Chapter 12 Principles of Steady-State Heat Transfer Nehal I. Abu-Lail Thank you Jordan University of Science and Technology • Dr. Sameer Al-Asheh • Dr. Mohammad Al-Saleh • Dr. Fahmi Abu Al-Rub • Google images Heat Transfer The basic requirement for heat transfer is the presence of a temperature difference. The rate of heat transfer in a certain direction depends on the magnitude of the temperature gradient in that direction. The larger the temperature gradient, the higher the rate of heat transfer. Application Areas of Heat Transfer More Applications of Heat Transfer Drying of fruits Fuel burning Drying of wood Evaporation of water & weather control Cooking Distillation Heat Transfer Mechanisms Heat can be transferred in three basic modes: Conduction, Convection, Radiation. All modes of heat transfer require the existence of a temperature difference. All modes are from the high-temperature medium to a lowertemperature one. 12.3A Unsteady State Heat Balance Making an unsteady-state heat balance for the x direction only on the element of volume or control volume in Fig. 12.3-1 by using Eqs. (12.3-1) and (12.32), with the cross-sectional area being A m2 and the consumption term ignored 𝐴𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 𝐼𝑛 − 𝑂𝑢𝑡 + 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 − 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝜕𝐸 = 𝑞𝑥 𝑥 − 𝑞𝑥 𝑥+∆𝑥 + 𝑞 ∆𝑥. 𝐴 𝜕𝑡 12.3-4 12.3-1 12.3A Unsteady State Heat Balance When only considering energy changes caused by a difference in temperature (ΔE = ΔH = mcpΔT), thus ignoring potential and kinetic energy changes, 𝜕𝐸 𝜕𝑇 = 𝜌𝐶𝑃 ∆𝑥. 𝐴 𝜕𝑡 𝜕𝑡 12.3-5 Substituting back into Eq. (12.3-4), 𝜌𝐶𝑃 𝜕𝑇 ∆𝑥. 𝐴 𝜕𝑡 = 𝑞𝑥 𝑥 − 𝑞𝑥 𝑥+∆𝑥 + 𝑞 ∆𝑥. 𝐴 12.3-6 where 𝑞 is rate of heat generated per unit volume. Assuming no heat generation and also assuming steady-state heat transfer, where the rate of accumulation is zero, Eq. (12.3-6) becomes 𝑞𝑥 𝑥 = 𝑞𝑥 𝑥+∆𝑥 12.3-7 This means the rate of heat input by conduction = the rate of heat output by conduction; or qx is a constant with time for steady-state heat transfer. 12.3A Unsteady State Heat Balance When only considering energy changes caused by a difference in temperature (ΔE = ΔH = mcpΔT), thus ignoring potential and kinetic energy changes, 𝜕𝐸 𝜕𝑇 = 𝜌𝐶𝑃 ∆𝑥. 𝐴 𝜕𝑡 𝜕𝑡 12.3-5 Substituting back into Eq. (12.3-4), 𝜌𝐶𝑃 𝜕𝑇 ∆𝑥. 𝐴 𝜕𝑡 = 𝑞𝑥 𝑥 − 𝑞𝑥 𝑥+∆𝑥 + 𝑞 ∆𝑥. 𝐴 12.3-6 where q is rate of heat generated per unit volume. Assuming no heat generation and also assuming steady-state heat transfer, where the rate of accumulation is zero, Eq. (12.3-6) becomes 𝑞𝑥 𝑥 = 𝑞𝑥 𝑥+∆𝑥 12.3-7 This means the rate of heat input by conduction = the rate of heat output by conduction; or qx is a constant with time for steady-state heat transfer. In this chapter, we are concerned with a control volume where the rate of accumulation of heat is zero and we have steady-state heat transfer. The rate of heat transfer is then constant with time, and the temperatures at various points in the system do not change with time. To solve problems in steady-state heat transfer, various mechanistic expressions in the form of differential equations for the different modes of heat transfer such as Fourier’s law are integrated. Expressions for the temperature profile and heat flux are then obtained in this chapter. In Chapter 14, the conservation-of-energy equations (4.2-2) and (12.3-6) will be used again when the rate of accumulation is not zero and unsteady-state heat transfer occurs. The mechanistic expression for Fourier’s law in the form of a partial differential equation will be used where the temperature at various points and the rate of heat transfer change with time. In Section 14.6, a general differential equation of energy change will be derived and integrated for various specific cases to determine the temperature profile and heat flux. 12.3 Conduction as a Basic Mechanism of Heat Transfer Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles. Conduction can take place in solids, liquids, or gases. Consider a gas occupy the two surfaces shown in the drawing with no bulk motion at any point: