Thermodynamics Yunus Solns

Heat Transfer Mechanisms 1-44C The thermal conductivity of a material is the rate of heat transfer through a unit thickness of the material per unit area and per unit temperature difference. The thermal conductivity of a material is a measure of how fast heat will be conducted in that material. 1-45C The mechanisms of heat transfer are conduction, convection and radiation. Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles. Convection is the mode of energy transfer between a solid surface and the adjacent liquid or gas which is in motion, and it involves combined effects of conduction and fluid motion. Radiation is energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules. 1-46C In solids, conduction is due to the combination of the vibrations of the molecules in a lattice and the energy transport by free electrons. In gases and liquids, it is due to the collisions of the molecules during their random motion. 1-47C The parameters that effect the rate of heat conduction through a windowless wall are the geometry and surface area of wall, its thickness, the material of the wall, and the temperature difference across the wall.

dT 1-48C Conduction is expressed by Fourier's law of conduction as Q cond = − kA where dT/dx is the temperature dx gradient, k is the thermal conductivity, and A is the area which is normal to the direction of heat transfer. Convection is expressed by Newton's law of cooling as Q = hA (T − T ) where h is the convection conv

s

s

∞

heat transfer coefficient, As is the surface area through which convection heat transfer takes place, Ts is the surface temperature and T∞ is the temperature of the fluid sufficiently far from the surface. Radiation is expressed by Stefan-Boltzman law as Q rad = εσAs (Ts 4 − Tsurr 4 ) where ε is the emissivity of surface, As is the surface area, Ts is the surface temperature, Tsurr is average surrounding surface temperature and σ = 5.67 × 10 −8 W / m 2 .K 4 is the Stefan-Boltzman constant. 1-49C Convection involves fluid motion, conduction does not. In a solid we can have only conduction. 1-50C No. It is purely by radiation. 1-51C In forced convection the fluid is forced to move by external means such as a fan, pump, or the wind. The fluid motion in natural convection is due to buoyancy effects only. 1-52C Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a blackbody at the same temperature. Absorptivity is the fraction of radiation incident on a surface that is absorbed by the surface. The Kirchhoff's law of radiation states that the emissivity and the absorptivity of a surface are equal at the same temperature and wavelength. 1-53C A blackbody is an idealized body which emits the maximum amount of radiation at a given temperature and which absorbs all the radiation incident on it. Real bodies emit and absorb less radiation than a blackbody at the same temperature.

1-54C No. Such a definition will imply that doubling the thickness will double the heat transfer rate. The equivalent but “more correct” unit of thermal conductivity is W.m/m2.°C that indicates product of heat transfer rate and thickness per unit surface area per unit temperature difference. 1-55C In a typical house, heat loss through the wall with glass window will be larger since the glass is much thinner than a wall, and its thermal conductivity is higher than the average conductivity of a wall. 1-56C Diamond is a better heat conductor. 1-57C The rate of heat transfer through both walls can be expressed as T −T T −T Q wood = k wood A 1 2 = (0.16 W/m. °C) A 1 2 = 1.6 A(T1 − T2 ) L wood 0 .1 m T −T T −T Q brick = k brick A 1 2 = (0.72 W/m. °C) A 1 2 = 2.88 A(T1 − T2 ) L brick 0.25 m where thermal conductivities are obtained from table A-5. Therefore, heat transfer through the brick wall will be larger despite its higher thickness. 1-58C The thermal conductivity of gases is proportional to the square root of absolute temperature. The thermal conductivity of most liquids, however, decreases with increasing temperature, with water being a notable exception. 1-59C Superinsulations are obtained by using layers of highly reflective sheets separated by glass fibers in an evacuated space. Radiation heat transfer between two surfaces is inversely proportional to the number of sheets used and thus heat loss by radiation will be very low by using this highly reflective sheets. At the same time, evacuating the space between the layers forms a vacuum under 0.000001 atm pressure which minimize conduction or convection through the air space between the layers. 1-60C Most ordinary insulations are obtained by mixing fibers, powders, or flakes of insulating materials with air. Heat transfer through such insulations is by conduction through the solid material, and conduction or convection through the air space as well as radiation. Such systems are characterized by apparent thermal conductivity instead of the ordinary thermal conductivity in order to incorporate these convection and radiation effects. 1-61C The thermal conductivity of an alloy of two metals will most likely be less than the thermal conductivities of both metals. 1-62 The inner and outer surfaces of a brick wall are maintained at specified temperatures. The rate of heat transfer through the wall is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal properties of the wall are constant. Properties The thermal conductivity of the wall is given to be k = 0.69 W/m⋅ °C. Analysis Under steady conditions, the rate of heat transfer through the wall is (20 − 5)°C ∆T Q cond = kA = (0.69W/m ⋅ °C)(5 × 6m 2 ) = 1035W L 0.3m

