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3-1 An arctic explorer builds a temporary shelter from wind-pack snow. The shelter is roughly hemispherical with an inside radius of 1.5 meters. After completing the shelter, the explorer crawls inside and closes off the entrance with a block of snow. Assume the shelter is now air tight and loses negligible heat by conduction through the walls. If the air temperature when the explorer completes the shelter is -10 o C, how long will it take before the air temperature inside reaches 10 o C? Assume the explorer does not freeze to death or suffocate, but sits patiently waiting for the temperature to rise. The explorer generates body heat at a rate of 300 kJ/h. Approach: Use the first law to find the change in temperature. Assumptions: 1. The air behaves like an ideal gas under these conditions. 2. The shelter is perfectly insulated and air-tight. 3. The specific heat of the air is constant. Solution: Let the system be the air inside the shelter. From the first law v Q U mc T =∆ = The mass of air can be determined from the ideal gas law ( ) ( ) ( ) ( )( ) 3 3 1 4 101kPa π 1.5 m 28.97 kg kmol 2 3 9.46 kg 8.314 kJ kmol K 10 + 273 K PVM m RT ⎞⎛ ⎟⎜ ⎠⎝ = = = The only heat added to the air is body heat. The walls are assumed to be thick and highly insulating. The rate of heat transfer is related to the total heat transferred by v Q Qt mc T = ∆= Solving for elapsed time ( ) ( ) ( ) ( ( ) ) 2 1 9.46 kg 0.717 kJ kg K 10 10 K 300kJ h v mc T T t Q t ∆= −− ∆= where c v may be found in Table A-8. Evaluating 0.45h 27 min t = = Answer Comments: In actuality, there must be some air entering and leaving the shelter or the explorer would be unable to breathe; therefore, the rise in temperature may not be as rapid as calculated. 3 - 1

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3-1

An arctic explorer builds a temporary shelter from wind-pack snow. The shelter is roughly hemispherical with an inside radius of 1.5 meters. After completing the shelter, the explorer crawls inside and closes off the entrance with a block of snow. Assume the shelter is now air tight and loses negligible heat by conduction through the walls. If the air temperature when the explorer completes the shelter is -10oC, how long will it take before the air temperature inside reaches 10oC? Assume the explorer does not freeze to death or suffocate, but sits patiently waiting for the temperature to rise. The explorer generates body heat at a rate of 300 kJ/h.

Approach:Use the first law to find the change in temperature.

Assumptions:1. The air behaves like an ideal gas under these conditions. 2. The shelter is perfectly insulated and air-tight. 3. The specific heat of the air is constant.

Solution:Let the system be the air inside the shelter. From the first law Q = U = mcv T The mass of air can be determined from the ideal gas law 1 4 3 (101kPa ) (1.5 ) m3 ( 28.97 kg kmol ) PVM 2 3 m= = = 9.46 kg RT ( 8.314 kJ kmol K )( 10 + 273) K The only heat added to the air is body heat. The walls are assumed to be thick and highly insulating. The rate of heat transfer is related to the total heat transferred by Q = Qt = mcv T Solving for elapsed time mc (T T ) t = v 2 1 Q300 kJ h where cv may be found in Table A-8. Evaluatingt =0.45 h = 27 min t =

( 9.46 kg )( 0.717 kJ

kg K ) (10 ( 10 ) ) K

Comments:In actuality, there must be some air entering and leaving the shelter or the explorer would be unable to breathe; therefore, the rise in temperature may not be as rapid as calculated.

3- 1

3-2

A well-insulated room with a volume of 60 m3 contains air initially at 100 kPa and 25oC. A 100-W light bulb is turned on for 3 hours. Assuming the room is air-tight, estimate the final temperature.

Approach:Use the first law to find the change in temperature.

Assumptions:1. The air behaves like an ideal gas under these conditions. 2. The room is perfectly insulated and air-tight. 3. The specific heat of the air is constant.

Solution:Let the system be the air and the light bulb. From the first law U = Q W The room is well-insulated, so Q = 0 . The electrical work done is 3600 s W = (100 W)(3h) 1h = 1.08 106 J Work is negative since it is done on the air. Because the air is assumed to be an ideal gas, the change in internal energy may be written in terms of the specific heat. Using this in the first law produces: mcV (T2 T1 ) = W To find the mass of air, m, use the ideal gas law PVM m= RT kg 1000 Pa (100 kPa)(60 m3 ) 28.97 kmol 1kPa m= kJ 1000 J 8.314 (25 + 273)K kmol K 1kJ = 70.2 kg where the molecular weight of air from Table A-1 has been used. For cV, use Table A-8 at 300K. Solving the first law for final temperature: W T2 = + T1 mcV

1kJ (1.08 106 )J 1000 J + 25 o C T2 = kJ (70.2 kg) 0.721 kg K = 46.3 o C

Comments:In actuality, there would be some air entering and leaving the room through doors, windows, etc. and some conduction through the walls. Therefore, the room air would not reach this elevated temperature.

3- 2

3-3

An elevator is required to carry 8 people to the top of a 12-story building in less than 1 min. A counterweight is used to balance the mass of the empty elevator cage. Assume that an average person weighs 155 lbf and that each story has a height of 12 ft. What is the minimum size of motor (in hp) that can be used in this application?

Approach:Write the first law in rate form. Eliminate all terms except work and potential energy.

Assumptions:1. The motor speed is constant. 2. The elevator is isothermal. 3. The elevator is adiabatic.

Solution:Take the elevator to be the system under study. From the first law: dKE dPE dU + + = Q W dt dt dt The elevator starts and stops at rest. Thus there is no net change in kinetic energy. The elevator is isothermal and its internal energy does not change. Furthermore, there is no heat transfer to or from the elevator. With these considerations, the first law reduces to dPE = W dt For a 1-minute time period (assuming a constant speed elevator) mg z = W t Each person has a mass of 155 lbm. Soft (8)(155lbm) 32.2 2 (12)(12 ft) s = W 2 60s 32.2 lbm ft/s (1min) 1lbf 1 min ft lbf 1hp W = 2976 s 550 ft lbf s = 5.41 hp

Comments:Work is negative because work is being done on the elevator cage.

3- 3

3-4

A climate controlled room in a semiconductor factory contains a conveyor belt. Electric power is supplied to the motor of the conveyor belt at 220 V and a current which varies linearly with time as: I = 1.0 t, where I is in amps when t is in minutes. An air conditioner removes heat from the room at a constant rate of 2 kW. The volume of air in the room is 600 m3. At t = 0, the air is at 25C and 101 kPa. Assume the mass of air is constant during this process and assume constant specific heats. a. Find the mass of air in the room (in kg). b. Find the air temperature after 30 min. (in C); ignore any temperature change of the motor or conveyor belt.

Approach:Find mass from the ideal gas law. Write the first law in rate form. Express rate of work as voltage times current, and substitute this expression in the first law. Integrate over time.

Assumptions:1. Air behaves like an ideal gas under these conditions. 2. The temperature of the conveyor belt and motor is unchanged. 3. No air enters or leaves the room. 4. The specific heat is constant.

Solution:a) From the ideal gas law: 1000 J kmol K )( 298 K ) RT 1kg = 0.847 m3 kg v= = PM 1000 Pa (101Pa ) ( 28.97 kg kmol ) 1kPa The mass of air is then 600 m3 Answer m= = 709 kg 0.847 m3 kg b) From the first law in rate form dE d Q W = = ( PE + KE + U ) dt dt The kinetic and potential energy of the air are unchanged and may be eliminated. Integrating the first law Q W dt = dU = U 2 U1

(8.314 kJ

(

)

Heat is removed by the air conditioner and work is added in the form of electrical work. The electrical work is W = VI = Vt ( 2 + 220t ) dt = m ( u2 u1 )t2 = m ( u2 u1 ) = mcv (T2 T1 ) 2 Solving for T2, (use Table A-8, at 300 K to get cv = 0.718 kJ kg K ) 2t + 220W 2 220 min ( 30 min ) 60s 2kW ( 30min ) + 220t 2 1000 W kW 2 2t + min 2 o = 25 C + = 29.6 o C T2 = T1 + 709 kg ( 0.718 kJ kg K ) mcv

3- 4

3-5

An interplanetary probe of volume 300 ft3 contains air at 14.7 psia and 77oF. The heaters fail and the air begins to cool. Assume heat is dissipated from the outside of the spacecraft by radiation at a steady rate of 60 Btu/h. On board electronics generate 12 W on average. Estimate the time required for the air to cool to 30oF.

Approach:Use the first law in rate form and the given values of heat and work to find the cooling time.

Assumptions:1. Air behaves like an ideal gas under these conditions. 2. The specific heat of the air is constant.

Solution:Let the system be the air. From the first law in rate form: dU du =m = Q W dt dt Assuming constant specific heat dT mc p = Q W dt To find the mass of air present, use the ideal gas law in the form: lbm (14.7 psia)(300 ft 3 ) 28.97 PVM lbmol = = 22.2 lbm m= 3 RT psia ft 10.73 (77 + 460)R lbmol R Separate variables in preparation for integration: Q W dT = dt mcv Both the heat rate and the power are constants; therefore, this expression integrates to Q W T = t mcv Solve for t and substitute values, using the specific heat of air in Table B-8 to get Btu o ( 22.2 lbm ) 0.171 ( 30 77 ) F mcv T lbm R t = = Btu Q W 3.412 h Btu 60 ( 12W ) h 1W t = 21.3h Answer

Comments:Note that heat is negative because heat leaves the system (the air). Work is also negative because electrical work is done on the system.

3- 5

3-6 A fan is installed in a 35 m3 sealed box containing air at 101 kPa and 20oC. The exterior of the boxis perfectly insulated. The fan does 250 W of work in stirring the air and operates for one hour. Find the final temperature and pressure of the air. Ignore the temperature change of any fan parts.

Approach:Use the first law in rate form to find the change in temperature. Then calculate final pressure from the ideal gas law.

Assumptions:1. The air behaves like an ideal gas under these conditions. 2. The box is perfectly insulated and air-tight. 3. The specific heat of the air is constant. 4. The mass of the fan parts is negligible compared to the mass of air in the box.

Solution:Let the system be the air in the box. From the first law dU du =m = Q W dt dt Assuming constant specific heat dT mc p = Q W dt To find the mass of air present, use the ideal gas law in the form: kg 1000 Pa (101kPa)(35 m3 ) 28.97 PVM kmol 1 kPa m= 1 = = 42 kg RT1 1000 J kJ 8.314 ( 20 + 273) K kmol K 1kJ Separating variables in preparation for integration: Q W dT = dt mcv Since the rates of work and heat are constants in this case, this expression integrates to Q W T = t mcv Using values of cP from Table A-8 3600 s 0 (250W)(1h) 1 h = 29.9o C T = kJ 1000 J (42 kg) 0.717 kg K 1kJ T = T2 T1 = T2 20 o C

T2 = 49.9 o C To find P2 , again use the ideal gas lawkJ (42 kg) 8.314 (49.9 + 273)K mRT2 kmol K = = 111 kPa P2 = kg VM (35 m 2 )(28.97 ) kmol P2 = 111 kPa

3- 6

3-7

A room contains 4 single-pane windows of size 5 ft by 2.5 ft. The thickness of the glass is in. If the inside glass surface is at 60oF and the outside surface is at 30oF, estimate the heat loss through the windows.

Approach:Use the one-dimensional conduction equation.

Assumptions:1. The thermal conductivity of the glass is constant.