T1>T2 Energy T T Kinetic energy Molecular energy collisions of neighbouring molecules Transfer of energy from more energetic to less energetic https://www.youtube.com /watch?v=vuBPgTMKUc0 Energy is related to the random translation of motion of https://www.youtube.c om/watch?v=9joLYfaye internal energy + vibrational motion of the molecules Thus, energy transfers in the direction of decreasing temperature e8 • Examples of conduction: • Walls of homes • Heat exchangers • Freezing of ground in winter • Heat treatment of steel forgings • Keeping constant skin temperature Frostbite is a disease of skin when exposed to freezing temperatures 12.3C Fourier’s Law of Heat Conduction qx dT k A dx W W K 2 m m.K m Assumptions: 1. Homogeneous media 2. 1Dimensional flow 12.3-3 btu btu 0 F 2 hr. ft hr. ft.0 F ft qx: Heat transfer rate by conduction, A: Area normal to flow, K: thermal conductivity dT/dx= Temperature gradient • For steady state heat transfer, constant cross sectional area of a large plate wall, qx is constant and we can integrate Fourier’s law for transfer between two points (x1, T1) and (x2 and T2), separation of variables: x2 x1 T 2 qx dx k dT ,12.3-13 A T1 qx T T k 1 2 A x2 x1 12.3-14 Note that T is a linear profile in X (similar to velocity profile we have seen for a moving plate). The rate of transport of heat can be thought of as: Rate of a transfer process Driving force Re sis tan ce 12.3-8 This equation states what we know intuitively: that in order to transfer a property such as heat or mass, we need a driving force to overcome a resistance. qx dT k A dx • • • 12.3-3 The units in Eq. (12.3-3) may also be expressed in the cgs system, with qx in cal/s, A in cm2, k in cal/s · °C · cm, T in °C, and x in cm. In the English system, qx is in btu/h, A in ft2, T in °F, x in ft, k in btu/h · °F · ft, and qx/A in btu/h · ft2. From Appendix A.1, the conversion factors for thermal conductivity are: 𝑏𝑡𝑢 𝑐𝑎𝑙 3 12.3-9 1 = 4.1365 × 10 2 ℎ𝑟. 𝑓𝑡 . ℉ 𝑠. 𝑐𝑚. ℃ 𝑏𝑡𝑢 𝑊 1 = 1.73073 2 ℎ𝑟. 𝑓𝑡 . ℉ 𝑚. 𝐾 12.3-10 Units of Heat Flux For heat flux and power, 𝑏𝑡𝑢 𝑊 1 = 3.1546 2 2 ℎ𝑟. 𝑓𝑡 𝑚 𝑏𝑡𝑢 1 = 0.29307 𝑊 ℎ𝑟 12.3-11 12.3-12 Question? What will happen to the rate of heat transfer, q, through the wall if: T across the wall doubles? A normal to the direction of heat transfer is doubled? Wall thickness is doubled? T q x kA x Example: The wall of an industrial furnace is constructed from 0.15 m thick fireclay brick having a thermal conductivity of 1.7 W/m.K. Measurements made during steady state operation reveal temperatures of 1400 and 1150 K at the inner and outer surfaces, respectively. What is the rate of heat loss through a wall which is 0.5 m by 3 m on a side? Solution Schematic: Always try to provide it Knowns: Steady state; L = 0.15 m; A = 0.5 × 3 m2; k = 1.7 W/m.K; T1 = 1400 L; T2 = 1150 K. Assumptions: 1. Steady state, 2. One dimensional through the wall, 3. Constant properties. Analysis: heat transfer by conduction Use Fouriers’law q/ A k (1400 1150) K T1 T2 1.7 W/m.K L 0.15 m = 2833 W/m2 (Flux) Heat transfer rate: q (q / A) A = (0.5 m × 3 m) 2833 W/m2 = 4250 W Example 12.3-1. Heat Loss through an Insulating Wall Calculate the heat loss per m2 of surface area for an insulating wall composed of 25.4-mm-thick fiber insulating board, where the inside temperature is 352.7 K and the outside temperature is 297.1 K. Solution of Example 12.3-1. Heat Loss Through an Insulating Wall Calculate the heat loss per m2 of surface area for an insulating wall composed of 25.4-mm-thick fiber insulating board, where the inside temperature is 352.7 K and the outside temperature is 297.1 K. From Appendix A.3, the thermal conductivity of fiber insulating board is 0.048 W/m·K. The thickness x2 – x1 = 0.0254 m. Substituting into Eq. (12.3-14): 𝑞 𝑘 0.048 𝑊 = 𝑇1 − 𝑇2 = 352.7 − 297.1 = 105.1 2 𝐴 𝑥2 − 𝑥1 0.0254 𝑚 12.3D Thermal Conductivity-k The thermal conductivity of a material is a measure of the ability of the material to conduct heat. High value for thermal conductivity Good heat conductor Low value Poor heat conductor or insulator 12.3D Thermal Conductivity - k It is a property