Brick wall 0.3 m

30 cm 20°C

5°C

1-63 The inner and outer surfaces of a window glass are maintained at specified temperatures. The amount of heat transfer through the glass in 5 h is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Thermal properties of the glass are constant. Properties The thermal conductivity of the glass is given to be k = 0.78 W/m⋅ °C. Analysis Under steady conditions, the rate of heat transfer through the glass by conduction is (10 − 3)°C ∆T Q cond = kA = (0.78 W/m ⋅ °C)(2 × 2 m 2 ) = 4368 W L 0.005m

Glass

Then the amount of heat transfer over a period of 5 h becomes Q = Q ∆t = (4.368 kJ/s)(5 × 3600 s) = 78,620 kJ cond

If the thickness of the glass doubled to 1 cm, then the amount of heat transfer will go down by half to 39,310 kJ. 10°C

3°C 0.5 cm

1-76 The heat generated in the circuitry on the surface of a 3-W silicon chip is conducted to the ceramic substrate. The temperature difference across the chip in steady operation is to be determined. Assumptions 1 Steady operating conditions exist. 2 Thermal properties of the chip are constant. Properties The thermal conductivity of the silicon chip is given to be k = 130 W/m⋅ °C. Analysis The temperature difference between the front and back surfaces of the chip is A = (0.006 m)(0.006 m) = 0.000036 m 2

Q L (3 W)(0.0005 m) ∆T Q = kA  → ∆T = = = 0.32°C L kA (130 W/m.°C)(0.00003 6 m 2 )

 Q 3W Ceramic substrate

Chip 6 × 6 × 0.5 mm

1-79 A hollow spherical iron container is filled with iced water at 0°C. The rate of heat loss from the sphere and the rate at which ice melts in the container are to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Heat transfer through the shell is one-dimensional. 3 Thermal properties of the iron shell are constant. 4 The inner surface of the shell is at the same temperature as the iced water, 0°C. Properties The thermal conductivity of iron is k = 80.2 W/m⋅ °C (Table A-3). The heat of fusion of water is given to be 333.7 kJ/kg. Analysis This spherical shell can be approximated as a plate of thickness 0.4 cm and area A = πD² = π(0.2 m)² = 0.126 m² Then the rate of heat transfer through the shell by conduction is (5 − 0)°C ∆T Q cond = kA = (80.2W/m ⋅ °C)(0.126 m 2 ) = 12,632 W L 0.004 m

5°C

Considering that it takes 333.7 kJ of energy to melt 1 kg of ice at 0°C, the rate at which ice melts in the container can be determined from Q 12.632 kJ / s m ice = = = 0.038 kg / s hif 333.7 kJ / kg

Iced water 0°C

0.4 cm

Discussion We should point out that this result is slightly in error for approximating a curved wall as a plain wall. The error in this case is very small because of the large diameter to thickness ratio. For better accuracy, we could use the inner surface area (D = 19.2 cm) or the mean surface area (D = 19.6 cm) in the calculations.

