1、4.1CHAPTER 4HEAT TRANSFERHeat Transfer Processes . 4.1Thermal Conduction 4.3Thermal Radiation 4.11Thermal Convection 4.17Heat Exchangers . 4.22Heat Transfer Augmentation. 4.24Symbols . 4.31EAT transfer is energy transferred because of a temperatureH difference. Energy moves from a higher-temperature
2、 region toa lower-temperature region by one or more of three modes:conduction, radiation, and convection. This chapter presents ele-mentary principles of single-phase heat transfer, with emphasis onHVAC applications. Boiling and condensation are discussed inChapter 5. More specific information on he
3、at transfer to or frombuildings or refrigerated spaces can be found in Chapters 14 to 19,23, and 27 of this volume and in Chapter 24 of the 2014 ASHRAEHandbookRefrigeration. Physical properties of substances can befound in Chapters 26, 28, 32, and 33 of this volume and in Chapter19 of the 2014 ASHRA
4、E HandbookRefrigeration. Heat transferequipment, including evaporators, condensers, heating and coolingcoils, furnaces, and radiators, is covered in the 2016 ASHRAE Hand-bookHVAC Systems and Equipment. For further information onheat transfer, see the Bibliography.1. HEAT TRANSFER PROCESSESConduction
5、Consider a wall that is 10 m long, 3 m tall, and 100 mm thick (Fig-ure 1A). One side of the wall is maintained at ts1= 25C, and theother is kept at ts2= 20C. Heat transfer occurs at rate q through thewall from the warmer side to the cooler. The heat transfer mode isconduction (the only way energy ca
6、n be transferred through a solid).If ts1is raised from 25 to 30C while everything else remains thesame, q doubles because ts1 ts2doubles.If the wall is twice as tall, thus doubling the area Acof the wall, qdoubles.If the wall is twice as thick, q is halved.From these relationships,q where means “pro
7、portional to” and L = wall thickness. However,this relation does not take wall material into account; if the wall werefoam instead of concrete, q would clearly be less. The constant ofproportionality is a material property, thermal conductivity k.Thus,q = k (1)where k has units of W/(mK). The denomi
8、nator L/(kAc) can be con-sidered the conduction resistance associated with the drivingpotential (ts1 ts2). This is analogous to current flow through an elec-trical resistance, I = (V1 V2)/R, where (V1 V2) is driving potential,R is electrical resistance, and current I is rate of flow of chargeinstead
9、 of rate of heat transfer q.Thermal resistance has units K/W. A wall with a resistance of5 K/W requires (ts1 ts2) = 5 K for heat transfer q of 1 W. The ther-mal/electrical resistance analogy allows tools used to solve electricalcircuits to be used for heat transfer problems.ConvectionConsider a surf
10、ace at temperature tsin contact with a fluid at t(Figure 1B). Newtons law of cooling expresses the rate of heattransfer from the surface of area Asasq = hcAs(ts t) = (2)where hcis the heat transfer coefficient (Table 1) and has unitsof W/(m2K). The convection resistance 1/(hcAs) has units of K/W.If
11、t ts, heat transfers from the fluid to the surface, and q is writ-ten as just q = hcAs(t ts). Resistance is the same, but the sign of thetemperature difference is reversed.For heat transfer to be considered convection, fluid in contactwith the surface must be in motion; if not, the mode of heat tran
12、sferis conduction. If fluid motion is caused by an external force (e.g.,fan, pump, wind), it is forced convection. If fluid motion resultsfrom buoyant forces caused by the surface being warmer or coolerthan the fluid, it is free (or natural) convection.The preparation of this chapter is assigned to
13、TC 1.3, Heat Transfer andFluid Flow.Fig. 1 (A) Conduction and (B) Convectionts1ts2AcL-Table 1 Heat Transfer Coefficients by Convection TypeConvection Type hc, W/(m2K)Free, gases 2 to 25Free, liquids 10 to 1000Forced, gases 25 to 250Forced, liquids 50 to 20 000Boiling, condensation 2500 to 100 000ts1
14、ts2AcL-ts1ts2LkAc-=tst1 hcAs-4.2 2017 ASHRAE HandbookFundamentals (SI)RadiationMatter emits thermal radiation at its surface when its temperatureis above absolute zero. This radiation is in the form of photons ofvarying frequency. These photons leaving the surface need nomedium to transport them, un
15、like conduction and convection (inwhich heat transfer occurs through matter). The rate of thermalradiant energy emitted by a surface depends on its absolute tempera-ture and its surface characteristics. A surface that absorbs all radia-tion incident upon it is called a black surface, and emits energ
16、y atthe maximum possible rate at a given temperature. The heat emis-sion from a black surface is given by the Stefan-Boltzmann law:qemitted, black = AsWb= AsTs4where Wb= Ts4 is the blackbody emissive power in W/m2; Tsisabsolute surface temperature, K; and = 5.67 108W/(m2K4) isthe Stefan-Boltzmann co
