The CRC Handbook of Mechanical Engineering. Chapter 4. Heat and Mass Transfer (776127), страница 35
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This equation in nondimensional formis given by111=+NTU ntu h (Ch Cmin ) ntu c (Cc Cmin )(4.5.64)12é 2 g Dp ùGi = ê c úë Deno û i(4.5.65)i = 1, 2whereé f ntu 2 3 æ 1 öæ 11ö11ù- ÷ + 1 - s 2 + KcDeno i = êPr ç ÷ + 2ç- 1 - s 2 - Keúriro úûè rø mè r o ri øêë j hoi(V=)A1A= 2a1 a 2()(4.5.66)(4.5.67)In the iterative solutions, the first time one needs ntuh and ntuc to start the iterations.
These can be eitherdetermined from the past experience or by estimations. If both fluids are gases or both fluids are liquid,one could consider that the design is “balanced,” i.e., that the thermal resistances are distributedapproximately equally on the hot and cold sides. In that case, Ch = Cc, andntu h » ntu c » 2 NTU(4.5.68)Alternatively, if we have liquid on one side and gas on the other side, consider 10% thermal resistanceon the liquid side, i.e.,11 ö=0.10æè UA ø ( ho hA)liq(4.5.69)Then, from Equations (4.5.63) and (4.5.64) with Cgas = Cmin, we can determine the ntu values on eachside as follows:ntu gas = 1.11NTU,ntu liq = 10C * NTU(4.5.70)Also note that initial guesses of ho and j/f are needed for the first iteration to solve Equation (4.5.66).For a good design, consider ho = 0.80 and determine an approximate value of j/f from the plot of j/f vs.© 1999 by CRC Press LLC4-153Heat and Mass TransferRe curve for the known j and f vs.
Re characteristics of each fluid side surface. The specific step-bystep design procedure is as follows:1. In order to compute the fluid bulk mean temperature and the fluid thermophysical properties oneach fluid side, determine the fluid outlet temperatures from the specified heat duty( ) (T˙ pq = mc2.3.4.5.6.7.hh,i) ( ) (T˙ p- Th,o = mccc ,o- Tc,i)(4.5.71)or from the specified exchanger effectiveness using Equation (4.5.52) and (4.5.53). For the firsttime, estimate the values of cp.For exchangers with C* ³ 0.5, the bulk mean temperature on each fluid side will be thearithmetic mean of inlet and outlet temperatures on each side. For exchangers with C* < 0.5, thebulk mean temperature on the Cmax side will be the arithmetic mean of the inlet and outlettemperatures on that side and the bulk mean temperature on the Cmin side will be the log-meanaverage as given by Equation (4.5.54).
With these bulk mean temperatures, determine cp anditerate one more time for the outlet temperatures if warranted. Subsequently, determine m, cp, k,Pr, and r on each fluid side.Calculate C* and e (if q is given), and determine NTU from the e-NTU expression, tables, orgraphical results for the selected flow arrangement (in this case, it is unmixed–unmixed crossflow, Table 4.5.4).
The influence of longitudinal heat conduction, if any, is ignored in the firstiteration since we don’t know the exchanger size yet.Determine ntu on each side by the approximations discussed with Equations (4.5.68) and (4.5.70)unless it can be estimated from past experience.For the selected surfaces on each fluid side, plot j/f vs. Re curve from the given surface characteristics and obtain an approximate value of j/f.
If fins are employed, assume ho = 0.80 unless abetter value can be estimated.Evaluate G from Equation (4.5.65) on each fluid side using the information from Steps 1 to 4and the input value of Dp.Calculate Reynolds number Re, and determine j and f on each fluid side from the given designdata for each surface.Compute h, hf, and ho using Equations (4.5.57) to (4.5.59).
For the first iteration, determine U1on Fluid 1 side from the following equation derived from Equations (4.5.6) and (4.5.67).a aa a111=++ 1 2 + 1 2U1 ( ho h)(ho hs )1 (ho hs )2 (ho h)21(4.5.72)where a1/a2 = A1/A2, a = A/V, V is the exchanger total volume, and subscripts 1 and 2 denoteFluid 1 and 2 sides. For a plate-fin exchanger, a terms are given by Shah (1981) and Kays andLondon (1984):a1 =b1b1b1 + b2 + 2 aa2 =b2b 2b1 + b2 + 2 a(4.5.73)Note that the wall thermal resistance in Equation (4.5.72) is ignored in the first iteration.