1-88 A 4-m diameter spherical tank filled with liquid nitrogen at 1 atm and -196°C is exposed to convection with ambient air. The rate of evaporation of liquid nitrogen in the tank as a result of the heat transfer from the ambient air is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is disregarded. 3 The convection heat transfer coefficient is constant and uniform over the surface. 4 The temperature of the thin-shelled spherical tank is nearly equal to the temperature of the nitrogen inside. Properties The heat of vaporization and density of liquid nitrogen at 1 atm are given to be 198 kJ/kg and 810 kg/m3, respectively. Analysis The rate of heat transfer to the nitrogen tank is Vapor As = πD 2 = π (4 m) 2 = 50.27 m 2 Air Q = hAs (Ts − Tair ) = (25 W/m 2 .°C) (50.27 m 2 )[ 20 − (−196 )]°C 20°C 1 atm = 271,430 W Liquid N2 Then the rate of evaporation of liquid nitrogen in the tank is determined to be -196°C Q 271.430 kJ/s Q = m h fg  → m = = = 1.37 kg/s h fg 198 kJ/kg

 Q

1-93 A sealed electronic box dissipating a total of 100 W of power is placed in a vacuum chamber. If this box is to be cooled by radiation alone and the outer surface temperature of the box is not to exceed 55°C, the temperature the surrounding surfaces must be kept is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection is disregarded. 3 The emissivity of the box is constant and uniform over the exposed surface. 4 Heat transfer from the bottom surface of the box to the stand is negligible. Properties The emissivity of the outer surface of the box is given to be 0.95. Analysis Disregarding the base area, the total heat transfer area of the electronic box is

As = (0.4 m)(0.4 m) + 4 × (0.2 m )(0.4 m ) = 0.48 m 2 The radiation heat transfer from the box can be expressed as Q = εσA (T 4 − T 4 ) rad

s

s

surr −8

100 W = (0.95)(5.67 × 10

[

W/m 2 .K 4 )(0.48 m 2 ) (55 + 273 K ) 4 − T surr 4

]

100 W ε = 0.95 Ts =55°C

which gives Tsurr = 296.3 K = 23.3°C. Therefore, the temperature of the surrounding surfaces must be less than 23.3°C.

1-102 The convection heat transfer coefficient for heat transfer from an electrically heated wire to air is to be determined by measuring temperatures when steady operating conditions are reached and the electric power consumed. Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Radiation heat transfer is negligible. Analysis In steady operation, the rate of heat loss from the wire equals the rate of heat generation in the wire as a result of resistance heating. That is, Q = E = VI = (110 V)(3 A) = 330 W 240°C generated

The surface area of the wire is

As = (πD ) L = π (0.002 m)(1.4 m) = 0.00880 m 2

L = 1.4 m The Newton's law of cooling for convection heat transfer is expressed as Q = hA (T − T ) s

s

D =0.2 cm

Q Air, 20°C

∞

Disregarding any heat transfer by radiation , the convection heat transfer coefficient is determined to be Q 330 W h= = = 170.5 W/m 2 .°C As (T1 − T∞ ) (0.00880 m 2 )( 240 − 20 )°C Discussion If the temperature of the surrounding surfaces is equal to the air temperature in the room, the value obtained above actually represents the combined convection and radiation heat transfer coefficient.

1-107 A spherical tank located outdoors is used to store iced water at 0°C. The rate of heat transfer to the iced water in the tank and the amount of ice at 0° C that melts during a 24-h period are to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal properties of the tank and the convection heat transfer coefficient is constant and uniform. 3 The average surrounding surface temperature for radiation exchange is 15°C. 4 The thermal resistance of the tank is negligible, and the entire steel tank is at 0°C. Properties The heat of fusion of water at atmospheric pressure is hif = 333. 7 kJ / kg . The emissivity of the outer surface of the tank is 0.6. Analysis (a) The outer surface area of the spherical tank is

As = πD 2 = π (3.02 m) 2 = 28.65 m 2 Then the rates of heat transfer to the tank by convection and radiation become Q = hA (T − T ) = (30 W/m 2 .°C)(28.65 m 2 )( 25 − 0)°C = 21,488 W conv

s

∞

s

4 Q rad = εAs σ (Tsurr − Ts4 ) = (0.6)( 28.65 m 2 )(5.67 × 10 -8 W/m 2 .K 4 )[( 288 K) 4 − (273 K ) 4 ] = 1292 W Q total = Q conv + Q rad = 21,488 + 1292 = 22,780 W

(b) The amount of heat transfer during a 24-hour period is Q = Q ∆t = (22.78 kJ/s)(24 × 3600 s) = 1,968,000 kJ

Air 25°C

Then the amount of ice that melts during this period becomes Q = mhif  → m =

Q 1,968,000 kJ = = 5898 kg hif 333 .7 kJ/kg

 Q

0°C Iced water 0°C

Discussion The amount of ice that melts can be reduced to a small fraction by insulating the tank.