Solution:The surface area of the glass perpendicular to the direction of heat flow is A = 4 ( 5ft )( 2.5ft ) = 50 ft 2 where all four windows have been accounted for. The rate of heat conduction is given by Fouriers law as kA (T1 T2 ) Q= L Using the thermal conductivity of glass in Table B-3, ( 0.8 Btu h ft R ) ( 50 ft 2 ) ( 60 30 ) o F Q= = 57, 600 Btu h Answer 1ft ( 0.25in.) 12in.

3-8

An L-shaped extrusion made of Aluminum alloy 2024-T6 is well insulated on all sides, as shown in the figure. Heat flows axially in the extrusion at a rate of 35 W. If the cool end is at 25oC, find the temperature at the hot end.

Approach:Use the one-dimensional conduction equation.

Assumptions:1. The thermal conductivity of the aluminum is constant.

Solution:Heat flows axially in the bar. The surface area of the bar perpendicular to the direction of heat flow is A = ( 4 )(1.5 ) + ( 4 )( 2 ) =14 cm 2 From the one-dimensional conduction equation kA (T1 T2 ) Q= L Solving for T1, and using thermal conductivity from Table A-2, ( 35) W (1.3 m ) QL + 25C = 209 o C T1 = + T2 = W 14 kA m2 177 m K 10, 000

3- 7

3-9

The wall of a furnace is a large surface of fire clay brick, which is 6.5 cm thick. The outer surface of the brick is measured to be at 35 C. The inner surface receives a heat flux of 2.3 W/cm2. Estimate the temperature of the inner surface of the brick.

Approach:Use the one-dimensional conduction equation.

Assumptions:1. The thermal conductivity of the brick is constant.

Solution:For one-dimensional steady-state conduction, kA (T1 T2 ) Q= L The wall receives a heat flux of 2.3 W/cm2, so k (T1 T2 ) Q W = 2.3 2 = A cm L Solving for T1, LQ T1 = + T2 k A The thermal conductivity should be evaluated at the average of T1 and T2; however, T1 is unknown. We begin by evaluating k at a reasonable temperature and iterating if necessary. From Table A-3, k = 1.0 W/ ( m K ) at T = 478 K. Using this value, 1m 4 2 100 cm 2.3 W 10 cm + 35o C = 1530 o C = 1803K T1 = 2 2 W cm 1 m 1.0 mK The average wall temperature is T + T2 1803 + 308 Tave = 1 = = 1056 K 2 2 This is much higher then the temperature at which k was evaluated, and Table A-3 shows that k varies with temperature. So find a new estimate for k by interpolating in Table A-3. At T = 1056 K, k = 1.572. The revised estimate of T1 is then 1 ( 6.5) 100 2.3 104 + 308 K = 1259 K T1 = ( ) 1.572 This is much different than the first estimate of T1 = 1803 K. Use the new value of 1259 K to find a new average temperature and a new value of k. Iterate until T1 no longer changes. The first few iterations are:

( 6.5 cm )

k 1.000 1.572 1.344 1.435 1.395

T1 K 1803 1259 1420 1350 1380

For greater accuracy, continue the iteration. At the last iteration, T = 1380 K

3- 8

3-10

A tungsten filament in a 60 W light bulb has a diameter of 0.04 mm and an electrical resistivity of 90 cm. The filament loses heat to the environment, which is at 20oC, by thermal radiation. The emissivity of the filament is 0.32 and the voltage across it is 115 V. Find the length of the filament and the filament surface temperature. (Electrical resistance equals electrical resistivity times filament length divided by filament cross-sectional area.)

Approach:Calculate the resistance of the wire from the known Joule heating losses and the applied voltage. Use resistance to find the wire length. Then equate the Joule heating to the loss by radiation to determine the wire surface temperature.

Assumptions:1. The only mode of heat transfer is radiation. 2. The surface of the filament is gray and diffuse. 3. There are no reflecting surfaces near the filament.

Solution:The electrical resistance is related to the electrical resistivity by L e L Re = e = Ax ( D / 2 )2 where Ax is the cross-sectional area. The power dissipated in the filament by Joule heating isRe Solving for resistance: 2 (115 V) 2 Re = = = 220 60 W Q Therefore the filament length is2 .04 1m D (220 ) mm Re 2 1000 mm 2 = L= e 1 1m (90 cm) 6 10 100 cm L = 0.308 m = 30.8 cm For radiation from a gray surface to a black environment Q = As (Ts 4 Ta 4 ) 2 2

Q = i =

2

where As is surface area, Ts is surface temperature, and Ta is the ambient temperature. Q 4 Q 4 + Ta 4 = + Ta 4 Ts = As DL 4 60 W Ts = + (20 + 273) 4 K 1m W -8 (0.32) 5.67 10 (0.380 m) (.04 mm) 2 m K 1000 mm 1 1 1

Ts = 3040 K

Comments:Tungsten glows brightly at this high temperature. The filament is coiled so it fits in the bulb.

3- 9

3-11

On a cold winter day, the interior walls of a room are at 55oF. A man standing in the room loses heat to the walls by thermal radiation. The mans surface area in 16 ft2, his clothing has an emissivity of 0.93 and his surface temperature is 70oF. He generates 300 Btu/h of body heat. What percentage of the mans body heat is transferred by radiation to the walls?

Approach:

4 Use Q = A (Ts4 Tsurr ) to determine radiation heat transfer

from the man and compare this to his body heat.

Assumptions:1. The mans clothing is gray and diffuse. 2. There are no reflecting surfaces near the man. 3. The mans temperature is uniform.

Solution:The heat lost by radiation is Qrad = A (Ts 4 Tsurr 4 ) Converting temperatures to absolute and substituting values Btu 4 Qrad = (0.93) 0.171 108 (16 ft 2 ) ( 70 + 460 ) (55 + 460) 4 R 4 2 4 h ft R Btu Qrad = 218 h To find the percent transferred by radiation: Qrad 218 = = 0.726 = 72.6% Qbody 300

Comments:Typically, radiation is an important mode of heat transfer when the only other mode is natural convection in air.

3 - 10

3-12

The sun can be approximated as a spherical black body with a surface temperature of 5762 K. The irradiation from the sun as measured by a satellite in earth orbit is 1353 W/m2. The distance from the earth to the sun is approximately 1.5 x 1011 m. Assuming that the sun radiates evenly in all directions, estimate the diameter of the sun.

Approach: Construct an imaginary sphere around the sun with a radius equal to the sun-earth distance. The total energy, in watts, falling on this sphere equals the total energy emitted by the sun. Use this total energy and the Stefan-Boltzman law to find the effective solar diameter. Assumptions: 1. The sun radiates uniformly in all directions. 2. The sun radiates like a black body.

Solution:If the sun had a radius, r, then the total heat radiated by the sun would be Q = As Ts 4Q = 4 r 2 Ts 4

where Q is in Watts. This heat is spread evenly over an imaginary sphere whose radius is the sun-earth distance, R. Therefore W Q = 1353 2 ( 4 R 2 ) m 2 m Q = (1353)(4 )(1.5 1011 ) 2= 3.83 1026 W The effective radius of the sun is

r= r=

Q 4 Ts 4 3.83 1026 W W 4 5.67 108 2 4 (5762 K) 4 m K

r = 6.98 108 m D = 2r = 1.40 109 m

3 - 11

3-13

A high torque motor has an approximately cylindrical housing 9.5 in. long and 6 in. in diameter. The motor delivers 1/8 hp in steady operation and has an efficiency of 0.72. All the heat generated by motor losses is removed by natural convection and radiation from the outer surface of the housing. The convective coefficient is 1.68 Btu/hft2 oF and the housing emissivity is 0.91. If the surroundings are at 58oF, what is the housings outer surface temperature?

Approach: Use the definition of efficiency to calculate the electrical work input to the motor. Then determine the rate of heat lost to the environment and set this rate equal to the sum of the radiative and convective heat losses. Assumptions: 1. The emissivity is not a function of temperature. 2. The heat transfer coefficient is not a function of temperature. 3. The emissivity is uniform over the surface of the housing. 4. The heat transfer coefficient is uniform over the surface of the housing. 5. The rate of heat loss is uniform over the surface of the housing. 6. The surface of the housing is gray and diffuse.

Solution:Because the motor has an efficiency of 0.72, 72% of the input electric power is converted into mechanical power, 1 0.72W = hp 8 Solve for electric power and convert units to Btu/h, 1 Btu hp 2544 8 h = 442 Btu W= h 0.72 1 hp The energy lost as heat from the outer casing is Btu Btu Qloss = (1 0.72 ) 442 =123.8 h h The surface area of the housing is A = 2 r 2 + 2 rl= 2 ( 3in.) + 2 ( 3in.)( 9.5in.) = 236in 2 = 1.64 ft 22

The surrounding temperature is Tsurr = 58 + 460 = 518 R Heat is lost by both convection and radiation, that is 4 Qloss = Qrad + Qconv = A (Ts4 Tsurr ) + hA (Ts Tsurr ) Substituting values Btu Btu Btu 2 4 4 2 124 = ( 0.91) 0.1714 108 (1.64 ft )(Ts 518 ) + 1.68 (1.64 ft ) (Ts 518 ) h h ft 2 R 4 h ft 2 R Solving Ts = 547 = 87 o F Answer

3 - 12

3-14

A flat plate solar collector 6 ft by 12 ft is mounted on the roof of a house. The outer surface of the collector is at 110oF and its emissivity is 0.9. The outside air is at 70oF and the sky has an effective temperature for radiation of 45oF. The collector transfers heat by natural convection to the air with a heat transfer coefficient of 3.2 Btu/hft2 oF and also transfers heat by radiation to the sky. Calculate the total heat lost from the solar collector.

Approach:Use the fundamental rate equations for convective and radiative heat transfer and add to get the total heat transfer.

Assumptions:1. The emissivity is not a function of temperature. 2. The heat transfer coefficient is not a function of temperature. 3. The emissivity is uniform over the surface of the collector. 4. The heat transfer coefficient is uniform over the surface of the collector. 5. The surface of the collector is gray and diffuse.

Solution:The surface area of the collector is A = (6 ft)(12 ft) = 72 ft 2 The heat lost by convection is Qconv = hA (Ts Tair )Btu 2 o = 3.2 (72 ft )(110 70) F h ft 2 F Btu = 9216 h The heat lost by radiation is 4 Qrad = A (Ts4 Tsky ) Qrad = (0.9)(0.171 108 ) = 4490 Btu 4 4 (110 + 460 ) ( 45 + 460 ) h ft 2 R 4

Btu h The total heat transferred is Qtot = Qconv + Qrad = 9216 + 4490 = 13, 700 Btu h

Comments:The effective temperature of the sky depends on the cloud cover. A clear sky will be much colder than an overcast one.

3 - 13

3-15

A CPU chip with a footprint of 3 cm by 2 cm is mounted on a circuit board. The chip generates 0.31 W/cm2 and rejects heat to the environment at 28oC by convection and radiation. The outer casing of the chip has an emissivity of 0.88 and the heat transfer coefficient is 48 W/m2K. Neglecting the thickness of the chip and any conduction into the circuit board, calculate the chip surface temperature.

Approach:Equate the heat generated to the sum of the heat removed by convection and radiation. Express the rates of convection and radiation in terms of surface temperature and solve for surface temperature.

Assumptions:1. No heat is removed from the chip by conduction. 2. All heat leaves from the top of the chip; the surface area of the sides of the chip is very small and will be neglected. 3. The surface of the chip is gray and diffuse. 4. The heat transfer coefficient and emissivity are not functions of temperature. 5. The heat transfer coefficient and emissivity are uniform over the surface of the chip.

