The CRC Handbook of Mechanical Engineering. Chapter 4. Heat and Mass Transfer (776127), страница 5
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Isothermal Vertical Channels, Figure 4.2.4A and B. Figure 4.2.4A shows an open cavity boundedby vertical walls and open at the top and bottom. The large opposing plates are isothermal, attemperatures T1 and T2, respectively, and the spacing between these plates is small. DT is theaverage temperature difference between the plates and T¥, as shown in Figure 4.2.4A, but T1 andT2 must not straddle T¥. For this caseæ æ Ra ö m14Nu = ç ç÷ + C1Cl Raè è f Re ø(© 1999 by CRC Press LLC)mö÷ø1mRa £ 10 5(4.2.15)4-19Heat and Mass Transferwhere f Re is the product of friction factor and Reynolds number for fully developed flow through,and C1 is a constant that accounts for the augmentation of heat transfer, relative to a vertical flatplate (Case 1), due to the chimney effect.
The fRe factor accounts for the cross-sectional shape(Elenbaas, 1942a). Symbols are defined in Figure 4.2.4A and B; in the Nu equation, q is the totalheat transferred to the ambient fluid from all heated surfaces.For the parallel plate channel shown in Figure 4.2.4(A), use f Re = 24, m = –1.9, and for gasesC1 » 1.2.
It should be noted, however, that C1 must approach 1.0 as Pr increases or as the platespacing increases. For channels of circular cross section (Figure 4.2.4B) fRe - 16, m = –1.03, andfor gases C1 » 1.17. For other cross-sectional shapes like the square (fRe = 14.23), hexagonal(fRe = 15.05), or equilateral triangle (fRe = 13.3), use Equation (4.2.15) with the appropriate fRe,and with m = –1.5, and C1 » 1.2 for gases.The heat transfer per unit cross-sectional area, q/Ac, for a given channel length H and temperaturedifference, passes through a maximum at approximately Ramax, whereRa maxæ f ReC1Cl ö=ç÷1mè 2ø43(4.2.16)Ramax provides the value of hydraulic radius r = 2Ac/P at this maximum.13. Isothermal Triangular Fins, Figure 4.2.4C.
For a large array of triangular fins (Karagiozis et al.,1994) in air, for 0.4 < Ra < 5 ´ 1053é3.26 ùNu = Cl Ra 1 4 ê1 + æ 0.21 ö úë è Ra ø û-1 30.4 < Ra < 5 ´ 10 5(4.2.17)In this equation, b is the average fin spacing (Figure 4.2.4C), defined such that bL is the crosssectional flow area between two adjacent fin surfaces up to the plane of the fin tips. For Ra <0.4, Equation (4.2.17) underestimates the convective heat transfer. When such fins are mountedhorizontally (vertical baseplate, but the fin tips are horizontal), there is a substantial reduction ofthe convective heat transfer (Karagiozis et al., 1994).14. U-Channel Fins, Figure 4.2.4C. For the fins most often used as heat sinks, there is uncertaintyabout the heat transfer at low Ra. By using a conservative approximation applying for Ra < 100(that underestimates the real heat transfer), the following equation may be used:é Ra -2Nu = êæ ö + C1Cl Raëè 24 ø()-2ùúû-0.5(4.2.18)For air C1 depends on aspect ratio of the fin as follows (Karagiozis, 1991):HùéC1 = ê1 + æ ö , 1.16úèøbû minë(4.2.19)Equation (4.2.18) agrees well with measurements for Ra > 200, but for smaller Ra it falls wellbelow data because the leading term does not account for heat transfer from the fin edges andfor three-dimensional conduction from the entire array.15.
Circular Fins on a Horizontal Tube, Figure 4.24D. For heat transfer from an array of circularfins (Edwards and Chaddock, 1963), for H/Di = 1.94, 5 < Ra < 104, and for air,© 1999 by CRC Press LLC4-20Section 4137 ö ùéNu = 0.125Ra 0.55 ê1 - expæ è Ra ø úûë0.294(4.2.20)A more general, but also more complex, relation is reported by Raithby and Hollands (1985).16. Square Fins on a Horizontal Tube, Figure 4.2.4D. Heat transfer (Elenbaas, 1942b) from the squarefins (excluding the cylinder that connects them) is correlated for gases by[(m1m14 m) + (0.62Ra )Nu = Ra 0.89 18](4.2.21)m = -2.7Heat Transfer in EnclosuresThis section deals with cavities where the bounding walls are entirely closed, so that no mass can enteror leave the cavity. The fluid motion inside the cavity is driven by natural convection, which enhancesthe heat transfer among the interior surfaces that bound the cavity.17.