of the material. It is a measure of the ability of the material to conduct heat. Determined experimentally (empirical constant). k depends on microscopic structure of the substance; • Generally: ksolids > kliq > kgas Why?? •kAl=202 > kBeef=1.35> kWater=0.569 > kair=0.0242 W/m.K • • • In Table 12.3-1, thermal conductivities of a few materials are given for the purpose of comparison. More-detailed data are given in Appendix A.3 for inorganic and organic materials and Appendix A.4 for food and biological materials. As seen in Table 12.3-1, gases have quite low values of thermal conductivity, liquids have intermediate values, and solid metals have very high values. Table 12.3-1: 1. Gases • Follows the kinetic theory of gases • Molecules are always moving • Energy transfers as molecules colloid from hotter to cooler molecules • Smaller molecules move faster and as such have higher conductivity k Tabsolute • k is independent of pressure up to few atm’s. • As P goes to vacuum pressures, k goes to 0. • Since gas theory is pretty accurate, we can use it to predict k for gases. Prediction of k for Gases • • Theories to predict thermal conductivities of gases are reasonably accurate. The predictive model developed by Chapman–Enskog for gases is given below: 𝑇 0.0829 𝑀𝑊 12.3-15 𝑘= 𝜎 2 Ω𝑘 Where k is thermal conductivity in W/m.K, T is temperature (K), MW is the molecular weight, s is the molecular diameter in (Å)(listed in A.6), and Ω𝑘 is the Lenard-Jones Collision integral. Ω𝑘 =1.16145 𝑇 ∗ −0.14874 +0.52487exp(−0.77329𝑇 ∗ )+2.16178exp(-2.43787𝑇 ∗ ) 𝑘 𝑇 12.3-16 Where 𝑇 ∗ = 𝐵 12.3-17 𝜀 e/kB=59.7, kB is Boltzmann constant and e is a scaling parameter of length specific for each molecule Example 12.3-2. Chapman–Enskog for the Prediction of Thermal Conductivity of H2 Use Eqs. 12.3-16, 12.3-17, and 12.3-18 to calculate the thermal conductivity of H2 at 273 K and 1 atm, and then compare that value to the one listed in Table 12.3-1. Solution of Example 12.3-2. Chapman–Enskog for the Prediction of Thermal Conductivity of H2 • H2 (MW = 2 g/gmol), T=273 K, P= 1 atm, s=2.827 Å from A.6. You are also given e/kB=59.7 K 0.0829 𝑘= • We need Ω𝑘 Ω𝑘 =1.16145 𝑇 ∗ 𝑇∗ = 𝑇 𝑀𝑊 𝜎 2 Ω𝑘 −0.14874 +0.52487exp(−0.77329𝑇 ∗ )+2.16178exp(-2.43787𝑇 ∗ ) 𝑘𝐵 𝑇 =273/59.7=4.6 𝜀 Ω𝑘 =1.16145 4.6 −0.14874 +0.52487exp(−0.77329 ∗ 4.6)+2.16178exp(2.43787*4.6) = 0.941 Solution of Example 12.3-2. Chapman–Enskog for the Prediction of Thermal Conductivity of H2 𝑘= 0.0829 273 2 2.827 2 (0.941) 𝑊 =0.129𝑚.𝐾 • The value listed in the Table 12.3-1 at 273 K is 0.167 W/m.K • The error between listed and calculated values is: Error = (0.167-0.129)×100% /0.167 =22.8% • The correlation should be used with caution 2. Liquids • Energy transfers as molecules colloid from hotter to cooler molecules. • Efficiency of energy transfer is better than in gases because molecules are highly backed together. • k’s are determined via empirical correlations. • k varies largely linearly with T k = a+bT with a and b being empirical constants 12.3-18 • For liquids, k is independent of P. • Water has very high k compared to organic solvents. • Dilute mixtures made largely with water have k values close to that of pure water (skim milk & apple sauce). 3. Solids • k for solids vary widely with homogeneous metals having very high k values compared to industrial solids such as rock, cardboard, wool or wood. • Mechanisms of heat transfer in metals can be divided into two categories: • Metals: Free electrons carry heat while they move within the metal lattice. • Heat is conducted by transmission of energy vibrations between adjacent atoms. • k of insulating materials like rock wool approaches k for air because they have lots of trapped air. • Super insulators (layers of composite materials filled with liquid H2) have lower k compared to air. • k for ice is much higher than that of water. k for frozen food is much higher than k for unfrozen food. Thermal diffusivity Heat conducted k Heat stored cp ( m2 s ) The thermal diffusivity represents how fast heat diffuses through a material compared to heat stored in it Appears in the transient heat conduction analysis (Chapter 14). A material that has a high thermal conductivity or a low heat capacity will have a large thermal diffusivity. The larger the thermal diffusivity, the faster the propagation of heat into the medium. 