Solution:The heat generated by the chip is W Qgen = 0.31 2 (3cm)(2 cm) = 1.86 W cm This heat is removed by convection and radiation; i.e. Qgen = Qconv + QradQgen = hA(Ts T f ) + A(Ts 4 Tsurr 4 )

The area of the top of the chip is 1m 2 A = (3cm)(2 cm) 4 2 = 0.0006 m 2 10 cm The surroundings are at the same temperature as the air. The air temperature is T f = 28 + 273 = 301K Substituting values: W W 2 -8 2 4 4 4 1.86 W= 48 2 (0.0006 m )(Ts 301)K+(0.88) 5.67 10 (0.0006 m )(Ts 301 )K 2 m K4 m K By trial and error or using an equation-solving programTs = 357 K = 84.2 o C

3 - 14

3-16

A metal plate 16 cm by 8 cm is placed outside on a clear night. The plate, which has an emissivity of 0.7, exchanges heat by radiation with the night sky, which is at 40oC. Air at 10oC flows over the top of the plate, cooling it with a heat transfer coefficient of 42 W/m2K. The plate is insulated on its underside and heated by an electric resistance heater. How much electric power must be supplied to maintain the plate at 55oC?

Approach:Calculate the radiation and convection from the fundamental rate equations. The sum of these rates equals the heat generated, by the first law.

Assumptions:1. The metal plate is gray and diffuse. 2. The plate is perfectly insulated on the back side. 3. Emissivity is uniform over the plate. 4. Emissivity is independent of temperature. 5. Heat transfer coefficient is uniform over the plate. 6. Heat transfer coefficient is independent of temperature.

Solution:The heat removed by radiation is Qrad = A (Tp 4 Ts 4 ) where Tp is the plate temperature and Ts is the sky temperature.Tp = 55 + 273 = 328 K Ts = 40 + 273 = 233K 1m 2 W Qrad = (16)(8)cm 2 4 2 (0.7) 5.67 108 2 4 (3284 2334 )K m K 10 cm Qrad = 4.38 W The heat removed by convection is Qconv = hA(TP Ta )

where Ta is air temperature. Ta = 10 + 273 = 263K 1m 2 W Qconv = 42 2 (16)(8)cm 2 4 2 (328 263)K m K 10 cm = 34.9 W Qtot = Qrad + Qconv = 4.38 + 34.9 = 39.3 W By the first law, dU = Q W dt The plate is in steady-state; therefore, there is no change in internal energy and the first law reduces to Q =W The electrical work (power to the heater) equals the heat lost from the plate W = Qtot = 39.3 W Answer

3 - 15

3-17

A home freezer is 1.8-m wide, 1-m high, and 1.2-m deep. The interior surface of the freezer must be kept at 10 C. The walls of the freezer are made of polystyrene insulation sandwiched between two thin layers of steel. The combined convective/radiative heat transfer coefficient on the exterior is 8.2 W/m2K and the ambient is at 25oC. If the power of the refrigeration unit is limited to 150 W, what thickness of polystyrene is needed? Assume the conduction resistance of the thin metal wall panels is very small and can be neglected and that the bottom of the freezer is perfectly insulated.

Approach:Use the resistance analogy to find the resistances due to conduction and convection. All resistances add in series.

Assumptions:1. Thermal conductivity is independent of temperature. 2. The resistance of the steel is negligible. 3. No heat is conducted into the floor. 4. The thickness of the insulation is small compared to the size of the freezer. 5. The heat transfer coefficient is uniform over the surface of the freezer and independent of temperature.

Solution:To maintain the freezer in steady state, the heat transfer into the freezer from the outside air must be no more than the freezer can remove. To find the heat transferred into the freezer, use the resistance analogy. The exposed surface area of the freezer (neglecting the bottom) is A = 2(1)(1.8) + 2(1)(1.2) + (1.2)(1.8) = 8.16 m 2 Assume the interior and exterior of the freezer have about the same area, since the insulation thickness is small compared to the size of the freezer. The heat entering the freezer is T T T T Q= a i = a i 1 L R1 + R2 + kA hA Solving for L T T 1 L = kA a i hA Q using k from Table A-4 o W 1 2 [ 25 ( 10) ] C L = 0.027 ( 8.16 m ) mK 250 W W 2 5.2 2 (8.16 m ) m K = 0.0481m = 4.81cm

3 - 16

3-18

The windshield of an automobile is heated on the inside by a flow of warm air. Cold air at 15F flows over the exterior of the windshield. The heat transfer coefficient on the inside is 16 Btu/hft2F and the heat transfer coefficient on the outside is 49 Btu/hft2F. The glass of the windshield has a thickness of 0.25 inch. What temperature should the inside air be so that the exterior surface temperature of the windshield is 3F?

Approach:Use the resistance analogy, adding all resistances in series.

Assumptions:1. Thermal conductivity is independent of temperature. 2. The heat transfer coefficient is uniform over the surface of the windshield and independent of temperature.

Solution:Heat is convected from the inside of the windshield with resistance R3, conducted through the glass with resistance R2 and convected from the outside of the windshield with resistance R1. All resistances add in series. Perform the calculation for a unit area of 1 ft2. The result is independent of area. The resistances are o 1 1 Fh = = 0.0204 R1 = Btu Btu h1 A 1 ft 2 ) 49 2 o ( h ft F 1 ft ( 0.25 in.) o L 12 in. = 0.026 F h R2 = = Btu Btu kA 2 0.8 (1 ft ) h ft o F o 1 1 Fh = = 0.0625 Btu Btu h2 A 2 16 (1 ft ) h ft 2 o F By the resistance analogy, heat is T Q= R From the resistance circuit, the heat through each resistance is the same, and T3 T1 T1 T0 = R2 + R3 R1 Solving for T3, ( 0.026 + 0.0625 ) ( 3 ( 15) ) ( R + R3 )(T1 T0 ) T3 = 2 + T1 = +3 0.0204 R1

R3 =

T3 = 81 o F

3 - 17

3-19

A copper busbar of length 40 cm carries electricity and produces 4.8 W in joule heating. The cross-section is square as shown in the figure and is covered with insulation of thermal conductivity 0.036 W/mK. All four sides are cooled by air at 20oC with an average heat transfer coefficient of 18 W/m2K. Assuming the copper is isothermal, estimate the maximum temperature of the insulation.

Approach:Use the resistance analogy, adding all resistances in series.

Assumptions:1. Thermal conductivity is independent of temperature. 2. The heat transfer coefficient is uniform over the surface of the insulation and independent of temperature. 3. Heat transfer is planar and onedimensional.

Solution:The heat transferred through each side of the bar is 4.8 W Q= = 12 W 4 The area of the inside of the insulation is A1 = ( 8cm )( 40 cm ) = 320 cm 2 The area of the outside of the insulation is A2 = (12 cm )( 40 cm ) = 480 cm 2 For conduction through the insulation, use the average of the inner and outer areas as an approximation 480 + 320 Aave = = 400 cm 2 2 The conduction resistance is L 2 cm R1 = = = 13.9 K W kAave 1m 2 ( 0.036 W m K ) ( 400 cm ) 100 cm The convection resistance is 1 1 R2 = = = 1.16 K W hA2 1m 2 2 2 (18 W m K )( 480 cm ) 10, 000 cm2 The total resistance is the sum of the two resistances in series Rtot = R1 + R2 = 15.1K W The maximum temperature of the insulation occurs next to the copper and is given by Tmax T = Rtot QTmax = (15.1K W )(1.2W ) + 20o C Tmax = 38.1 o C

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3-20

A freezer maintains one side of a slab of ice 3 cm thick at 10oC. The other side exchanges heat with the ambient at 15oC by combined natural convection and radiation. In steady state, the ice does not melt. Find the highest possible value of heat transfer coefficient on the ice surface exposed to the ambient.

Approach:Use the resistance analogy to relate the surface temperature to the heat transfer coefficient.

Assumptions:1. The combined radiative/convective heat transfer coefficient does not depend significantly on temperature. 2. The thermal conductivity is constant. 3. The heat transfer coefficient is uniform over the surface of the ice.

Solution:When the ice surface temperature, T2 , is 0 C , the heat transfer coefficient has its maximum value. If the heat transfer coefficient is higher than this value, the ice would begin to melt. Conduction through the ice occurs in series with combined convection/radiation at the outer surface, as shown in the figure. The conduction resistance is: L R1 = kA The heat conducted through the ice is T T kA(T2 T1 ) Q= 2 1 = R1 L For the combined convection and radiation, the resistance is 1 R2 = hA The heat conducted through the ice is equal to the heat arriving by convection and radiation; therefore, T T Q = 3 2 = hA(T3 T2 ) R2 Eliminating Q giveskA(T2 T1 ) = hA(T3 T2 ) L Solving for h, and using the thermal conductivity of ice from Table A-3, W 1.88 [ 0 (10) ] k (T2 T1 ) mK h= = (0.03 m) (15-0 ) L(T3 T2 ) = 41.8 W m oC2

3 - 19

3-21

The door of a kitchen oven contains a window made of a single pane of 1/4 in. thick Pyrex glass. The interior oven temperature is 550oF and the room air is at 68oF. The combined convective/radiative heat transfer coefficient on the oven interior is 1.7 Btu/hft2oF and on the oven exterior, it is 0.88 Btu/hft2oF. A toddler comes by and touches the window. Calculate the temperature of the surface that the child's hand contacts.

Approach:Use the resistance analogy, adding all resistances in series.

Assumptions:1. The combined radiative/convective heat transfer coefficient does not depend significantly on temperature. 2. The thermal conductivity is constant. 3. The heat transfer coefficient is uniform over the surface of the glass. 4. Heat transfer is one-dimensional.

Solution:From the resistance network in the figure, T T T T Q = win room = oven win R3 R1 + R2 Substituting values, with thermal conductivity from Table B-3, (Twin 68) o F (550 Twin ) o F = 1 1ft (0.25in.) Btu 1 12 in. A + 0.88 2 o h ft F Btu Btu A 0.7 1.7 A 2 o h ft o F h ft F Solving for the window temperature, Twin, yields Twin = 380 o F

3 - 20

3-22

An electronic device may be modeled as three plane layers, as shown in the figure. The entire package is cooled on both sides by air at 20C. Heat is generated in a very thin layer between two contacting surfaces at a rate of 500 W/m2, as shown. The heat transfer coefficient on both sides is 8.7 W/m2K. Assume the layers are very large in extent in the direction not shown. Using data in the figure below, calculate the temperature T2.

Approach:Use the resistance analogy, adding all resistances in series.

Assumptions:1. The combined radiative/convective heat transfer coefficient does not depend significantly on temperature. 2. The thermal conductivity is constant. 3. The heat transfer coefficient is uniform over the surface of the device. 4. Heat is generated in a layer of negligible thickness.

Solution:The resistance network that can be used to model this device is:

where T0 is the air temperature. Now calculate the resistances for an area of 1 m2. The resistances may be calculated as 0.003m K L R1 = 1 = = 0.00882 W W k1 A 2 0.34 (1m ) mK 1 1 K R2 = = = 0.1149 W hA W 2 8.7 2 (1m ) m K 0.0194 m K L R3 = 2 = = 0.1213 W W k2 A 2 0.16 (1m ) mK Another way to represent the resistance network is shown in the figure to the right, where R4 is given by K R4 = R1 + R2 + R3 = 0.245 W and R5 is K R5 = R1 + R2 = 0.1237 W For two resistors in parallel,

3 - 21

R4 R5 1 1 1 = + Rtot = or Rtot R4 R5 R4 + R5 Substituting values gives K Rtot = 0.0822 W From the resistance analogy, T = QRtot = T2 T0 Solving for T2, K T2 = ( 500 W ) 0.0822 + 20 o C W T2 = 61 o C

3 - 22

3-23

A cardboard box is used to ship flowers on a summer day when the ambient temperature is 80oF. The air inside the box is maintained at 45oF by the use of cold packs. The box is lined with a layer of styrofoam (ks = 0.015 Btu/hftR) one half inch thick. The cardboard itself is 1/8-inch thick and has kc = 0.13 Btu/hftR. The box measures 8 in. by 8 in. by 2.5 ft. Assume h on the inside is 2.0 Btu/hft2R and h on the outside is 9.3 Btu/hft2R. Calculate the rate of heat transfer into the box. Neglect heat transfer on the ends.

