1. CHAPTER 12 MECHANISMS OF HEAT TRANSFER

2. INTRODUCTION
Heat: The form of energy that can be transferred from one system to another as a result of temperature difference.
Thermodynamics deal with the amount of energy in form of heat, work and mass transport during the process and only considers a process from one equilibrium state to another, and it gives no indication about how long the process will take.
Heat Transfer deals with the determination of the rates of such energy transfers as well as variation of temperature(non equilibrium process).
The transfer of energy as heat is always from the higher-temperature medium to the lower-temperature one.
Heat transfer stops when the two mediums reach the same temperature.
Heat can be transferred in three different modes:
conduction, convection, radiation

3. CONDUCTION
Conduction: 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.
In gases and liquids, conduction (heat conduction) is due to the collisions and diffusion of the molecules during their random motion.
In solids, it (heat conduction) is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons.
A cold canned drink in a warm room, for example, eventually warms up to the room temperature as a result of heat transfer from the room to the drink through the aluminum can by conduction
Heat conduction through a large plane wall of thickness x and area A.
The rate of heat conduction through a plane layer is proportional to the temperature difference across the layer and the heat transfer area, but is inversely proportional to the thickness of the layer.

5. When Δx → 0
Thermal conductivity, k: A measure of the ability of a material to conduct heat.
Temperature gradient dT/dx: The slope of the temperature curve on a T-x diagram.
Heat is conducted in the direction of decreasing temperature, and the temperature gradient becomes negative when temperature decreases with increasing x. The negative sign in the equation ensures that heat transfer in the positive x direction is a positive quantity.
In heat conduction analysis, A represents the area normal to the direction of heat transfer.
The rate of heat conduction through a solid is directly proportional to its thermal conductivity.

6. Thermal Conductivity
Thermal conductivity: The rate of heat transfer through a unit thickness of the material per unit area per unit temperature difference.
The thermal conductivity of a material is a measure of the ability of the material to conduct heat.
A high value for thermal conductivity indicates that the material is a good heat conductor, and a low value indicates that the material is a poor heat conductor or insulator.
A simple experimental setup to determine the thermal conductivity of a material.

7. The range of thermal conductivity of various materials at room temperature.

8. The thermal conductivities of gases such as air vary by a factor of 104 from those of pure metals such as copper.
Pure crystals and metals have the highest thermal conductivities, and gases and insulating materials the lowest.
The mechanisms of heat conduction in different phases of a substance.

9. The variation of the thermal conductivity of various solids, liquids, and gases with temperature.

10. Thermal Diffusivity
cp Specific heat, J/kg · °C: Heat capacity per unit mass
cp Heat capacity, J/m3 · °C: Heat capacity per unit volume
 Thermal diffusivity, m2/s: Represents how fast heat diffuses through a material
A material that has a high thermal conductivity or a low heat capacity will obviously have a large thermal diffusivity.
The larger the thermal diffusivity, the faster the propagation of heat into the medium.
A small value of thermal diffusivity means that heat is mostly absorbed by the material and a small amount of heat is conducted further.

11. Example 12-1
A flat wall is composed of 20 cm of brick having a thermal conductivity kt = 0.72 W/mK. The right face temperature of the brick is 900C, and the left face temperature of the brick is 20C. Determine the rate of heat conduction through the wall per unit area of wall.
Tright = 900C
Tleft = 20C
20 cm

12. Example 12-2
The roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m thick, and is made of a flat layer of concrete whose thermal conductivity is k = 0.8 W/m.°C. The temperatures of the inner and the outer surfaces of the roof one night are are measured to be 15°C and 4°C, respectively, for a period of 10 hours. Determine (a) the rate of heat loss through that roof that night and (b) the cost of that heat loss to the home owner if the cost of electricity is $0.08/kWh.

