The CRC Handbook of Mechanical Engineering. Chapter 4. Heat and Mass Transfer (776127), страница 9
Текст из файла (страница 9)
(1979):1 < Re d Pr < 100Nu d = 1.125(Re d Pr )0.413(4.2.66)For more information on heat transfer with flow over cylinders, refer to Morgan (1975) and Zukauskas(1987).Spheres: For flows over spheres (Figure 4.2.16) use one of the following two correlations.FIGURE 4.2.16 A fluid flowing over a sphere.1. Whitaker (1972): Evaluate properties at T¥ except ms at Ts.3.5 < Re d < 76, 000© 1999 by CRC Press LLC0.71 < Pr < 3801 < m m s < 3.24-37Heat and Mass Transfer(12Nu d = 2.0 + 0.4 Re d + 0.06 Re23d) Pr25æ möçm ÷è sø14(4.2.67)2.
Achenbach (1978): Evaluate properties at (Ts + T¥)/2:100 < Re d < 2 ´ 10 5Pr = 0.71(1.6Nu d = 2 + 0.25Re d + 3 ´ 10 -4 Re d4 ´ 10 5 < Re d < 5 ´ 10 612)(4.2.68)Pr = 0.7123Nu d = 430 + 5 ´ 10 -3 Re d + 0.25 ´ 10 -9 Re d - 3.1 ´ 10 -17 Re d(4.2.69)3. Liquid Metals: From experimental results with liquid sodium, Witte (1968) proposed3.6 ´ 10 4 < Re d < 1.5 ´ 10 512Nu d = 2 + 0.386(Re d Pr )(4.2.70)Other Geometries: For geometries other than cylinders and spheres, use Equation (4.2.71) with thecharacteristic dimensions and values of the constants given in the Table 4.2.1.m(4.2.71)Nu D = cRe DAlthough Equation (4.2.71) is based on experimental data with gases, its use can be extended to fluidswith moderate Prandtl numbers by multiplying Equation (4.2.71) by (Pr/0.7)1/3.Heat Transfer across Tube BanksWhen tube banks are used in heat exchangers, the flow over the tubes in the second subsequent rowsof tubes is different from the flow over a single tube.
Even in the first row the flow is modified by thepresence of the neighboring tubes. The extent of modification depends on the spacing between the tubes.If the spacing is very much greater than the diameter of the tubes, correlations for single tubes can beused. Correlations for flow over tube banks when the spacing between tubes in a row and a column isnot much greater than the diameter of the tubes have been developed for use in heat-exchanger applications. Two arrangements of the tubes are considered — aligned and staggered as shown in Figure4.2.17. The nomenclature used in this section is shown in the figure.For the average convective heat transfer coefficient with tubes at uniform surface temperature, fromexperimental results, Zukauskas (1987) recommends correlations of the form:pæ Pr ömaNu d = cæ ö Re d Pr n ç÷è bøè Prs ø0.25(4.2.72)In Equation (4.2.72) all properties are evaluated at the arithmetic mean of the inlet and exit temperaturesof the fluid, except Prs which is evaluated at the surface temperature Ts.
The values of the constants c,p, m, and n are given in Table 4.2.2 for in-line arrangement and in Table 4.2.3 for staggeredarrangement.© 1999 by CRC Press LLC4-38Section 4TABLE 4.2.1Values of c and m in Equation (4.2.71)From Jakob, 1949. With permission.FIGURE 4.2.17 Two arrangements of tube banks. In-line or aligned arrangement on the left and staggered arrangement on the right. (a = ST /d; b = SL/d.)© 1999 by CRC Press LLC4-39Heat and Mass TransferTABLE 4.2.2In-Line Arrangement — Values of Constants in Equation (4.2.72) (p = 0 in all cases)TABLE 4.2.3Redcmn1–100100–1000103–2 ´ 1052 ´ 105 – 2 ´ 1060.90.520.270.0330.40.50.630.80.360.360.360.4Staggered Arrangement — Values of Constants in Equation (4.2.72)Redcpmn1–500500–1000103–2 ´ 1052 ´ 105 – 2 ´ 1061.040.710.350.031000.20.20.40.50.60.80.360.360.360.36In computing Red, the maximum average velocity between tubes is used.
The maximum velocitiesfor the in-line and staggered arrangements are given byU max =In-line:Sd >Staggered:Sd <Staggered:U ¥ STST - dST + d2(4.2.73)U max =ST + d2U max =2éæS ö ùSd = êSL2 + ç T ÷ úè 2 ø úûêëU ¥ STST - d(4.2.74)U ¥ ST2( Sd - d )(4.2.75)12Equation (4.2.72) is for tube banks with 16 or more rows.
When there are fewer than 16 rows, theheat transfer coefficient given by Equation (4.2.72) is multiplied by the correction factor c1 defined byEquation (4.2.76) and given in Table 4.2.4.hN= c1h16(4.2.76)wherehN = heat transfer coefficient with N rows (fewer than 16)h16 = heat transfer coefficient with 16 or more rowsTABLE 4.2.4Correction Factor c1 to Be Used with Equation (4.2.76)Number of Rows (N)Tube ArrangementIn-lineStaggered© 1999 by CRC Press LLC12345710130.700.640.800.760.860.840.900.890.930.930.960.960.980.980.990.994-40Section 4Pressure Drop: With tube banks, pressure drop is a significant factor, as it determines the fan powerrequired to maintain the fluid flow.
