 W PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 9-90 9-106 A spherical tank made of stainless steel is used to store iced water The rate of heat transfer to the iced water and the amount of ice that melts during a 24-h period are to be determined Assumptions Steady operating conditions exist Air is an ideal gas with constant properties Thermal resistance of the tank is negligible The local atmospheric pressure is atm Ts = 0°C Properties The properties of air at atm and the film temperature of (Ts+T∞)/2 = (0+20)/2 = 10°C are (Table A-15) k = 0.02439 W/m.°C T∞ = 20°C ν = 1.426 ×10 −5 m /s Pr = 0.7336 1 β= = = 0.003534 K -1 (10 + 273)K Tf Iced water Di = m 0°C Q& 1.5 cm Analysis (a) The characteristic length in this case is Lc = Do = 6.03 m Then, Ra = gβ (T∞ − Ts ) D o3 ν2 Nu = + h= Pr = (9.81 m/s )(0.003534 K -1 )(20 − K )(6.03 m) 0.589 Ra / [1 + (0.469 / Pr ) ] / 16 / (1.426 × 10 −5 m /s) = 2+ 0.589(5.485 × 1011 )1 / [1 + (0.469 / 0.7336) ] / 16 / (0.7336) = 5.485 × 1011 = 394.5 k 0.02439 W/m.°C Nu = (394.5) = 1.596 W/m °C Do 6.03 m As = πDo2 = π (6.03 m) = 114.2 m and Q& = hAs (T∞ − Ts ) = (1.596 W/m °C)(114.2 m )(20 − 0)°C = 3646 W Heat transfer by radiation and the total rate of heat transfer are Q& rad = εAs σ (T s − Tsurr ) = (1)(114.2 m )(5.67 × 10 −8 W/m K )[(20 + 273 K ) − (0 + 273 K ) ] = 11,759 W Q& total = 3646 + 11,759 = 15,404 W ≅ 15.4 kW (b) The total amount of heat transfer during a 24-hour period is Q = Q& Δt = (15.4 kJ/s)(24 h/day × 3600 s/h ) = 1.331×10 kJ/day Then the amount of ice that melts during this period becomes Q = mhif ⎯ ⎯→ m = Q 1.331 × 10 kJ = = 3988 kg 333.7 kJ/kg hif PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 9-105 9-117 A hot part of the vertical front section of a natural gas furnace in a plant is considered The rate of heat loss from this section and the annual cost of this heat loss are to be determined Assumptions Steady operating conditions exist Air is an ideal gas with constant properties The local atmospheric pressure is atm Any heat transfer from other surfaces of the tank is disregarded Properties The properties of air at atm and the film temperature of (Ts+T∞)/2 = (110+25)/2 = 67.5°C are (Table A-15) k = 0.02863 W/m.°C ν = 1.97 × 10 −5 m /s Pr = 0.7183 1 β= = = 0.002937 K -1 Tf (67.5 + 273)K Plate on furnace 1.5 m × 1.5 Room 25°C ε = 0.7 Ts = 110°C Analysis The characteristic length in this case is the height of that section of furnace, Lc = L = 1.5 m Then, Ra = gβ (Ts − T∞ ) L3 ν2 ⎧ ⎪ ⎪ ⎪ Nu = ⎨0.825 + ⎪ ⎪ ⎪⎩ h= Pr = (9.81 m/s )(0.002937 K -1 )(110 − 25 K )(1.5 m) (1.97 × 10 −5 m /s) 2 ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ 1/ 0.387Ra ⎪ ⎪ = ⎨0.825 + / 27 ⎬ ⎪ ⎪ ⎡ ⎛ 0.492 ⎞ / 16 ⎤ ⎪ ⎪ ⎥ ⎢1 + ⎜ ⎟ ⎪⎩ ⎥⎦ ⎪⎭ ⎢⎣ ⎝ Pr ⎠ (0.7183) = 1.530 × 1010 ⎫ ⎪ 10 / ⎪ 0.387(1.530 × 10 ) ⎪ = 289.1 / 27 ⎬ ⎪ ⎡ ⎛ 0.492 ⎞ / 16 ⎤ ⎪ ⎥ ⎢1 + ⎜ ⎟ ⎥⎦ ⎪⎭ ⎢⎣ ⎝ 0.7183 ⎠ k 0.02863 W/m.