13. Assumption 1 Steady operating conditions exits during the entire night since the surface temperatures of the roof remain constant at the specified value. 2 Constant properties can be used for the roof.
Analysis (a) Noting that heat transfer through the roof is by conduction and the area of the roof is A = 6 m x 8 m = 48 m2, the steady rate of heat transfer through the roof is
(b) The amount of heat lost through the roof during a 10-hour period and its cost is
Cost = (Amount of energy)(Unit cost of energy)
= (16.9 kWh)($0.08/kWh) = $1.35

14. Then the amount of heat transfer over a period of 5 h becomes
Example 12-3
The inner and outer surfaces of a 0.5-cm thick 2-m x 2-m window glass in winter are 10oC and 3oC, respectively. If the thermal conductivity of the glass is 0.78 W/m. K, determine the amount of heat loss through the glass over a period of 5 h. What would your answer be if the glass were 1 cm thick?
Glass
3C
10C
0.5 cm
Then the amount of heat transfer over a period of 5 h becomes
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.

15. Analysis The heat transfer area is A =  r2 =  (0.075 m)2 = 0.0177 m2
Example 12-4
An aluminum pan whose thermal conductivity is 237 W/m . oC has a flat bottom with diameter 15 cm and thickness 0.4 cm. Heat is transferred steadily to boiling water in the pan through it bottom at a rate of 800 W. If the inner surface of the bottom of the pan is at 105oC, determine the temperature of the outer surface of the bottom of the pan.
Analysis The heat transfer area is
A =  r2 =  (0.075 m)2 = m2
Under steady conditions, the rate of heat transfer through the bottom of the pan by conduction is
105C
800 W
0.4 cm
Substituting,
which gives
T2 = C

16. CONVECTION
Convection: The mode of energy transfer between a solid surface and the adjacent liquid or gas that is in motion, and it involves the combined effects of conduction and fluid motion.
The faster the fluid motion, the greater the convection heat transfer.
In the absence of any bulk fluid motion, heat transfer between a solid surface and the adjacent fluid is by pure conduction.
Heat transfer from a hot surface to air by convection.

17. Forced convection: If the fluid is forced to flow over the surface by external means such as a fan, pump, or the wind.
Natural (or free) convection: If the fluid motion is caused by buoyancy forces that are induced by density differences due to the variation of temperature in the fluid.
The cooling of a boiled egg by forced and natural convection.
Heat transfer processes that involve change of phase of a fluid are also considered to be convection because of the fluid motion induced during the process, such as the rise of the vapor bubbles during boiling or the fall of the liquid droplets during condensation.

18. Newton’s law of cooling
h convection heat transfer coefficient, W/m2 · °C
As the surface area through which convection heat transfer takes place
Ts the surface temperature
T the temperature of the fluid sufficiently far from the surface.
The convection heat transfer coefficient h is not a property of the fluid.
It is an experimentally determined parameter whose value depends on all the variables influencing convection such as surface geometry.
the surface geometry
the nature of fluid motion
the properties of the fluid
the bulk fluid velocity

19. Example 12-5
A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at 15oC, as shown in figure above. Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be 152oC in steady operation. Also, the voltage drop and electric current through the wire are measured to be 60 V and 1.5 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
Assumptions 1 Steady operating conditions exit since the temperature readings do not change with time. 2 Radiation heat transfer is negligible.

20. The surface area of the wire is
Newton’s law of cooling for convection heat transfer is expressed as
Disregarding any heat transfer by radiation and thus assuming all the heat loss from the wire to occur by convection, the convection heat transfer coefficient is determined to be

21. Example 12-6
A transistor with a height of 0.4 cm and a diameter of 0.6 cm is mounted on a circuit board as shown in figure below. The transistor is cooled by air flowing over it with an average heat transfer coefficient of 30 W/m2.°C. The air temperature is 55°C and the transistor case temperature is not to exceed 70°C
Determine the amount of power this transistor can dissipate safely. Disregard any heat transfer from the transistor base

22. Analysis Disregarding the base area, the total heat transfer area of the transistor is
Then the rate of heat transfer from the power transistor at specified conditions is
Therefore, the amount of power this transistor can dissipate safely is W.