Approach:Use the resistance analogy, adding all resistances in series.

Assumptions:1. The heat transfer coefficient does not depend significantly on temperature. 2. The thermal conductivity of both materials is constant. 3. The heat transfer coefficient is uniform over the surface of the box.

Solution:The area of the inside of the box (neglecting the ends) is 6.75 2 A1 = ( 2.5ft ) ft ( 4 ) = 5.625ft 12 The area of the outside of the styrofoam is 7.75 2 A2 = ( 2.5ft ) ft ( 4 ) = 6.46 ft 12 The outer area of the box is 8 A3 = ( 2.5ft ) ft ( 4 ) = 6.67 ft 2 12 All resistances add in series. To find the inside convective resistance, use 1 1 R1 = = = 0.075 h o F Btu 2 o h1 A1 ( 2.0 Btu h ft F )( 5.625ft 2 ) For conduction through the foam, the area is different on inside and out. Estimate the appropriate area by taking the average of inner and outer areas. 0.5 ft L 12 R2 = = = 0.417 h o F Btu kA ( 0.015 Btu h ft R ) ( 6.04 ft 2 ) Likewise for the cardboard, take the average in inner and outer areas to get 1 1 ft L 8 12 = = 0.012 h o F Btu R3 = kA ( 0.13Btu h ft R ) ( 6.56 ft 2 ) The exterior convective resistance is 1 1 ft L 8 12 = = 0.012 h o F Btu R3 = kA ( 0.13Btu h ft R ) ( 6.56 ft 2 )Rtot = R1 + R2 + R3 + R4 = 0.577 h o F Btu Q= T 80 45 Btu = = 60.7 Rtot 0.577 h

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3-24

A living room floor 3 m by 4.5 m is constructed of a layer of oak planks 1.2 cm thick laid over plywood 2.0 cm thick. In winter, the basement air is at 15oC while the living room air is at 20oC. The heat transfer coefficient on the living room floor and the basement ceiling are 3.6 and 6.8 W/m2K, respectively. If the home is heated electrically and the cost of electricity is \$0.08 per kWh, estimate the cost per month of the energy lost through the floor. If the room is carpeted with wall-to-wall carpeting 1.6 cm thick (k = 0.06 W/mK), what would the energy cost be?

Approach:Use the resistance analogy, adding all resistances in series.

Assumptions:1. Neglect conduction through the floor joists (not shown). 2. Conduction is one-dimensional. 3. Thermal conductivity is constant.

Solution:First consider the case without the carpet. The thermal resistances for convection and conduction are: (using Table A-3 and A-5 for thermal conductivities)R1 = 1 1 K = = 0.0206 W hA W 3.6 2 (3)(4.5)m m K L 0.012m K = = 0.00523 W kA W 2 0.17 (3)(4.5)m mK L 0.02 m K = = 0.0124 W kA W 2 0.12 (3)(4.5)m mK

R2 =

R3 =

1 1 K = = 0.0109 W hA W 2 6.8 2 (3)(4.5)m m K All these resistances add in series R4 = Rtot = R1 + R2 + R3 + R4 = 0.0206 + 0.00523 + 0.0124 + 0.0109 = 0.049 K W

The heat transfer rate is given by T (20 15) o C Q= = = 102 W K Rtot 0.049 W 30 days 24 h \$.08 1kW cost = (1 month) (102 h) 1 month 1 day kWh 1000 W = \$5.87

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The carpet adds a resistance to the system. The resistance of the carpet is: L 0.016 cm R5 = = W kA 2 0.06 (3)(4.5)m mK K = 0.0198 W Adding this resistance to the previous total resistance gives a revised total resistance of K R*tot = Rtot + R5 = 0.049 + 0.0198 = 0.0688 W The new heat transfer rate is 20 15 = 72.7 W Q* = 0.0688 The ratio of the new cost to the old cost is new cost 72.7 W = old cost 102 W Solving for new cost: 72.7 new cost = (5.87) = \$4.19 102

Comments: The carpet may improve appearance or comfort in the room; however, it does not contribute substantially to energy savings.

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3-25

The wall of a furnace must be designed to transmit no more than 220 Btu/hft2. Two types of bricks are available for construction: one with a thermal conductivity of 0.38 Btu/hftR and a maximum allowable temperature of 1400oF and the other with a thermal conductivity of 0.98 Btu/hftR and a maximum allowable temperature of 2300oF. The inside wall of the furnace is at 2100oF and the outside wall is at 300oF. Both types of bricks have dimensions of 9 by 4.5 by 3 in. and both are the same cost per brick. If the bricks can be laid up in any manner, determine the most economical arrangement of bricks.

Approach:Use the resistance analogy to find the temperature drop across the layers of brick. Use as few high temperature bricks as possible since these have poor insulation properties.

Assumptions:1. The thermal conductivity of the bricks is not a function of temperature. 2. Heat conduction is one-dimensional. 3. The wall is very large.

Solution:First calculate the necessary thickness of the high temperature layer, which is made of high conductivity bricks. This thickness is found from T T Q= = L R kA Taking A = 1ft 2 and Q = 220 Btu/hBtu 2 (2200 1400) o F 0.98 (1ft ) h ft o F = 3.12 ft = 37.4in. Btu 220 h Because of the integral size of the bricks, we cannot design a layer with exactly this dimension. The actual layer must be at least 37.4 in. Given the size of the bricks, the layup which uses the fewest number of bricks and exceeds 37.4 in. by the least amount is: Tk1 A L= = Q

which gives an actual thickness of 39 in. The two-layer wall may be represented as:

To find the thickness of the second layer, L2 , use

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T (2100 300) o F h oF = = 8.18 Btu Q Btu 220 h The total resistance is the series combination of the resistances of each layer: L L Rtot = 1 + 2 k1 A k2 A Solving for L2, 1 (39) ft L1 Btu h oF 12 2 L2 = k2 A Rtot = 0.38 (1ft ) 8.18 k1 A Btu (0.98)(1 ft 2 ) h ft o F = 1.85ft = 22.2 in. The layup of the second layer must be:Rtot =

In summary, the most economical layup which satisfies all constraints is

Comments:In actual application, the mortar would provide additional insulating value. The bricks could also have been laid in the other direction with three bricks of edge length 3 in. spanning what is shown as one 9-in. brick in the diagram or two bricks of edge length 4.5 in. spanning what is shown as one 9-in. brick. This would result in the same total number of bricks when the third dimension is considered.

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3-26

A chemical reactor is in the shape of a long cylinder, as shown in the figure. The reactor is covered with a layer of insulation 17.7 cm thick. The reactor loses heat through the insulation at a rate of 15.3 W per meter of length. The thermal conductivity of the insulation is 0.04 W/mK. On the outside of the insulation, air at 26oC removes heat by forced convection, with a heat transfer coefficient of 32 W/m2K. Find the maximum temperature of the insulation. Neglect radiation.

Approach:Use the electrical resistivity analogy adding all resistances in series. The maximum temperature is at the inner radius of the insulation.

Assumptions:1. Heat transfer is one-dimensional. 2. The thermal conductivity of the insulation is constant. 3. The heat transfer coefficient is uniform over the surface of the insulation and independent of temperature.

Solution:The reactor is hot inside and the air is cool outside. The maximum temperature of the insulation will be at r = r1 1.3m r1 = 0.177 m = 0.473m 2 The outer radius of the insulation is r2 = r1 + 0.177 m = 0.65 m The resistance across the insulation is r 0.65 ln 2 ln r1 K 0.473 = =1.265 R1 = W 2 Lk W 2 (1m ) 0.04 mK where L=1m because the value of Q is given for a one-meter length. The resistance for convection is1 1 K = 0.00765 = W W hA 32 2 ( 2 )( 0.65 m )(1m ) m K K Rtot = R1 + R2 =1.265 + 0.00765 =1.273 W From the resistance analogy T = QRT = T1 T2 Rearranging T1 = T2 + QRTR2 =

K = 26C + (15.3W ) 1.273 W = 45.5C

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3-27

An insulated copper wire with a length of 1.2 m carries 20 A of current. The copper is 1 mm in diameter and the insulation (k = 0.13 W/moC) has a thickness of 0.8 cm. Air at 25oC blows in crossflow over the wire to produce an external convective heat transfer coefficient of 219 W/m2K. Assuming the copper is isothermal, find the copper temperature. Take the electrical resistivity of copper to be constant at 2.1 x 10-8 m.

Approach:Use the electrical resistivity of the wire to find the wires electrical resistance. Then calculate the power dissipated by Joule heating. Finally add the conductive resistance across the insulation to the convective resistance and use the total resistance to compute the wire temperature.

Assumptions:1. The copper is a perfect conductor and hence, isothermal. 2. The thermal conductivity and electrical resistivity are constant. 3. Heat transfer is one-dimensional. 4. The heat transfer coefficient is uniform over the surface of the wire and independent of temperature.

Solution:The radii of the inner and outer surfaces of the insulation are .5 mm r1 = = 0.0005 m (1000 mm/m ) 1m r2 = r1 + 0.8cm = 0.0005 + 0.008 = 0.0085 m 100 cm The cross-sectional area, Ax , of the copper isAx = r12 = (0.0005) 2 = 7.85 107 m 2 The electrical resistance of the wire is L (2.1 108 m)(1.2 m) Re = = = 0.0321 Ax 7.85 10-7 m 2 The power produced by Joule heating in the wire is Q = i 2 R = (20 A) 2 (0.0321 ) = 12.8 W The thermal resistances are ln ( r2 / r1 ) ln ( 0.0085/0.0005 ) K = = 2.89 R1 = W 2 Lk W 2 (1.2 m) 0.13 mK

1 1 K = 0.0712 = W W hAs 219 2 2 (0.0085 m)(1.2 m) m K where As is the surface area of the outside of the insulation. Temperature is found from T T Q= 3 1 or T3 = T1 + Q( R1 + R2 ) R1 + R2R2 = T3 = 25 C+(12.8 W)(2.89+0.0712)

K = 63 o C W

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3-28

The wall of a submarine is 1-in. thick stainless steel (AISI 304) insulated on the interior with a 1.5-in. layer of polyurethane foam (k = 0.017 Btu/hftF). The heat transfer coefficient on the interior is 3.7 Btu/hftoF. At full speed, the exterior heat transfer coefficient is 135 Btu/hftoF. The sub is approximately cylindrical with the length 240 ft and outer diameter 30 ft. If the seawater is a 40oF, at what rate must heat be added to the interior air to keep it at 70oF? As a first approximation, neglect heat transfer through the ends.

Approach:Use the resistance analogy, adding all resistances in series.

Assumptions:1. The combined radiative/convective heat transfer coefficient does not depend significantly on temperature. 2. The thermal conductivity is constant. 3. The heat transfer coefficient is uniform over the surface of the submarine. 4. Heat transfer is one-dimensional.

