Introduction to thermodynamics and heat transfer 2nd edition

http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS Chapter Basics of Heat Transfer Chapter BASICS OF HEAT TRANSFER Thermodynamics and Heat Transfer 1-1C Thermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state to another Heat transfer, on the other hand, deals with the rate of heat transfer as well as the temperature distribution within the system at a specified time 1-2C (a) The driving force for heat transfer is the temperature difference (b) The driving force for electric current flow is the electric potential difference (voltage) (a) The driving force for fluid flow is the pressure difference 1-3C The caloric theory is based on the assumption that heat is a fluid-like substance called the "caloric" which is a massless, colorless, odorless substance It was abandoned in the middle of the nineteenth century after it was shown that there is no such thing as the caloric 1-4C The rating problems deal with the determination of the heat transfer rate for an existing system at a specified temperature difference The sizing problems deal with the determination of the size of a system in order to transfer heat at a specified rate for a specified temperature difference 1-5C The experimental approach (testing and taking measurements) has the advantage of dealing with the actual physical system, and getting a physical value within the limits of experimental error However, this approach is expensive, time consuming, and often impractical The analytical approach (analysis or calculations) has the advantage that it is fast and inexpensive, but the results obtained are subject to the accuracy of the assumptions and idealizations made in the analysis 1-6C Modeling makes it possible to predict the course of an event before it actually occurs, or to study various aspects of an event mathematically without actually running expensive and time-consuming experiments When preparing a mathematical model, all the variables that affect the phenomena are identified, reasonable assumptions and approximations are made, and the interdependence of these variables are studied The relevant physical laws and principles are invoked, and the problem is formulated mathematically Finally, the problem is solved using an appropriate approach, and the results are interpreted 1-7C The right choice between a crude and complex model is usually the simplest model which yields adequate results Preparing very accurate but complex models is not necessarily a better choice since such models are not much use to an analyst if they are very difficult and time consuming to solve At the minimum, the model should reflect the essential features of the physical problem it represents 1-1 Chapter Basics of Heat Transfer Heat and Other Forms of Energy 1-8C The rate of heat transfer per unit surface area is called heat flux q It is related to the rate of heat transfer by Q   qdA A 1-9C Energy can be transferred by heat, work, and mass An energy transfer is heat transfer when its driving force is temperature difference 1-10C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in daily life 1-11C For the constant pressure case This is because the heat transfer to an ideal gas is mCpT at constant pressure and mCpT at constant volume, and Cp is always greater than Cv 1-12 A cylindrical resistor on a circuit board dissipates 0.6 W of power The amount of heat dissipated in 24 h, the heat flux, and the fraction of heat dissipated from the top and bottom surfaces are to be determined Assumptions Heat is transferred uniformly from all surfaces Analysis (a) The amount of heat this resistor dissipates during a 24-hour period is Q  Q t  (0.6 W)(24 h)  14.4 Wh = 51.84 kJ (since Wh = 3600 Ws = 3.6 kJ) Resistor 0.6 W (b) The heat flux on the surface of the resistor is As  q s  D  DL   (0.4 cm) Q   (0.4 cm)(1.5 cm)  0.251  1.885  2.136 cm Q 0.60 W   0.2809 W/cm As 2.136 cm (c) Assuming the heat transfer coefficient to be uniform, heat transfer is proportional to the surface area Then the fraction of heat dissipated from the top and bottom surfaces of the resistor becomes Qtop  base Qtotal  Atop  base Atotal  0.251  0.118 or (11.8%) 2136 Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side surface 1-2 Chapter Basics of Heat Transfer 1-13E A logic chip in a computer dissipates W of power The amount heat dissipated in h and the heat flux on the surface of the chip are to be determined Assumptions Heat transfer from the surface is uniform Analysis (a) The amount of heat the chip dissipates during an 8-hour period is Q  Q t  (3 W)(8 h)  24 Wh  0.024 kWh Logic chip Q  W (b) The heat flux on the surface of the chip is q s  Q 3W   37.5 W/in As 0.08 in 1-14 The filament of a 150 W incandescent lamp is cm long and has a diameter of 0.5 mm The heat flux on the surface of the filament, the heat flux on the surface of the glass bulb, and the annual electricity cost of the bulb are to be determined Assumptions Heat transfer from the surface of the filament and the bulb of the lamp is uniform Analysis (a) The heat transfer surface area and the heat flux on the surface of the filament are As  DL   (0.05 cm )(5 cm )  0.785 cm q s  Q 150 W   191 W/cm  1.91 10 W/m As 0.785 cm Lamp 150 W Q (b) The heat flux on the surface of glass bulb