Chapter 6 Heat Transfer Objectives • Describe the basic mechanisms of heat transfer, which are conduction, convection, and radiation, and Fourier's law of heat conduction, Newton's law of cooling, and the Stefan–Boltzmann law of radiation. • Identify the mechanisms of heat transfer that occur simultaneously in practice. • Explain the concept of thermal resistance, and develop thermal resistance networks for practical heat conduction problems. • Solve steady conduction problems that involve multilayer rectangular geometries. • Solve various heat transfer problems encountered in practice. Introduction › Heat as the form of energy that can be transferred from one system to another as a result of temperature difference. › A thermodynamic analysis is concerned with the amount of heat transfer as a system undergoes a process from one equilibrium state to another. › The science that deals with the determination of the rates of such energy transfers is the heat transfer. › The transfer of energy as heat is always from the higher-temperature medium to the lowertemperature one, and heat transfer stops when the two mediums reach the same temperature. › Heat can be transferred in three basic modes: – Conduction. – Convection. – Radiation. All modes of heat transfer require the existence of a temperature difference. 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 is due to the collisions and diffusion of the molecules during their random motion. › In solids, it is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons. › 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. Area Temperature difference Rate of heat conduction ∝ Thickness παΆ cond = ππ΄ π1 − π2 Δπ = −ππ΄ Δπ₯ Δπ₯ W Heat conduction through a large plane wall of thickness Δx and area A. Conduction 2 Thermal conductivity, k [W/mοK] A measure of the ability of a material to conduct heat. Fourier’s law of heat conduction ππ παΆ cond = −ππ΄ ππ₯ ππ/ππ₯ : Temperature gradient 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. Conduction 4 › Thermal conductivity (k): The rate of heat transfer through a unit thickness of the material per unit area (1 m2) per unit temperature difference. › 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. π1 − π2 αΆ πcond = ππ΄ Δπ₯ πΏ π= παΆ π΄ π1 − π2 A simple experimental setup to determine the thermal conductivity of a material. Conduction Material k, W/mοK* Diamond 2300 Silver 429 Copper 401 Gold 317 Aluminum 237 Iron 80.2 Mercury (l) 8.54 Glass 0.78 Brick 0.72 Water (l) 0.607 Human skin 0.37 Wood (oak) 0.17 Helium (g) 0.152 Soft rubber 0.13 Glass fiber 0.043 Air (g) 0.026 Urethane, rigid foam 0.026 Thermal conductivities @ room temperature Conduction • In solids, heat conduction is due to two effects: (1) the lattice vibrational waves induced by the vibrational motions of the molecules positioned at relatively fixed positions in a periodic manner called a lattice and (2) the energy transported via the free flow of electrons in the solid. • Gas and liquid molecules undergo molecular collisions and molecular diffusions. • The thermal conductivities of gases such as air vary by a 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. Crystalline solids such as diamond and semiconductors such as silicon are good heat conductors but poor electrical conductors. As a result, such materials find widespread use in the electronics industry. Application: Thermal Paste Conduction The thermal conductivity of an alloy is usually much lower than the thermal conductivity of either metal of which it is composed Pure Metal or Alloy k, W/m.K, at 300 k Copper 401 Nickel 91 Constantan (55% cu, 45% Ni) 23 Copper 401 Aluminum 237 Commercial bronze (90% Cu, 10% A1) 52 Conduction › c p › ρc Specific heat, [J/kg·°C] p β Heat capacity per unit mass Heat capacity, [J/m3 ⋅β C] β Heat capacity per unit volume › α Thermal diffusivity, [m2 /s] β Represents how fast heat diffuses through a material Heat conduction π › πΌ= = m2 /s Heat storage πππ › 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. Conduction (Thermal Diffusivities @ Room Temperature) Discussion The roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m thick, and it is made of a flat layer of concrete whose thermal conductivity is k = 0.8 W/m⋅K. The temperatures of the inner and the outer surfaces of the roof one night