The CRC Handbook of Mechanical Engineering. Chapter 4. Heat and Mass Transfer (776127), страница 61
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Since the air inlet temperatureis usually lower than the water inlet temperature, the water is cooled both by evaporation and by sensibleheat loss. For usual operating conditions the evaporative heat loss is considerably larger than the sensibleheat loss. Figure 4.8.7b shows a mechanical draft cross-flow unit. Figure 4.8.8 shows a natural-draftcross-flow tower for a power plant.Packing Thermal PerformanceCounterflow units. Merkel’s method (Merkel, 1925) for calculating the number of transfer units requiredto cool the water stream, for specified inlet and outlet water temperatures and inlet air condition is (Mills,1995)Ntu =g mS=m˙ LòhL ,outhL ,indhLhs - hG(hG = hG,in + (m˙ L m˙ G ) hL - hL,out(4.8.12))hs ( P, Ts ) = hs ( P, TL )(4.8.13)(4.8.14)It is imperative that the usual enthalpy datum states be used, namely, zero enthalpy for dry air and liquidwater at 0°C. Table 4.8.1 gives enthalpy data for 1 atm pressure.
The important assumptions requiredto obtain this result include1.2.3.4.A Lewis number of unity;Low mass transfer rate theory is valid;The liquid-side heat transfer resistance is negligible, that is, Ts . TL;The amount of water evaporated is small compared with the water and air flow rates.© 1999 by CRC Press LLC4-251Heat and Mass Transfer(a)(b)FIGURE 4.8.7 (a) A natural-draft counterflow cooling tower for a power plant. (b) A cross-flow cooling tower foran air-conditioning system.The method is accurate up to temperatures of about 60°C; comparisons with more exact results areusually within 3 to 5%, and seldom show errors greater than 10%.
Notice that the method does notgive the outlet state of the air; however, in situations encountered in practice, the outlet air can beassumed to be saturated for the purposes of calculating its density. It is possible to extend Merkel’smethod to include a finite liquid-side heat transfer resistance, but such refinement is seldom warranted.For typical operating conditions the bulk liquid temperature is seldom more than 0.3 K above theinterface temperature.Cross-flow units. Figure 4.8.9 shows a schematic of a cross-flow packing. If we assume that both theliquid and gas streams are unidirectional, and that there is no mixing in either stream, then use ofMerkel’s assumptions leads to the following pair of differential equations (Mills, 1995):© 1999 by CRC Press LLC4-252Section 4FIGURE 4.8.8 A natural-draft cross-flow cooling tower for a power plant.¶hGg a= m (hs - hG )¶xG(4.8.15)g a¶hL= - m (hs - hG )¶yL(4.8.16)Also hs = hs(hL) for a negligible liquid-side heat transfer resistance and the required boundary conditionsare the inlet enthalpies of both streams.
Equations (4.8.15) and (4.8.16) are solved numerically and thesolution used to evaluate the average enthalpy of the outlet liquid,hL,out =1XXòh0L ,outdx(4.8.17)Substituting in an exchanger energy balance on the liquid stream gives the heat transfer as(q = m˙ L hL,in - hL,out© 1999 by CRC Press LLC)(4.8.18)4-253Heat and Mass TransferTABLE 4.8.1Thermodynamic Properties of Water Vapor-Air Mixtures at 1 atmSpecific Volume, m3/kgEnthalpya,b kJ/kgTemp.,°CSaturationMass FractionDry AirSaturatedAirLiquidWaterDry AirSaturatedAir101112131415161718192021222324252627282930313233343536373839404142434445464748490.0076080.0081360.0086960.0092890.0099180.010580.011290 012040 012830.013660.014550.015480.016470.017510.018610.019780.021000.022290.023660.025090.026600.028200.029870.031640.033500.035450.037510.039670.041940.044320.046830.049460.052220.055120.058170.061370.064720.068420.071930.075800.80180.80460.80750.81030.81310.81600.81880.82170.82450.82730.83020.83300.83590.83870.84150.84440.84720.85000.85290.85570.85860.86140.86420.86710.86990.87280.87560.87840.88130.88410.88700.88980.89260.89550.89830.90120.90400.90680.90970.91250.80540.80860.81170.81480.81800.82120.82440.82760.83090.83410.83740.84080.84410.84750.85100.85440.85790.86150.86500.86860.87230.87600.87980.88360.88740.89140.89530.89940.90350.90770.91190.91620.92060.92510.92970.93430.93910.94390.94890.953942.1346.3250.5254.7158.9063.0867.2771.4575.6479.8283.9988.1792.3596.53100.71104.89109.07113.25117.43121.61125.79129.97134.15138.32142.50146.68150.86155.04159.22163.40167.58171.76175.94180.12184.29188.47192.65196.83201.01205.1910.05911.06512.07113.07714.08315.08916.09517.10118.10719.11320.12021.12822.13423.14024.14725.15326.15927.16628.17229.17830.18531.19132.19833.20434.21135.21836.22437.23138.23839.24540.25241.25942.26643.27344.28045.28746.29447.30148.30849.31629.14531.48133.89836.40138.99541.68444.47347.36750.37253.49356.73660.10763.61267.25971.05475.00479.11683.40087.86292.51197.357102.408107.674113.166118.893124.868131.100137.604144.389151.471158.862166.577174.630183.037191.815200.980210.550220.543230.980241.881abThe enthalpies of dry air and liquid water are set equal to zero at a datum temperature of 0°C.The enthalpy of an unsaturated water vapor-air mixture can be calculated as h = hdry air +(m1/m1,sat)(hsat – hdry air).Sample calculation.