is As  D   (8 cm)  201.1 cm q s  Q 150 W   0.75 W/cm  7500 W/m As 201.1 cm (c) The amount and cost of electrical energy consumed during a one-year period is Electricity Consumption  Q t  (015 kW)(365  h / yr)  438 kWh / yr Annual Cost = (438 kWh / yr)($0.08 / kWh)  $35.04 / yr 1-15 A 1200 W iron is left on the ironing board with its base exposed to the air The amount of heat the iron dissipates in h, the heat flux on the surface of the iron base, and the cost of the electricity are to be determined Assumptions Heat transfer from the surface is uniform Analysis (a) The amount of heat the iron dissipates during a 2-h period is Q  Q t  (12 kW)(2 h)  2.4 kWh (b) The heat flux on the surface of the iron base is Q base  (0.9)(1200 W) = 1080 W Q 1080 W q  base   72,000 W / m Abase 0.015 m (c) The cost of electricity consumed during this period is Cost of electricity = (2.4 kWh)  ($0.07 / kWh)  $0.17 1-3 Iron 1200 W Chapter Basics of Heat Transfer 1-16 A 15 cm  20 cm circuit board houses 120 closely spaced 0.12 W logic chips The amount of heat dissipated in 10 h and the heat flux on the surface of the circuit board are to be determined Assumptions Heat transfer from the back surface of the board is negligible Heat transfer from the front surface is uniform Analysis (a) The amount of heat this circuit board dissipates during a 10-h period is Chips, 0.12 W Q  (120)(0.12 W)  14.4 W Q  Q t  (0.0144 kW)(10 h)  0.144 kWh Q (b) The heat flux on the surface of the circuit board is As  (0.15 m )(0.2 m )  0.03 m 15 cm Q 14.4 W q s    480 W/m As 0.03 m 20 cm 1-17 An aluminum ball is to be heated from 80C to 200C The amount of heat that needs to be transferred to the aluminum ball is to be determined Assumptions The properties of the aluminum ball are constant Properties The average density and specific heat of aluminum are given to be  = 2,700 kg/m3 and C p  0.90 kJ/kg.C Metal ball Analysis The amount of energy added to the ball is simply the change in its internal energy, and is determined from Etransfer  U  mC (T2  T1 ) where E m  V   D3   (2700 kg / m3 )(015 m)  4.77 kg Substituting, Etransfer  (4.77 kg)(0.90 kJ / kg  C)(200 - 80) C = 515 kJ Therefore, 515 kJ of energy (heat or work such as electrical energy) needs to be transferred to the aluminum ball to heat it to 200C 1-18 The body temperature of a man rises from 37°C to 39°C during strenuous exercise The resulting increase in the thermal energy content of the body is to be determined Assumptions The body temperature changes uniformly Properties The average specific heat of the human body is given to be 3.6 kJ/kg.°C Analysis The change in the sensible internal energy content of the body as a result of the body temperature rising 2C during strenuous exercise is U = mCT = (70 kg)(3.6 kJ/kg.C)(2C) = 504 kJ 1-4 Chapter Basics of Heat Transfer 1-19 An electrically heated house maintained at 22°C experiences infiltration losses at a rate of 0.7 ACH The amount of energy loss from the house due to infiltration per day and its cost are to be determined Assumptions Air as an ideal gas with a constant specific heats at room temperature The volume occupied by the furniture and other belongings is negligible The house is maintained at a constant temperature and pressure at all times The infiltrating air exfiltrates at the indoors temperature of 22°C Properties The specific heat of air at room temperature is C p = 1.007 kJ/kg.C (Table A-15) Analysis The volume of the air in the house is V  ( floor space)(height)  (200 m2 )(3 m)  600 m3 Noting that the infiltration rate is 0.7 ACH (air changes per hour) and thus the air in the house is completely replaced by the outdoor air 0.724 = 16.8 times per day, the mass flow rate of air through the house due to infiltration is m air   PoVair Po (ACH  V house )  RTo RTo (89.6 kPa)(16.8  600 m / day) (0.287 kPa.m /kg.K)(5 + 273.15 K) 0.7 ACH 22C AIR 5C  11,314 kg/day Noting that outdoor air enters at 5C and leaves at 22C, the energy loss of this house per day is Q infilt  m air C p (Tindoors  Toutdoors )  (11,314 kg/day)(1.007 kJ/kg.C)(22  5)C  193,681 kJ/day = 53.8 kWh/day At a unit cost of $0.082/kWh, the cost of this electrical energy lost by infiltration is Enegy Cost = (Energy used)(Unit cost of energy)  (53.8 kWh/day)($0.082/kWh )  $4.41/day 1-5 Chapter Basics of Heat Transfer 1-20 A house is heated from 10C to 22C by an electric heater, and some air escapes through the cracks as the heated air in the house expands at constant pressure The amount of heat transfer to the air and its cost are to be determined Assumptions Air as an ideal gas with a constant specific heats at room temperature The volume occupied by the furniture and other belongings is negligible The pressure in the house remains constant at all times Heat loss from the house to the outdoors is negligible during heating The air leaks out at 22C Properties The specific heat of air at room temperature is C p = 1.007 kJ/kg.C (Table A-15) 22C Analysis The volume and mass of the air in the house are V  ( floor space)(height)  (200 m2 )(3 m)  600 m3 m 10C AIR PV (1013 kPa)(600 m3 )   747.9 kg RT (0.287 kPa.m3 / kg.K)(10 + 273.15 K) Noting that the pressure in the house remains constant during heating, the amount of heat that must be transferred to the air in the house as it is heated from 10 to 22C