3-23 Thermal Contact Resistance 3-40C The resistance that an interface offers to heat transfer per unit interface area is called thermal contact resistance, Rc The inverse of thermal contact resistance is called the thermal contact conductance 3-41C The thermal contact resistance will be greater for rough surfaces because an interface with rough surfaces will contain more air gaps whose thermal conductivity is low 3-42C 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 3-43C 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 3-44C Heat transfer through the voids at an interface is by conduction and radiation Evacuating the interface eliminates heat transfer by conduction, and thus increases the thermal contact resistance 3-45C Thermal contact resistance can be minimized by (1) applying a thermally conducting liquid on the surfaces before they are pressed against each other, (2) by replacing the air at the interface by a better conducting gas such as helium or hydrogen, (3) by increasing the interface pressure, and (4) by inserting a soft metallic foil such as tin, silver, copper, nickel, or aluminum between the two surfaces 3-46 The thickness of copper plate whose thermal resistance is equal to the thermal contact resistance is to be determined Properties The thermal conductivity of copper is k = 386 W/m⋅°C Analysis Noting that thermal contact resistance is the inverse of thermal contact conductance, the thermal contact resistance is determined to be Rc = 1 = = 5.556 × 10 −5 m °C/W hc 18,000 W/m °C L where L is the thickness of k the plate and k is the thermal conductivity Setting R = R c , the equivalent thickness is determined from the relation above to be For a unit surface area, the thermal resistance of a flat plate is defined as R = L = kR = kRc = (386 W/m ⋅ °C)(5.556 ×10 −5 m ⋅ °C/W) = 0.0214 m = 2.14 cm Therefore, the interface between the two plates offers as much resistance to heat transfer as a 2.14 cm thick copper Note that the thermal contact resistance in this case is greater than the sum of the thermal resistances of both plates 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 3-24 3-47 Six identical power transistors are attached on a copper plate For a maximum case temperature of 75°C, the maximum power dissipation and the temperature jump at the interface are to be determined Assumptions Steady operating conditions exist Heat transfer can be approximated as being onedimensional, although it is recognized that heat conduction in some parts of the plate will be twodimensional since the plate area is much larger than the base area of the transistor But the large thermal conductivity of copper will minimize this effect All the heat generated at the junction is dissipated through the back surface of the plate since the transistors are covered by a thick plexiglass layer Thermal conductivities are constant Properties The thermal conductivity of copper is given to be k = 386 W/m⋅°C The contact conductance at the interface of copper-aluminum plates for the case of 1.17-1.4 μm roughness and 10 MPa pressure is hc = 49,000 W/m2⋅°C (Table 3-2) Analysis The contact area between the case and the plate is given to be cm2, and the plate area for each transistor is 100 cm2 The thermal resistance network of this problem consists of three resistances in series (contact, plate, and convection) which are determined to be R contact = R plate = 1 = = 0.0227 °C/W hc Ac (49,000 W/m ⋅ °C)(9 × 10 − m ) L 0.012 m = = 0.0031 °C/W kA (386 W/m ⋅ °C)(0.01 m ) Rconvection = Plate L 1 = = 3.333 °C/W ho A (30 W/m ⋅ °C)(0.01 m ) Q& The total thermal resistance is then R total = Rcontact + Rplate + Rconvection = 0.0227 + 0.0031 + 3.333 = 3.359 °C/W Note that the thermal resistance of copper plate is very small and can be ignored all together Then the rate of heat transfer is determined to be (75 − 23)°C ΔT = = 15.5 W Q& = R total 3.359 °C/W Rcontact Rplate Rconv Tcase T∞ Therefore, the power transistor should not be operated at power levels greater than 15.5 W if the case temperature is not to exceed 75°C The temperature jump at the interface is determined from ΔTinterface = Q& Rcontact = (15.5 W)(0.0227 °C/W) = 0.35°C which is not very large Therefore, even if we eliminate the thermal contact resistance at the