Solution:The radii that will be needed in the calculation are r3 = 15 ft r2 = 15 1/ 12 = 14.917 ft r1 = r2 1.5 /12 = 14.792 ft Heat transfer from the 70F interior air to the 40F exterior water can be calculated using the following resistance network:

The convective resistance on the interior is 1 h oF 1 1 R1 = = = = 1.21 105 Btu h1 A1 h1 2 r1 L Btu 3.7 2 (14.792ft)(240 ft) h ft 2 o F The conductive resistance through the polyurethane foam is r 14.917 ln 2 ln r1 h oF 14.792 = 3.28 104 R2 = = Btu Btu 2 Lk1 2 (240 ft) 0.017 h ft o F The conductive resistance through the stainless steel (with thermal conductivity from Table B-2) is r ln 3 ln 15 r h oF 14.917 = 4.30 107 R3 = 2 = Btu 2 Lk2 2 (240)(8.6) Finally, the convective resistance on the exterior is 1 h oF 1 R4 = = = 3.27 107 h3 2 r3 L (135)2 (15)(240) BtuRtot = R1 + R2 + R3 + R4 = 3.49 105

From the resistance analogy Q =

T 70 40 Btu = = 8.79 104 4 Rtot 3.41 10 h

3 - 30

3-29

An aluminum wire 2.5 m long conducts 12 A with an imposed voltage of 1.5 V. The wire, which has a diameter of 2.4 mm, is covered with a layer of insulation 2 mm thick. The thermal conductivity of the insulation is 0.15 W/moC. Air at 40oC flows over the exterior of the wire to give a convective heat transfer coefficient of 32 W/m2oC. Assume the aluminum is isothermal and compute the temperature on the inside surface and also on the outside surface of the insulation.

Approach:Calculate the power dissipated by Joule heating. Use the thermal resistance analogy to find the unknown temperatures.

Assumptions:1. The aluminum is a perfect conductor and hence, isothermal. 2. The thermal conductivity is constant. 3. Heat transfer is one-dimensional. 4. The heat transfer coefficient is uniform over the surface of the wire and independent of temperature.

Solution:The radii of the inner and outer surfaces of the insulation are 1m r1 = 1.2 mm = 0.0012 m 1000 mm 1m r2 = r1 + 2 mm = 0.0012 + 0.002 = 0.0032 m 1000 mm The thermal resistances are r 0.0032 ln 2 ln r1 K 0.0012 = = 0.416 R1 = W W 2 kL 2 0.15 (2.5 m) mK 1 K 1 1 R2 = = = = 0.622 W W hA h(2 r2 L) 32 2 2 (0.0032 m)(2.5 m) m K The heat generated by Joule heating is Q = i = (1.5 V)(12 A) = 18 W From the thermal resistance analogy, T T Q= 3 2 R1 + R2K = 58.7 o C Answer W This is the temperature at the interior of the insulation. Similarly, at the exterior, temperature is K T2 = T1 + QR2 = 40 o C + (18 W ) (0.622) = 51.2 o C Answer WT3 = T1 + Q( R1 + R2 ) = 40 o C + (18 W ) (0.416+0.622)

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3-30

An insulated steel pipe carries hot water at 80oC. The outer surface of the insulation loses heat to the environment by convection and radiation. For convection, assume hconv = 5.8 W/m2oC. The emissivity of the insulation is 0.88. The surroundings are at 30oC. Assume the inner surface of the insulation is at the water temperature. What is the surface temperature of the insulation? Use data on the figure below.

Approach:Use the thermal resistance analogy to find the unknown temperature.

Assumptions:1. The thermal conductivity is constant. 2. Heat transfer is one-dimensional. 3. The heat transfer coefficient is uniform over the surface of the pipe and independent of temperature. 4. The insulation is gray and diffuse.

Solution:The total heat transfer coefficient is the sum of radiative and convective heat transfer coefficients, thus htot = hconv + hrad = hconv + (Ts + Tsurr )(Ts 2 + Tsurr 2 ) From the resistance circuit Tw Ts Ts Tsurr = R1 R2 which becomes Tw Ts T T = s surr r2 1 ln r1 2htot r2 L 2 Lk It is not possible to solve this equation directly for the unknown temperature Ts because htot is a function of Ts. By trial and error, the solution isTs = 42.8 o C

Comments:Equation-solving software is very useful in problems of this type. Remember to use absolute temperauture (Kelvin) in the computations.

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3-31

A cylinder of radius r1 is covered with a layer of insulation of thermal conductivity, k. A fluid flows over the outside of the insulation, exchanging heat with a heat transfer coefficient, h. Let r2 be the radius at the outer surface of the insulation. Cooling of the cylinder is controlled by the combination of conduction and convection resistances. If r2 is small, the conduction resistance is small. As r2 increases, the conduction resistance increases, but the surface area of exposed insulation also increases, and this results in a decrease in convective resistance. As a result, there is an optimal value of r2 which produces the largest possible total resistance to heat transfer. Derive an expression for the optimum value of r2 as a function of r1, k, and h.

Approach:Write an equation for the total resistance due to conduction through the insulation and convection from the outer surface. To find the maximum value of resistance, take the derivative with respect to the outer radius, r2 and set it equal to zero.

Assumptions:1. Thermal conductivity is constant. 2. Heat transfer is one-dimensional. 3. The heat transfer coefficient is uniform over the surface of the insulation and independent of temperature.

Solution:We assume the length, L, of the cylinder is very large compared to the radius, so that heat transfer is onedimensional. In that case, the total resistance is Rtot = Rcond + Rconv where Rcond is conduction resistance and Rconv is convection resistance. These resistances are given by1 hA 2 Lk Expressing the total resistance explicitly as a function of r2 gives ln ( r2 / r1 ) 1 + Rtot (r2 ) = h2 r2 L 2 LkRtot =

ln ( r2 / r1 )

+

To find the optimum value of r2 , for which Rtot is maximized, take the derivative of Rtot with respect to r2 and set the result equal to zero. r1 1 2 dRtot r2 r1 ( r2 ) = + dr2 2 Lk 2 hL Let r0 be the optimal value of r2 ; thenr0 1 r 2 0 2 Lk 2 hL 1 1 0= k r0 h Solving for r0, k r0 = h

0=

Note that r0 is independent of r 1 , but that r0 > r1 . If the cylinder radius is greater than r0 , there is no optimal value of outer radius, and increasing insulation thickness always increases total resistance.

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3-32

A frozen pipe is filled with ice at 0oC. A heating tape wrapped around the pipe provides 90 W per meter of pipe length. Insulation is placed over the heating tape. The insulation has a thickness of 0.5 cm and a thermal conductivity of 0.082 W/moC. Convection and radiation occur from the outside of the insulation to the environment, which is at 15oC. The heat transfer coefficient is 7.7 W/m2oC and the emissivity is 0.94. The pipe wall remains at 0oC during the heating and the heating tape is very thin. The pipe has as I.D. of 3 cm and a wall thickness of 4 mm. How much time is required to completely melt the ice? (heat of fusion = 3.34 x 105 J/kg, density = 921 kg/m3)

Approach:Use a thermal resistance network. The heat from the heating tape is a source term added at the node for the pipe wall, which is held at a constant temperature of 0C. Determine how much heat conducts through the insulation and subtract this from the heat added by the tape. The result is the rate of heat transfer into the ice. Using the latent heat of fusion, the time to melt the ice can be calculated.

Assumptions:1. The pipe wall is a perfect conductor and hence, isothermal. 2. The thermal conductivity of the insulation is constant. 3. Heat transfer is one-dimensional. 4. The heat transfer coefficient is uniform over the surface of the insulation and independent of temperature. 5. The insulation is gray and diffuse.

Solution:The process may be modeling using the following thermal resistance network:

where Q3 is the heat supplied by the heating tape and Ts is the surface temperature of the insulation. The heat added by the tape splits into two directions part travels through the insulation ( Q1 ) and the other part ( Q2 ) acts to melt the ice. First find Q1 . We will need the following radii:3cm = 1.5cm = 0.015 m 2 4 r2 = 0.015 m + m = 0.019 m 1000 0.5 m = 0.024 m r3 = 0.019 m + 100r1 =

3 - 34

where L = 1m because the heat rate of the tape is given per meter of pipe length. The resistance to conduction is r 0.024 ln 3 ln r2 K 0.019 = = 0.453 R1 = W 2 Lk W 2 (1m ) 0.082 mK The resistance to convection is 1 1 1 K R2 = = = = 0.861 W hconv A hconv 2 r3 L W 7.7 2 2 ( 0.024 m )(1m ) m K The radiative resistance is 1 1 R3 = = where hrad = (Ts + Tair )(Ts 2 + Tair 2 ) hrad A hrad 2 r3 L The surface temperature, Ts, is unknown, so hrad cannot be determined. It is necessary to do an iterative calculation. We will assume a value of Ts = 266 K, which happens to be the right answer. Of course, it is very unlikely that a person would pick that exact number as an initial guess; however, any reasonable guess (something between 0C and 15C) will give a result close to the correct answer. The guess can be adjusted and the calculation repeated if greater accuracy is desired. With the assumed temperature, W W hrad = ( 0.94 ) 5.67 108 2 4 ( 266 + 258 ) K ( 2662 + 2582 ) K 2 = 3.84 2 m K m K R3 =

1

W 3.84 2 2 ( 0.024 m )(1m ) m K Define R23 as the parallel combination of R2 and R3, that is RR ( 0.861)(1.73) = 0.575 K R23 = 2 3 = R2 + R3 0.861 + 1.73 WRtot = R1 + R23 = 0.453 + 0.575 = 1.03

= 1.73

K W

K W

The heat traveling through the insulation is T T ( 273 258) K = 14.5 W Q1 = ice air = K Rtot 1.03 W The heat available to melt the ice is Q2 = Q3 Q1 = 90 14.5 = 75.5 W The total amount of heat needed to melt the ice is J kg 2 5 Q2 = hif Vice ice = hif r12 L ice = 3.34 105 ( 0.015 m ) (1m ) 921 3 = 2.17 10 J kg m Q2 = Q2 t Q2 2.17 105 J = = 2878s = 48 min 75.5 W Q2

Thereforet =

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3-33

Show that the conduction thermal resistance of a spherical shell of inner radius r1 and outer radius r2 is given by:Rsphere = r2 r1 4 r1r2 k

Approach:Start with Fouriers law. Write surface area, A, in terms of the radius of the sphere and integrate from the inner radius to the outer radius of the spherical shell. Compare the resulting equation to a resistive circuit.

Assumptions:1. Thermal conductivity is constant. 2. Heat transfer is one-dimensional.

Solution:From Fouriers law dT Q = kA dr Separate variables and integrate from the inner to the outer radius r2 Q T2 r1 Adr = k T1 dT To perform this integration, we need A in terms of r ( Q is a constant). Substituting the formula for the surface area of a sphere and integrating the right hand side, r2 dr Q = k (T2 T1 ) r1 4 r 2Q ( r 1 ) rr21 = k (T2 T1 ) 4 Q 1 1 = k (T2 T1 ) 4 r2 r1 r1 r 2 = k (T2 T1 ) r1r2 r1r2 Compare this result to a resistive circuit to obtain T T r r Rsphere = 1 2 = 2 1 4 kr1r2 Q Q 4

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3-34

A hollow sphere made of pure aluminum has an inner radius of 3 cm and an outer radius of 18 cm. The temperature at the inner radius is maintained at 0oC. The outer surface is exposed to air at 25oC. The convective heat transfer coefficient is 65 W/m2K, and radiation may be neglected. Calculate the rate of heat transfer and the temperature of the outer surface of the sphere.

Approach:Use the resistance analogy to determine the heat transfer rate and the surface temperature. Calculate the conductive and convective thermal resistances and add them to find the total resistance.

Assumptions:1. The thermal conductivity is constant. 2. Heat transfer is one-dimensional. 3. The heat transfer coefficient is uniform over the surface of the sphere and independent of temperature.