is determined to be Q  mC p (T2  T1 )  (747.9 kg)(1.007 kJ/kg.C)(22  10)C  9038 kJ Noting that kWh = 3600 kJ, the cost of this electrical energy at a unit cost of $0.075/kWh is Enegy Cost = (Energy used)(Unit cost of energy)  (9038 / 3600 kWh)($0.075/kWh)  $0.19 Therefore, it will cost the homeowner about 19 cents to raise the temperature in his house from 10 to 22C 1-21E A water heater is initially filled with water at 45F The amount of energy that needs to be transferred to the water to raise its temperature to 140F is to be determined Assumptions Water is an incompressible substance with constant specific heats at room temperature No water flows in or out of the tank during heating Properties The density and specific heat of water are given to be 62 lbm/ft and 1.0 Btu/lbm.F Analysis The mass of water in the tank is  ft    497.3 lbm m  V  (62 lbm/ft )(60 gal)  7.48 gal    Then, the amount of heat that must be transferred to the water in the tank as it is heated from 45 to140F is determined to be Q  mC (T2  T1 )  (497.3 lbm)(1.0 Btu/lbm F)(140  45)F  47,250 Btu The First Law of Thermodynamics 1-6 140F 45F Water Chapter Basics of Heat Transfer 1-22C Warmer Because energy is added to the room air in the form of electrical work 1-23C Warmer If we take the room that contains the refrigerator as our system, we will see that electrical work is supplied to this room to run the refrigerator, which is eventually dissipated to the room as waste heat 1-24C Mass flow rate m is the amount of mass flowing through a cross-section per unit time whereas the volume flow rate V is the amount of volume flowing through a cross-section per unit time They are related to each other by m  V where  is density 1-25 Two identical cars have a head-on collusion on a road, and come to a complete rest after the crash The average temperature rise of the remains of the cars immediately after the crash is to be determined Assumptions No heat is transferred from the cars All the kinetic energy of cars is converted to thermal energy Properties The average specific heat of the cars is given to be 0.45 kJ/kg.C Analysis We take both cars as the system This is a closed system since it involves a fixed amount of mass (no mass transfer) Under the stated assumptions, the energy balance on the system can be expressed as E E inout   Net energy transfer by heat, work, and mass E system    Change in internal, kinetic, potential, etc energies  U cars  KE cars  (mCT ) cars  [m(0  V ) / 2]cars That is, the decrease in the kinetic energy of the cars must be equal to the increase in their internal energy Solving for the velocity and substituting the given quantities, the temperature rise of the cars becomes T  mV / V / (90,000 / 3600 m/s) /  kJ/kg       0.69C mC C 0.45 kJ/kg.C  1000 m /s  1-26 A classroom is to be air-conditioned using window air-conditioning units The cooling load is due to people, lights, and heat transfer through the walls and the windows The number of 5-kW window air conditioning units required is to be determined Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room Analysis The total cooling load of the room is determined from Q cooling  Q lights  Q people  Q heat gain 1-7 kJ/h 15,000 Room 40 people 10 bulbs · Qcool Chapter Basics of Heat Transfer where Q lights  10  100 W  kW Q people  40  360kJ/h  14,400 kJ/h  4kW Q heat gain  15,000 kJ/h  4.17 kW Substituting, Q cooling    4.17  9.17 kW Thus the number of air-conditioning units required is 9.17 kW  1.83   units kW/unit 1-27E The air in a rigid tank is heated until its pressure doubles The volume of the tank and the amount of heat transfer are to be determined Assumptions Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -141C and 3.77 MPa The kinetic and potential energy changes are negligible, pe  ke  Constant specific heats at room temperature can be used for air This assumption results in negligible error in heating and air-conditioning applications Properties The gas constant of air is R = 0.3704 psia.ft3/lbm.R = 0.06855 Btu/lbm.R (Table A-1) Analysis (a) We take the air in the tank as our system This is a closed system since no mass enters or leaves The volume of the tank can be determined from the ideal gas relation, V mRT1 (20lbm)(0.3704 psia  ft /lbm  R)(80 + 460R)   80.0ft P1 50 psia (b) Under the stated assumptions and observations, the energy balance becomes E E inout  E system     Net energy transfer by heat, work, and mass Change in internal, kinetic, potential, etc energies Qin  U   Qin  m(u2  u1 )  mCv (T2  T1 ) The final temperature of air is PV PV  T1 T2   T2  P2 T1   (540 R)  1080 R P1 The specific heat of air at the average temperature of Tave = (540+1080)/2= 810 R = 350F is Cv,ave = Cp,ave – R = 0.2433 - 0.06855 = 0.175 Btu/lbm.R Substituting, Q = (20 lbm)( 0.175 Btu/lbm.R)(1080 - 540) R = 1890 Btu Air 20 lbm 50 psia 80F Q 1-8 Chapter Basics of Heat Transfer 1-28 The hydrogen gas in a rigid tank is cooled until its temperature drops to 300 K The final pressure in the tank and the amount of heat transfer are to be determined Assumptions Hydrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -240C and 1.30 MPa The kinetic and potential energy changes are negligible, ke  pe  Properties The gas constant of hydrogen is R = 4.124 kPa.m3/kg.K (Table A-1) Analysis (a) We take the