23. Example 12-7
Four power transistors, each dissipating 15 W, are mounted on a thin vertical plate 22 cm x 22 cm in size. The heat generated by the transistors is to be dissipated by both surfaces of the plate to the surrounding air at 25oC, which is blown over the plate by a fan. The entire plate can be assumed to be nearly isothermal, and the exposed surface area of the transistor can be taken to be equal to its base area. If the average convection heat transfer coefficient is 25 W/m2 . oC, determine the temperature of the aluminum plate. Disregard any radiation effects. Ts
15 W
Disregarding any radiation effects, the temperature of the aluminum plate is determined to be

24. RADIATION
Radiation: The energy emitted by matter in the form of electromagnetic waves a result of the changes in the electronic configurations of the atoms or molecules.
Unlike conduction and convection, the transfer of heat by radiation does not require the presence of an intervening medium.
In fact, heat transfer by radiation is fastest (at the speed of light) and it will pass through vacuum (a volume of space that is essentially empty of matter, such that its gaseous pressure is much less than atmospheric pressure) . This is how the energy of the sun reaches the earth.
In heat transfer studies we are interested in thermal radiation, which is the form of radiation emitted by bodies because of their temperature. It differs from other forms of electromagnetic radiation such as x-rays, gamma rays, microwaves, radio waves, and television waves that are not related to temperature.
All bodies at a temperature above absolute zero(0 Kelvin) emit thermal radiation.

25.  = 5.670  108 W/m2 · K4 Stefan–Boltzmann constant
Radiation is a volumetric phenomenon, and all solids, liquids, and gases emit, absorb, or transmit radiation to varying degrees.
However, radiation is usually considered to be a surface phenomenon for solids.
The maximum rate of radiation that can be emitted from a surface at a thermodynamic temperature Ts ( in K or R) is given by the Stefan-Boltzmann law as
 =  108 W/m2 · K4 Stefan–Boltzmann constant
Blackbody radiation represents the maximum amount of radiation that can be emitted from a surface at a specified temperature.

26. The radiation emitted by all real surfaces is less than the radiation emitted by a blackbody at the same temperature, and is expressd as
Emissivity  : A measure of how closely a surface approximates a blackbody for which  = 1 of the surface. 0   1.

27. Example 12-8
Consider a person standing in a room maintained at 22°C at all times. The inner surfaces of the walls, floors, and the ceiling of the house are observed to be at an average temperature of 10°C in winter and 25°C in summer. Determine the rate of radiation heat transfer between this person and the surrounding surfaces if the exposed surface area and the average outer surface temperature of the person are 1.4 m2 and 30°C, respectively.

28. and

29. STEADY HEAT CONDUCTION IN PLANE WALLS
Heat transfer through the wall of a house can be modeled as steady and one-dimensional.
The temperature of the wall in this case depends on one direction only (say the x-direction) and can be expressed as T(x).
Heat transfer through a wall is one-dimensional when the temperature of the wall varies in one direction only.
Heat transfer is the only energy interaction involved in this case and there is no heat generation, the balance for the wall is

30. Fourier’s law of heat conduction
for steady operation, since there is no change in the temperature of the wall with at time at any point
Therefore, the rate of heat transfer into the wall must be equal to the rate of heat transfer out of it. So, the rate of heat transfer through the wall must be constant,
Fourier’s law of heat conduction
Under steady conditions, the temperature distribution in a plane wall is a straight line: dT/dx = const.

31. Performing the integrations and rearranging gives:
The rate of heat conduction through a plane wall is proportional to the average thermal conductivity, the wall area, and the temperature difference, but is inversely proportional to the wall thickness.
Once the rate of heat conduction is available, the temperature T(x) at any location x can be determined by replacing T2 by T, and L by x.

32. Thermal Resistance ConceptOr
Where
Conduction resistance of the wall: Thermal resistance of the wall against heat conduction.
Thermal resistance of a medium depends on the geometry and the thermal properties of the medium.
This equation for heat transfer is analogous to the relation for electric current flow I, expressed as
Electrical resistance

33. rate of heat transfer  electric current
thermal resistance  electrical resistance temperature difference  voltage difference
Newton’s law of cooling
Convection resistance of the surface: Thermal resistance of the surface against heat convection.
Schematic for convection resistance at a surface.