are measured to be 15° C and 4° C, respectively, for a period of 10 h. Determine the rate of heat loss through the roof that night. Ans: 1.69 kW 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 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. Heat transfer from a hot surface to air by convection. Convection 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. Newton’s law of cooling παΆ conv = βπ΄π ππ − π∞ W h convection heat transfer coefficient, [W/m2 ⋅β C] not a property of the fluid A s the surface area through which convection heat transfer takes place. T s the surface temperature. T ∞ the temperature of the fluid sufficiently far from the surface (e.g.: Tats) Convection h is determined parameter: Convection heat transfer coefficient • The surface geometry Type of Convection h, W/m2·K* • The nature of fluid motion Free convection of gases 2-25 Free convection of liquids 10-1000 Forced convection of gases 25-250 Forced convection of liquids 50-20,000 Boiling and condensation 2500-100,000 • The properties of the fluid • The bulk fluid velocity *Multiply by 0.176 to convert to Btu/h⋅ft2 ⋅°F. Discussion A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at 15° C. Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be 152° C in steady operation. The circuit is connected for 20s, and 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. Ans: 34.9 W/ m 2 ·K Radiation • Radiation: The energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules. • 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 suffers no attenuation in a vacuum. This is how the energy of the sun reaches the earth. • In heat transfer studies, the focus is in thermal radiation, which is the form of radiation emitted by bodies because of their temperature. • All bodies at a temperature above absolute zero emit thermal radiation. • 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. Radiation Stefan–Boltzmann law π = παΆ emit,max = ππ΄π ππ 4 W Stefan–Boltzmann constant π = 5.670 × 10−8 W/m2 ⋅ πΎ 4 Blackbody radiation represents the maximum amount of radiation that can be emitted from a surface at a specified temperature. Blackbody: The idealized surface that emits radiation at the maximum rate. Radiation Radiation emitted by real surfaces παΆ emit = πππ΄π ππ 4 W Emissivity ε : A measure of how closely a surface (grey body) approximates a blackbody for which ε = 1 of the surface. 0 ≤ ε ≤ 1. Radiation › Absorptivity α: The fraction of the radiation energy incident on a surface that is absorbed by the surface. 0 ≤ α ≤ 1. › A blackbody absorbs the entire radiation incident on it (α = 1). › Kirchhoff’s law: The emissivity and the absorptivity of a surface at a given temperature and wavelength are equal. παΆ absorbed = πΌ παΆ incident (W) The absorption of radiation incident on an opaque surface of absorptivity α. Radiation Net radiation heat transfer: The difference between the rates of radiation emitted by the surface and the radiation absorbed. The determination of the net rate of heat transfer by radiation between two surfaces is a complicated matter since it depends on: (1)The properties of the surfaces. (2)Their orientation relative to each other. (3)The interaction of the medium between the surfaces with radiation. 4 παΆ rad = πππ΄π ππ 4 − ππ π’ππ W Radiation › When radiation and convection occur simultaneously between a surface and a gas 4 παΆ total = παΆ conv + παΆ rad = βconv π΄π ππ − πsurr + πππ΄π ππ 4 − πsurr παΆ total = βcombined π΄π ππ − π∞ W Radiation is usually significant relative to conduction or natural convection, but negligible relative to forced convection. βcombined = βconv + βrad 2 = βconv + ππ ππ + πsurr ππ 2 + πsurr Combined heat transfer coefficient hcombined Discussion 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 at an average temperature of 10°C in winter and 25°C in summer. Determine the rate of radiation heat transfer between this lady 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. Ans: 40.9W, 152 W Discussion It is a common experience to feel chilly in winter and warm in summer in our homes even when the thermostat setting is kept the same. This is due to the so-called radiation effect resulting from radiation heat exchange between our bodies and the surrounding surfaces of the walls and the ceiling. Simultaneous Heat