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Download 3-30 An exposed hot surface of an industrial natural gas furnace is
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Transcript
3-30 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss through that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to be determined. Also, the length of time it will take for the insulation to pay for itself from the energy it saves will be determined. Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2 Thermal conductivities are constant. 3 The furnace operates continuously. 4 The given heat transfer coefficient accounts for the radiation effects. Properties The thermal conductivity of the glass wool insulation is given to be k = 0.038 W/m°C. Analysis The rate of heat transfer without insulation is A (2 m)(1.5 m) 3 m2 Insulation Q hA(Ts T ) (10 W / m2 . C)(3 m2 )(80 30) C 1500 W In order to reduce heat loss by 90%, the new heat transfer rate and thermal resistance must be Q 010 . 1500 W 150 W T T (80 30) C Q Rtotal 0.333 C / W Rtotal 150 W Q Rinsulation Ro T Ts L and in order to have this thermal resistance, the thickness of insulation must be 1 L Rtotal Rconv Rinsulation hA kA 1 L 0.333 C/W 2 2 (10 W/m .C)(3 m ) (0.038 W/m. C)(3 m 2 ) L 0.034 m 3.4 cm Noting that heat is saved at a rate of 0.91500 = 1350 W and the furnace operates continuously and thus 36524 = 8760 h per year, and that the furnace efficiency is 78%, the amount of natural gas saved per year is Q t (1.350 kJ/s)(8760 h) 3600 s 1 therm Energy Saved saved 517 .4 therms Efficiency 0.78 1 h 105,500 kJ The money saved is Money saved ( Energy Saved)(Cost of energy) (517.4 therms)($0.55 / therm) $284.5 (per year) The insulation will pay for its cost of $250 in Money spent $250 Payback period 0.88 yr Money saved $284.5 / yr which is less than one year. 3-55 A wall is constructed of two layers of sheetrock spaced by 5 cm 12 cm wood studs. The space between the studs is filled with fiberglass insulation. The thermal resistance of the wall and the rate of heat transfer through the wall are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation heat transfer. Properties The thermal conductivities are given to be k = 0.17 W/m°C for sheetrock, k = 0.11 W/m°C for wood studs, and k = 0.034 W/m°C for fiberglass insulation. Analysis (a) The representative surface area is A 1 0.65 0.65 m2 . The thermal resistance network and the individual thermal resistances are Ri T1 R2 R1 R3 R4 R5 T2 1 1 0.185 C/W 2 hi A (8.3 W/m .C) (0.65 m 2 ) L 0.01 m R1 R 4 R sheetrock 0.090 C/W kA (0.17 W/m. C) (0.65 m 2 ) Ri L 0.12 m 21 .818 C/W kA (0.11 W/m. C) (0.05 m 2 ) L 0.12 m R3 R fiberglass 5.882 C/W kA (0.034 W/m. C) (0.60 m 2 ) R 2 R stud ( 1 1 0.045 C/W 2 o ho A (34 W/m . C) (0.65 m 2 ) 1 1 1 1 R mid 4.633 C/W R 2 R3 21 .818 5.882 Ro 1 R mid Rtotal Ri R1 R mid R 4 Ro 0.185 0.090 4.633 0.090 0.045 4.858 C/W (for a 1 m 0.65 m section) T T [20 (5)]C Q 1 2 5.15 W Rtotal 4.858 C/W b) Then steady rate of heat transfer through entire wall becomes (12 m) (5 m) Q total (5.15 W) 475 W 0.65 m 2 3-41C An interface acts like a very thin layer of insulation, and thus the thermal contact resistance has significance only for highly conducting materials like metals. Therefore, the thermal contact resistance can be ignored for two layers of insulation pressed against each other. 3-42C An interface acts like a very thin layer of insulation, and thus the thermal contact resistance is significant for highly conducting materials like metals. Therefore, the thermal contact resistance must be considered for two layers of metals pressed against each other. 3-48 A thin copper plate is sandwiched between two epoxy boards. The error involved in the total thermal resistance of the plate if the thermal contact conductances are ignored is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since the plate is large. 