Extensive Horizontal Layers, Figure 4.2.5A with q = 0°. If the heated plate, in a horizontalparallel-plate cavity, is on the top, heat transfer is by conduction alone, so that Nu = 1. For heattransfer from below (Hollands, 1984):1- ln ( Ra1 3 k2 ) ù··é13æ Ra 1 3 ö1708 ù êéú + éæ Ra ö - 1ù+Nu = 1 + ê1 k2êúç k ÷ú ëè 5830 øRa úû ê 1ëè 2 øûêëúû(4.2.22)where[ x ]· = ( x, 0) maxk1 =1.441 + 0.018 Pr + 0.00136 Pr 2(k2 = 75 exp 1.5Pr-1 2)(4.2.23)The equation has been validated for Ra < 1011 for water, Ra < 108 for air, and over a smaller Rarange for other fluids.
Equation (4.2.22) applies to extensive layers: W/L ³ 5. Correlations fornonextensive layers are provided by Raithby and Hollands (1985).FIGURE 4.2.5 Nomenclature for enclosure problems.© 1999 by CRC Press LLC4-21Heat and Mass Transfer18. Vertical Layers, Figure 4.2.5(A), with q = 90°. W/L > 5. For a vertical, gas-filled (Pr » 0.7) cavitywith H/L ³ 5, the following equation closely fits the data, for example that of Shewen et al. (1996)for Ra(H/L)3 £ 5 ´ 1010 and H/L ³ 40.2é æö ùê çú0.0665Ra 1 3 ÷ úNu1 = ê1 + ç÷1.4ê çæ 9000 ö ÷ úê ç1+úè Ra ø ÷ø úêë èû120.273LNu 2 = 0.242æ Ra öèHø[Nu = Nu1 , Nu 2]max(4.2.24)For Pr ³ 4, the following equation is recommended (Seki et al., 1978) for Ra(H/L)3 < 4 ´ 10120.360.1éùLLNu = ê1,0.36Pr 0.051 æ ö Ra 0.25 , 0.084 Pr 0.051 æ ö Ra 0.3 úèøèøHHëû max(4.2.25a)and for Ra (H/L)3 > 4 ´ 1012Nu = 0.039Ra 1 3(4.2.25b)19.
Tilted Layers, Figure 4.25A, with 0 £ q £ 90°, W/L > 8. For gases (Pr » 0.7), 0 £ q £ 60° andRa £ 105 (Hollands et al., 1976), use·1.613ùé1708 ù é 1708(sin 1.8q) ù éæ Ra cosq öNu = 1 + 1.44 ê1 1+- 1úêúêúèøRa cosqúû ë 5830ë Ra cosq û êëû·(4.2.26)See equation (4.2.23) for definition of [x]°. For 60° £ q £ 90° linear interpolation is recommendedusing Equations (4.2.24) for q = 90° and (4.2.26) for q = 60°.20.
Concentric Cylinders, Figure 4.2.5B. For heat transfer across the gap between horizontal concentric cylinders, the Nusselt number is defined as Nu = q¢ ln(Do/Di)/2pDT where q¢ is the heattransfer per unit length of cylinder. For Ra £ 8 ´ 107, 0.7 £ Pr £ 6000, 1.15 £ D/Di £ 8 (Raithbyand Hollands, 1975)éln( Do Di )Ra 1 4êNu = ê0.603Cl3535êL Di ) + ( L Do )(ë[ùú5 4 ,1úúû max(4.2.27)]For eccentric cylinders, see Raithby and Hollands (1985).21. Concentric Spheres, Figure 4.2.5B. The heat transfer between concentric spheres is given by thefollowing equation (Raithby and Hollands, 1975) for Ra £ 6 ´ 108, 5 £ Pr £ 4000, 1.25 < Do /Di£ 2.5,é14æ LöêqLRa 1 4= ê1.16Cl ç ÷Nu =Di Do kDT êè Di ø ( D D )3 5 + ( D D ) 4 5iooië[For eccentric spheres, see Raithby and Hollands (1985).© 1999 by CRC Press LLCùú5 4 ,1úúû max](4.2.28)4-22Section 4Example CalculationsProblem 1: Heat Transfer from Vertical Plate, Figure 4.2.6A.