17、nstant. If a surface is not black, the emissionper unit time per unit area isW = Wb= Ts4where W is emissive power, and is emissivity, where 0 1. Fora black surface, = 1.Nonblack surfaces do not absorb all incident radiation. Theabsorbed radiation isqabsorbed= AsGwhere absorptivity is the fraction of
18、 incident radiation absorbed,and irradiation G is the rate of radiant energy incident on a surfaceper unit area of the receiving surface. For a black surface, = 1.A surfaces emissivity and absorptivity are often both functionsof the wavelength distribution of photons emitted and absorbed,respectivel
19、y, by the surface. However, in many cases, it is reason-able to assume that both and are independent of wavelength. Ifso, = (a gray surface).Two surfaces at different temperatures that can “see” each othercan exchange energy through radiation. The net exchange ratedepends on the surfaces (1) relativ
20、e size, (2) relative orientation andshape, (3) temperatures, and (4) emissivity and absorptivity.However, for a small area Asin a large enclosure at constant tem-perature tsurr, the irradiation on Asfrom the surroundings is theblackbody emissive power of the surroundings Wb,surr. So, if tstsurr, net
21、 heat loss from gray surface Asin the radiation exchangewith the surroundings at Tsurrisqnet= qemitted qabsorbed= AsWbs AsWb,surr= As(Ts4 T4surr)(3)where = for the gray surface. If tsL/5Corner of three adjoining walls (inner surface at T1and outer surface at T2)0.15LL WL d, W, HThin isothermal recta
22、ngular plate buried in semi-infinite mediumd = 0, W Ld WW Ld 2WW LCylinder centered inside square of length LL WW 2RIsothermal cylinder buried in semi-infinite medium L RL Rd 3Rd RL dHorizontal cylinder of length L midway between two infinite, parallel, isothermal surfacesL dIsothermal sphere in sem
23、i-infinite mediumIsothermal sphere in infinite medium 4R2.756L1dW-+ln0.59-Hd-0.078Wln 4WL-2Wln 4WL-2Wln 2dL-2Lln 0.54WR-2Lcosh1dR-2Lln 2dR-2LlnLR- 1ln L 2dln LR-2Lln4dR-4R1 R 2d-4.6 2017 ASHRAE HandbookFundamentals (SI)Extended SurfacesHeat transfer from a surface can be increased by attaching finso
24、r extended surfaces to increase the area available for heat transfer.A few common fin geometries are shown in Figures 5 to 8. Fins pro-vide a large surface area in a low volume, thus lowering materialcosts for a given performance. To achieve optimum design, fins aregenerally located on the side of t
25、he heat exchanger with lower heattransfer coefficients (e.g., the air side of an air-to-water coil).Equipment with extended surfaces includes natural- and forced-convection coils and shell-and-tube evaporators and condensers.Fins are also used inside tubes in condensers and dry expansionevaporators.
26、Fin Efficiency. As heat flows from the root of a fin to its tip, tem-perature drops because of the fin materials thermal resistance. Thetemperature difference between the fin and surrounding fluid istherefore greater at the root than at the tip, causing a correspondingvariation in heat flux. Therefo
27、re, increases in fin length result in pro-portionately less additional heat transfer. To account for this effect,fin efficiency is defined as the ratio of the actual heat transferredfrom the fin to the heat that would be transferred if the entire finwere at its root or base temperature: = (6)where q
28、 is heat transfer rate into/out of the fins root, teis tempera-ture of the surrounding environment, tris temperature at fin root,and Asis surface area of the fin. Fin efficiency is low for long or thinfins, or fins made of low-thermal-conductivity material. Fin effi-ciency decreases as the heat tran
29、sfer coefficient increases because ofincreased heat flow. For natural convection in air-cooled condensersand evaporators, where the air-side h is low, fins can be fairly largeand fabricated from low-conductivity materials such as steel insteadof from copper or aluminum. For condensing and boiling, w
30、herelarge heat transfer coefficients are involved, fins must be very shortfor optimum use of material. Fin efficiencies for a few geometriesare shown in Figures 5 to 8. Temperature distribution and fin effi-ciencies for various fin shapes are derived in most heat transfertexts.Fig. 6 Efficiency of A
31、nnular Fins with Constant Metal Area for Heat FlowFig. 7 Efficiency of Several Types of Straight Fins Fig. 8 Efficiency of Four Types of SpinesqhAstrte-Heat Transfer 4.7Constant-Area Fins and Spines. For fins or spines with constantcross-sectional area e.g., straight fins (option A in Figure 7), cy-