hydrogen in the tank as our system This is a closed system since no mass enters or leaves The final pressure of hydrogen can be determined from the ideal gas relation, P1V P2V T 300 K    P2  P1  (250 kPa)  178.6 kPa T1 T2 T1 420 K (b) The energy balance for this system can be expressed as E E inout  Net energy transfer by heat, work, and mass  E system    Change in internal, kinetic, potential, etc energies  Qout  U Qout   U   m(u2  u1 )  mCv (T1  T2 ) where (250 kPa)(1.0 m ) PV m   0.1443 kg RT1 (4.124 kPa  m /kg  K)(420 K) H2 250 kPa 420 K Q Using the Cv (=Cp – R) = 14.516 – 4.124 = 10.392 kJ/kg.K value at the average temperature of 360 K and substituting, the heat transfer is determined to be Qout = (0.1443 kg)(10.392 kJ/kg·K)(420 - 300)K = 180.0 kJ 1-9 Chapter 15 Cooling of Electronic Equipment 15-133 "!PROBLEM 15-133" "GIVEN" "Q_dot_total=3.5 [W], parameter to be varied" T_ambient=50 "[C]" T_chip=95 "[C]" A=0.8 "[cm^2]" "ANALYSIS" q_dot=Q_dot_total/A Q_dot_total=h*A*Convert(cm^2, m^2)*(T_chip-T_ambient) Q_dot_total=(T_chip-T_ambient)/R_ChipFluid Qtotal [W] 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10 q [W/cm2] 2.5 3.125 3.75 4.375 5.625 6.25 6.875 7.5 8.125 8.75 9.375 10 10.63 11.25 11.88 12.5 h [W/m2-C] 555.6 694.4 833.3 972.2 1111 1250 1389 1528 1667 1806 1944 2083 2222 2361 2500 2639 2778 15-70 RChipFluid [C/W] 22.5 18 15 12.86 11.25 10 8.182 7.5 6.923 6.429 5.625 5.294 4.737 4.5 Chapter 15 Cooling of Electronic Equipment 13 3000 10.8 2500 8.6 2] 2000 q h 6.4 m c / W [ q ] C 1500 m / W [ 4.2 2 1000 Q total [W ] h 500 10 22.5 18.5 14.5 ] W / C [ 10.5 d i u l F p i h C 6.5 R 2.5 Q total [W ] 15-71 10 Chapter 15 Cooling of Electronic Equipment 15-134 A computer chip is to be cooled by immersion in a dielectric fluid The minimum heat transfer coefficient and the appropriate type of cooling mechanism are to be determined Assumptions Steady operating conditions exist Analysis The average heat transfer coefficient over the surface of the chip is determined from Newton's law of cooling to be 10C Q  hAs (Tchip  T fluid ) Ts = 55C Q 5W h  As (Tchip  T fluid ) (0.4  10 -4 m )(55  10)C 5W  2778 W/m C which is rather high An examination of Fig 15-62 reveals that we can obtain such heat transfer coefficients with the boiling of fluorocarbon fluids Therefore, a suitable cooling technique in this case is immersion cooling in such a fluid 15-135 A chip is cooled by boiling in a dielectric fluid The surface temperature of the chip is to be determined 45C Assumptions The boiling curve in Fig 15-63 is prepared for a chip having a surface area of 0.457 cm2 being cooled in FC86 maintained at 5C The chart can be used for similar cases with reasonable accuracy Analysis The heat flux in this case is q  Q 3W   15 W/cm 2 As 0.2 cm The temperature of the chip surface corresponding to this value is determined from Fig 15-63 to be Tchip  T fluid  63C   Tchip  (T fluid  63)C = (45 + 63)C = 108C 15-72 Chip 3W Chapter 15 Cooling of Electronic Equipment 15-136 A chip is cooled by boiling in a dielectric fluid The maximum power that the chip can dissipate safely is to be determined Assumptions The boiling curve in Fig 15-63 is prepared for a chip having a surface area of 0.457 cm2 being cooled in FC86 maintained at 5C The chart can be used for similar cases with reasonable accuracy Analysis The temperature difference between the 35C chip surface and the liquid is Tchip  T fluid  (60  35) C  25 C Ts = 60C Using this value, the heat flux can be determined from Fig 15-63 to be q  3.3 W/cm Then the maximum power that the chip can dissipate safely becomes Q  qA  (3.3 W/cm )(0.3 cm ) = 0.99 W s 15-137 An electronic device is to be cooled by immersion in a dielectric fluid It is to be determined if the heat generated inside can be dissipated to the ambient air by natural convection and radiation as well as the heat transfer coefficient at the surface of the electronic device Assumptions Steady operating conditions exist 60C Analysis Assuming the surfaces of the cubic enclosure to be at the temperature of the boiling dielectric fluid at 60  C , the rate at which heat can be dissipated to the ambient air at Tair = 20C 20  C by combined natural convection and radiation is determined from kW  Q  hAs (Ts  Tair )  h(6 a )(Ts  Tair )  (10 W/m C)[6(1 m) ](60 - 20)C = 2400 W = 2.4 kW Therefore, the heat generated inside the cubic enclosure can be dissipated by natural convection and radiation The heat transfer coefficient at the surface of the electronic device is Q 2000 W Q  hAs (Ts  T fluid )   h    8333 W/m C As (Ts  T fluid ) (0.012 m )(80  60)C Review Problems 15-138C For most effective cooling, (1) the transistors must be mounted directly over the cooling lines, (2) the thermal contact resistance between the transistors and the cold plate must be minimized by attaching them tightly with a thermal grease, and (3) the thickness of the plates and the tubes should be as small as possible to minimize the thermal resistance between the transistors and the tubes 15-139C There is no such thing as heat rising Only heated fluid rises because of lower density due to buoyancy Heat conduction in a solid is due to the molecular vibrations and electron movement, and gravitational force has no effect on it Therefore, the orientation of the bar is irrelevant 15-73 