12.4 Convection 12.4A Convection as a Basic Heat Transfer Mechanism The transfer of heat by convection implies the transfer of heat by bulk transport and the mixing of macroscopic elements of warmer portions with cooler portions of a gas or liquid. It also often refers to the energy exchange between a solid surface and a fluid. Examples of heat transfer by convection are heat loss from a car radiator where air is circulated by a fan, stirring foods being cooked in a vessel, blowing across the surface of a hot cup of coffee to cool it, and so on. 12.4A Convection as a Basic Heat Transfer Mechanism Convection is commonly classified into two sub-modes: Forced convection: Where a fluid is forced to flow past a solid surface by a pump, fan, or other mechanical means, and Natural (or free) convection: where warmer or cooler fluid next to the solid surface causes a circulation because of a density difference resulting from the temperature differences in the fluid. A hot egg will cool faster when we blow air on it. Natural and Forced Convection: The forces used to create convection currents in fluids are of two types: Forced convection (a): If the currents are set in motion by action of a mechanical device, such as a pump or agitator blower. The flow is independent of the density. Natural convection (b): If the hot plate exposed to ambient air without external sources of motion, i.e. the movement of air is due to density gradient near the plate (i.e. difference in density causes the T in the fluid). Generally: hfree < hforced Convection Examples: Loss of heat from a car radiator when air is circulated by a fan Cooling of food in vessels being stirred Cooling of a hot cup of coffee by blowing over it Sweating Hot, less dense Water rises Water cools, become more dense sinks 12.4B Convective Heat Transfer Coefficient The rate of convection heat transfer is expressed by Newton’s law of cooling as: q hA(Tw T f ) 12.4-1 Tw: Wall temperature, Tf: Bulk Fluid temperature, A: Area normal to flow h is the convection heat transfer coefficient in W/m2. °C. Often called also as film coefficient as when fluid passes over a solid, there will be a stationary thin film of fluid adjacent to the surface of solid and this layer represents the most resistance to heat flow. h depends on variables such as: The surface geometry, h can be predicted empirically The nature of fluid motion, Water has high h value. The properties of the fluid, To convert h from English to SI units: 1 btu/hr.ft2.oF= The bulk fluid velocity, 5.6783 W/m2. K Temperature gradient. Table 12.4-1 Forced convection Natural convection Forced convection Example: Water at 300 K flows over both sides of a plate of 1m × 2m in area, maintained at 400 K. If the convective heat transfer coefficient is 200 W/m2.K, calculate the heat transfer. q 2hA(Ts T ) Why 2? q= (2) (1) (2) m2 (200) (W/m2.K)(400 – 300) K q = 80,000 W Water 300 K Solid plate 400 k 2m 1m Example 12.4-1. Heat Loss Through Convection The forced convective heat-transfer coefficient of flowing hot water (393 K) over a cold steel surface (283 K) is 1000. Determine the heat transfer rate per unit surface area (W/m2) from the water to the steel surface. Solution of Example 12.4-1. Heat Loss Through Convection The forced convective heat-transfer coefficient of flowing hot water (393 K) over a cold steel surface (283 K) is 1000. Determine the heat transfer rate per unit surface area (W/m2) from the water to the steel surface. Given: h=1000 W/m2.K Tf= 393 K Tw=283 K 𝑞 = ℎ(𝑇𝑓 − 𝑇𝑤 )=1000(393-283)=110,000 W/m2 𝐴 12.5 Radiation 12.5A Radiation as a Basic Heat Transfer Mechanism • • • • Radiation differs from heat transfer by conduction and convection in that no physical medium is needed for its propagation. Radiation is the transfer of energy through space by means of electromagnetic waves in much the same way as electromagnetic light waves transfer light. The same laws that govern the transfer of light govern the radiant transfer of heat. Solids and liquids tend to absorb the radiation being transferred through them, so that radiation is important primarily in transfer through space or gases. Examples of Radiation Radiation from sun to earth. Cooking food when passed below red-hot electric heater. Heating of fluids inside coils of a combustion furnace. Radiation therapy in cancer treatment. 