The CRC Handbook of Mechanical Engineering. Chapter 4. Heat and Mass Transfer (776127), страница 34
Текст из файла (страница 34)
Shah (1988a) presented further refinements in these procedures as well as step-by-stepprocedures for two-pass cross-counterflow plate-fin exchangers, and single-pass crossflow and two-passcross-counterflow tube-fin exchangers. Also, step-by-step solution procedures for the rating and sizingproblems for rotary regenerators (Shah, 1988b), heat pipe heat exchangers (Shah and Giovannelli, 1988)and PHEs (Shah and Wanniarachchi, 1991) are available. As an illustration, the step-by-step solutionprocedures will be covered here for a single-pass crossflow exchanger.Rating Problem for a Crossflow Plate-Fin Exchanger.Following is a step-by-step procedure for rating a crossflow plate-fin exchanger. Inputs to the ratingproblem for a two-fluid exchanger are the exchanger construction, flow arrangement and overall dimensions, complete details on the materials and surface geometries on both sides including their nondimensional heat transfer and pressure drop characteristics (j and f vs.
Re), fluid flow rates, inlet temperatures,and fouling factors. The fluid outlet temperatures, total heat transfer rate, and pressure drops on eachside of the exchanger are then determined as the rating problem solution.1. Determine the surface geometric properties on each fluid side. This includes the minimum freeflow area Ao, heat transfer surface area A (both primary and secondary), flow lengths L, hydraulicdiameter Dh, heat transfer surface area density b, the ratio of minimum free-flow area to frontalarea s, fin length lf , and fin thickness d for fin efficiency determination, and any specializeddimensions used for heat transfer and pressure drop correlations.2. Compute the fluid bulk mean temperature and fluid thermophysical properties on each fluid side.Since the outlet temperatures are not known for the rating problem, they are estimated initially.Unless it is known from past experience, assume an exchanger effectiveness as 60 to 75% formost single-pass crossflow exchangers, or 80 to 85% for single-pass counterflow exchangers.
Forthe assumed effectiveness, calculate the fluid outlet temperatures.()(4.5.52)()(4.5.53)Th, o = Th, i - e(Cmin Ch ) Th, i - Tc, iTc, o = Tc, i - e(Cmin Cc ) Th, i - Tc, iInitially, assume Cc/Ch = m˙ c / m˙ h for a gas-to-gas exchanger, or Cc/Ch = m˙ c c p,c / m˙ h c p,h for a gasto-liquid exchanger with very approximate values of cp for the fluids in question.For exchangers with C* > 0.5 (usually gas-to-gas exchangers), the bulk mean temperatures oneach fluid side will be the arithmetic mean of the inlet and outlet temperatures on each fluid side(Shah, 1981). For exchangers with C* > 0.5 (usually gas-to-gas exchangers), the bulk meantemperature on the Cmax side will be the arithmetic mean of inlet and outlet temperatures; thebulk mean temperature on the Cmin side will be the log-mean average temperature obtained asfollows:Tm,Cmin = Tm,Cmax ± DTlm(4.5.54)where DTlm is the LMTD based on the terminal temperatures (see Equation 4.5.19).
Use the plussign if the Cmin side is hot; otherwise, use the negative sign.Once the bulk mean temperature is obtained on each fluid side, obtain the fluid properties fromthermophysical property books or from handbooks. The properties needed for the rating problemare m, cp, k, Pr, and r. With this cp, one more iteration may be carried out to determine Th,o or Tc,ofrom Equation 4.5.52 or 4.5.53 on the Cmax side, and subsequently Tm on the Cmax side, and refinefluid properties accordingly.© 1999 by CRC Press LLC4-150Section 43. Calculate the Reynolds number Re = GDh/m and/or any other pertinent dimensionless groups(from the basic definitions) needed to determine the nondimensional heat transfer and flow frictioncharacteristics (e.g., j or Nu and f) of heat transfer surfaces on each side of the exchanger.Subsequently, compute j or Nu and f factors.
Correct Nu (or j) for variable fluid property effects(Shah, 1981) in the second and subsequent iterations from the following equations.For gases:For liquids:éT ùNu=ê wúNu cp ë Tm ûn¢ém ùNu=ê wúNu cp ë m m ûéT ùf=ê wúfcp ë Tm ûn¢m¢( 4.5.55)ém ùf=ê wúfcp ë m m ûm¢( 4.5.56)where the subscript cp denotes constant properties, and m¢ and n¢ are empirical constants providedin Table 4.5.7. Note that Tw and Tm in Equations (4.5.55) and (4.5.56) and in Tables 4.5.7a and band are absolute temperatures.TABLE 4.5.7a Property Ratio Method Exponents of Equations (4.5.55) and(4.5.56) for Laminar FlowFluidHeatingGasesn¢ = 0.00, m¢ = 1.00for 1 < Tw /Tm < 3n¢ = –0.14, m¢ = 0.58for mw/mm < 1LiquidsCoolingn¢ = 0.0, m¢ =0.81for 0.5 < Tw /Tm < 1n¢ = –0.14, m¢ = 0.54for mw/mm > 1Source: Shah, R.K., in Heat Exchangers: Thermal-Hydraulic Fundamentals and Design,S.