Transfer Mechanisms › Heat transfer is only by conduction in opaque solids, but by conduction and radiation in semitransparent solids. › A solid may involve conduction and radiation but not convection. A solid may involve convection and/or radiation on its surfaces exposed to a fluid or other surfaces. › Heat transfer is by conduction and possibly by radiation in a still fluid (no bulk fluid motion) and by convection and radiation in a flowing fluid. › In the absence of radiation, heat transfer through a fluid is either by conduction or convection, depending on the presence of any bulk fluid motion. › Convection = Conduction + Fluid motion › Heat transfer through a vacuum is by radiation. › Most gases between two solid surfaces do not interfere with radiation. › Liquids are usually strong absorbers of radiation. Simultaneous Heat Transfer Mechanisms (1) Heat transfer is only by conduction in opaque solids, but by conduction and radiation in semitransparent solids. (2) A solid may involve conduction and radiation but not convection. A solid may involve convection and/or radiation on its surfaces exposed to a fluid or other surfaces. (1) In the absence of radiation, heat transfer through a fluid is either by conduction or convection, depending on the presence of any bulk fluid motion or still fluid. (2) Convection = Conduction + Fluid motion (1) Heat transfer through a vacuum is by radiation. Discussion Consider a person standing in a breezy room at 20° C. Determine the total rate of heat transfer from this person if the exposed surface area and the average outer surface temperature of the person are 1.6m2 and 29°C, respectively, and the convection heat transfer coefficient is 6 W/ m 2 ⋅K Ans: 168 W Steady Heat Conduction in Plane Walls › Heat transfer through the wall of a house can be modeled as steady and one-dimensional (One Direction) ππ Fourier’s law of heat conduction παΆ cond, wall = −ππ΄ ππ₯ (W) › The temperature of the wall in this case depends on one direction only (say the x-direction) and can be expressed as T(x). ENERGY BALANCE (Rate of heat transfer into the wall) − (Rate of heat transfer out of the wall) = (Rate of change of the energy of the wall) Heat transfer through a wall is onedimensional when the temperature of the wall varies in one direction only. παΆ in − παΆ out = ππΈwall ππ‘ Steady Heat Conduction in Plane Walls Steady operation ππΈwall /ππ‘ = 0 παΆ cond, wall = −ππ΄ ππ ππ₯ β πΏ ΰΆ± παΆ cond, wall ππ₯ = − ΰΆ± π₯=0 παΆ cond, wall = ππ΄ παΆ cond, wall ππ₯ = −ππ΄ ππ π2 ππ΄ ππ π=π1 π1 − π2 πΏ (W) › 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. Under steady conditions, the temperature distribution in a plane wall is a straight line.. › 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. Thermal Resistance (Rwall): Conduction Thermal Resistance Concept παΆ cond, wall = ππ΄ Electrical Resistance Concept π1 − π2 πΏ π1 − π2 αΆ πcond, wall = π wall πΏ π wall = (°C/W) ππ΄ V1 − V2 πΌ= π π (W) Conduction resistance of the wall: Thermal resistance of the wall against heat conduction. Thermal resistance of a medium depends on the geometry (area and thickness) and the thermal properties (k) of the medium. π π = πΏ/ππ π΄ Analogy between thermal and electrical resistance concepts. › rate of heat transfer (π)αΆ → electric current (πΌ) › thermal resistance (π π€πππ ) → electrical resistance (π ) › temperature difference (π) → voltage difference (π) Conduction in Composite Wall What happens when you have multiple layers together? Steady operation ππΈwall /ππ‘ = 0 If we know the temperatures at the interface (T2 and T3) then we could use the equation state. But, most of the time, we only know T1 and T4, which are the surface temperatures. 