1. Nature of Radiant Heat Transfer • In conduction, heat is transferred from one part of a body to another, and the intervening material is heated. • In convection, heat is transferred by the actual mixing of materials and by conduction. • In radiant heat transfer, the medium through which the heat is transferred usually is not heated. • Again, radiation heat transfer is the transfer of heat by electromagnetic radiation. 1. Nature of Radiant Heat Transfer • Thermal radiation is a form of electromagnetic radiation similar to X rays, light waves, gamma rays, and so on, differing only in wavelength. • It obeys the same laws as light: it travels in straight lines, can be transmitted through space and vacuum, and so on. • It is an important mode of heat transfer and is especially important where large temperature differences occur, as, for example, in a furnace with boiler tubes, in radiant dryers, or in an oven baking food. • Radiation often occurs in combination with conduction and convection. Ch. 17 will cover radiation in details. 1. Nature of Radiant Heat Transfer In an elementary sense, the mechanism of radiant heat transfer is composed of three distinct steps or phases: • The thermal energy of a hot source, such as the wall of a furnace at T1, is converted into energy in the form of electromagnetic-radiation waves. • These waves travel through the intervening space in straight lines and strike a cold object at T2, such as a furnace tube containing water to be heated. • The electromagnetic waves that strike the body are absorbed by it and converted back to thermal energy or heat. 2. Absorptivity and Black Bodies • When thermal radiation (such as light waves) falls upon a body, a portion is absorbed by the body in the form of heat, a portion is reflected back into space, and a portion may actually be transmitted through the body. • For most cases in process engineering, bodies are opaque to transmission, so this will be neglected. Hence, for opaque bodies, 𝛼 + 𝜌 = 1 12.5-1 Where is absorptivity or fraction absorbed and ρ is reflectivity or fraction reflected. 2. Absorptivity and Black Bodies • • • • • • • • A black body is defined as one that absorbs all radiant energy and reflects none. Hence, ρ = 0 and α = 1.0 for a black body. Actually, in practice there are no perfect black bodies, but a close approximation is a small hole in a hollow body, as shown in Fig. 12.5-1. The inside surface of the hollow body is blackened by charcoal. The radiation enters the hole and impinges on the rear wall; a portion is absorbed there and a portion is reflected in all directions. The reflected rays impinge again, a portion is absorbed, and the process continues. Hence, essentially all of the energy entering is absorbed and the area of the hole acts as a perfect black body. The surface of the inside walls is “rough” and rays are scattered in all directions, unlike a mirror, where they are reflected at a definite angle. 2. Absorptivity and Black Bodies • As stated previously, a black body absorbs all radiant energy falling on it and reflects none. • Such a black body also emits radiation, depending on its temperature, and does not reflect any. • The ratio of the emissive power of a surface to that of a black body is called emissivity ε and is 1.0 for a black body. • Kirchhoff ’s law states that at the same temperature T1, α1 and ε1 of a given surface are the same, or 𝛼1 = 𝜀1 12.5-2 Eq. (12.5-2) holds for any black or nonblack solid surface. 