34. When the convection heat transfer coefficient is very large (h → ), the convection resistance becomes zero and Ts  T∞.
That is, the surface offers no resistance to convection, and thus it does not slow down the heat transfer process.
This situation is approached in practice at surfaces where boiling and condensation occur.

35. Thermal Resistance Network
(rate of heat convection into the wall)= (rate of heat conduction through the wall)= (rate of heat convection from the wall)

36. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.
Temperature drop
Temperature drop across any layer is equal to the rate of heat transfer times the thermal resistance across that layer.
The temperature drop across a layer is proportional to its thermal resistance.

37. Multilayer Plane Walls
Plane wall that consist of several layers of different materials. Consider a plane wall that consists of two layers (such as a brick wall with a layer of insulation). The rate of steady heat transfer through this two-layer composite wall can be expressed as
The thermal resistance network for heat transfer through a two-layer plane wall subjected to convection on both sides.

38. The rate of steady heat transfer through a multilayer medium is constant, and thus it must be the same through each layer. Once is known, an unknown surface temperature Tj at any surface or interface j can be determined from
The interface temperature T2 between the two walls can be determined from

39. GENERALIZED THERMAL RESISTANCE NETWORKS
Thermal resistance network for two parallel layers.

40. Two assumptions in solving complex multidimensional heat transfer problems by treating them as one-dimensional using the thermal resistance network are
any plane wall normal to the x-axis is isothermal (i.e., to assume the temperature to vary in the x-direction only)
any plane parallel to the x-axis is adiabatic (i.e., to assume heat transfer to occur in the x-direction only)
Do they give the same result?
Thermal resistance network for combined series-parallel arrangement.

41. Example 12-9
Consider a 3-m-high, 5-m-wide, and 0.3-m-thick wall whose thermal conductivity is k = 0.9 W/m.K. On a certain day, the temperatures of the inner and the outer surfaces of the wall are measured to be 16oC and 2oC, respectively. Determine the rate of heat loss through the wall on that day.

42. Example 12-10
Consider a 0.8-m-high and 1.5-m-wide double-pane window consisting of two 8 mm thick layers of glass (k = 0.78 W/m.oC). Determine the steady rate of heat transfer through this glass window and the temperature of its inner surface for a day during which the room is maintained at 20oC while the temperature of the outdoors is -10oC. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h1=10W/m2.oC and h2 = 40 W/m2 .oC, which includes the effects of radiation.

43. Example 12-11
Consider a 0.8-m-high and 1.5-m-wide double-pane window consisting of two 4-mm-thick layers of glass (k = 0.78 W/m.K) separated by a 10-mm-wide stagnant air space (k = W/m.K). Determine the steady rate of heat transfer through this double-pane window and the temperature of its inner surface for a day during which the room is maintained at 20oC while the temperature of the outdoors is -10oC. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h1=10W/m2.oC and h2 = 40 W/m2 .oC, which includes the effects of radiation.

44. Example 12-12
A 3-m-high and 5-m-wide consists of long 16-cm x 22-cm cross section horizontal bricks (k = 0.72 W/m.oC) separated by 3-cm-thick plaster layer (k = 0.22 W/m.oC). There are also 2-cm-thick plaster layer on each side of the brick and a 3-cm-thick rigid foam (k = W/m.oC) on the inner side of the wall. The indoor and the outdoor temperatures are 20oC and -10oC, respectively., and convection heat transfer coefficient on the inner and outer side are h1 = 10 W/m2.oC and h2 = 25 W/m2.oC , respectively. Assuming one-dimensional heat transfer and disregarding radiation, determine the rate of heat transfer through the wall.

45. Example 12-13
Consider a 5 m high, 8 m long and 0.22 m thick wall whose representative cross section is as shown in the Figure below. The thermal conductivities of various materials used, in W/m·ͦC, are kA = kF = 2, kB = 8, kC = 20, kD = 15, and kE = 35. The left and right surfaces of the wall are maintained at uniform temperature of 300oC and 100oC, respectively. Assuming heat transfer through the wall to be one-dimensional, determine (a) the rate of heat transfer through the wall; (b) the temperature at the point where the section B, D, and E meet; and (c) the temperature drop across the section F. Disregard any contact resistances at the interfaces.