interface completely, we will lower the operating temperature of the transistor in this case by less than 1°C 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 3-25 3-48 Two cylindrical aluminum bars with ground surfaces are pressed against each other in an insulation sleeve For specified top and bottom surface temperatures, the rate of heat transfer along the cylinders and the temperature drop at the interface are to be determined Assumptions Steady operating conditions exist Heat transfer is one-dimensional in the axial direction since the lateral surfaces of both cylinders are wellinsulated Thermal conductivities are constant Interface Bar Properties The thermal conductivity of aluminum bars is given to be k = 176 W/m⋅°C The contact conductance at the interface of aluminum-aluminum plates for the case of ground surfaces and of 20 atm ≈ MPa pressure is hc = 11,400 W/m2⋅°C (Table 3-2) Analysis (a) The thermal resistance network in this case consists of two conduction resistance and the contact resistance, and they are determined to be Rcontact = R plate = Ri T1 Bar Rglass Ro T2 1 = = 0.0447 °C/W hc Ac (11,400 W/m ⋅ °C)[π (0.05 m) /4] 0.15 m L = = 0.4341 °C/W kA (176 W/m ⋅ °C)[π (0.05 m) /4] Then the rate of heat transfer is determined to be (150 − 20)°C ΔT ΔT = = = 142.4 W Q& = R total Rcontact + R bar (0.0447 + × 0.4341) °C/W Therefore, the rate of heat transfer through the bars is 142.4 W (b) The temperature drop at the interface is determined to be ΔTinterface = Q& Rcontact = (142.4 W)(0.0447 °C/W) = 6.4°C 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 3-26 3-49 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 Steady operating conditions exist Heat transfer is one-dimensional since the plate is large 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/m2⋅°C Analysis The thermal resistances of different layers for unit surface area of m2 are Rcontact = R plate = Repoxy Copper plate Epoxy Epoxy 1 = = 0.00017 °C/W hc Ac (6000 W/m ⋅ °C)(1 m ) 0.001 m L = = 2.6 × 10 −6 °C/W kA (386 W/m ⋅ °C)(1 m ) Q& 0.005 m L = = = 0.01923 °C/W kA (0.26 W/m ⋅ °C)(1 m ) mm mm The total thermal resistance is R total = Rcontact + R plate + Repoxy = × 0.00017 + 2.6 × 10 − + × 0.01923 = 0.03880 °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 Rcontact × 0.00017 %Error = × 100 = × 100 = 0.88% 0.03880 R total Rplate Repoxy Repoxy T1 Rcontact T2 Rcontact which is negligible 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 3-27 Generalized Thermal Resistance Networks 3-50C Parallel resistances indicate simultaneous heat transfer (such as convection and radiation on a surface) Series resistances indicate sequential heat transfer (such as two homogeneous layers of a wall) 3-51C The thermal resistance network approach will give adequate results for multi-dimensional heat transfer problems if heat transfer occurs predominantly in one direction 3-52C Two approaches used in development of the thermal resistance network in the x-direction for multidimensional problems are (1) to assume any plane wall normal to the x-axis to be isothermal and (2) to assume any plane parallel to the x-axis to be adiabatic 3-53 A typical section of a building wall is considered The average heat flux through the wall is to be determined Assumptions Steady operating conditions exist Properties The thermal conductivities are given to be k23b = 50 W/m⋅K, k23a = 0.03 W/m⋅K, k12 = 0.5 W/m⋅K, k34 = 1.0 W/m⋅K Analysis We consider m2 of wall area The thermal resistances are R12 = t12 0.01 m = = 0.02 m ⋅ °C/W k12 (0.5 W/m ⋅ °C) R 23a = t 23 La k 23a ( La + Lb ) 0.6 m = 2.645 m ⋅ °C/W (0.03 W/m ⋅ °C)(0.6 + 0.005) Lb k 23b ( La + Lb ) = (0.08 m) R 23b = t 23 = (0.08 m) R34 = 0.005 m = 1.32 × 10 −5 m ⋅ °C/W (50 W/m ⋅ °C)(0.6 + 0.005) t 34 0.1 m = = 0.1 m ⋅ °C/W k 34 (1.0 W/m ⋅ °C) The total thermal resistance and the rate of heat transfer are ⎛ R R ⎞ R total = R12 + ⎜⎜ 23a 23b ⎟⎟ + R34 ⎝ R 23a + R 23b ⎠ ⎛ 1.32 × 10 −5 = 0.02 + 2.645⎜⎜ −5 ⎝ 2.645 + 1.32 × 10 q& = ⎞ ⎟ + 0.1 = 0.120 m ⋅ °C/W ⎟ ⎠ T4 − T1 (35 − 20)°C = = 125 W/m 2 R total 0.120 m ⋅ C/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 3-28 3-54 A wall consists of horizontal bricks separated by plaster layers There are also plaster layers on each side of the wall, and a rigid foam on the