Consider a counterflow unit that is required to cool water from 40 to 26°C whenthe inlet air is at 10°C, 1 atm, and saturated. We will calculate the number of transfer units required forbalanced flow, that is, m˙ G / m˙ L = 1. Equation (4.8.12) is to be integrated numerically, with hG obtainedfrom Equation 4.8.13. The required thermodynamic properties can be obtained from Table 4.8.1. UsingTable 4.8.1, hG,in = hsat (10°C) = 29.15 kJ/kg, hL,out = hL(26°C) = 109.07 kJ/kg. Substituting in Equation(4.8.13),hG = 29.15 + (hL - 109.07)© 1999 by CRC Press LLC4-254Section 4TL, °ChL, kJ/kghG, kJ/kghs, kJ/kghs – hg, kJ/kg1hs - h g2628303234363840109.07117.43125.79134.15142.50150.86159.22167.5829.1537.5145.8754.2362.5870.9479.3087.6679.1287.8697.36107.67118.89131.10144.39158.8649.9750.3551.4953.4456.3160.1665.0971.200.020010.019860.019420.018710.017760.016620.015360.01404Choosing 2°C intervals for convenient numerical integration, the above table is constructed, with hLand hs = hs(TL) also obtained from Table 4.8.1.Using the trapezoidal rule,òhL ,inhL ,outdhL8.36=[0.02001 + 2(0.01986 + 0.01942 + 0.01871 + 0.01776hs - hG2+ 0.01662 + 0.01536) + 0.01404]= 1.043From Equation (4.8.12), Ntu = 1.043.
Also, by using Table 4.8.1, TG,out = 27.9° for saturated outlet air.FIGURE 4.8.9 Schematic of a cross-flow cooling tower packing showing the coordinate system.Thermal-Hydraulic Design of Cooling TowersThe thermal-hydraulic design of a mechanical-draft cooling tower is relatively straightforward.
The flowrate ratio m˙ L / m˙ G can be specified and varied parametrically to obtain an optimal design, for which thesize and cost of the packing is balanced against fan power requirements and operating cost. Data arerequired for mass transfer conductances and friction for candidate packings. Tables 4.8.2a and b givecorrelations for a selection of packings. In Table 4.8.2b, the mass transfer conductance is correlated asgma/L, where a is the transfer area per unit volume and L = m˙ L / A fr is the superficial mass velocity ofthe water flow (also called the water loading on the packing). Similarly, we define G = m˙ G / A fr . Typicalwater loadings are 1.8 to 2.7 kg/m2 sec, and superficial air velocities fall in the range 1.5 to 4 m/sec.No attempt is made to correlate gm and a separately.
The number of transfer units of a packing of heightH is then© 1999 by CRC Press LLC4-255Heat and Mass TransferNtu =g m S g m aH=m˙ LL(4.8.19)TABLE 4.8.2a Packings for Counterflow and Cross-Flow Cooling Towers:Designations and DescriptionsCounterflow Packings1.2.3.4.5.6.7.8.9.10.11.12.Flat asbestos sheets, pitch 4.45 cmFlat asbestos sheets, pitch 3.81 cmFlat asbestos sheets, pitch 3.18 cmFlat asbestos sheets, pitch 2.54 cm60° angle corrugated plastic, Munters M12060, pitch 1.17 in.60° angle corrugated plastic, Munters M19060, pitch 1.8 in.Vertical corrugated plastic, American Tower Plastics Coolfilm, pitch 1.63 in.Horizontal plastic screen, American Tower Plastics Cooldrop, pitch 8 in. 2 in.
gridHorizontal plastic grid, Ecodyne shape 10, pitch 12 in.Angled corrugated plastic, Marley MC67, pitch 1.88 in.Dimpled sheets, Toschi Asbestos-Free Cement, pitch 0.72 in.Vertical plastic honeycomb, Brentwood Industries Accu-Pack, pitch 1.75 in.Cross-Flow Packings1.2.3.4.5.6.7.8.Doron V-bar, 4 ´ 8 in. spacingDoron V-bar, 8 ´ 8 in. spacingEcodyne T-bar, 4 ´ 8 in. spacingEcodyne T-bar, 8 ´ 8 in. spacingWood lath, parallel to air flow, 4 ´ 4 in. spacingWood lath, perpendicular to air flow, 4 ´ 4 in.
spacingMarley a-bar, parallel to air flow, 16 ´ 4 in. spacingMarley ladder, parallel to air flow, 8 ´ 2 in. spacingThe correlations are in terms of dimensionless mass velocities L+ and G+, and a hot water correction+. The hot water correction accounts for a number of factors, such as errors associated with Merkel’sTHWmethod, deviations from low mass transfer rate theory at higher values of Ts, and fluid property dependence on temperature. Frictional resistance to air flow through the packings is correlated as a losscoefficient N = DP/(rV2/2) per unit height or depth of packing, as a function of L+ and G+.
The velocityV is superficial gas velocity. No hot water correction is required.In a natural-draft tower, the thermal and hydraulic performance of the tower are coupled, and the flowrate ratio m˙ L / m˙ G cannot be specified a priori. The buoyancy force producing the air flow depends onthe state of the air leaving the packing which in turn depends on m˙ L / m˙ G and the inlet air and waterstates. An iterative solution is required to find the operating point of the tower. The buoyancy forceavailable to overcome the shell and packing pressure drops is()DP B = g r a - rG,out H(4.8.20)where ra is the ambient air density and H is usually taken as the distance from the bottom of the packingto the top of the shell.
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