Solution:Approach this problem by using the resistance analogy. The conduction resistance is r r Rcond = 2 1 4 r1r2 k The value for thermal conductivity of aluminum is available in Table A-2. Using this and given values: (0.18 0.03) m K = 0.00933 Rcond = W W 4 (0.18)(0.03)m 2 237 mK The convection resistance is 1 1 1 K Rconv = = = = 0.0378 2 W W hA h ( 4 r2 ) 2 2 65 2 4 (0.18) m m K Heat is first conducted through the aluminum and then convected from the surface. The conduction and convection occur in series; therefore, add the resistances to get total resistance. The heat transfer becomes: T T T1 T (0 25) o C Q= 1 = = Rtot Rcond + Rconv (0.00933 + 0.0378) K W Q = 1811W Answer Heat rate is negative because heat is flowing in the negative r-direction (from outside to inside). To find surface temperature, T2 , note that the total heat is equal to the heat removed by convection, so thatT2 T Rconv Solving for T2: T2 = QRconv + T Q= K T2 = (1811W) 0.0378 + 25 o C W o = 4.95 C

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3-35

A bathosphere of inside diameter 3.4 m is at an ocean depth where the water temperature is 5oC. The wall of the bathosphere is made of 5 cm thick steel. The convective heat transfer coefficient between the air and the inside wall is 9.2 W/m2K and that between the water and the outside wall is 860 W/m2K. After the divers return to the surface they complain to the designer that the bathosphere was chilly. If the maximum power of the heater is 2.5 kW, estimate the air temperature inside the bathosphere.

Approach:Use the thermal resistance analogy.

Assumptions:1. Neglect body heat generated by the divers. 2. The bathosphere is spherical in shape. 3. Neglect heat dissipated by any on-board electronics. 4. Heat transfer is one-dimensional. 5. Thermal conductivity is constant. 6. The heat transfer coefficients are uniform over the sphere and independent of temperature.

Solution:The inside and outside radii of the bathosphere are r1 = 1.7 m r2 = 1.7 + 0.05 = 1.75 m The inside and outside surface areas, are, respectively A1 = 4 r12 = 4 (1.7) 2 m 2 = 32.3 m 2A2 = 4 r2 2 = 4 (1.75) 2 m 2 = 38.5 m 2 The thermal conductivity of steel is available in Table A-2. The thermal resistances are: 1 1 K R1 = = = 0.00299 W W h1 A1 2 9.2 2 (32.3m ) m K (r r ) (1.75 1.7)m K R2 = 2 1 = = 2.21 105 W W 4 r1r2 k 4 (1.7 m)(1.75 m) 60.5 mK 1 1 K R3 = = = 3.02 105 W W h2 A2 2 860 2 ( 38.5 m ) m K K Rtot = R1 + R2 + R3 = 0.00305 W Note that the only resistance that matters is the interior convection resistance. The rate of heat transfer is T T Q = air water Rtot Solving for the bathosphere air temperature Tair = QRtot + Twater K o 1000 W = 2.5 kW 0.00305 + 5 C W 1 kW = 12.6 o C = 54.7 o F

Comment: It is rather chilly in the bathosphere.

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3-36

A high-pressure chemical reactor contains a gas mixture at 1000oF. The reactor is made of AISI 1010 carbon steel and is spherical, with an inner diameter of 3.2 ft and a wall thickness of 0.75 in. The outer wall of the reactor is encased in a 2.5 in. thick layer of insulation (k = 0.03 Btu/hftR). The convective heat transfer coefficient on the inside wall of the reactor is 8.3 Btu/hft2oF and on the outside of the insulation, the combined convective/radiative heat transfer coefficient is 11.7 Btu/hft2oF. If the ambient is at 80oF, find the rate of heat transfer from the reactor to the surroundings.

Approach:Use the resistance analogy to find the resistances due to conduction and convection. All resistances add in series.

Assumptions:1. Thermal conductivity is independent of temperature. 2. The heat transfer coefficient is uniform over the surface of the reactor and independent of temperature. 3. Heat transfer is one-dimensional.

Solution:The various radii needed to solve the problem, as shown in the figure, are: 3.2 ft 0.75 r1 = = 1.6 ft r2 = r1 + L1 = 1.6 + = 1.66 ft 2 12 2.5 r3 = r2 + L2 = 1.66 + = 1.87 ft 12 The convective resistance on the inside is o 1 1 1 Fh R1 = = = = 0.00374 2 Btu Btu h1 A1 h1 4 r1 2 8.3 4 (1.6 ft) h ft 2 o F For conduction, the resistances are (using k for steel at 1440 R from Table B-2) o r r (1.66 1.6)ft Fh R2 = 2 1 = = 0.000165 Btu 2 r1r2 k1 Btu 2 (1.66)(1.6)ft 2 22.7 h ft o F o r r 1.87 1.66 Fh R3 = 3 2 = = 0.356 2 r2 r3 k2 2 (1.87)(1.66)(0.03) Btu The convective resistance on the outside is o 1 1 1 Fh R4 = = = = 0.00194 h2 A2 h2 4 r32 (11.7)4 (1.87) 2 Btu As you can see, the conduction resistance through the insulation dominates. The total resistance is o Fh Rtot = R1 + R2 + R3 + R4 = 0.00374 + 0.000165 + 0.356 + 0.00194 = 0.361 Btu The heat transferred is T (1000 80) o F Btu Q= = = 2550 Answer o Fh Rtot h 0.361 Btu

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3-37

A novelty drink container is made of plastic in the shape of a sphere. The container has an outside diameter of 6.5 cm and a wall thickness of 2.5 mm. The container is initially filled with soda and crushed ice. The ice occupies 30% of the volume of the drink. The plastic has a conductivity of 0.07 W/mK and an emissivity of 0.92. The inside surface of the container may be assumed to be at the freezing temperature of water. The heat transfer coefficient due to convection on the outside of the container is 9.4 W/m2K. Ambient temperature is 18C. The latent heat of fusion of water is 333.7 kJ/kg and the density of ice is 921 kg/m3. Neglecting any transient effects, estimate the time until all the ice has just melted.

Approach:Use the thermal resistance analogy to determine the heat transfer to the soda and ice mixture. Since the surface temperature is unknown and radiation depends on this temperature, it will be necessary to iterate. As a final step, calculate the mass of ice and use the heat of fusion of the ice to determine the melting time.

Assumptions:1. Soda has the properties of water. 2. Heat transfer is one-dimensional. 3. Thermal conductivity is constant. 4. The heat transfer coefficients are uniform over the sphere and independent of temperature. 5. The outer surface is gray and diffuse.

Solution:First find the steady-state rate of heat transfer from the ambient to the soda and ice. The relevant radii are 6.5 r2 = = 0.0325 m 2 )(100 ) (2.5 = 0.03m 1000 The conductive resistance through the plastic is given by r r 0.0325 0.03 K = 2.91 R1 = 2 1 = 4 r1r2 k 4 ( 0.03)( 0.0325 )( 0.07 ) W r1 = r2

The area of the outside is A = 4 r22 = 0.0133 m 2 The resistance for the convection and radiation on the outside of the container is, 1 R2 = ( hconv + hrad ) A The resistance R2 depends on the radiative heat transfer coefficient, which is hrad = (T2 + T3 )(T2 2 + T32 ) This cannot be calculated until T2 is known or assumed. Since the heat that is transferred to the outer surface by convection and radiation must conduct through the plastic, T3 T2 T2 T 1 = R2 R1 The last three equations are solved simultaneously for T2, R2 and hrad (iteration is necessary). All temperatures must be expressed in Kelvin. The result is

3 - 40

W K R2 = 5.346 T2 = 279.4 K 2 m K W The rate of heat transfer into the soda and ice mixture is thus T T 279.4 273 Q= 2 1 = = 2.18 W R1 2.915 The volume of the drink is 4 4 3 V = r13 = ( 0.03) = 1.13 104 m3 3 3 The volume of ice is 30% of total volume, so Vice = ( 0.3) V = 3.39 105 m3 hrad = 4.69 kg mice = Vice = 921 3 ( 3.39 105 m3 ) = 0.03125 kg m The first law for the ice is Q = H = mh Qt = mhif

where hif is the heat of fusion of the ice. Solving for time kJ ( 0.0132 kg ) 333.7 m hif kg t = = Q 1 kJ ( 2.18 W ) 1000 J = 4786s = 1.33h

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3-38

A small rod made of pure copper is 0.5 cm in diameter and 1.4 cm long. The rod is initially at 10oC. It is then exposed to a hot air flow at 30oC. The heat transfer coefficient between the rod and the air is 25 W/m2oC. What will the rod temperature be after 45 seconds?

Approach:Calculate the Biot number to see if the lumped system approximation is valid. Then use the lumped system approximation to determine final temperature.

Assumptions:1. The lumped system approximation is valid. 2. The heat transfer coefficient is uniform over the surface of the rod and independent of temperature. 3. The specific heat and thermal conductivity of the copper are constant.

Solution:Check the Biot number to see if the lumped-system approximation is valid. A representative length is ( 0.25cm )(1.4 cm ) = .106 cm V R2 L RL Lchar = = = = 2 A ( 2 RL + 2 R ) 2 ( L + R ) (2)(0.25+1.4) cm The thermal conductivity of pure copper is found in Table A-2 1m W 25 (0.106 cm) hLchar m 2 o C 100 cm = 6.61 105 = Bi = W k 401 mK Since Bi 0.1 , the lumped-system approximation is valid. The temperature of the rod as a function of time is ht +T T = (Ti T f ) exp c L f p char Using values for the properties of copper from Table A-2 W 25 2 o (45s) m C + 30 o C T = (10 30) o C exp 1m kg J 8933 3 385 (0.106 cm) m kg K 100 cm T = 15.3 o C Answer

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3-39

A slab of aluminum (2024-T6), which measures 16 cm 16 cm 1.5 cm is initially at 750 K. The slab is then annealed by a water spray at 15C, which strikes both sides of the slab. The convective heat transfer coefficient is estimated to be 1500 W/m2K. How much time is required to cool the slab to 320 K? Neglect convection off the edges of the slab since almost all the surface area is on the two 16 cm x 16 cm sides.

Approach:Calculate the Biot number to see if the lumped system approximation is valid. Then use the lumped system approximation to determine the time required.

Assumptions:1. The lumped system approximation is valid. 2. The heat transfer coefficient is uniform over the surface of the slab and independent of temperature. 3. The specific heat and thermal conductivity of the aluminum are constant. 4. Heat transfer on the edges is negligible.

Solution:Check the Biot number to see if the lumped-system approximation is valid. A representative length is V sA s (1.5cm ) Lchar = = s = = = 0.75cm A 2 As 2 (2) where s is the thickness of the slab and As is the surface area of one 16 cm 16 cm side. The thermal conductivity of 2024-T5 aluminum alloy depends on temperature, as shown in Table A-2. Assume an average slab temperature of Tave = ( 750 + 320 ) K / 2 = 535 K. The Biot number then becomes 1m W 1500 2 (0.75cm) m K hL 100 cm = 6.05 102 Bi = char = W k 186 mK Since Bi 0.1 , the lumped-system approximation is valid. The time to cool the slab is given by c p Lchar T ( t ) T f ln t= h Ti T f Using values for the properties of 2024-T5 aluminum alloy from Table A-2 at 535 K (interpolating for cp), and expressing final temperature in Kelvin, 1m kg J 2770 3 1004 ( 0.75 cm ) m kg K 100 cm ln 320 288 t= W 750 288 1500 2 m K t = 37.1 s Answer

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3-40

Buckshot initially at 450oF is quenched in an oil bath at 85oF. The buckshot is spherical with a diameter of 0.2 in. and is made of lead. The shot falls through the bath, reaching the bottom after 20 s. The convective heat transfer coefficient between buckshot and oil is 36 Btu/hft2oF. Calculate the temperature of the shot just as it reaches the bottom of the bath.