inner side of the wall The rate of heat transfer through the wall is to be determined Assumptions Heat transfer is steady since there is no indication of change with time Heat transfer through the wall is one-dimensional Thermal conductivities are constant Heat transfer by radiation is disregarded Properties The thermal conductivities are given to be k = 0.72 W/m⋅°C for bricks, k = 0.22 W/m⋅°C for plaster layers, and k = 0.026 W/m⋅°C for the rigid foam Analysis We consider m deep and 0.33 m high portion of wall which is representative of the entire wall The thermal resistance network and individual resistances are R3 Ri R1 R2 T∞1 R4 R6 R7 R5 T∞2 1 = = 0.303 °C/W h1 A (10 W/m ⋅ °C)(0.33 × m ) L 0.02 m = = = 2.33 °C/W kA (0.026 W/m ⋅ °C)(0.33 × m ) Ri = Rconv,1 = R1 = R foam L 0.02 m = = 0.275 °C/W kA (0.22 W/m ⋅ °C)(0.33 × m ) L 0.18 m = = = 54.55 °C/W ho A (0.22 W/m ⋅ °C)(0.015 × m ) R = R6 = R plaster = side R3 = R5 = R plaster center L 0.18 m = = 0.833 °C/W kA (0.72 W/m ⋅ °C)(0.30 × m ) 1 = = = 0.152 °C/W h2 A (20 W/m ⋅ °C)(0.33 × m ) R = Rbrick = Ro = Rconv, R mid = 1 1 1 + + = + + ⎯ ⎯→ R mid = 0.81 °C/W R3 R R5 54.55 0.833 54.55 Rtotal = Ri + R1 + R + R mid + Ro = 0.303 + 2.33 + 2(0.275) + 0.81 + 0.152 = 4.145 °C/W The steady rate of heat transfer through the wall per 0.33 m2 is T −T [(22 − (−4)]°C Q& = ∞1 ∞ = = 6.27 W 4.145°C/W Rtotal Then steady rate of heat transfer through the entire wall becomes ( × 6) m Q& total = (6.27 W) = 456 W 0.33 m 2 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 3-29 3-55 EES Prob 3-54 is reconsidered The rate of heat transfer through the wall as a function of the thickness of the rigid foam is to be plotted Analysis The problem is solved using EES, and the solution is given below "GIVEN" A=4*6 [m^2] L_brick=0.18 [m] L_plaster_center=0.18 [m] L_plaster_side=0.02 [m] L_foam=2 [cm] k_brick=0.72 [W/m-C] k_plaster=0.22 [W/m-C] k_foam=0.026 [W/m-C] T_infinity_1=22 [C] T_infinity_2=-4 [C] h_1=10 [W/m^2-C] h_2=20 [W/m^2-C] "ANALYSIS" R_conv_1=1/(h_1*A_1) A_1=0.33*1 "[m^2]" R_foam=(L_foam*Convert(cm, m))/(k_foam*A_1) "L_foam is in cm" R_plaster_side=L_plaster_side/(k_plaster*A_1) A_2=0.30*1 "[m^2]" R_plaster_center=L_plaster_center/(k_plaster*A_3) A_3=0.015*1 "[m^2]" R_brick=L_brick/(k_brick*A_2) R_conv_2=1/(h_2*A_1) 1/R_mid=2*1/R_plaster_center+1/R_brick R_total=R_conv_1+R_foam+2*R_plaster_side+R_mid+R_conv_2 Q_dot=(T_infinity_1-T_infinity_2)/R_total Q_dot_total=Q_dot*A/A_1 Qtotal [W] 634.6 456.2 356.1 292 247.4 214.7 189.6 169.8 153.7 140.4 700 600 Qtotal [W] Lfoam [cm] 10 500 400 300 200 100 Lfoam [cm] 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 10 3-30 3-56 A wall is to be constructed of 10-cm thick wood studs or with pairs of 5-cm thick wood studs nailed to each other The rate of heat transfer through the solid stud and through a stud pair nailed to each other, as well as the effective conductivity of the nailed stud pair are to be determined Assumptions Heat transfer is steady since there is no indication of change with time Heat transfer can be approximated as being one-dimensional since it is predominantly in the x direction Thermal conductivities are constant The thermal contact resistance between the two layers is negligible Heat transfer by radiation is disregarded Properties The thermal conductivities are given to be k = 0.11 W/m⋅°C for wood studs and k = 50 W/m⋅°C for manganese steel nails Analysis (a) The heat transfer area of the stud is A = (0.1 m)(2.5 m) = 0.25 m2 The thermal resistance and heat transfer rate through the solid stud are L 0.1 m = = 3.636 °C/W kA (0.11 W/m ⋅ °C)(0.25 m ) 8°C ΔT = = 2.2 W Q& = R stud 3.636 °C/W R stud = Stud L Q& (b) The thermal resistances of stud pair and nails are in parallel ⎡ π (0.004 m) ⎤ = 50⎢ ⎥ = 0.000628 m 4 ⎥⎦ ⎣⎢ 0.1 m L = = = 3.18 °C/W kA (50 W/m ⋅ °C)(0.000628 m ) 0.1 m L = = = 3.65 °C/W kA (0.11 W/m ⋅ °C)(0.25 − 0.000628 m ) 1 1 = + = + ⎯ ⎯→ Rtotal = 1.70 °C/W R stud R nails 3.65 3.18 Anails = 50 R nails R stud Rtotal πD 2 T1 T2 Rstud T1 T2 8°C ΔT = 4.7 W Q& = = R stud 1.70 °C/W (c) The effective conductivity of the nailed stud pair can be determined from (4.7 W)(0.1 m) Q& L ΔT ⎯ ⎯→ k eff = = = 0.235 W/m.