°C Nu = (289.1) = 5.518 W/m °C L 1.5 m As = (1 m)(1.5 m) = 1.5 m and Q& = hAs (Ts − T∞ ) = (5.518 W/m °C)(1.5 m )(110 − 25)°C = 703.5 W The radiation heat loss is Q& rad = εAs σ (Tsurr − Ts ) = (0.7)(1.5 m )(5.67 × 10 −8 W/m K )[(110 + 273 K ) − (25 + 273 K ) ] = 812 W Q& total = 703.5 + 812 = 1515 W The amount and cost of natural gas used to overcome this heat loss per year is Q& (1.515 kJ/s) Q gas = Q& gas Δt = total Δt = (310 days/yr × 10 hr/day × 3600 s/hr) = 2.14 × 10 kJ 0.79 0.79 Cost = (2.14 × 10 / 105,500 therm)($1.20/therm) = $243 PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 9-106 9-118 A group of 25 transistors are cooled by attaching them to a square aluminum plate and mounting the plate on the wall of a room The required size of the plate to limit the surface temperature to 50°C is to be determined Assumptions Steady operating conditions exist Air is an ideal gas with constant properties The local atmospheric pressure is atm Any heat transfer from the back side of the plate is negligible Properties The properties of air at atm and the film temperature of (Ts+T∞)/2 = (50+30)/2 = 40°C are (Table A-15) Plate L×L k = 0.02662 W/m.°C Transistors, 25×1.5 W ν = 1.702 × 10 −5 m /s Pr = 0.7255 β= 1 = = 0.003195 K -1 Tf (40 + 273)K Room 30°C ε = 0.9 Ts = 50°C Analysis The Rayleigh number can be determined in terms of the characteristic length (length of the plate) to be Ra = gβ (T∞ − Ts ) L3c ν2 Pr = (9.81 m/s )(0.003195 K -1 )(50 − 30 K )( L) (1.702 × 10 −5 m /s) (0.7255) = 1.571× 10 L3 The Nusselt number relation is ⎧ ⎪ ⎪ ⎪ Nu = ⎨0.825 + ⎪ ⎪ ⎪⎩ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ 1/ ⎪ 1/ 0.387(1.571× 10 L ) ⎪ 0.387Ra ⎪ ⎪ = ⎨0.825 + / 27 ⎬ / 27 ⎬ ⎪ ⎪ ⎪ ⎡ ⎛ 0.492 ⎞ / 16 ⎤ ⎡ ⎛ 0.492 ⎞ / 16 ⎤ ⎪ ⎪ ⎥ ⎪ ⎢1 + ⎜ ⎥ ⎢1 + ⎜ ⎟ ⎟ ⎥⎦ ⎪⎭ ⎢⎣ ⎝ 0.7255 ⎠ ⎪⎩ ⎥⎦ ⎪⎭ ⎢⎣ ⎝ Pr ⎠ The heat transfer coefficient is k 0.02662 W/m.°C Nu = Nu L L As = L2 h= Noting that both the surface and surrounding temperatures are known, the rate of convection and radiation heat transfer are expressed as 0.02662 W/m.°C Q& conv = hAs (Ts − T∞ ) = NuL2 (50 − 30)°C L Q& rad = εAs σ (Ts − Tsky ) = (0.9) L2 (5.67 ×10 −8 W/m K )[(50 + 273) − (30 + 273) ]K = 125.3L2 The rate of total heat transfer is expressed as Q& = Q& + Q& total conv rad 0.02662 W/m.°C 25 × (1.5 W) = NuL2 (50 − 30)°C + 125.3L2 L Substituting Nusselt number expression above into this equation and solving for L, the length of the plate is determined to be L = 0.426 m PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 9-107 9-119 A group of 25 transistors are cooled by attaching them to a square aluminum plate and positioning the plate horizontally in a room The required size of the plate to limit the surface temperature to 50°C is to be determined for two cases Assumptions Steady operating conditions exist Air is an ideal gas with constant properties The local atmospheric pressure is atm Any heat transfer from the back side of the plate is negligible Properties The properties of air at atm and the film temperature of (Ts+T∞)/2 = (50+30)/2 = 40°C are (Table A-15) k = 0.02662 W/m.°C Plate Transistors, L×L ν = 1.702 × 10 −5 m /s 25×1.5 W Pr = 0.7255 β= 1 = = 0.003195 K -1 Tf (40 + 273)K Room 30°C ε = 0.9 Ts = 50°C Analysis The characteristic length and the Rayleigh number for the horizontal case are determined to be A L2 L = Lc = s = p 4L Ra = gβ (T∞ − Ts ) L3c Pr = (9.81 m/s )(0.003195 K -1 )(50 − 30 K )( L / 4) −5 2 (0.7255) = 2.454 × 10 L3 ν (1.702 × 10 m /s) Noting that both the surface and surrounding temperatures are known, the rate of radiation heat transfer is determined to be Q& = εA σ (T − T ) = (0.9) L2 (5.67 ×10 −8 W/m K )[(50 + 273) − (30 + 273) ]K = 125.3L2 rad s s sky (a) Hot surface facing up: We assume Ra < 107 and thus L