Approach:Calculate the Biot number to see if the lumped system approximation is valid. Then use the lumped system approximation to determine final temperature.

Assumptions:1. The lumped system approximation is valid. 2. The temperature of the oil is constant. 3. The specific heat and thermal conductivity of the lead is constant. 4. The heat transfer coefficient is independent of temperature and uniform over the surface of the shot.

Solution:Check the Biot number to see if the lumped-system approximation is valid. A representative length is 1ft 4 ( 0.2in.) R3 V 3 R 12 in. Lchar = = = = = .00278ft A 4 R 2 3 (2)(3) The thermal conductivity of lead is found in Table B-2 Btu 36 (0.00278ft) hLchar h ft 2 o F = Bi = Btu k 20.4 h ft o F Bi = 0.0049 Since Bi 0.1 , the lumped-system approximation is valid. The temperature of the buckshot is ht +T T = (Ti T f ) exp c L f p Using values for the properties of lead from Table B-2Btu 1h - 36 h ft 2 o F (20s) 3600 s + 85 o F T = (450 85) F exp lbm Btu 708 ft 3 0.03 lbm R (0.00278ft) = 97.3 o Fo

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3-41

A thermocouple is a temperature-measuring device which relies on quantum mechanical effects. Thermocouples are constructed of two thin wires of different metals welded together to form a spherical bead. Consider a thermocouple 0.1 mm in diameter which is suddenly immersed in ice water. Ideally, the thermocouple bead should immediately drop to 0oC, but in practice, there is a time delay. The heat transfer coefficient between the thermocouple and the ice water is 32 W/m2oC. The density, specific heat, and thermal conductivity of the bead are 8925 kg/m3, 385 J/kgK, and 23 W/mK, respectively. Assuming no conduction in the thermocouple wires and an initial thermocouple temperature of 25oC, estimate the time required for the bead to reach 0.1oC.

Approach:Calculate the Biot number to see if the lumped system approximation is valid. Then use the lumped system approximation to determine the unknown time.

Assumptions:1. The lumped system approximation is valid. 2. There is no conduction in the thermocouple wires. 3. The thermophysical properties of the thermocouple are constant.

Solution:Check the Biot number to see if the lumped-system approximation is valid. The characteristic length is 4 3 r V 3 r Lchar = = = A 4 r 2 3Lchar = 0.05 mm 1m 5 = 1.67 10 m 3 1000 mm

The Biot number is given by W 32 (1.6710-5m ) hLchar m 2 K Bi = = = 2.32 105 W k 23 mK Since Bi 0.1, the lumped system approximation is valid. The time to reach a specified temperature ist= cP L T (t ) T f ln T T h f i

kg J 5 (1.67 10 m) 8925 3 385 m kg K 0.1-0 ln t= W 25-0 32 2 o m C = 9.88 s

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3-42

A long uninsulated Nichrome wire of diameter 1/16 in. is cooled convectively by air at 70oF. The heat transfer coefficient is 6.6 Btu/hft2oF. Current runs through the wire, generating heat at a rate of 1.9 W per foot. a. Find the steady state temperature of the wire. b. Assume the wire is initially at 70oF. After the current is turned on, how long will it take for the wire temperature to rise to 90% of the difference between its initial temperature and its steady-state temperature?

Approach:For part a, set heat generated equal to heat convected and solve for temperature. For part b, use the lumped system approximation. Start from the first law and include both generation and convection in the heat term. The first law becomes a differential equation which must be solved to find the required time.

Assumptions:1. The wire is very long so that there is no conduction down the wire. 2. The specific heat and thermal conductivity are constant. 3. The heat transfer coefficient is uniform over the surface of the thermocouple and independent of temperature.

Solution:a) In steady state, the heat generated equals the heat convected from the wire, or Qgen = hA (Ts T f ) For a 1-ft length of wire 1 1ft 2 A = dL = in. (1ft) = 0.0164 ft 16 12 in. Solving for surface temperature Btu 3.41 W h 1.9 ft 1 W Qgen Ts T f = Answer = + 70 F = 130 o F Btu hA 2 6.6 (0.0164 ft ) h ft 2 o F b) Check to see if the lumped system approximation is valid by finding the Biot number for the wire. The characteristic length is 1 1ft in. 2 V r L r 16 12 in. = 0.0013ft = = Lchar = = 4 A 2 rL 2 Btu 6.6 (0.0013ft) hLchar h ft 2 o F = = 0.00125 Bi = Btu K 6.9 h ft o F where the thermal conductivity of Nichrome is found in Table B-2. The Biot number is less than 0.1; therefore, the lumped system approximation is reasonable. Apply the first law to the wire dU = Q W dt By the lumped system approximation, the entire wire is at the same temperature so that internal energy is only a function of time and not a function of space. The heat term includes the heat generated and the

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heat convected from the wire. Since we account for the current in the wire as generated heat, we set W = 0. With these simplifications, the first law becomes dT mc p = Qgen hA(T T f ) dt Separating variables dT dt = Qgen hA(T T f ) mc p Integrating both sides T t dt dT Ti Q hA (T T ) = 0 mc p gen f Let

= Qgen hA(T T f )

d = hAdT Using these substitutions on the left hand side and integrating the right hand side produces T d t Ti ( hA ) = mc pln T Ti

=

hAt mc pT

Evaluating at the limits:ln Qgen hA(T T f ) = Ti

hAt mcP

Qgen hA(T T f ) hAt ln = Qgen hA(Ti T f ) mc p

We are asked for the time t at which T = 0.9 Ts = (0.9)(130) = 117 o F . Solving for t Qgen hA(T T f ) ln hA Qgen hA(Ti T f ) If L is the length of wire ( L =1ft)t= mc p t=

L r 2 c p Qgen hA(T T f ) ln h2 rL Qgen hA(Ti T f )

1 1ft 2 A = 2 rL = 2 in. (1ft) = 0.0164 ft 32 12in. using the density and specific heat of Nichrome from Table B-2, Btu 3.41 h Btu 2 o 1.9W 6.6 (0.0164 ft )(117 70) F 2 lbm 1 1ft Btu h ft F 1W 524 2 in. 0.1 ft 32 12 in. lbm R ln t= Btu (1.9)(3.41) (6.6)(0.0164)(70 70) 2 6.6 2 o h ft F

t = 0.0158 h = 56.9 s

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3-43 A copper sphere 3 cm in diameter is painted black so that it has an emissivity very close to 1. Thesphere is heated to 700oC and then placed in a vacuum chamber whose walls are very cold. How long will it take for the sphere to cool to 300oC? Use the lumped system model.

Approach:Calculate the Biot number to see if the lumped system approximation is valid. Then use the first law to express the energy balance is terms of temperatures. It will be necessary to integrate to find the final temperature.

Assumptions:1. The lumped system approximation is valid. 2. The painted copper is a black body. 3. The thermal conductivity and specific heat of the sphere are constant.

Solution:Check the Biot number to see if the lumped-system approximation is valid. The radiation heat transfer coefficient is (with Ts = 700 + 273 = 973K and Tsurr = 300 + 273 = 573K )W hrad = (Ts + Tsurr )(Ts 2 + Tsurr 2 ) = (1) 5.67 108 2 4 ( 973 + 573) K ( 9732 + 5732 ) K 2 m K W hrad = 112 2 m K The representative length for use in the Biot number is 1m 4 3 ( 3.0 cm ) r V 3 r 100 cm Lchar = = = = = .005 m A 4 r 2 3 (2)(3) The thermal conductivity of copper at the average temperature of 500oC is found in Table A-2. With this W 112 2 (0.005 m) hLchar m K Bi = = = 0.00145 k W 386 m K Since Bi 0.1 , the lumped-system approximation is valid. From the first law: dU = Q W dt Since W = 0 dT = AT 4 mc p dt Separating variables T dT t Adt Ti T 4 = 0 mc p

Integrating both sidesT 3 At = 3) T mc p (i

T

Evaluating at the limits 1 1 3 At 3 = 3 T Ti mc p Solving for t

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4 r 3c p mc p 1 1 1 1 3 t= = 3 A T 3 Ti 3 3 ( 4 r 2 ) T 3 Ti 3 which simplifies to rc p 1 1 t= 3 3 9 T Ti

Using properties of pure copper from Table A-2 at the average temperature of 500 K, kg .03 J 8933 3 m 385 m 2 kg K 1 1 t= 3 3 W ( 300 + 273) ( 700 + 273) 9 5.67 10-8 2 4 m K = 428s = 7.13 min Answer

Comments:The radiative heat transfer coefficient changes as the sphere cools, since it depends on temperature. Evaluating hrad at the final sphere temperature of 300oC gives a Biot number of 0.000553. Thus the Biot number is small enough to justify using the lumped system approximation throughout the entire cooling process.

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3-44

The roof of a house is partially covered with snow. The roof is made from plywood covered with shingles (ksh = 0.4 Btu/hftoF). In the attic space, the heat transfer coefficient between the air and the plywood is 3.1 Btu/hft2oF. The heat transfer coefficient over the snow and the exposed shingles on the outside of the roof is 7.6 Btu/hft2oF. Assume the snow has a density of 12 lbm/ft3 and it covers 64% of the roof area. Calculate the thermal resistance from the attic air to the outside air for the 30-ft by 60-ft roof panel shown below.

Approach:Use the thermal resistance analogy, adding resistances in parallel and series as necessary.

Assumptions:1. The roof is very large so edge effects may be neglected. 2. The plywood and shingles are isothermal. 3. The heat transfer coefficient is uniform and independent of temperature. 4. Thermal conductivity is constant.

Solution:A thermal resistance model for transport from the attic air at Tin to the ambient air at Tout is:

The area of the roof panel is A = ( 30 ft )( 60 ft ) = 1800 ft 2 With properties from Tables B-4 and B-5, the resistances are 1 1 Ro = = = 1.79 104 ( h o F ) ( Btu ) hA 3.1Btu ( h ft o F ) (1800 ft 2 )

(

)

0.75 ft L 12 = = 4.96 104 ( h o F ) Btu R1 = kA ( 0.07 Btu ( h ft R ) ) (1800 ft 2 ) 0.5 L 12 = = 5.79 105 ( h o F ) Btu R2 = kA ( 0.4 )(1800 )

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3 L 12 = = 48.2 104 h o F Btu R3 = kA ( 0.045 )(1800 )( 0.64 ) R4 = R5 = 1 1 = = 1.14 104 h o F Btu hA ( 7.6 )(1800 )( 0.64 )

1 1 = = 2.03 104 h o F Btu hA ( 7.6 )(1800 )( 0.36 ) Adding resistances in series and parallel gives R ( R + R4 ) Rtot = Ro + R1 + R2 + 5 3 = 9.28 104 h o F Btu R5 + R3 + R4

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3-45

A dining area has a glass ceiling built of square units. Each unit consists of two glass panes supported by a steel frame, as shown below. The space between the panes contains a gas. The heat transfer coefficients, as shown on the figure, are h1 = 4.11 W/m2oC (inside the room) h2 = 3.63 W/m2oC (between the panes) h3 = 7.45 W/m2oC (outside the room) The air in the room is at 26oC and the exterior air is at 15oC. The glass has a thermal conductivity of 1.4 W/mK and the steel has a thermal conductivity of 37.7 W/mK. Using dimensions on the figure, find the total heat loss through one unit.

Approach:Use the thermal resistance analogy, adding resistances in parallel and series as necessary.