°C Q& = k eff A ΔTA (8°C)(0.25 m ) L 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 3-31 3-57 A wall is constructed of two layers of sheetrock spaced by 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 Heat transfer is steady since there is no indication of change with time Heat transfer through the wall is one-dimensional Thermal conductivities are constant 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 m The thermal resistance network and the individual thermal resistances are R2 Ri R1 R4 R5 T∞1 T∞2 R3 1 = = 0.185 °C/W hi A (8.3 W/m ⋅ °C)(0.65 m ) 0.01 m L R1 = R = R sheetrock = = = 0.090 °C/W kA (0.17 W/m ⋅ °C)(0.65 m ) Ri = 0.16 m L = = 29.091 °C/W kA (0.11 W/m ⋅ °C)(0.05 m ) 0.16 m L R3 = R fiberglass = = = 7.843 °C/W kA (0.034 W/m ⋅ °C)(0.60 m ) R = R stud = 1 = = 0.045 °C/W o ho A (34 W/m ⋅ C)(0.65 m ) 1 1 = + = + ⎯ ⎯→ R mid = 6.178 °C/W R R3 29.091 7.843 Ro = R mid Rtotal = Ri + R1 + R mid + R + Ro = 0.185 + 0.090 + 6.178 + 0.090 + 0.045 = 6.588 °C/W (for a m × 0.65 m section) T −T [20 − (−9)]°C Q& = ∞1 ∞ = = 4.40 W 6.588 °C/W Rtotal (b) Then steady rate of heat transfer through entire wall becomes (12 m)(5 m) Q& total = (4.40 W) = 406 W 0.65 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 3-32 3-58E A wall is to be constructed using solid bricks or identical size bricks with square air holes There is a 0.5 in thick sheetrock layer between two adjacent bricks on all four sides, and on both sides of the wall The rates of heat transfer through the wall constructed of solid bricks and of bricks with air holes are to be determined Assumptions Heat transfer is steady since there is no indication of change with time Heat transfer through the wall is one-dimensional Thermal conductivities are constant Heat transfer coefficients account for the radiation heat transfer Properties The thermal conductivities are given to be k = 0.40 Btu/h⋅ft⋅°F for bricks, k = 0.015 Btu/h⋅ft⋅°F for air, and k = 0.10 Btu/h⋅ft⋅°F for sheetrock Analysis (a) The representative surface area is A = (7.5 / 12)(7.5 / 12) = 0.3906 ft The thermal resistance network and the individual thermal resistances if the wall is constructed of solid bricks are R2 Ri R1 T∞1 R3 R5 Ro R4 T∞2 1 = = 1.7068 h °F/Btu hi A (1.5 Btu/h ⋅ ft ⋅ °F)(0.3906 ft ) L 0.5 / 12 ft R1 = R5 = R plaster = = = 1.0667 h ⋅ °F/Btu kA (0.10 Btu/h ⋅ ft ⋅ °F)(0.3906 ft ) Ri = L / 12 ft = = 288 h ⋅ °F/Btu kA (0.10 Btu/h ⋅ ft ⋅ °F)[(7.5 / 12) × (0.5 / 12)]ft L / 12 ft = = = 308.57 h ⋅ °F/Btu kA (0.10 Btu/h ⋅ ft⋅ o F)[(7 / 12) × (0.5 / 12)]ft R = R plaster = R3 = R plaster L / 12 ft = = 5.51 h ⋅ °F/Btu kA (0.40 Btu/h ⋅ ft ⋅ °F)[(7 / 12) × (7 / 12)]ft 1 = = 0.64 h ⋅ °F/Btu Ro = ho A (4 Btu/h ⋅ ft ⋅ °F)(0.3906 ft ) 1 1 1 = + + = + + ⎯ ⎯→ R mid = 5.3135 h ⋅ °F/Btu R mid R R3 R 288 308.57 5.51 R = Rbrick = Rtotal = Ri + R1 + R mid + R5 + Ro = 1.7068 + 1.0667 + 5.3135 + 1.0667 + 0.64 = 9.7937 h ⋅ °F/Btu T −T (80 − 30)°F Q& = ∞1 ∞ = = 5.1053 Btu/h 9.7937 h ⋅ °F/Btu Rtotal Then steady rate of heat transfer through entire wall becomes (30 ft)(10 ft) Q& total = (5.1053 Btu/h) = 3921 Btu/h 0.3906 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 3-33 (b) The thermal resistance network and the individual thermal resistances if the wall is constructed of bricks with air holes are R2 Ri R1 T∞1 R3 R6 Ro R4 T∞2 R5 Aairholes = 9(1.5 / 12) × (1.55 / 12) = 0.1406 ft Abricks = (7 / 12 ft) − 0.1406 = 0.1997 ft L / 12 ft = = 355.62 h ⋅ °F/Btu kA (0.015 Btu/h ⋅ ft ⋅ °F)(0.1406 ft ) L / 12 ft = = = 9.389 h ⋅ °F/Btu kA (0.40 Btu/h ⋅ ft ⋅ °F)(0.1997 ft ) R = R airholes = R5 = Rbrick 1 1 1 1 = + + + = + + + ⎯ ⎯→ R mid = 8.618 h ⋅ °F/Btu R mid R R3 R R5 288 308.57 355.62 9.389 Rtotal = Ri + R1 + R mid + R6 + Ro = 1.7068 + 1.0667 + 8.618 + 1.0677 + 0.64 = 13.0992 h ⋅ °F/Btu T −T (80 − 30)°F Q& = ∞1 ∞ = = 3.817 Btu/h 13.0992 h ⋅ °F/Btu Rtotal Then steady rate of heat transfer through entire wall becomes (30 ft)(10 ft) Q& total = (3.817 Btu/h) = 2932 Btu/h 0.3906 ft 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 3-34 3-59 A composite wall consists of several horizontal and vertical layers The left and right surfaces of the wall are maintained at uniform temperatures The rate of heat transfer through the wall, the interface temperatures, and the temperature drop across the section F are to be determined Assumptions Heat transfer is steady since there is no indication of change