32、lindrical spines (option D in Figure 8), the efficiency can be cal-culated as = (7)wherem =P = fin perimeterAc= fin cross-sectional areaWc= corrected fin/spine length = W + Ac/PAc/P = d/4 for a cylindrical spine with diameter d= a/4 for an a a square spine= yb= /2 for a straight fin with thickness E
33、mpirical Expressions for Fins on Tubes. Schmidt (1949) pres-ents approximate, but reasonably accurate, analytical expressions(for computer use) for the fin efficiency of circular, rectangular, andhexagonal arrays of fins on round tubes, as shown in Figures 5, 9,and 10, respectively. Rectangular fin
34、arrays are used for an in-linetube arrangement in finned-tube heat exchangers, and hexagonalarrays are used for staggered tubes. Schmidts empirical solution isgiven by = (8)where rbis tube radius, m = , = fin thickness, and Z isgiven by Z = (re /rb) 11 + 0.35 ln(re /rb)where reis the actual or equiv
35、alent fin tip radius. For circular fins,re /rbis the actual ratio of fin tip radius to tube radius. For rectangu-lar fins (Figure 9),re /rb = 1.28 = M/rb = L/M 1where M and L are defined by Figure 9 as a/2 or b/2, depending onwhich is greater. For hexagonal fins (Figure 10),re /rb = 1.27 where and a
36、re defined as previously, and M and L are defined byFigure 10 as a/2 or b (whichever is less) and 0.5 ,respectively.For constant-thickness square fins on a round tube (L = M in Fig-ure 9), the efficiency of a constant-thickness annular fin of the samearea can be used. For more accuracy, particularly
37、 with rectangularfins of large aspect ratio, divide the fin into circular sectors asdescribed by Rich (1966).Other sources of information on finned surfaces are listed in theReferences and Bibliography.Surface Efficiency. Heat transfer from a finned surface (e.g., atube) that includes both fin area
38、As and unfinned or prime area Apisgiven byq = (hpAp+ hsAs)(tr te)(9)Assuming the heat transfer coefficients for the fin and prime sur-faces are equal, a surface efficiency scan be derived ass= (10)where A = As+ Apis the total surface area, the sum of the fin andprime areas. The heat transfer in Equa
39、tion (8) can then be written asq = shA(tr te) = (11)where 1/(shA) is the finned surface resistance.Example 3. An aluminum tube with k = 186 W/(mK), ID = 45 mm, andOD = 50 mm has circular aluminum fins = 1 mm thick with an outerdiameter of Dfin= 100 mm. There are N = 250 fins per metre of tubelength.
40、 Steam condenses inside the tube at ti= 200C with a large heattransfer coefficient on the inner tube surface. Air at t= 25C isheated by the steam. The heat transfer coefficient outside the tube is40 W/(m2K). Find the rate of heat transfer per metre of tube length.Solution: From Figure 5s efficiency
41、curve, the efficiency of these cir-cular fins isThe fin area for L = 1 m isAs= 250 2(Dfin2 OD2)/4 = 2.945 m2The unfinned area for L = 1 m isAp= OD L(1 N) = (0.05 m)(1 m)(1 250 0.001) = 0.118 m2and the total area A = As+ Ap= 3.063 m2. Surface efficiency ismWctanhmWc-hP kAcmrbZtanhmrbZ-2hkFig. 9 Recta
42、ngular Tube Array 0.2 0.3a 22b2+ApAs+A-trte1 shA-Fig. 10 Hexagonal Tube ArrayWDfinOD2 0.10 0.052 0.025 m= =XeXb0.10 0.05 2.0Whk 2-0.02540 W/(m2K)186 W mK0.0005 m-0.52 0.89=4.8 2017 ASHRAE HandbookFundamentals (SI)s= = 0.894and resistance of the finned surface isRs= = 9.13 103K/WTube wall resistance
43、isThe rate of heat transfer is thenq = = 18 981 WHad Schmidts approach been used for fin efficiency,m = = 20.74 m1rb= OD/2 = 0.025 mZ = (Dfin/OD) 11 + 0.35 ln(Dfin/OD) = 1.243 = = 0.88the same as given by Figure 5.Contact Resistance. Fins can be extruded from the prime surface(e.g., short fins on tu
44、bes in flooded evaporators or water-cooled con-densers) or can be fabricated separately, sometimes of a differentmaterial, and bonded to the prime surface. Metallurgical bonds areachieved by furnace-brazing, dip-brazing, or soldering; nonmetallicbonding materials, such as epoxy resin, are also used.
45、 Mechanicalbonds are obtained by tension-winding fins around tubes (spiral fins)or expanding the tubes into the fins (plate fins). Metallurgical bond-ing, properly done, leaves negligible thermal resistance at the jointbut is not always economical. Contact resistance of a mechanicalbond may or may n
46、ot be negligible, depending on the application,quality of manufacture, materials, and temperatures involved. Testsof plate-fin coils with expanded tubes indicate that substantial lossesin performance can occur with fins that have cracked collars, but neg-ligible contact resistance was found in coils with continuous collarsand properly expanded tubes (Dart 1959).Contact resistance at an interface between two solids is largely afunction of the surface properties and characteristics of the solids,contact pressure, and fluid in the interface, if any. Eckels (1977)modeled the infl
