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 33 ft long, 10 ft tall, and 0.3 ft thick (Figure1A). One side of the wall is maintained at ts1= 77F, and the otheris kept at ts2= 68F. Heat transfer occurs at rate q through the wallfrom the warmer side to the cooler. The heat transfer mode is con-duction (the only way energy
6、 can be transferred through a solid).If ts1is raised from 77 to 86F 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 “
7、proportional 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 Btu/hftF. The d
8、enominator L/(kAc) can beconsidered 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 chargeinst
9、ead of rate of heat transfer q.Thermal resistance has units hF/Btu. A wall with a resistance of3 hF/Btu requires (ts1 ts2) = 3F for heat transfer q of 1 Btu/h. Thethermal/electrical resistance analogy allows tools used to solve elec-trical circuits to be used for heat transfer problems.ConvectionCon
10、sider a surface 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 Btu/hft2F. The convection resistance 1/(hcAs) has un
11、its ofhF/Btu.If 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 m
12、ode of heat transferis 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
13、 is assigned to TC 1.3, Heat Transfer andFluid Flow.Fig. 1 (A) Conduction and (B) Convectionts1ts2AcL-Table 1 Heat Transfer Coefficients by Convection TypeConvection Type hc, Btu/hft2FFree, gases 0.35 to 4.5Free, liquids 1.8 to 180Forced, gases 4.5 to 45Forced, liquids 9 to 3500Boiling, condensation
14、 450 to 18,000ts1ts2AcL-ts1ts2LkAc-=tst1 hcAs-4.2 2017 ASHRAE HandbookFundamentals 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 trans
15、port them, unlike 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, an
16、d emits energy 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 Btu/hft2; Tsisabsolute surface temperature, R; and = 0.1712 108Btu/hft2R4is th
17、e Stefan-Boltzmann constant. If a surface is not black, the emis-sion per 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 absorpti
18、vity is the fraction of 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 a
19、nd absorbed,respectively, 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 t
20、he surfaces (1) relative 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,su
21、rr. So, if tstsurr, net 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,
22、 HThin isothermal rectangular 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 dI
23、sothermal sphere in semi-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 Extended SurfacesHeat transfer from a surface can be increased
24、 by attaching finsor 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 locat
25、ed on the side of the 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 ex
26、pansionevaporators.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
27、 heat flux. Therefore, 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 temper
28、ature: = (6)where q 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 decreas
29、es as the heat transfer 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 conden
30、sing and boiling, wherelarge 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
31、. 6 Efficiency of Annular 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
32、A in Figure 7), cy-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 f
33、in with thickness Empirical 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, respectivel
34、y. Rectangular fin 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
35、the actual or equivalent 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
36、 = 1.27 where and are 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 acc
37、uracy, particularly 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 incl
38、udes both fin area 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 he
39、at transfer in Equation (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 = 1290 Btuin/hft2F, ID = 1.8 in.,and OD = 2 in. has circular aluminum fins = 0.04 in. thick with anouter diameter of Dfin= 3.9 in. There are N = 76
40、 fins per foot of tubelength. Steam condenses inside the tube at ti= 392F with a large heattransfer coefficient on the inner tube surface. Air at t= 77F isheated by the steam. The heat transfer coefficient outside the tube is7 Btu/hft2F. Find the rate of heat transfer per foot of tube length.Solutio
41、n: From Figure 5s efficiency curve, the efficiency of these cir-cular fins isThe fin area for L = 1 ft isAs= NL 2(Dfin2 OD2)/4 = 1338 in2= 9.29 ft2The unfinned area for L = 1 ft isAp= OD L(1 N) = (2/12) ft 1 ft(1 76 0.04/12) = 0.39 ft2and the total area A = As+ Ap= 9.68 ft2. Surface efficiency ismWc
42、tanhmWc-hP kAcmrbZtanhmrbZ-2hkFig. 9 Rectangular Tube Array 0.2 0.3a 22b2+ApAs+A-trte1 shA-Fig. 10 Hexagonal Tube ArrayWDfinOD2 3.9 22 0.95 in.=XeXb3.9 222-1.95 in.=Whk 2- 0 . 9 5 i n .7 Btu/hft2F1290 Btuin/hft2F0.02 in.- 0 . 4 9 0.89=4.8 2017 ASHRAE HandbookFundamentals s= = 0.894and resistance of
43、the finned surface isRs= = 0.0165 hF/BtuTube wall resistance isThe rate of heat transfer is thenq = = 18,912 Btu/hHad Schmidts approach been used for fin efficiency,m = = 6.25 ft1rb= OD/2 = 1 in. = 0.0833 ftZ = (Dfin/OD) 11 + 0.35 ln(Dfin/OD) = 1.172 = = 0.89the same as given by Figure 5.Contact Res
44、istance. Fins can be extruded from the prime surface(e.g., short fins on tubes 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 s
45、oldering; nonmetallicbonding materials, such as epoxy resin, are also used. 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
46、 not always economical. Contact resistance of a mechanicalbond may or may not 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 surfa