Chapter 15 Cooling of Electronic Equipment 15-140 A multilayer circuit board consisting of four layers of copper and three layers of glass-epoxy sandwiched together is considered The magnitude and location of the maximum temperature that occurs in the PCB are to be determined Assumptions Steady operating conditions exist Thermal properties are constant There is no direct heat dissipation from the surface of the PCB, and thus all the heat generated is conducted by the PCB to the heat sink Heat sink 11.25 W 10.5 W 9W 7.5 W 6W 4.5 W 3W 1.5 W Center 0.5 cm cm cm Glass-epoxy Analysis The effective thermal conductivity of the board is determined from Copper (k1t1 ) copper  4[(386 W/m.C)(0.0001 m)] = 0.1544 W/ C (k t ) epoxy  3[(0.26 W/m.C)(0.0005 m)] = 0.00039 W/ C k eff  (k 1t1 ) copper  (k t ) epoxy t1  t  81.5 W/m.C  (0.1544  0.00039)W/ C 4(0.0001 m )  3(0.0005 m ) 15 cm 15 cm The maximum temperature will occur in the middle of the board which is farthest away from the heat sink We consider half of the board because of symmetry, and divide the region in 1-cm thick strips, starting at the mid-plane Then from Fourier’s law, the temperature difference across a strip can be determined from T Q  k eff A L   T   QL k eff A where L = cm = 0.01 m (except it is 0.5 cm for the strip attached to the heat sink), and the heat transfer area for all the strips is A  (015 m)[4(0.0001 m) + 3(0.0005 m)]  0.000285 m Then the temperature at the center of the board is determined by adding the temperature differences across all the strips as  (Q  Q  Q  Q  Q  Q  Q  Q / 2) L QL Tcenter  heat sink   k eff A k eff A   and (15   4.5   7.5   10.5  1125 / W)(0.01 m) (815 W / m  C)(0.000285 m )  20.5 C Tcenter  Theat sink  Tcenter  heat sink  35 C + 20.5 C  55.5 C Discussion This problem can also be solved approximately by using the “average” heat transfer rate for the entire half board, and treating it as a constant The heat transfer rate in each half changes from at the center to 22.5/2 = 11.25 W at the heat sink, with an average of 11.25/2 = 5.625 W Then the center temperature becomes T T Q L (5.625 W)(0.075 m) Q ave  k eff A   Tcenter  Theat sink  ave  35 C +  53.2 C L k eff A (81.5 W / m  C)(0.000285 m ) 15-74 Chapter 15 Cooling of Electronic Equipment 15-141 A circuit board consisting of a single layer of glass-epoxy is considered The magnitude and location of the maximum temperature that occurs in the PCB are to be determined Assumptions Steady operating conditions exist Thermal properties are constant There is no direct heat dissipation from the surface of the PCB, and thus all the heat generated is conducted by the PCB to the heat sink Glass-epoxy 1.5 mm 15 cm 15 cm Heat sink 11.25 W 10.5 W 9W 7.5 W 6W 4.5 W 3W 1.5 W Center 0.5 cm cm cm Analysis In this case the board consists of a 1.5-mm thick layer of epoxy Again the maximum temperature will occur in the middle of the board which is farthest away from the heat sink We consider half of the board because of symmetry, and divide the region in 1-cm thick strips, starting at the mid-plane Then from Fourier’s law, the temperature difference across a strip can be determined from T Q  k eff A L   T   QL k eff A where L = cm = 0.01 m (except it is 0.5 cm for the strip attached to the heat sink), and the heat transfer area for all the strips is A  (015 m)(0.0015 m) = 0.000225 m Then the temperature at the center of the board is determined by adding the temperature differences across all the strips as Tcenter  heat sink   and  (Q  Q  Q  Q  Q  Q  Q  Q / 2) L QL  k eff A eff A k (15   4.5   7.5   10.5  1125 / W)(0.01 m) (0.26 W / m  C)(0.000225 m )  8141 C Tcenter  Theat sink  Tcenter  heat sink  35 C + 8141 C  8176 C Discussion This problem can also be solved approximately by using the “average” heat transfer rate for the entire half board, and treating it as a constant The heat transfer rate in each half changes from at the center to 22.5/2 = 11.25 W at the heat sink, with an average of 11.25/2 = 5.625 W Then the center temperature becomes T T Q L (5.625 W)(0.075 m) Q ave  k eff A   Tcenter  Theat sink  ave  35 C +  7247 C L k eff A (0.26 W / m  C)(0.000225 m ) 15-75 Chapter 15 Cooling of Electronic Equipment 15-142 The components of an electronic system located in a horizontal duct of rectangular cross-section are cooled by forced air flowing through the duct The heat transfer from the outer surfaces of the duct by natural convection, the average temperature of the duct and the highest component surface temperature in the duct are to be determined Assumptions Steady operating conditions exist Air is an ideal gas The local atmospheric pressure is atm Air Properties We use the properties of air at 30C (30+45)/2 = 37.5C are (Table A-15) Air duct 10 cm  10 cm   1.136 kg/m C p  1007 J/kg.C 45C 150 W Pr  0.726 k  0.0264 W/m.C L=1m v  1.68  10 5 m / s Air 30C 50 m/min Analysis (a) The volume and the mass flow rates of air are V  Ac V = (0.1 m)(0.1 m)(50/60 m/s) = 0.008333 m /s m  V  (1.136 kg/m )(0.008333 m /s) = 0.009466 kg/s The rate of heat transfer to the air flowing through the duct is Q forced conv  m C p (Tin  Tout )  (0.009466 kg/s)(1007 J/kg.C)(45 - 30)C = 143.0 W Then the rate of heat loss from the outer surfaces of the duct to the ambient air by natural convection becomes Q conv  Q total  Q forced