ASHRAE FUNDAMENTALS IP CH 4-2013 Heat Transfer.pdf

1、4.1CHAPTER 4HEAT TRANSFERHeat Transfer Processes . 4.1Thermal Conduction 4.3Thermal Radiation 4.11Thermal Convection 4.17Heat Exchangers . 4.21Heat 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 2010 ASHRAEHandbookRefrigeration. Physical properties of substances can befound in Chapters 26, 28, 32, and 33 of this volume and in Chapter19 of the 2010 ASHRA

4、E HandbookRefrigeration. Heat transferequipment, including evaporators, condensers, heating and coolingcoils, furnaces, and radiators, is covered in the 2012 ASHRAE Hand-bookHVAC Systems and Equipment. For further information onheat transfer, see the Bibliography.HEAT TRANSFER PROCESSESConductionCon

5、sider 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 ca

6、n 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 “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 Btu/hftF. The deno

8、minator 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 chargeinstead

9、 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.ConvectionConsid

10、er 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 units

11、 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 mode

12、 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 is

13、 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 45

14、0 to 18,000ts1ts2AcL-ts1ts2LkAc-=tst1 hcAs-4.2 2013 ASHRAE HandbookFundamentalsRadiationMatter 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

15、 them, unlike conduction and convection (inwhich heat transfer occurs through matter). The rate of thermalradiant energy emitted by a surface depends on its absolute temper-ature and its surface characteristics. A surface that absorbs all radi-ation incident upon it is called a black surface, and em

16、its 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 the St

17、efan-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 absorptivity

18、 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 and a

19、bsorbed,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 the s

20、urfaces (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,surr.

21、So, if tstsurr, net heat loss from gray surface Asin the radiation exchangewith the surroundings at Tsurrisqnet= qemitted qabsorbed= AsWbs AsWb,surr= As(Ts4 T4sum)(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, HThi

22、n 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 mediumL RL Rd 3Rd RL dHorizontal cylinder of length L midway between two infinite, parallel, isothermal surfacesL dIsother

23、mal 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 2013 ASHRAE HandbookFundamentalsa given performance. To achieve optimum design, fins are generallyloc

24、ated on the side of the heat exchanger with lower heat transfercoefficients (e.g., the air side of an air-to-water coil). Equipmentwith extended surfaces includes natural- and forced-convectioncoils and shell-and-tube evaporators and condensers. Fins are alsoused inside tubes in condensers and dry e

25、xpansion evaporators.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

26、in 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 temp

27、erature: = (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 decre

28、ases 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 cond

29、ensing 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.C

30、onstant-Area Fins and Spines. For fins or spines with constantcross-sectional area e.g., straight fins (option A in Figure 7), cy-lindrical spines (option D in Figure 8), the efficiency can be cal-culated asFig. 6 Efficiency of Annular Fins with Constant Metal Area for Heat FlowFig. 7 Efficiency of

31、Several Types of Straight FinsqhAstrte-Fig. 8 Efficiency of Four Types of SpinesHeat Transfer 4.7 = (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

32、fin 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, respective

33、ly. 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

34、 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 /r

35、b = 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 ac

36、curacy, 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 inc

37、ludes 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 h

38、eat 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 = 7

39、6 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.Soluti

40、on: 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 ismW

41、ctanhmWc-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 . 9 5 i n .=Whk 2- 0 . 9 5 i n .7 Btu/hft2F1290 Btuin/hft2F0.02 in.- 0 . 4 9 0.89=4.8 2013 ASHRAE HandbookFundamentalss= = 0.894and resistan

42、ce of 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,mi = = 6.25 ft1rb= OD/2 = 1 in. = 0.0833 ftZ = (Dfin/OD) 1 1 + 0.35 ln(Dfin/OD) = 1.172 = = 0.89the same as given by Figure 5.Con

43、tact Resistance. Fins can be extruded from the prime sur-face (e.g., short fins on tubes in flooded evaporators or water-cooledcondensers) or can be fabricated separately, sometimes of a differ-ent material, and bonded to the prime surface. Metallurgical bondsare achieved by furnace-brazing, dip-bra

44、zing, or soldering; nonme-tallic bonding materials, such as epoxy resin, are also used.Mechanical bonds are obtained by tension-winding fins aroundtubes (spiral fins) or expanding the tubes into the fins (plate fins).Metallurgical bonding, properly done, leaves negligible thermalresistance at the jo

45、int but is not always economical. Contact resis-tance of a mechanical bond may or may not be negligible, depend-ing on the application, quality of manufacture, materials, andtemperatures involved. Tests of plate-fin coils with expanded tubesindicate that substantial losses in performance can occur w

46、ith finsthat have cracked collars, but negligible contact resistance wasfound in coils with continuous collars and 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 influence of fin density, fin thickness, and tube diameteron contact pressure and compared it to data for wet and dry coils.Shlykov (1964) showed that the range