with time Heat transfer through the wall is one-dimensional Thermal conductivities are constant Thermal contact resistances at the interfaces are disregarded Properties The thermal conductivities are given to be kA = kF = 2, kB = 8, kC = 20, kD = 15, kE = 35 W/m⋅°C B Analysis (a) The representative surface area is A = 0.12 × = 0.12 m The thermal resistance network and the individual thermal resistances are R2 R5 R1 R7 R3 T1 T2 R4 0.01 m R6 ⎛ L ⎞ R1 = R A = ⎜ ⎟ = = 0.04 °C/W ⎝ kA ⎠ A (2 W/m ⋅ °C)(0.12 m ) 0.05 m ⎛ L ⎞ R = R = RC = ⎜ ⎟ = = 0.06 °C/W ⎝ kA ⎠ C (20 W/m ⋅ °C)(0.04 m ) 0.05 m ⎛ L ⎞ = 0.16 °C/W R3 = R B = ⎜ ⎟ = ⎝ kA ⎠ B (8 W/m ⋅ °C)(0.04 m ) 0.1 m ⎛ L ⎞ R5 = R D = ⎜ ⎟ = = 0.11 °C/W ⎝ kA ⎠ D (15 W/m⋅ o C)(0.06 m ) m ⎛ L ⎞ R6 = R E = ⎜ ⎟ = = 0.05 o C/W ⎝ kA ⎠ E (35 W/m ⋅ °C)(0.06 m ) 0.06 m ⎛ L ⎞ = 0.25 °C/W R7 = R F = ⎜ ⎟ = ⎝ kA ⎠ F (2 W/m ⋅ °C)(0.12 m ) Rmid ,1 = 1 1 1 + + = + + ⎯ ⎯→ Rmid ,1 = 0.025 °C/W R2 R3 R4 0.06 0.16 0.06 1 1 = + = + ⎯ ⎯→ Rmid , = 0.034 °C/W Rmid , R5 R6 0.11 0.05 Rtotal = R1 + Rmid ,1 + Rmid , + R7 = 0.04 + 0.025 + 0.034 + 0.25 = 0.349 °C/W T − T∞ (300 − 100)°C = = 572 W (for a 0.12 m × m section) Q& = ∞1 0.349 °C/W Rtotal Then steady rate of heat transfer through entire wall becomes (5 m)(8 m) Q& total = (572 W) = 1.91 × 10 W 0.12 m (b) The total thermal resistance between left surface and the point where the sections B, D, and E meet is Rtotal = R1 + R mid ,1 = 0.04 + 0.025 = 0.065 °C/W Then the temperature at the point where the sections B, D, and E meet becomes T −T Q& = ⎯ ⎯→ T = T1 − Q& Rtotal = 300°C − (572 W)(0.065 °C/W) = 263°C Rtotal (c) The temperature drop across the section F can be determined from ΔT Q& = → ΔT = Q& R F = (572 W)(0.25 °C/W) = 143°C RF 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 3-35 3-60 A composite wall consists of several horizontal and vertical layers The left and right surfaces of the wall are maintained at uniform temperatures The rate of heat transfer through the wall, the interface temperatures, and the temperature drop across the section F are to be determined Assumptions Heat transfer is steady since there is no indication of change with time Heat transfer through the wall is one-dimensional Thermal conductivities are constant Thermal contact resistances at the interfaces are to be considered Properties The thermal conductivities of various materials used are given to be kA = kF = 2, kB = 8, kC = 20, kD = 15, and kE = 35 W/m⋅°C B Analysis The representative surface area is A = 0.12 × = 0.12 m R2 R5 R1 R3 R4 R7 R8 R6 (a) The thermal resistance network and the individual thermal resistances are 0.01 m ⎛ L ⎞ R1 = R A = ⎜ ⎟ = = 0.04 °C/W kA ⎝ ⎠ A (2 W/m ⋅ °C)(0.12 m ) 0.05 m ⎛ L ⎞ R = R = RC = ⎜ ⎟ = = 0.06 °C/W ⎝ kA ⎠ C (20 W/m ⋅ °C)(0.04 m ) 0.05 m ⎛ L ⎞ = 0.16 °C/W R3 = R B = ⎜ ⎟ = ⎝ kA ⎠ B (8 W/m ⋅ °C)(0.04 m ) 0.1 m ⎛ L ⎞ R5 = R D = ⎜ ⎟ = = 0.11 °C/W ⎝ kA ⎠ D (15 W/m⋅ o C)(0.06 m ) 0.1 m ⎛ L ⎞ R6 = R E = ⎜ ⎟ = = 0.05 o C/W ⎝ kA ⎠ E (35 W/m ⋅ °C)(0.06 m ) 0.06 m ⎛ L ⎞ = 0.25 °C/W R7 = R F = ⎜ ⎟ = kA ⎝ ⎠ F (2 W/m ⋅ °C)(0.12 m ) R8 = 0.00012 m ⋅ °C/W 0.12 m R mid ,1 = = 0.001 °C/W 1 1 1 + + = + + ⎯ ⎯→ R mid ,1 = 0.025 °C/W R R3 R 0.06 0.16 0.06 1 1 = + = + ⎯ ⎯→ R mid , = 0.034 °C/W R mid , R5 R6 0.11 0.05 Rtotal = R1 + R mid ,1 + R mid , + R7 + R8 = 0.04 + 0.025 + 0.034 + 0.25 + 0.001 = 0.350 °C/W T −T (300 − 100)°C = 571 W (for a 0.12 m × m section) Q& = ∞1 ∞ = 0.350 °C/W Rtotal Then steady rate of heat transfer through entire wall becomes (5 m)(8 m) Q& total = (571 W) = 1.90 × 10 W 0.12 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 3-36 (b) The total thermal resistance between left surface and the point where the sections B, D, and E meet is Rtotal = R1 + R mid ,1 = 0.04 + 0.025 = 0.065 °C/W Then the temperature at the point where The sections B, D, and E meet becomes T −T Q& = ⎯ ⎯→ T = T1 − Q& Rtotal = 300°C − (571 W)(0.065 °C/W) = 263°C Rtotal (c) The temperature drop across the section F can be determined from ΔT Q& = ⎯ ⎯→ ΔT = Q& R F = (571 W)(0.25 °C/W) = 143°C RF 3-61 A coat is made of layers of 0.1 mm thick synthetic fabric separated by 1.5 mm thick air space The rate of heat loss through the jacket is to be determined, and the result is to be compared to the heat loss through a jackets without the air space Also, the