For the vertical isothermal surface inFigure 4.2.6A with Ts = 40°C, H1 = 1 m, H2 = 1 m, W1 = 1 m, W2 = 1 m and for an ambient air temperatureof T¥ = 20°C (at 1 atm), find the heat transfer from one side of the plate.FIGURE 4.2.6 Sketches for example problems.Properties: At Tf = (Tw + T¥)/2 = 30°C and atmospheric pressure for air: n = 1.59 ´ 10–5 m2/sec, Pr= 0.71, k = 0.0263 W/mK. At T¥, b » 1/T¥ = 1(273 + 20) = 0.00341 K–1.Solution: For the geometry shown in Figure 4.2.6A:H öæAs = ( H1 + H2 )W1 + ç H1 + 2 ÷ W2 = 3.5 m 2è2 øòW1 + W2034S 3 4 dc = ( H1 + H2 ) W1 +14L1 4 = ( H1 + H2 )L1 4Ra =ò744 W2( H1 + H2 ) - H17 4 = 3.03 m 7 47 H2[W1 + W2S 3 4 dc=Asgb ¥ L3 (Tw - T¥ )na=]= 1.19 m1 4 (see comments below)0C1 =(plate surface area)1.19 ´ 3.03= 1.033.59.81 ´ 0.00341 ´ 2 3 ´ (40 - 20)= 1.50 ´ 10101.59 ´ 10 -5 ´ 2.25 ´ 10 -5Cl = 0.514 from Equation (4.2.3); Ct = CtV = 0.103 from Equation (4.2.4).
NuT = C1 Cl Ra1/4 = 185from Equation (4.2.11).Nu l =2.0= 186ln 1 + 2.0 Nu T(Nu t = CtV Ra 1 3üïïý (from Equation (4.2.7))ï1 + 1.4 ´ 10 9 Pr Ra = 238ïþ)(Nu =© 1999 by CRC Press LLC)qL= Nu 6l + Nu t6ADTk(16)= 2464-23Heat and Mass Transferfrom Equation (4.2.6) with m = 6.q=As DTkNu 3.5 ´ 20 ´ 0.0263 ´ 246== 226WL2Comments on Problem 1: Since Nul < Nut, the heat transfer is primarily turbulent. Do not neglectradiation.
Had the surface been specified to be at constant heat flux, rather than isothermal, the equationsin this section can be used to find the approximate average temperature difference between the plate andfluid.Problem 2: Heat Transfer from Horizontal Strip, Figure 4.2.6B. Find the rate of heat loss per unit lengthfrom a very long strip of width W = 0.1 m with a surface temperature of Ts = 70°C in water at T¥ = 30°C.Properties: At Tf = (Ts + T¥)1/2 = 50°Cv = 5.35 ´ 10 -7 m 2 /seca = 1.56 ´ 10 -7 m 2 /seck = 0.645 W/mKb = 2.76 ´ 10 -4 K -1Pr = 3.42Solution: This problem corresponds to Case 3 and Figure 4.2.1C.CtH = 0.14from Equation 4.2.5 and Cl = 0.563 from Equation (4.2.3).WH ö W== 0.05 mL* = lim æH ®¥è 2W + 2 H ø2from Figure 4.2.1C.Ra =gbDTL*3= 1.62 ´ 10 8vaNu l =Nu T = 0.835Cl Ra 1 4 = 53.51.4= 54.2ln 1 + 1.4 Nu T(Nu =)Nu t = CtH Ra 1 3 = 76.3qL*10= Nu10l + Nu tWHDT kq H=()0.1= 76.5WDTkNu= 3950 W/m-lengthL*Comments: Turbulent heat transfer is dominant.
Radiation can be ignored (since it lies in the farinfrared region where it is not transmitted by the water).Problem 3: Heat Loss across a Window Cavity, Figure 4.2.6C. The interior glazing is at temperature T1= 10°C, the exterior glazing at T2 = –10°C, the window dimensions are W = 1 m, H = 1.7 m, and theair gap between the glazings is L = 1 cm and is at atmospheric pressure. Find the heat flux loss acrossthe window.© 1999 by CRC Press LLC4-24Section 4Properties: At T = T1 + T2/2 = 0°C = 273Kv = 1.35 ´ 10 -5 m 2 /seca = 1.89 ´ 10 -5 m 2 /seck = 0.024 W/mKb = 1 273 = 3.66 ´ 10 -3 K -1Pr = 0.71Solution: The appropriate correlations are given in Case 18 and by Equation (4.2.24).Ra =gb(T1 - T2 ) L3va3=9.81 ´ 3.66 ´ 10 -3 ´ 20 ´ (0.01)= 2.81 ´ 10 31.35 ´ 10 -5 ´ 1.89 ´ 10 -52é ìü ùê ïï úï 0.0665Ra 1 3 ï úNu1 = ê1 + í1.4 ýêæ 9000 ö ï úê ï1 +úêë ïî è Ra ø ïþ úûLNu 2 = 0.242æ Ra öèHøNu =q / WH =0.27312= 1.010.01ö= 0.242æ 2.81 ´ 10 3 ´è1.7 ø0.273= 0.520qL= (Nu1 , Nu 2 ) max = 1.01WH (T1 - T2 )kN (T1 - T2 )kL=1.01 ´ 20 ´ 0.24= 48.5 W/m 20.01Comments: For pure conduction across the air layer, Nu = 1.0.
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