32 Conduction in Composite Wall Current: rate of heat transfer (π)αΆ Voltage: temperature difference (π) Resistance: thermal resistance (π π€πππ ) βπ₯ π wall(π‘β) = π π΄ βπ₯π΄ βπ₯π΅ βπ₯πΆ π π(π΄π΅πΆ) = + + ππ΄ π΄ ππ΅ π΄ ππΆ π΄ παΆ π(π΄π΅πΆ) π1 − π4 = π π(π΄π΅πΆ) 33 Conduction in Composite Wall Current: rate of heat transfer (π)αΆ Voltage: temperature difference (π) Resistance: thermal resistance (π π€πππ ) παΆ π(π΄π΅πΆ) = βππ΄ βππ΅ βππΆ = = π π΄ π π΅ π πΆ 34 Discussion Find the heat transfer per m2 through a wall comprising of 250mm brick (k = 0.69 W/mK) and an outer layer of concrete 50mm wide (k = 0.93 W/mK). The temperature of the inside surface of the brick, T1 is 30°C and the outside temperature of the concrete, T3 is 5°C. T3 Ans: 60 W/m2 Thermal Resistance: Convection Newton’s law of cooling παΆ conv = βπ΄π ππ − π∞ ππ − π∞ αΆ πconv = π conv (W) W 1 π conv = βπ΄π (°C/W) › Convection resistance of the surface (π conv ): Thermal resistance of the surface against heat convection › When the convection heat transfer coefficient is very large (h → ∞), the convection resistance becomes zero and ππ ≈ π. › 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. Thermal Resistance: Radiation 4 παΆ rad = πππ΄π ππ 4 − πsurr ππ − πsurr αΆ πrad = βrad π΄π ππ − πsurr = π rad παΆ rad βrad = π΄π (ππ − πsurr ) 2 βrad = ππ(ππ 2 + πsurr )(ππ + πsurr ) (W/m2 ⋅ K) Radiation resistance of the surface: Thermal resistance of the surface against radiation. 1 π rad = (K/W) βrad π΄π If πsurr ≈ π∞ H combined = h conv + h rad Combined heat transfer coefficient Thermal Resistance Network (Rate of heat convection into the wall) = (Rate of heat conduction through the wall) = (Rate of heat convection from the wall) παΆ = π∞1 − π1 π1 − π2 π2 − π∞2 π∞1 − π1 π1 − π2 π2 − π∞2 = = = = = 1/β1 π΄ πΏ/ππ΄ 1/β2 π΄ π conv, 1 π wall π conv, 2 Thermal Resistance Network π∞1 − π∞2 παΆ = π total π total = π conv,1 + π wall + π conv,2 1 πΏ 1 π total = + + [°C/W] β1 π΄ ππ΄ β2 π΄ Thermal Resistance Network Temperature drop, (Δπ) αΆ Δπ = ππ °C παΆ = ππ΄ Δπ [W] ππ΄ = 1 π total [°C/K] U overall heat transfer coefficient Once παΆ is evaluated, the surface temperature T1 can be determined from π −π π −π παΆ = ∞1 1 π conv, 1 = ∞1 1/β1 π΄ 1 Temperature drop proportional thermal resistance Multilayer Plane Walls παΆ = π∞1 − π∞2 π total π total = π conv, 1 + π wall, 1 +π wall, 2 + π conv, 2 1 πΏ1 πΏ2 1 π total = + + + β1 π΄ π1 π΄ π2 π΄ β2 π΄ The thermal resistance network for heat transfer through a two-layer plane wall subjected to convection on both sides. π∞1 − π2 π∞1 − π2 παΆ = = 1 πΏ1 π conv,1 + π Wall,1 + β1 π΄ π1 π΄ Discussion Consider a 0.8-m-high and 1.5-m-wide glass window with a thickness of 8 mm and a thermal conductivity of k = 0.78 W/m⋅K. 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 20°C while the temperature of the outdoors is −10°C. Take the heat transfer coefficients on the inner and outer surfaces of the window to be h1= 10 W/m2⋅K and h2 = 40 W/m2 ⋅K, which includes the effects of radiation. Ans: 266 W Discussion Consider a 0.8-m-high and 1.5-m-wide double-pane window consisting of two 4-mmthick layers of glass (k = 0.78 W/m⋅K) separated by a 10-mm-wide stagnant airspace (k = 0.026 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 20°C while the temperature of the outdoors is − 10°C. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h1=10 W/m2⋅K and h2=40 W/ m2⋅K , which includes the effects of radiation. Ans: 69.2 W Generalized Thermal Resistance Networks παΆ = παΆ 1 + παΆ 2 = παΆ = π1 − π2 π1 − π2 + = π1 − π2 π 1 π 2 1 1 + π 1 π 2 π1 − π2 π total 1 1 1 = + π total π 1 π 2 π 1 π 2 π total = π 1 + π 2 Thermal resistance network for two parallel layers. Generalized Thermal Resistance Networks παΆ = π1 − π∞ π total πΏ1 π 1 = π1 π΄1 π 3 = πΏ3 π3 π΄3 πΏ2 π 2 = π2 π΄2 1 π conv = βπ΄3 π 1 π 2 π total = π 12 + π 3 + π conv = + π 3 + π conv π 1 + π 2 › Two assumptions in solving complex multidimensional heat transfer problems by treating them as onedimensional using the thermal resistance network are: 1) Any plane wall normal to the x-axis is isothermal (i.e., to assume the temperature to vary in the x-direction only). 2) Any plane parallel to the x-axis is adiabatic (i.e., to assume heat transfer to occur in the x-direction only). Discussion A 3-m-high and 5-m-wide wall consists of long 16-cm × 22-cm cross section horizontal bricks (k = 0.72 W/m⋅K) separated by 3-cm-thick plaster layers (k = 0.22 W/m⋅K). There are also 2-cm-thick plaster layers on each side of the brick and a 3-cm-thick rigid foam (k = 0.026 W/m⋅K) on the inner side of the wall. The indoor and the outdoor temperatures are 20° C and − 10° C, respectively, and the convection heat transfer coefficients on the inner and the outer sides are h1=10 W/m2⋅K and h2=25W/m2⋅K, respectively. Assuming onedimensional heat transfer and disregarding radiation, determine the rate of heat transfer through the wall. Ans: 263 W Discussion