3. Radiation From a Body and Emissivity. • The basic equation for heat transfer by radiation from a perfect black body with an emissivity ε = 1.0 is 𝑞 = 𝐴𝜎𝑇 4 12.5-3 where q is heat flow in W, A is m2 surface area of the body, σ is a constant 5.676 × 10–8 W/m2 · K4 (0.1714 × 10–8 btu/h·ft2·°R4), and T is temperature of the black body in K (°R). • • For a body that is not a black body and has an emissivity ε < 1.0 the emissive power is reduced by ε, or 𝑞 = 𝐴𝜖𝜎𝑇 4 12.5-4 Substances that have emissivities of less than 1.0 are called gray bodies when the emissivity is independent of the wavelength. All real materials have an emissivity ε < 1. Table 12.5-1. Total Emissivity, ε, of Various Surfaces Surface T(K) Polished aluminum T(°F) Emissivity, ε Polished iron Oxidized iron Polished copper Asbestos board Oil paints, all colors 500 850 450 373 353 296 373 440 1070 350 212 176 74 212 0.039 0.057 0.052 0.74 0.018 0.96 0.92–0.96 Water 273 32 0.95 3. Radiation From a Body and Emissivity. • Since the emissivity ε and absorptivity α of a body are equal at the same temperature, the emissivity, like absorptivity, is low for polished metal surfaces and high for oxidized metal surfaces. • Typical values are given in Table 12.5-1, but they do vary some with temperature. • Most nonmetallic substances have high values. Additional data are tabulated in Appendix A.3. 12.5B Radiation to a Small Object from Its Surroundings • In the case of a small gray object of area A1 m2 at temperature T1 in a large enclosure at a higher temperature T2, there is a net radiation to the small object. • The small body emits an amount of radiation to the enclosure given by Eq. (12.5-4) as 𝐴1 𝜀1 𝜎𝑇14 . The emissivity ε1 of this body is taken at T1. • The small body also absorbs an amount of energy from its surroundings at T2 given by 𝐴2 α12 𝜎𝑇24 • The α12 is the absorptivity of body 1 for radiation from the enclosure at T2. The value of α12 is approximately the same as the emissivity of this body at T2. 12.5B Radiation to a Small Object from Its Surroundings • The net heat of absorption is then, by the Stefan–Boltzmann equation, 𝑞 = 𝐴1 𝜀1 𝜎𝑇14 − 𝐴1 𝛼12 𝜎𝑇24 = 𝐴1 𝜎 𝜀1 𝑇14 − 𝛼12 𝑇24 12.5-5 • A further simplification of Eq. (12.5-5) is usually made for engineering purposes by using only one emissivity for the small body, at temperature T2. Thus, 𝑞 = 𝐴1 𝜎𝜀 𝑇14 − 𝑇24 12.5-6 Example 12.5-1. Radiation to a Metal Tube A small oxidized horizontal metal tube with an OD of 0.0254 m (1 in.), 0.61 m (2 ft) long, and with a surface temperature at 588 K (600°F) is in a very large furnace enclosure with firebrick walls and the surrounding air at 1088 K (1500°F). The emissivity of the metal tube is 0.60 at 1088 K and 0.46 at 588 K. Calculate the heat transfer to the tube by radiation using SI and English units. Solution of Example 12.5-1. Radiation to a Metal Tube A small oxidized horizontal metal tube with an OD of 0.0254 m (1 in.), 0.61 m (2 ft) long, and with a surface temperature at 588 K (600°F) is in a very large furnace enclosure with firebrick walls and the surrounding air at 1088 K (1500°F). The emissivity of the metal tube is 0.60 at 1088 K and 0.46 at 588 K. Calculate the heat transfer to the tube by radiation using SI and English units. • Since the large-furnace surroundings are very large compared to the small enclosed tube, the surroundings, when viewed from the position of the small body, appear black even if they are actually gray, and Eq. (12.5-6) is applicable. Solution of Example 12.5-1. Radiation to a Metal Tube • Substituting given values into Eq. (12.5-6) with an ε of 0.6 at 1088 K, 𝑞 = 𝐴1 𝜎𝜀 𝑇14 − 𝑇24 A1=pDL=p(0.0254 m)(0.61 m) = 0.049 m2 𝑞 = 𝐴1 𝜎𝜀 𝑇14 − 𝑇24 𝑊 q=0.049 m2(0.6)(5.676×108 2 4)[(588)4K4-(1088)4K4]=-2125 W 𝑚 .𝐾 • You can do the same for English units Other examples of small objects in large enclosures that occur in the process industries are: • A loaf of bread in an oven receiving radiation from the walls around it, • A package of meat or food radiating heat to the walls of a freezing enclosure, • A hot ingot of solid iron cooling and radiating heat in a large room, and • A thermometer measuring the temperature in a large duct. All images were taken from our textbook or from Google.com