3 Thermal conductivities are constant. Properties The thermal conductivities are given to be k = 386 W/m°C for copper plates and k = 0.26 W/m°C for epoxy boards. The contact conductance at the interface of copper-epoxy layers is given to be hc = 6000 W/m2C. Analysis The thermal resistances of different layers for unit surface area of 1 m2 are Copp er 1 1 Rcontact 0.00017 C/W Epox 2 2 plate Epox hc Ac (6000 W/m .C)(1 m ) y y Rplate L 0.001 m 2.6 106 C / W kA (386 W / m. C)(1 m2 ) Repoxy L 0.005 m 0.01923 C / W kA (0.26 W / m. C)(1 m2 ) Q 5 mm 5 mm The total thermal resistance is R total 2 Rcontact R plate 2 Repoxy 2 0.00017 2.6 10 6 2 0.01923 0.03914 C/W Then the percent error involved in the total thermal resistance of the plate if the thermal contact resistances are ignored is determined to be %Error 2Rcontact 2 0.00017 100 100 0.87% Rtotal 0.03914 which is negligible. 3-68 A steam pipe covered with 3-cm thick glass wool insulation is subjected to convection on its surfaces. The rate of heat transfer per unit length and the temperature drops across the pipe and the insulation are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. Properties The thermal conductivities are given to be k = 15 W/m°C for steel and k = 0.038 W/m°C for glass wool insulation Analysis The inner and the outer surface areas of the insulated pipe per unit length are Ai Di L (0.05 m)(1 m) 0157 . m2 Ao Do L (0.055 0.06 m)(1 m) 0.361 m2 Ri The individual thermal resistances are R1 R2 T1 Ro T2 1 1 Ri 0.08 C/W 2 hi Ai (80 W/m .C) (0.157 m 2 ) ln( r2 / r1 ) ln( 2.75 / 2.5) R1 R pipe 0.00101 C/W 2k1 L 2 (15 W/m. C) (1 m) R 2 Rinsulation ln( r3 / r2 ) ln(5.75 / 2.75 ) 3.089 C/W 2k 2 L 2 (0.038 W/m. C) (1 m) 1 1 0.1847 C/W ho Ao (15 W/m 2 .C) (0.361 m 2 ) Ri R1 R 2 Ro 0.08 0.00101 3.089 0.1847 3.355 C/W Ro Rtotal Then the steady rate of heat loss from the steam per m. pipe length becomes T T (320 5) C Q 1 2 93.9 W Rtotal 3.355 C / W The temperature drops across the pipe and the insulation are Tpipe QR pipe (93.9 W)(0.00101 C / W) 0.095 C Tinsulation QR insulation (93.9 W)(3.089 C / W) 290 C 3-108 A commercially available heat sink is to be selected to keep the case temperature of a transistor below 90C in an environment at 20 C . Assumptions 1 Steady operating conditions exist. 2 The transistor case is isothermal at 90 C . 3 The contact resistance between the transistor and the heat sink is negligible. Analysis The thermal resistance between the transistor attached to the sink and the ambient air is determined to be R Ts T T T (90 20) C T Q Rcase ambient transistor 1.75 C / W Rcase ambient 40 W Q The thermal resistance of the heat sink must be below 175 . o C / W . Table 3-4 reveals that HS6071 in vertical position, HS5030 and HS6115 in both horizontal and vertical position can be selected. 3-111E The handle of a stainless steel spoon partially immersed in boiling water extends 7 in. in the air from the free surface of the water. The temperature difference across the exposed surface of the spoon handle is to be determined. Assumptions 1 The temperature of the submerged portion of the spoon is equal to the water temperature. 2 The temperature in the spoon varies in the axial direction only (along the spoon), T(x). 3 The heat transfer from the tip of the spoon is negligible. 4 The heat transfer coefficient is constant and uniform over the entire spoon surface. 5 The thermal properties of the spoon are constant. 6 The heat transfer coefficient accounts for the effect of radiation from the spoon.. Properties The thermal conductivity of the spoon is given to be k = 8.7 Btu/hft°F. Analysis Noting that the cross-sectional area of the spoon is constant and measuring x from the free surface of water, the variation of temperature along the spoon can be expressed as T ( x) T cosh a ( L x) Tb T cosh aL h, T D where p 2(0.5 / 12 ft 0.08 / 12 ft ) 0.0967 ft Tb Ac (0.5 / 12 ft)(0.08 / 12 ft ) 0.000278 ft 2 a hp kAc (3 Btu / h.ft 2 . F)(0.0967 ft ) (8.7 Btu / h.ft. F)(0.000278 ft 2 ) L = 7 in 10.95 ft -1 Noting that x = L = 7/12=0.583 ft at the tip and substituting, the tip temperature of the spoon is determined to be 0 . 