1 cm

1-122 The duct of an air heating system of a house passes through an unheated space in the attic. The rate of heat loss from the air in the duct to the attic and its cost under steady conditions are to be determined. Assumptions 1 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. 2 Steady operating conditions exist since there is no change with time at any point and thus ∆mCV = 0 and ∆E CV = 0 . 3 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0 . 4 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and airconditioning applications. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The specific heat of air at room temperature is Cp = 1.007 kJ/kg·°C (Table A-15). Analysis We take the heating duct as the system. This is a control volume since mass crosses the system boundary during the process. There is only one inlet and one exit and thus m 1 = m 2 = m . Then the energy balance for this steady-flow system can be expressed in the rate form as E − E = ∆E system 0 (steady) = 0 → E in = E out inout     Rate of net energy transfer by heat, work, and mass

Rate of change in internal, kinetic, potential, etc. energies

 1 = Q out + mh  2 (since ∆ke ≅ ∆pe ≅ 0) mh  p (T1 − T2 ) Q out = mC

3 m/s

AIR

· Q

The density of air at the inlet conditions is determined from the ideal gas relation to be P 100 kPa ρ= = = 1031 . kg / m 3 RT (0.287 kPa.m 3 / kg.K)(65+ 273)K The cross-sectional area of the duct is

Ac = πD 2 / 4 = π (0.20 m)2 / 4 = 0.0314 m 2 Then the mass flow rate of air through the duct and the rate of heat loss become

m = ρAc V = (1031 . kg / m 3 )(0.0314 m2 )(3 m / s) = 0.0971 kg / s and Q loss = m C p (Tin − Tout ) = (0.0971 kg/s)(1.0 07 kJ/kg. °C)(65 − 60 )°C = 0.489 kJ/s or 1760 kJ/h. The cost of this heat loss to the home owner is (Rate of heat loss)(Unit cost of energy input) Cost of Heat Loss = Furnace efficiency =

(1760 kJ/h)($0.5 8/therm) 0.82

 1 therm     105,500 kJ 

= $0.012/h Discussion The heat loss from the heating ducts in the attic is costing the homeowner 1.2 cents per hour. Assuming the heater operates 2,000 hours during a heating season, the annual cost of this heat loss adds up to $24. Most of this money can be saved by insulating the heating ducts in the unheated areas.

1–44C Define thermal conductivity and explain its significance in heat transfer. 1–45C What are the mechanisms of heat transfer? How are they distinguished from each other? 1–46C What is the physical mechanism of heat conduction in a solid, a liquid, and a gas? 1–47C Consider heat transfer through a windowless wall of a house in a winter day. Discuss the parameters that affect the rate of heat conduction through the wall. 1–48C Write down the expressions for the physical laws that govern each mode of heat transfer, and identify the variables involved in each relation. 1–49C How does heat conduction differ from convection? 1–50C Does any of the energy of the sun reach the earth by conduction or convection? 1–51C How does forced convection differ from natural convection? 1–52C Define emissivity and absorptivity. What is Kirchhoff’s law of radiation? 1–53C What is a blackbody? How do real bodies differ from blackbodies? 1–54C Judging from its unit W/m · °C, can we define thermal conductivity of a material as the rate of heat transfer through the material per unit thickness per unit temperature difference? Explain. 1–55C Consider heat loss through the two walls of a house on a winter night. The walls are identical, except that one of them has a tightly fit glass window. Through which wall will the house lose more heat? Explain. 1–56C Which is a better heat conductor, diamond or silver? 1–57C Consider two walls of a house that are identical except that one is made of 10-cm-thick wood, while the other is made of 25-cm-thick brick. Through which wall will the house lose more heat in winter? 1–58C How do the thermal conductivity of gases and liquids vary with temperature? 1–59C Why is the thermal conductivity of super insulation orders of magnitude lower than the thermal conductivity of ordinary insulation? 1–60C Why do we characterize the heat conduction ability of insulators in terms of their apparent thermal conductivity instead of the ordinary thermal conductivity? 1–61C Consider an alloy of two metals whose thermal conductivities are k1 and k2. Will the thermal conductivity of the alloy be less than k1, greater than k2, or between k1 and k2? 1–62 The inner and outer surfaces of a 5-m _ 6-m brick wall of thickness 30 cm and thermal conductivity 0.69 W/m · °C are maintained at temperatures of 20°C and 5°C, respectively. Determine the rate of heat transfer through the wall, in W. Answer: 1035 W