equivalent thickness of a wool coat is to be determined Assumptions Heat transfer is steady since there is no indication of change with time Heat transfer through the jacket is one-dimensional Thermal conductivities are constant Heat transfer coefficients account for the radiation heat transfer Properties The thermal conductivities are given to be k = 0.13 W/m⋅°C for synthetic fabric, k = 0.026 W/m⋅°C for air, and k = 0.035 W/m⋅°C for wool fabric Analysis The thermal resistance network and the individual thermal resistances are R1 R2 R3 R4 R5 R6 R7 R8 R9 Ro T∞2 Ts1 0.0001 m L = = 0.0006 °C/W kA (0.13 W/m ⋅ °C)(1.25 m ) L 0.0015 m R air = R = R = R6 = R8 = = = 0.0462 °C/W kA (0.026 W/m ⋅ °C)(1.25 m ) 1 Ro = = = 0.0320 °C/W hA (25 W/m ⋅ °C)(1.25 m ) Rtotal = R fabric + R air + Ro = × 0.0006 + × 0.0462 + 0.0320 = 0.2198 °C/W R fabric = R1 = R3 = R5 = R7 = R9 = and T −T (28 − 0)°C = 127 W Q& = s1 ∞ = Rtotal 0.2198 °C/W If the jacket is made of a single layer of 0.5 mm thick synthetic fabric, the rate of heat transfer would be T − T∞ Ts1 − T∞ (28 − 0)°C Q& = s1 = = = 800 W Rtotal × R fabric + Ro (5 × 0.0006 + 0.0320) °C/W The thickness of a wool fabric that has the same thermal resistance is determined from L R total = R wool + Ro = + kA hA fabric 0.2198 °C/W = L (0.035 W/m ⋅ °C)(1.25 m ) + 0.0320 ⎯ ⎯→ L = 0.0082 m = 8.2 mm 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 3-37 3-62 A coat is made of layers of 0.1 mm thick cotton fabric separated by 1.5 mm thick air space The rate of heat loss through the jacket is to be determined, and the result is to be compared to the heat loss through a jackets without the air space Also, the equivalent thickness of a wool coat is to be determined Assumptions Heat transfer is steady since there is no indication of change with time Heat transfer through the jacket is one-dimensional Thermal conductivities are constant Heat transfer coefficients account for the radiation heat transfer Properties The thermal conductivities are given to be k = 0.06 W/m⋅°C for cotton fabric, k = 0.026 W/m⋅°C for air, and k = 0.035 W/m⋅°C for wool fabric Analysis The thermal resistance network and the individual thermal resistances are R1 R2 R3 R4 R5 R6 R7 R8 R9 Ro T∞2 T1 L 0.0001 m = = 0.00133 °C/W kA (0.06 W/m ⋅ °C)(1.25 m ) L 0.0015 m R air = R = R = R6 = R8 = = = 0.0462 °C/W kA (0.026 W/m⋅ o C)(1.25 m ) 1 Ro = = = 0.0320 °C/W hA (25 W/m ⋅ °C)(1.25 m ) Rtotal = 5R fabric + R air + Ro = × 0.00133 + × 0.0462 + 0.0320 = 0.2235 °C/W Rcot ton = R1 = R3 = R5 = R = R9 = and T −T (28 − 0)°C = 125 W Q& = s1 ∞ = Rtotal 0.2235 °C/W If the jacket is made of a single layer of 0.5 mm thick cotton fabric, the rate of heat transfer will be T − T∞ Ts1 − T∞ (28 − 0)°C Q& = s1 = = = 724 W Rtotal × R fabric + Ro (5 × 0.00133 + 0.0320) °C/W The thickness of a wool fabric for that case can be determined from R total = R wool + Ro = fabric 0.2235 °C/W = L + kA hA L (0.035 W/m ⋅ °C)(1.25 m ) + 0.0320 ⎯ ⎯→ L = 0.0084 m = 8.4 mm 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 3-38 3-63 A kiln is made of 20 cm thick concrete walls and ceiling The two ends of the kiln are made of thin sheet metal covered with 2-cm thick styrofoam For specified indoor and outdoor temperatures, the rate of heat transfer from the kiln is to be determined Assumptions Heat transfer is steady since there is no indication of change with time Heat transfer through the walls and ceiling is one-dimensional Thermal conductivities are constant Heat transfer coefficients account for the radiation heat transfer Heat loss through the floor is negligible Thermal resistance of sheet metal is negligible Properties The thermal conductivities are given to be k = 0.9 W/m⋅°C for concrete and k = 0.033 W/m⋅°C for styrofoam insulation Analysis In this problem there is a question of which surface area to use We will use the outer surface area for outer convection resistance, the inner surface area for inner convection resistance, and the average area for the conduction resistance Or we could use the inner or the outer surface areas in the calculation of all thermal resistances with little loss in accuracy For top and the two side surfaces: Ri Rconcrete Ro Tin Tout 1 = = 0.0071× 10 − °C/W hi Ai (3000 W/m ⋅ °C)[(40 m)(13 − 1.2) m] L 0.2 m = = = 4.480 × 10 − °C/W kAave (0.9 W/m ⋅ °C)[(40 m)(13 − 0.6) m] Ri = Rconcrete Ro = 1 = = 0.769 × 10 − °C/W ho Ao (25 W/m ⋅ °C)[(40 m)(13 m)] Rtotal = Ri + Rconcrete + Ro = (0.0071 + 4.480 + 0.769) × 10 − = 5.256 × 10 − °C/W and T − Tout [40 − (−4)]°C = = 83,700 W Q& top + sides = in Rtotal 5.256 × 10 − °C/W Heat loss through the end surface of the kiln with styrofoam: Ri Rstyrofoam Ro Tin Tout 1 = = 0.201× 10 − °C/W hi Ai (3000 W/m ⋅ °C)[(4 − 0.4)(5 − 0.4) m ] 0.02 m L = = = 0.0332 °C/W kAave (0.033 W/m ⋅ °C)[(4 − 0.2)(5 − 0.2) m ] Ri = R styrofoam Ro = 1 = = 0.0020 °C/W ho Ao (25 W/m ⋅ °C)[4 × m ] Rtotal = Ri + R styrpfoam + Ro = 0.201 × 10 − + 0.0332 + 0.0020 = 0.0352 °C/W and T − Tout [40 − (−4)]°C = = 1250 W Q& end surface = in 0.0352 °C/W Rtotal Then the total rate of heat transfer from the kiln becomes Q& total = Q& top + sides + 2Q& side = 83,700 + × 1250 = 86,200 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 3-39 3-64 EES Prob 3-63 is reconsidered The effects of the thickness of the wall and the convection heat transfer coefficient on the outer surface of the rate of heat loss from the kiln are to be investigated Analysis The problem is solved using EES, and the solution is given below "GIVEN" width=5 [m] height=4 [m] length=40 [m] L_wall=0.2 [m] k_concrete=0.9 [W/m-C] T_in=40 [C] T_out=-4 [C] L_sheet=0.003 [m] L_styrofoam=0.02 [m] k_styrofoam=0.033 [W/m-C] h_i=3000 [W/m^2-C] h_o=25 [W/m^2-C] "ANALYSIS" R_conv_i=1/(h_i*A_1) A_1=(2*height+width-6*L_wall)*length R_concrete=L_wall/(k_concrete*A_2) A_2=(2*height+width-3*L_wall)*length R_conv_o=1/(h_o*A_3) A_3=(2*height+width)*length R_total_top_sides=R_conv_i+R_concrete+R_conv_o Q_dot_top_sides=(T_in-T_out)/R_total_top_sides "Heat loss from top and the two side surfaces" R_conv_i_end=1/(h_i*A_4) A_4=(height-2*L_wall)*(width-2*L_wall) R_styrofoam=L_styrofoam/(k_styrofoam*A_5) A_5=(height-L_wall)*(width-L_wall) R_conv_o_end=1/(h_o*A_6) A_6=height*width R_total_end=R_conv_i_end+R_styrofoam+R_conv_o_end Q_dot_end=(T_in-T_out)/R_total_end "Heat loss from one end surface" Q_dot_total=Q_dot_top_sides+2*Q_dot_end Lwall [m] 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 Qtotal [W] 151098 131499 116335 104251 94395 86201 79281 73359 68233 63751 59800 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 3-40 ho [W/m2.C] 10 15 20 25 30 35 40 45 50 Qtotal [W] 54834 70939 78670 83212 86201 88318 89895 91116 92089 92882 160000 Qtotal [W] 140000 120000 100000 80000 60000 0.08 0.12 0.16 0.2 0.24 0.28 0.32 Lwall [m] 95000 90000 Qtotal [W] 85000 80000 75000 70000 65000 60000 55000 50000 10 15 20 25 30 35 40 45 50 ho [W/m -C] 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 3-41 3-65E The thermal resistance of an epoxy glass laminate across its thickness is to be reduced by planting cylindrical copper fillings throughout The thermal resistance of the epoxy board for heat conduction across its thickness as a result of this modification is to be determined Assumptions Steady operating conditions exist Heat transfer through the plate is one-dimensional Thermal conductivities are constant Properties The thermal conductivities are given to be k = 0.10 Btu/h⋅ft⋅°F for epoxy glass laminate and k = 223 Btu/h⋅ft⋅°F for copper fillings Analysis The thermal resistances of copper fillings and the epoxy board are in parallel The number of copper fillings in the board and the area they comprise are Atotal = (6 / 12 ft)(8 / 12 ft) = 0.333 ft n copper = 0.33 ft = 13,333 (number of copper fillings) (0.06 / 12 ft)(0.06 / 12 ft) Acopper = n πD = 13,333 Aepoxy = Atotal − Acopper π (0.02 / 12 ft) = 0.0291 ft = 0.3333 − 0.0291 = 0.3042 ft Rcopper The thermal resistances are evaluated to be 0.05 / 12 ft L = = 0.00064 h ⋅ °F/Btu kA (223 Btu/h ⋅ ft ⋅ °F)(0.0291 ft ) 0.05 / 12 ft L = = = 0.137 h ⋅ °F/Btu kA (0.10 Btu/h ⋅ ft ⋅ °F)(0.3042 ft ) Rcopper = Repoxy Repoxy Then the thermal resistance of the entire epoxy board becomes 1 1 = + = + ⎯ ⎯→ Rboard = 0.00064 h ⋅ °F/Btu Rboard Rcopper Repoxy 0.00064 0.137 Heat Conduction in Cylinders and Spheres 3-66C When the diameter of cylinder is very small compared to its length, it can be treated as an infinitely long cylinder Cylindrical rods can also be treated as being infinitely long when dealing with heat transfer at locations far from the top or