Zukauskas (1987) recommends that the pressure drop be computedfrom the relationDp = pi - pe = N c2rU maxf2(4.2.77)where pi and pe are the fluid pressures at inlet and exit of the tube banks. The values of c and f arepresented in Figure 4.2.18A. In Figure 4.2.18A the friction factor f for in-line arrangement is presentedfor different values of b (SL/d) for SL = ST . For values of SL/ST other than 1, the correction factor c isgiven in the inset for different values of (a – 1)/(b – 1). Similarly, the friction factor for staggeredarrangement (for equilateral triangle arrangement) and a correction factor for different values of a/b arealso given in Figure 4.2.18b.
The value of f is for one row of tubes; the total pressure drop is obtainedby multiplying the pressure drop for one row by the number of rows, N.The temperature of the fluid varies in the direction of flow, and, therefore, the value of the convectiveheat transfer coefficient (which depends on the temperature-dependent properties of the fluid) also variesin the direction of flow. However, it is common practice to compute the total heat transfer rate with theassumption of uniform convective heat transfer coefficient evaluated at the arithmetic mean of the inletand exit temperatures of the fluid.
With such an assumption of uniform convective heat transfer coefficient, uniform surface temperature and constant specific heat (evaluated at the mean fluid temperature),the inlet and exit fluid temperatures are related byæ T - Te öhA=- slnç s÷˙ pmcè Ts - Ti ø(4.2.78)The heat transfer rate to the fluid is related by the equation˙ p (Ti - Te )q = mc(4.2.79)ExampleA heat exchanger with aligned tubes is used to heat 40 kg/sec of atmospheric air from 10 to 50°C withthe tube surfaces maintained at 100°C.
Details of the heat exchanger areDiameter of tubes25 mmNumber of columns20Length of each tube3mSL = ST75 mmDetermine the number of rows required.Solution: Average air temperature = (Ti + Te)/2 = 30°C. Properties of atmospheric air (from Suryanarayana, 1995):© 1999 by CRC Press LLCr = 1.165 kg m 3c p = 1007 J kg Km = 1.865 ´ 10 -5 Nsec m 3k = 0.0264 W mKPr = 0.712Prs (at 100°C) = 0.7054-41Heat and Mass Transferf86Re=10341010421.50xa=bb = 1.2510610.60.41.25105 , 1060.210.10.06 0.10.20.41(a-1)/(b-1)2462.002.500.13101102103104Re(a)10510680Re=1021.6105 , 1061.410310St = Sd1.2a=1.251.0104104103f1021050.90.40.60.8124a/b11.502.502.000.12101102103Re(b)104105FIGURE 4.2.18 Friction factors for tube banks.
(a) In-line arrangement; (b) Staggered arrangement.To find Umax we need the minimum area of cross section for fluid flow (Figure 4.2.19).H = 20 ´ 0.075 = 1.5 mAmin = 20(0.075 - 0.025) ´ 3 = 3 m 2U max =© 1999 by CRC Press LLCm˙40== 11.44 m secrAmin 1.165 ´ 31064-42Section 4FIGURE 4.2.19 Aligned tube heat exchanger (only a few of the 20 columns and rows are shown).Re d =rU max d 1.165 ´ 11.44 ´ 0.025== 17, 865m1.865 ´ 10 -5With values from Table 4.2.2,0.712 öNu d = 0.27 ´ 17, 8650.63 ´ 0.712 0.36 æè 0.705 øh=0.25= 114.3114.3 ´ 0.0264= 120.7 W m 2 K0.025From Equation 4.2.78,120.7 ´ As100 - 50 ö=lnæè 100 - 10 ø40 ´ 1007As = p ´ 0.025 ´ 3 ´ 20 ´ NN = number of rows = 42Fan Power: From the first law of thermodynamics (see Chapter 2), the fan power isæppv2 öW˙ F = m˙ ç i + e + e ÷2øè ri r epi and pe are the pressures at inlet and exit of the heat exchanger and ve is the fluid velocity at exit.Assuming constant density evaluated at (Ti + Te)/2 the pressure drop is found from Figure 4.2.18a.Re p = 17, 865:a = b = ST d = 75 25 = 3In Figure 4.2.18, although the friction factor is available for values of b up to 2.5, we will estimate thevalue of f for b = 3.
From Figure 4.2.18, f » 0.11. The correction factor c = 1.pi - pe = N c2rU max1.165 ´ 11.44 2f = 42 ´ 1´ 0.11 = 352.2 kPa22ve =© 1999 by CRC Press LLC11.44 ´ 50= 7.63 m/sec754-43Heat and Mass Transferæ7.632 öW˙ F = 40ç 352.2 +÷ = 15, 250 W2 øèHeat Transfer with Jet ImpingementJet impingement (Figure 4.2.20) on a heated (or cooled) surface results in high heat transfer rates, andis used in annealing of metals, tempering of glass, cooling of electronic equipment, internal combustionengines, and in a wide variety of industries — textiles, paper, wood, and so on. Usually, the jets arecircular, issuing from a round nozzle of diameter d, or rectangular, issuing from a slot of width w. Theymay be used singly or in an array.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