1–63 The inner and outer surfaces of a 0.5-cm-thick 2-m _ 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass is 0.78 W/m · °C, determine the amount of heat loss, in kJ, through the glass over a period of 5 hours. What would your answer be if the glass were 1 cm thick? Answers: 78,624 kJ, 39,312 kJ 1–76 The heat generated in the circuitry on the surface of a silicon chip (k _ 130 W/m · °C) is conducted to the ceramic substrate to which it is attached. The chip is 6 mm _ 6 mm in size and 0.5 mm thick and dissipates 3Wof power. Disregarding any heat transfer through the 0.5-mm-high side surfaces, determine the temperature difference between the front and back surfaces of the chip in steady operation. 1–79 A hollow spherical iron container with outer diameter 20 cm and thickness 0.4 cm is filled with iced water at 0°C. If the outer surface temperature is 5°C, determine the approximate rate of heat loss from the sphere, in kW, and the rate at which ice melts in the container. The heat from fusion of water is 333.7 kJ/kg. 1–88 The boiling temperature of nitrogen at atmospheric pressure at sea level (1 atm) is 196°C. Therefore, nitrogen is commonly used in low temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at 196°C until the liquid nitrogen in the tank is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of 198 kJ/kg and a density of 810 kg/m3 at 1 atm. Consider a 4-m-diameter spherical tank initially filled with liquid nitrogen at 1 atm and 196°C. The tank is exposed to 20°C ambient air with a heat transfer coefficient of 25 W/m2 · °C. The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Disregarding any radiation heat exchange, determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air. 1–93 Consider a sealed 20-cm-high electronic box whose base dimensions are 40 cm 40 cm placed in a vacuum chamber. The emissivity of the outer surface of the box is 0.95. If the electronic components in the box dissipate a total of 100 W of power and the outer surface temperature of the box is not to exceed 55°C, determine the temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible.

1–102 A1.4-m-long, 0.2-cm-diameter electrical wire extends across a room that is maintained at 20°C. Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be 240°C in steady operation. Also, the voltage drop and electric current through the wire are measured to be 110 V and 3 A, respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room. Answer: 170.5 W/m2 · °C 1–107 A3-m-internal-diameter spherical tank made of 1-cmthick stainless steel is used to store iced water at 0°C. The tank is located outdoors at 25°C. Assuming the entire steel tank to be at 0°C and thus the thermal resistance of the tank to be negligible, determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0°C that melts during a 24-hour period. The heat of fusion of water at atmospheric pressure is hif 333.7 kJ/kg. The emissivity of the outer surface of the tank is 0.6, and the convection heat transfer coefficient on the outer surface can be taken to be 30 W/m2 · °C. Assume the average surrounding surface temperature for radiation exchange to be 15°C. Answer: 5898 kg

1–122 A4-m-long section of an air heating system of a house passes through an unheated space in the attic. The inner diameter of the circular duct of the heating system is 20 cm. Hot air enters the duct at 100 kPa and 65°C at an average velocity of 3 m/s. The temperature of the air in the duct drops to 60°C as a result of heat loss to the cool space in the attic. Determine the rate of heat loss from the air in the duct to the attic under steady conditions. Also, determine the cost of this heat loss per hour if the house is heated by a natural gas furnace having an efficiency of 82 percent, and the cost of the natural gas in that area is $0.58/therm (1 therm 105,500 kJ). Answers: 0.488 kJ/s, $0.012/h