conv  150 W - 143 W = 57 W (b) The average surface temperature can be determined from Q conv  hAs (Ts ,duct  Tambient ) air But we first need to determine convection heat transfer coefficient Using the Nusselt number relation from Table 15-2, As  (4 )(0.1 m)(1 m) = 0.4 m Re  VD h (50 / 60 m/s)(0.1 m)   4960 v 1.68  10 5 m /s Nu  0.102 Re0.675 Pr1 /  (0.102)(4960)0.675 (0.726 )1 /  28.6 h k 0.0264 W/m.C Nu  (28.6 )  7.56 W/m C Dh m Then the average surface temperature of the duct becomes Q 57 W Q conv  hAs (Ts  Tambient )   Ts  Tambient  conv  30C +  48.9C hAs (7.56 W/m C)(0.4 m ) (c) The highest component surface temperature will occur at the exit of the duct From Newton's law relation at the exit, q conv  h(Ts ,max  Tair ,exit )   Ts ,max  Tair ,exit  Q conv / As 57 W  45C +  63.8C h (7.56 W/m C)(0.4 m ) 15-76 Chapter 15 Cooling of Electronic Equipment 15-143 Two power transistors are cooled by mounting them on the two sides of an aluminum bracket that is attached to a liquid-cooled plate The temperature of the transistor case and the fraction of heat dissipation to the ambient air by natural convection and to the cold plate by conduction are to be determined Assumptions Steady operating conditions exist Conduction heat transfer is one-dimensional We assume the ambient temperature is 25C Analysis The rate of heat transfer by conduction is Q conduction  (0.80)(12 W) = 9.6 W Assuming heat conduction in the plate to be one-dimensional for simplicity, the thermal resistance of the aluminum plate and epoxy adhesive are L 0.02 m   0.938 C / W kA (237 W / m  C)(0.003 m)(0.03 m) L 0.0002 m    1235 C / W kA (1.8 W / m  C)(0.003 m)(0.03 m) Ralu um  Repoxy The total thermal resistance of the plate and the epoxy is R plate  epoxy  Repoxy  R plate  1235  0.938  2.173 C / W Heat generated by the transistors is conducted to the plate, and then it is dissipated to the cold plate by conduction, and to the ambient air by convection Denoting the plate temperature where the transistors are connected as Ts,max and using the heat transfer coefficient relation from Table 15-1 for a vertical plate, the total heat transfer from the plate can be expressed as Q total  Q cond  Q conv   Tplate R plate  epoxy Ts , max  Tcold plate R plate  epoxy  hAside (Ts , ave  Tair ) T )   (T  1.42 s , ave air  L   0.25 Aside (Ts , max  Tair ) where Ts, ave = (Ts, max + Tcold plate )/2, L 0.03 m, and Aside = 2(0.03 m)(0.03 m) = 0.00018 m2 Substituting the known quantities gives 20 W  Ts,max  40 2.173 C / W  142 (0.00018) Liquid cooled plate cm [(Ts,max  40) /  25]1.25 Transistor (0.03) 0.25 cm Solving for Ts, max gives Ts, max = 83.3C Then the rate of heat transfer by natural convection becomes Q conv  142 (0.00018) [(83.3  40) /  25]1.25 (0.03) 0.25 Aluminum bracket Plastic washer  0.055 W which is 0.055/20 = 0.00275 or 0.3% of the total heat dissipated The remaining 99.7% of the heat is transferred by conduction Therefore, heat transfer by natural convection is negligible Then the surface temperature of the transistor case becomes  Tcase  Ts,max  QR plastic washer  83.3 C + (10 W)(2 C / W) = 103.3 C 15-77 Chapter 15 Cooling of Electronic Equipment 15-144E A plastic DIP with 24 leads is cooled by forced air Using data supplied by the manufacturer, the junction temperature is to be determined for two cases Assumptions Steady operating conditions exist Analysis The junction-to-ambient thermal resistance of the device with 24 leads corresponding to an air velocity of 500  0.3048 = 152.4 m/min is determined from Fig 15-23 to be Air 70F 500 ft/min R junction  ambient  50  C / W = (50  1.8) + 32 = 122  F / W 1.5 W Then the junction temperature becomes Q  T junction  Tambient R junction  ambient   T junction  Tambient  Q R junction  ambient  70F + (1.5 W)(122 F/W) = 253F When the fan fails the total thermal resistance is determined from Fig 15-23 by reading the value at the intersection of the curve at the vertical axis to be R junction  ambient  66  C / W = (66  1.8) + 32 = 150.8  F / W which yields Q  T junction  Tambient R junction  ambient   T junction  Tambient  Q R junction  ambient  70F + (1.5 W)(150.8 F/W) = 296F 15-78 Chapter 15 Cooling of Electronic Equipment 15-145 A circuit board is to be conduction-cooled by aluminum core plate sandwiched between two epoxy laminates The maximum temperature rise between the center and the sides of the PCB is to be determined Assumptions Steady operating conditions exist Thermal properties are constant There is no direct heat dissipation from the surface of the PCB., and thus all the heat generated is conducted by the PCB to the heat sink T9 PCB 15 cm  18 cm Epoxy adhesive 1W Aluminum core Repoxy Radhesive Cold plate 9W 8W 7W 6W 5W 4W 3W 2W Cold plate 1W Raluminum  Center Analysis Using the half thickness of the aluminum frame because of symmetry, the thermal resistances against heat flow in the vertical direction for a 1-cm wide strip are L 0.0006 m Raluminum,    0.00169  C / W kA (237 W / m  C)(0.01 m)(015 m) L 0.0005 m Repoxy    128205 C / W kA (0.26 W / m  C)(0.01 m)(015 m) L 0.0001 m Radhesive    0.03703  C / W kA (18 W / m  C)(0.01 m)(0.15 m) Rvertical  Raluminum,  Repoxy  Radhesive  0.00169  128205  0.03704  1321 C / W We assume heat conduction along the epoxy and adhesive in the horizontal direction to be negligible, and