Kakaç et al., Eds., Hemisphere Publishing, Washington, D.C., 1981. With permission.TABLE 4.5.7b Property Ratio Method Correlations of Exponents ofEquations (4.5.55) and (4.5.56) for Turbulent FlowFluidGasesLiquidsHeatingNu = 5 + 0.012 Re0.83 (Pr + 0.29) (Tw/Tm)nn = –[log10(Tw/Tm)]1/4 + 0.3for 1 < Tw/Tm < 5, 0.6 < Pr < 0.9,104< Re < 106, and L/Dh > 40m¢ = –0.1for 1 < Tw/Tm < 2.4n¢ = –0.11afor 0.08 < mw/mm < 1f/fcp = (7 – mm/mw)/6b or m¢ = 0.25for 0.35 < mw/mm < 1Coolingn¢ = 0m¢ = –0.1 (tentative)n¢ = –0.25afor 1 < mw/mm < 40m¢ = 0.24bfor 1 < mw/mm < 2aValid for 2 £ Pr £ 140, 104 £ Re £ 1.25 ´ 105.Valid for 1.3 £ Pr £ 10, 104 £ Re £ 2.3 ´ 105.Source: Shah, R.K., in Heat Exchangers: Thermal-Hydraulic Fundamentals andDesign, S. Kakaç et al., Eds., Hemisphere Publishing, Washington, D.C., 1981.With permission.b4.
From Nu or j, compute the heat transfer coefficients for both fluid streams.h = Nu k Dh = jGc p Pr -2 3Subsequently, determine the fin efficiency hf and the extended surface efficiency ho© 1999 by CRC Press LLC(4.5.57)4-151Heat and Mass Transferhf =tanh mlmlwherem2 =hP˜k f Ak(4.5.58)where P̃ is the wetted perimeter of the fin surface.ho = 1 -AfA(1 - h )f(4.5.59)Also calculate the wall thermal resistance Rw = d/Awkw. Finally, compute the overall thermalconductance UA from Equation (4.5.6) knowing the individual convective film resistances, wallthermal resistances, and fouling resistances, if any.5. From the known heat capacity rates on each fluid side, compute C* = Cmin/Cmax. From the knownUA, determine NTU = UA/Cmin.
Also calculate the longitudinal conduction parameter l. With theknown NTU, C*, l, and the flow arrangement, determine the exchanger effectiveness e from eitherclosed-form equations of Table 4.5.4 or tabular/graphical results from Kays and London (1984).6. With this e, finally compute the outlet temperatures from Equations (4.5.52) and (4.5.53). If theseoutlet temperatures are significantly different from those assumed in Step 2, use these outlettemperatures in Step 2 and continue iterating Steps 2 to 6, until the assumed and computed outlettemperatures converge within the desired degree of accuracy.
For a gas-to-gas exchanger, mostprobably one or two iterations will be sufficient.7. Finally, compute the heat duty from(q = eCmin Th,i - Tc,i)(4.5.60)8. For the pressure drop calculations, first we need to determine the fluid densities at the exchangerinlet and outlet (ri and ro) for each fluid. The mean specific volume on each fluid side is thencomputed from Equation (4.5.30).Next, the entrance and exit loss coefficients, Kc and Ke, are obtained from Figure 4.5.14 forknown s, Re, and the flow passage entrance geometry.The friction factor on each fluid side is corrected for variable fluid properties using Equation(4.5.55) or (4.5.56). Here, the wall temperature Tw is computed fromTw,h = Tm,h - Rh + Rs,h q()(4.5.61)()(4.5.62)Tw,c = Tm,c + Rc + Rs,c qwhere the various resistance terms are defined by Equation (4.5.6).The core pressure drops on each fluid side are then calculated from Equation (4.5.29).
This thencompletes the procedure for solving the rating problem.Sizing Problem for a Crossflow Plate-Fin Exchanger.As defined earlier, we will concentrate here to determine the physical size (length, width, and height)of a single-pass crossflow exchanger for specified heat duty and pressure drops. More specifically, inputsto the sizing problem are surface geometries (including their nondimensional heat transfer and pressuredrop characteristics), fluid flow rates, inlet and outlet fluid temperatures, fouling factors, and pressuredrops on each side.For the solution to this problem, there are four unknowns — two flow rates or Reynolds numbers (todetermine correct heat transfer coefficients and friction factors) and two surface areas — for the two-© 1999 by CRC Press LLC4-152Section 4fluid crossflow exchanger. The following four equations (Equations (4.5.63), (4.5.65), and (4.5.67) areused to solve iteratively the surface areas on each fluid side: UA in Equation (4.5.63) is determined fromNTU computed from the known heat duty or e and C*; G in Equation (4.5.65) represents two equations,for Fluids 1 and 2 (Shah, 1988a); and the volume of the exchanger in Equation (4.5.67) is the samebased on the surface area density of Fluid 1 or Fluid 2.111»+UA ( ho hA)hAh()coh(4.5.63)Here we have neglected the wall and fouling thermal resistances.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