Assumptions:1. All heat flow is perpendicular to the ceiling. 2. The heat transfer coefficients are uniform and independent of temperature. 3. Thermal conductivity is constant.

Solution:The heat transport may be modeled with the resistance network shown, where R1 Convection on inside steel surface R2 Conduction in the steel R3 Convection on outside steel surface R4 Convection on inside glass surface R5 Conduction in the glass R6 Convection in interior gas space R7 Convection of outside glass surface The areas needed for the calculation are A1 = (0.73m) 2 = 0.533m 2A2 = 0.73 + 2 ( 0.041) A1 2

= 0.1264 m 2

The thermal resistances are 1 1 C R1 = = =1.92 W W h1 A2 2 4.11 2 o ( 0.1264 m ) m C R2 = 0.084 m L C = = 0.0176 W W k2 A2 2 37.7 o ( 0.1264 ) m m C

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R3 = R4 = R5 = R6 = R7 =

1 1 C = =1.06 h3 A2 ( 7.45 )( 0.1264 ) W 1 1 C = = 0.457 h1 A1 ( 4.11)( 0.533) W L 0.084 C = = 0.00804 k1 A1 (1.4 )( 0.533) W 1 1 C = = 0.517 h2 A1 ( 3.63)( 0.533) W 1 1 C = = 0.252 h3 A1 ( 7.45 )( 0.533) W

We now combine the three resistances on the left leg of the circuit into C R8 = R1 + R2 + R3 = 3.003 W We also combine the resistances on the right leg to get C R9 = R4 + 2 ( R5 + R6 ) + R7 =1.76 W The parallel combination of the last two resistances is RR ( 3)(1.76 ) =1.11 C RT = 8 9 = R8 + R9 3 + 1.76 W Finally, the rate of heat loss may be determined as o T ( 26 15 ) C Q= = = 9.92 W C Rtot 1.11 W

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3-46

A man is wearing a shirt and a jacket which is unzipped in front. His skin temperature is 70oF. The convective and radiative heat transfer coefficients on the outside of the jacket and exposed part of the shirt are estimated as 0.8 and 0.43 Btu/hft2oF respectively. Model the mans torso as a cylinder of diameter 1.3 ft and height 2 ft. Assume the shirt is a layer of cloth of thickness 0.05 in. with ks = 0.12 Btu/hftR. The jacket is 0.4-in. thick with kj = 0.094 Btu/hftR. Assume the jacket covers half the mans torso. The surroundings are at 45oF. Calculate the total rate of heat loss from the mans torso. Neglect the thermal resistance due to any air layers between the shirt and the skin, or the shirt and the jacket.

Approach:Use the thermal resistance analogy, adding resistances in parallel and series as necessary.

Assumptions:1. The resistance due to trapped air layers is small. 2. The heat transfer coefficients are uniform and independent of temperature. 3. Thermal conductivity is constant. 4. The jacket and shirt are gray and diffuse.

Solution:The radii that will be needed, as shown in the figure above, are r1 = 0.65ft0.05 = 0.65416 ft 12 0.4 r3 = r2 + = 0.6875ft 12 A thermal resistance network is shown to the right. The resistances are: r2 = 0.65 +

R1 conduction in the shirt R2 convection and radiation from the exposed part of the shirt R3 conduction in the jacket R4 convection and radiation from the jacket

r 0.65416 ln 2 ln r1 0.65 = = 0.00424 R h Btu R1 = 2 Lk s 2 ( 2 ft )( 0.12 Btu h ft R ) R2 = 1 htot A = 1 2 where D2 is the diameter at radius r2. Substituting values 1 R2 = = 0.198 R h Btu ( 0.43 + 0.8)( )( 0.65416 )( 2 )

D2 L

In evaluating the conduction in the jacket, we recognize that the jacket covers only half the body, so the area perpendicular to the direction of heat flow is only half that of a cylindrical shell. The resistance to conduction in the jacket is, therefore, twice, the resistance of a cylindrical shell (less area implies higher resistance). This resistance is

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r3 0.687 ln ln 0.65416 r2 = 0.0842 R h Btu = 2 R3 = 2 2 Lk 2 ( 2 )( 0.094 ) 1 1 1 = = = 0.188 R h Btu D3 L ( 0.43 + 0.8 ) ( 0.687 ) 2 htot A ( hvrad + hcon ) 2 The thermal circuit may now be solved as R ( R + R4 ) 0.198 ( 0.0842 + 0.188 ) RT = R1 + 2 3 = 0.00424 + = 0.119 R h Btu R2 + R3 + R4 0.198 + 0.0842 + 0.188 The rate of heat transfer from the man is T 70 45 = = 210 Btu h Q= Answer R 0.119 R4 =

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3-47

The inside wall of a machine is covered with acoustic tile 3.5 cm thick for noise abatement. The tile increases the thermal resistance of the wall, and, as a result, the interior air temperature rises to unacceptable levels. An engineer suggests drilling holes in the tile and welding steel rods 3.5 cm long and 1.8 cm in diameter to the wall, so as to increase its effective thermal conductivity, as shown in the figure. The rods are in a square array on 10 cm centers. The machine dissipates 150 W per square meter of wall area through its outer wall. The heat transfer coefficients on the interior and exterior are 4.6 and 11.4 W/m2K respectively. If the exterior air temperature is 25oC, calculate the interior air temperature with and without the rods.

Approach:Use the thermal resistance analogy, adding resistances in parallel and series as necessary.

Assumptions:1. The inside wall of the machine is very large so that edge effects may be neglected. 2. The steel wall is isothermal. 3. Thermal conductivity is constant.

Solution:Select a unit cell of wall which contains one rod, as shown to the right. Defining q as the heat transfer per unit area and A3 as the surface area of the unit cell, the rate of heat transfer through the unit cell is 1m 2 (10 cm)(10 cm) 4 2 150 W 10 cm = 1.5 W Q = qA3 = 2 2 1m 1m Note that there are 100 of these unit cells in 1 square meter. Assuming that the steel wall is isothermal, the thermal resistance network is

Ro inside convective resistance R1 conduction through the tile R2 conduction through the rod

R3 conduction in the steel wall R4 outside convective resistance

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The areas across which heat flows are: 1 m2 A3 = (10 cm)(10 cm) 4 2 = 0.01m 2 10 cm A2 = r 2 = (0.009 m) 2 = 2.54 104 m 2 A1 = A3 A2 = 0.00975 m 2 With these areas, the resistances may be calculated as: 1 1 K R = = = 21.7 hin A3 W W 2 4.6 2 (0.01m ) m K L1 0.035 m K = = 61.0 W W k1 A1 2 0.058 (0.00975 m ) mK L1 0.035 m K R2 = = = 2.27 W W k2 A2 4 2 60.5 (2.54 10 m ) mK L 0.0075 m K R3 = 2 = = 0.0124 W W k2 A3 2 60.5 (0.01m ) mK 1 1 K R4 = = = 8.77 W W hout A3 2 11.4 2 (0.01m ) m K where the thermal conductivity of steel and acoustic tile are found in Tables A-2 and A-5, respectively. For the parallel combination of R1 and R2: ( 61.9 )( 2.27 ) = 2.19 K RR R5 = 1 2 = R1 + R2 61.9 + 2.27 W This resistance now combines in series with the other three resistances, i.e., Rtot = R + R5 + R3 + R4 R1 = = 21.7 + 2.19 + 0.0124 + 8.77 = 32.7 K W

The basic rate equation is T Tin Tout Q= = Rtot Rtot Solving for Tin, Tin = Tout + QRtot Substituting values: 32.7 K o Tin = 25 o C + 1.5 W with rods Answer = 74.1 C W To find the interior temperature without the steel rods, repeat the analysis, but set the thermal conductivity of the rods to that of acoustic tile. The result is Tin = 161 o C without rods Answer

Comments:The rods make a big difference. The engineer might also think about thinner tiles.

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3-48

In a certain localized area, the earth can be approximately represented by areas of stone, soil, and iron ore, as shown below. Using data on the figure and assuming the geometry is two-dimensional, find the effective thermal conductivity in the vertical direction. This is the conductivity that the earth would have if it were all made of the same material.

Approach:Use the thermal resistance analogy, adding resistances in parallel and series as necessary.

Assumptions:1. All heat flow is in the vertical direction. 2. Thermal conductivity is constant.

Solution:The heat transport may be modeled with the resistance network shown. For the soil (for a 1 ft. depth of earth), L 4 ft hR R1 = = = 2.42 Btu Btu kA 0.3 (1.5 + 2.5 + 1.5 ) ft (1ft ) h ft R For the iron core, L 4 ft R2 = = = 0.32 Btu kA 25 0.25 + 0.25 ) ft (1ft ) ( h ft R For the stone, L 0.5ft R3 = = = 0.052 Btu kA 1.6 ( 6 ) ft (1ft ) h ft R The total resistance may be calculated as ( 2.42 )( 0.32 ) = 0.335 h R RR Rtot = R3 + 1 2 = 0.052 + Btu R1 + R2 2.42 + 0.32 The effective thermal conductivity must satisfy L Rtot = keff A Thereforekeff = L Rtot A = 4.5 Btu = 2.24 h ft R ( 0.335)( 6 )(1)

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3-49

A drinking glass with an outside diameter of 3.5 in. and a wall thickness of 0.125 in. is filled to the height of 6.2 in. with a mixture of soda and ice. The glass is placed in an insulated soft rubber sleeve 0.75 in thick. (cut-away view shown in figure) The exposed top surface of the drink is at 32F and gains heat by natural convection and radiation from the surroundings, which are at 85F. On the top surface, the heat transfer coefficients for convection and radiation are 1.6 Btu/hft2oF and 0.7 Btu/hft2oF, respectively. The natural convective heat transfer coefficient between the soda-ice mixture and the inside wall of the glass is 57 Btu/hft2oF. On the outside of the rubber sleeve, the heat transfer coefficients for convection and radiation are 2.3 Btu/hft2oF and 0.85 Btu/hft2oF, respectively. Assume no heat is transferred through the bottom of the glass. The initial mass of ice in the drink is 0.09 lbm. The latent heat of fusion of water is 143.5 Btu/lbm. Assuming a steady-state temperature profile in the glass wall and rubber, calculate the time required for the ice to completely melt. (Assume no one takes a sip from the glass.)

Approach:Use the thermal resistance analogy, adding resistances in parallel and series as necessary.

Assumptions:1. The heat transfer coefficients are uniform and independent of temperature. 2. Thermal conductivity is constant. 3. The bottom of the glass is perfectly insulated. 4. Steady-state conditions prevail. 5. The glass is a cylinder (not slightly conical as shown).

Solution:The heat transport may be modeled with the resistance network shown above. The resistances in this circuit are defined as R0 Convection between the soda and the glass R1 Conduction through the glass (not shown actually within the glass due to space constraints). R2 Conduction in the rubber sleeve R3 Convection on the outside of the rubber sleeve R4 Radiation on the outside of the rubber sleeve R5 Convection on the top of the soda R6 Radiation on the top of the soda To evaluate these resistances, the following radii will be needed (see the figure): 3.5 2 (.125 ) r0 = =1.625 in. 2 3.5 r1 = =1.75 in. 2 3.5 r2 = + 0.75 = 2.5 in. 2 To calculate R0, we need the lateral surface area of the inside of the glass A1 = D0 L = 3.5 2 (.125 ) ( 6.2 ) = 63.3 in.2

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in.2 1144 2 ft h oF 1 R0 = = = 0.04 Btu Btu h1 A1 63.3 in.2 ) 57 2 o ( h ft F r 1.75 ln 1 ln o r0 1.