00 . 8 5 i in n cosh a( L L) cosh aL cosh 0 1 = 75 F + (200 75) = 75 F + (200 75) = 75.4 F cosh(10 .95 0.583 ) 296 T ( L) T (Tb T ) Therefore, the temperature difference across the exposed section of the spoon handle is T Tb Ttip (200 75.4)F 124.6F 3-114 A circuit board houses 80 logic chips on one side, dissipating 0.04 W each through the back side of the board to the surrounding medium. The temperatures on the two sides of the circuit board are to be determined for the cases of no fins and 864 aluminum pin fins on the back surface. Assumptions 1 Steady operating conditions exist. 2 The temperature in the board and along the fins varies in one direction only (normal to the board). 3 All the heat generated in the chips is conducted across the circuit board, and is dissipated from the back side of the board. 4 Heat transfer from the fin tips is negligible. 5 The heat transfer coefficient is constant and uniform over the entire fin surface. 6 The thermal properties of the fins are constant. 7 The heat transfer coefficient accounts for the effect of radiation from the fins. Properties The thermal conductivities are given to be k = 20 W/m°C for the circuit board, k = 237 W/m°C for the aluminum plate and fins, and k = 1.8 W/m°C for the epoxy adhesive. Analysis (a) The total rate of heat transfer dissipated by the chips is Q 80 (0.04 W) 3.2 W 2 cm The individual resistances are Repoxy Rboard RAluminum Rconv T1 T2 T2 A (012 . m)(018 . m) 0.0216 m2 L 0.003 m 0.00694 C / W kA (20 W / m. C)(0.0216 m2 ) 1 1 0.9259 C / W hA (50 W / m2 . C)(0.0216 m2 ) Rboard Rconv Rtotal Rboard Rconv 0.00694 0.9259 0.93284 C / W The temperatures on the two sides of the circuit board are T T Q 1 2 T1 T2 QR total 40 C (3.2 W)(0.93284 C / W) 43.0 C Rtotal T T Q 1 2 T2 T1 QR board 43.0 C (3.2 W)(0.00694 C / W) 40.5 0.02 43.0 C Rboard Therefore, the board is nearly isothermal. (b) Noting that the cross-sectional areas of the fins are constant, the efficiency of the circular fins can be determined to be hp kAc a hD kD / 4 2 4h kD 4(50 W / m2 . C) 18.37 m-1 (237 W / m. C)(0.0025 m) tanh aL tanh(18.37 m-1 0.02 m) 0.957 aL 18.37 m-1 0.02 m The fins can be assumed to be at base temperature provided that the fin area is modified by multiplying it by 0.957 (or alternately use the method in parenthesis below). Then the various thermal resistances are fin L 0.0002 m 0.0051 C / W kA (18 . W / m. C)(0.0216 m2 ) L 0.002 m 0.00039 C / W kA (237 W / m. C)(0.0216 m2 ) Repoxy RAl Afinned fin nDL 0.957 864 (0.0025 m)(0.02 m) 0130 . m2 Aunfinned 0.0216 864 D 2 4 Atotal,with fins Afinned Aunfinned Rconv (or use Rcon v Rtotal 0.0216 864 (0.0025) 2 4 0130 . 0.017 0147 . m2 0.0174 m2 1 1 01361 . C / W hAtotal,with fins (50 W / m2 . C)(0147 . m2 ) 1 and overall NAf Qtotal 1 1 f ; you should get the same answer) hAt b At overall hAt Rboard Repoxy Raluminum Rconv 0.00694 0.0051 0.00039 01361 . 01484 . C / W Then the temperatures on the two sides of the circuit board becomes T T Q 1 2 T1 T2 QR . C / W) 40.5 C total 40 C (3.2 W)( 01484 Rtotal T T Q 1 2 T2 T1 QR board 40.5 C (3.2 W)(0.00694 C / W) 40.5 0.02 40.5 C Rboard 3-124 Hot and cold water pipes run parallel to each other in a thick concrete layer. The rate of heat transfer between the pipes is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the concrete is constant. T1 = 60C Properties The thermal conductivity of concrete is given to be k = 0.75 W/m°C. T2 = 15C Analysis The shape factor for this configuration is given in Table 3-5 to be S 2L 4 z D1 2 D 2 2 cosh 1 2 D D 1 2 2(8 m) 2 z = 40 cm 9.078 m 4(0.4 m) 2 (0.05 m) 2 (0.05 m) 2 cosh 2(0.05 m)( 0.05 m) Then the steady rate of heat transfer between the pipes becomes Q Sk (T T ) (9.078 m)(0.75 W/m.C)(60 15)C 306 W 1 1 2 D = 5 cm L=8m