heat to be conduction to the heat sink along the aluminum frame The thermal resistance of the aluminum frame against heat conduction in the horizontal direction for a 1-cm long strip is L 0.01 m Rframe  Raluminum,||    01953 C / W kA (237 W / m  C)(0.0012 m)(0.18 m) The temperature difference across a strip is determined from  T  QR aluminum,|| The maximum temperature rise across the cm distance between the center and the sides of the board is determined by adding the temperature differences across all the strips as  T  QR  (Q  Q  Q  Q  Q  Q  Q  Q 2) R horizontal  aluminum,|| aluminum,||  (1         W)(0.1953  C / W) = 8.8 C The temperature difference between the center of the aluminum core and the outer surface of the PCB is determined similarly to be  Tvertical  QR vertical,   (1 W)(1.321  C / W) = 1.3 C  Then the maximum temperature rise across the 9-cm distance between the center and the sides of the PCB becomes Tmax  Thorizontal  Tvertical  8.8  13  10.1 C 15-79 Chapter 15 Cooling of Electronic Equipment 15-146 Ten power transistors attached to an aluminum plate are cooled from two sides of the plate by liquid The temperature rise between the transistors and the heat sink is to be determined Assumptions Steady operating conditions exist Thermal properties are constant Analysis We consider only half of the plate because of symmetry It is stated that 70% of the heat generated is conducted through the aluminum plate, and this heat will be conducted across the 1-cm wide section between the transistors and the cooled edge of the plate (Note that the mid section of the plate will essentially be isothermal and thus there will be no significant heat transfer towards the midsection) The rate of heat conduction to each side is of the plate is Transistor Aluminum plate Cut-out section cm  cm cm Q cond,1-side  0.70  (10 W )  W Then the temperature rise across the 1-cm wide section of the plate can be determined from Q cond 1-side  kA cm ( T ) plate 40C L Solving for ( T ) plate and substituting gives (T ) plate  Q cond 1-side L (7 W)(0.01 m)   2.1C kA (237 W/m.C)(0.07  0.002 m ) 15-80 cm cm 40C Chapter 15 Cooling of Electronic Equipment 15-147 The components of an electronic system located in a horizontal duct are cooled by air flowing over the duct The total power rating of the electronic devices that can be mounted in the duct is to be determined for two cases Assumptions Steady operating conditions exist Thermal properties of air are constant The local atmospheric pressure is atm Properties The properties of air at the film temperature of (30+60)/2 = 45C are (Table A-15) Pr  0.724 k  0.0270 W/m.C v  1.75  10 5 m / s Air 30C 250 m/min Analysis The surface area of the duct is 150 W Ts < 60C As  2[(1.2 m)(0.1 m)]  [(1.2 m)(0.2 m)] = 0.72 m The duct is oriented such that air strikes the 10 cm high side normally Using the Nusselt number relation from Table 15-2 for a 10-cm by 10-cm cross-section square as an approximation, the heat transfer coefficient is determined to be Re  L = 1.2 m Air duct 10 cm  20 cm VD (250 / 60 m/s)(0.1 m)   23,810 v 1.75  10 5 m /s Nu  0.102 Re0.675 Pr1 /  (0.102)(23,810)0.675 (0.724 )1 /  82.4 h k 0.0270 W/m.C Nu  (82.4 )  22.3 W/m C Dh m Then the heat transfer rate (and thus the power rating of the components inside) in this orientation is determined from Q  hAs (Ts  T fluid )  (22.3 W/m C)(0.72 m )(60 - 30)C = 481 W We now consider the duct oriented such that air strikes the 20 cm high side normally Using the Nusselt number relation from Table 15-2 for a 20-cm by 20-cm cross-section square as an approximation, the heat transfer coefficient is determined to be Re  VD (250 / 60 m/s)(0.2 m)   47,619 v 1.75  10 5 m /s Nu  0.102 Re0.675 Pr1 /  (0.102)(47,619)0.675 (0.724 )1 /  131.6 h k 0.0270 W/m.C Nu  (131.6 )  17.8 W/m C Dh m Then the heat transfer rate (and thus the power rating of the components inside) in this orientation is determined from Q  hAs (Ts  T fluid )  (17.8 W/m C)(0.72 m )(60 - 30)C = 384 W 15-81 Chapter 15 Cooling of Electronic Equipment 15-148 The components of an electronic system located in a horizontal duct are cooled by air flowing over the duct The total power rating of the electronic devices that can be mounted in the duct is to be determined for two cases Assumptions Steady operating conditions exist Thermal properties of air are constant The local atmospheric pressure is atm Properties The properties of air at the film temperature of (30+60)/2 = 45C and 54.05 kPa are (Table A15) Pr  0.724 k  0.0270 W/m.C v 1.75  10 5 m / s  3.28  10 5 m / s 54.05/101.325 Analysis The surface area of the duct is Air 30C 250 m/min 150 W Ts < 60C As  2[(1.2 m)(0.1 m)]  [(1.2 m)(0.2 m)] = 0.72 m The duct is oriented such that air strikes the 10 cm high side normally Using the Nusselt number relation from Table 15-2 for a 10-cm by 10-cm cross-section square as an approximation, the heat transfer coefficient is determined to be Re  L = 1.2 m Air duct 10 cm  20 cm VD (250 / 60 m/s)(0.1 m)   12,703 v 3.28  10 5 m /s Nu  0.102 Re0.675 Pr1 /  (0.102)(12,703)0.675 (0.724 )1 /  53.9 h k 0.0270 W/m.C Nu  (53.9)  14.6 W/m C Dh m Then the heat transfer rate (and thus the power rating of the components inside) in this orientation is determined from Q  hAs (Ts  T fluid )  (14.6 W/m C)(0.72 m )(60 - 30)C = 315 