bottom surfaces However, it is not proper to use this model when finding temperatures near the bottom and the top of the cylinder 3-67C Heat transfer in this short cylinder is one-dimensional since there will be no heat transfer in the axial and tangential directions 3-68C No In steady-operation the temperature of a solid cylinder or sphere does not change in radial direction (unless there is heat generation) 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 3-42 3-69 Chilled water is flowing inside a pipe The thickness of the insulation needed to reduce the temperature rise of water to one-fourth of the original value is to be determined Assumptions Heat transfer is steady since there is no indication of any change with time Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction Thermal conductivities are constant The thermal contact resistance at the interface is negligible Properties The thermal conductivity is given to be k = 0.05 W/m⋅°C for insulation Insulation Analysis The rate of heat transfer without the insulation is Q& old = m& c p ΔT = (0.98 kg/s)(4180 J/kg ⋅ °C)(8 - 7)°C = 4096 W r2 The total resistance in this case is T − Tw Q& old = ∞ R total 4096 W = Water (30 − 7.5)°C ⎯ ⎯→ R total = 0.005493°C/W R total L R1 Ro Rins T∞2 T∞1 The convection resistance on the outer surface is Ro = r1 1 = = 0.004716 °C/W ho Ao (9 W/m ⋅ °C)π (0.05 m)(150 m) The rest of thermal resistances are due to convection resistance on the inner surface and the resistance of the pipe and it is determined from R1 = R total − Ro = 0.005493 − 0.004716 = 0.0007769 °C/W The rate of heat transfer with the insulation is Q& = m& c ΔT = (0.98 kg/s)(4180 J/kg ⋅ °C)(0.25 °C) = 1024 W new p The total thermal resistance with the insulation is T − Tw [30 − (7 + 7.25) / 2)]°C Q& new = ∞ ⎯ ⎯→ 1024 W = ⎯ ⎯→ R total, new = 0.02234°C/W R total, new R total, new It is expressed by R total,new = R1 + R o, new + Rins = R1 + 0.02234°C/W = 0.0007769 + ln( D / D1 ) + ho Ao 2πk ins L (9 W/m ⋅ °C)πD (150 m) + ln( D / 0.05) 2π (0.05 W/m ⋅ °C)(150 m) Solving this equation by trial-error or by using an equation solver such as EES, we obtain D = 0.1265 m Then the required thickness of the insulation becomes t ins = ( D − D1 ) / = (0.05 − 0.1265) / = 0.0382 m = 3.8 cm 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 3-43 3-70 Steam flows in a steel pipe, which is insulated by gypsum plaster The rate of heat transfer from the steam and the temperature on the outside surface of the insulation are be determined Assumptions Heat transfer is steady since there is no indication of any change with time Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction Thermal conductivities are constant The thermal contact resistance at the interface is negligible Properties (a) The thermal conductivities of steel and gypsum plaster are given to be 50 and 0.5 W/m⋅°C, respectively Insulation Analysis The thermal resistances are Ri Ti Rsteel Rins Steam Ro To 1 = = 0.0003316°C/W hi Ai (800 W/m ⋅ °C)π (0.06 m)(20 m) ln( D / D1 ) ln(8 / 6) = = = 0.0000458°C/W 2πk steel L 2π (50 W/m ⋅ °C)(20 m) L Ri = Rsteel Rins = Ro = ln( D3 / D ) ln(16 / 8) = = 0.011032°C/W 2πk ins L 2π (0.5 W/m ⋅ °C)(20 m) 1 = = 0.0004974°C/W ho Ao (200 W/m ⋅ °C)π (0.16 m)(20 m) The total thermal resistance and the rate of heat transfer are R total = Ri + Rsteel + Rins + Ro = 0.0003316 + 0.0000458 + 0.011032 + 0.0004974 = 0.011907°C/W T − To (200 − 10)°C = Q& = i = 15,957 W R total 0.011907 m ⋅ C/W (b) The temperature at the outer surface of the insulation is determined from (Ts − 10)°C T − To Q& = s ⎯ ⎯→ 15,957 W = ⎯ ⎯→ Ts = 17.9°C Ro 0.0004974 m ⋅ °C/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 ... indicate simultaneous heat transfer (such as convection and radiation on a surface) Series resistances indicate sequential heat transfer (such as two homogeneous layers of a wall) 3- 51C The thermal... using this Manual, you are using it without permission 3- 34 3- 59 A composite wall consists of several horizontal and vertical layers The left and right surfaces of the wall are maintained at uniform... There are also plaster layers on each side of the wall, and a rigid foam on the inner side of the wall The rate of heat transfer through the wall is to be determined Assumptions Heat transfer