W We now consider the duct oriented such that air strikes the 20 cm high side normally Using the Nusselt number relation from Table 15-2 for a 20-cm by 20-cm cross-section square as an approximation, the heat transfer coefficient is determined to be Re  VD (250 / 60 m/s)(0.2 m)   25,407 v 3.28  10 5 m /s Nu  0.102 Re0.675 Pr1 /  (0.102)(25,407)0.675 (0.724 )1 /  86.1 h k 0.0270 W/m.C Nu  (86.1)  11.6 W/m C Dh m Then the heat transfer rate (and thus the power rating of the components inside) in this orientation is determined from Q  hAs (Ts  T fluid )  (11.6 W/m C)(0.72 m )(60 - 30)C = 251 W 15-82 Chapter 15 Cooling of Electronic Equipment 15-149E A computer is cooled by a fan blowing air into the computer enclosure The fraction of heat lost from the outer surfaces of the computer case is to be determined Assumptions Steady operating conditions exist Thermal properties of air are constant The local atmospheric pressure is atm Analysis Using the proper relation from Table 15-1, the heat transfer coefficient and the rate of natural convection heat transfer from the vertical side surfaces are determined to be ft 12 24    20  (2) ft + ft  ft   3.67 ft 12  12   12 L Aside  T  hconv, side  1.42   L  Q h A conv , side conv , side 0.25  95  80   1.42   / 12  side (T s 0.25  3.32 Btu/h.ft F  T fluid )  (3.32 Btu/h.ft F)(3.67 ft )(95  80)F = 182.9 Btu/h Similarly, the rate of heat transfer from the horizontal top surface by natural convection is determined to be  20  24  4 ft  ft   12  12  L   1.82 ft p  20   24   ft    ft   12   12   20  24  Atop   ft  ft   3.33 ft 12 12    Atop 0.25 Air 80F 95F 80 W 0.25  T   95  80  hconv,top  1.32  1.32  2.24 Btu/h.ft F    L   1.82  Q conv,top  hconv,top Atop (Ts  T fluid )  (2.24 Btu/h.ft F)(3.33 ft )(95  80)F = 111.7 Btu/h The rate of heat transfer from the outer surfaces of the computer case by radiation is Q rad  As  (Ts  Tsurr )  (0.85)(3.67 ft  3.33 ft )(0.1714 Btu/h.ft R )[(95  460 R)  (80  273 R) ]  100.4 Btu/h Then the total rate of heat transfer from the outer surfaces of the computer case becomes Q  Q  Q  Q  182.9  1117  100.4  395 Btu / h total conv , side conv ,top rad Therefore, the fraction of the heat loss from the outer surfaces of the computer case is f  (395 / 3.41214) W  0.68  68% 170 W 15-150 15-152 Design and Essay Problems  15-83 Computer case Preface This manual is prepared as an aide to the instructors in correcting homework assignments, but it can also be used as a source of additional example problems for use in the classroom With this in mind, all solutions are prepared in full detail in a systematic manner, using a word processor with an equation editor The solutions are structured into the following sections to make it easy to locate information and to follow the solution procedure, as appropriate: Solution Assumptions Properties Analysis Discussion - The problem is posed, and the quantities to be found are stated The significant assumptions in solving the problem are stated The material properties needed to solve the problem are listed The problem is solved in a systematic manner, showing all steps Comments are made on the results, as appropriate A sketch is included with most solutions to help the students visualize the physical problem, and also to enable the instructor to glance through several types of problems quickly, and to make selections easily Problems designated with the CD icon in the text are also solved with the EES software, and electronic solutions complete with parametric studies are available on the CD that accompanies the text Comprehensive problems designated with the computer-EES icon [pick one of the four given] are solved using the EES software, and their solutions are placed at the Instructor Manual section of the Online Learning Center (OLC) at www.mhhe.com/cengel Access to solutions is limited to instructors only who adopted the text, and instructors may obtain their passwords for the OLC by contacting their McGraw-Hill Sales Representative at http://www.mhhe.com/catalogs/rep/ Every effort is made to produce an error-free Solutions Manual However, in a text of this magnitude, it is inevitable to have some, and we will appreciate hearing about them We hope the text and this Manual serve their purpose in aiding with the instruction of Heat Transfer, and making the Heat Transfer experience of both the instructors and students a pleasant and fruitful one We acknowledge, with appreciation, the contributions of numerous users of the first edition of the book who took the time to report the errors that they discovered All of their suggestions have been incorporated Special thanks are due to Dr Mehmet Kanoglu who checked the accuracy of most solutions in this Manual Yunus A Çengel July 2002 ... the heat transfer coefficient to be uniform, heat transfer is proportional to the surface area Then the fraction of heat dissipated from the top and bottom surfaces of the resistor becomes Qtop...Chapter Basics of Heat Transfer Chapter BASICS OF HEAT TRANSFER Thermodynamics and Heat Transfer 1-1C Thermodynamics deals with the amount of heat transfer as a system undergoes... Chapter Basics of Heat Transfer Heat and Other Forms of Energy 1-8C The rate of heat transfer per unit surface area is called heat flux q It is related to the rate of heat transfer by Q  