In secondand subsequent iterations, compute U1 fromA AA AdA1111=+++ 1 2 + 1 2U1 ( ho h)kAhhhhho h) 2()()(w wo s 1o s 21© 1999 by CRC Press LLC(4.5.74)4-154Section 4where the necessary geometry information A1/A2 and A1/Aw is determined from the geometrycalculated in the previous iteration.8. Now calculate the core dimensions. In the first iteration, use NTU computed in Step 2.
Forsubsequent iterations, calculate longitudinal conduction parameter l (and other dimensionlessgroups for a crossflow exchanger). With known e, C*, and l, determine the correct value of NTUusing either a closed-form equation or tabular/graphical results (Kays and London, 1984). Determine A1 from NTU using U1 from the previous step and known Cmin.A1 = NTU Cmin U1(4.5.75)A2 = ( A2 A1 ) A1 = (a 2 a 1 ) A1(4.5.76)and henceAo from known ṁ and G is given byAo,1 = (m˙ G)1Ao,2 = (m˙ G) 2(4.5.77)A fr ,1 = Ao,1 s1A fr ,2 = Ao,2 s 2(4.5.78)so thatwhere s1 and s2 are generally specified for the surface or can be computed for plate-fin surfacesfrom Shah (1981) and Kays and London (1984):s1 =b1b1 Dh,1 4s2 =b1 + b2 + 2db2b 2 Dh,2 4b1 + b2 + 2d(4.5.79)Now compute the fluid flow lengths on each side (see Figure 4.5.17) from the definition of thehydraulic diameter of the surface employed on each side.æ D AöL1 = ç h ÷è 4 Ao ø 1æ D AöL2 = ç h ÷è 4 Ao ø 2(4.5.80)Since Afr,1 = L2L3 and Afr,2 = L1L3, we can obtainL3 =A fr ,1L2orL3 =A fr ,2L1(4.5.81)Theoretically, L3 calculated from both expressions of Equation (4.5.81) should be identical.
Inreality, they may differ slightly because of the round-off error. In that case, consider an averagevalue for L3.9. Finally, compute the pressure drop on each fluid side, after correcting f factors for variable propertyeffects, in a manner similar to Step 8 of the rating problem for a Crossflow Plate Fin Exchanger.10. If the calculated values of Dp are within and close to input specifications, the solution to thesizing problem is completed. Finer refinements in the core dimensions, such as integer numbersof flow passages, etc., may be carried out at this time. Otherwise, compute the new value of Gon each fluid side using Equation (4.5.29) in which Dp is the input-specified value, and f, Kc, Ke,and geometric dimensions are from the previous iteration.© 1999 by CRC Press LLCHeat and Mass Transfer4-155FIGURE 4.5.17 A single-pass crossflow exchanger.11.
Repeat (iterate) Steps 6 to 10 until both transfer and pressure drops are met as specified. It shouldbe emphasized that since we have imposed no constraints on the exchanger dimensions, the aboveprocedure will yield L1, L2, and L3 for the selected surfaces such that the design will meet exactlythe heat duty and pressure drops on both fluid sides.Flow MaldistributionIn the previously presented heat transfer (e-NTU, MTD, etc. methods) and pressure drop analyses, it ispresumed that the fluid is uniformly distributed through the core.
In practice, flow maldistribution doesoccur to some extent and often severely, and may result in a significant reduction in exchanger heattransfer performance and an increase in the pressure drop. Hence, it may be necessary for the designerto take into account the effect of flow maldistribution causing undesirable performance deterioration upfront while designing a heat exchanger.Some maldistributions are geometry-induced (i.e., the result of exchanger fabrication conditions, suchas header design or manufacturing tolerances, or the duct geometry/structure upstream of the exchanger),and other maldistributions are the result of exchanger operating conditions.
Gross, passage-to-passageand manifold-induced flow maldistributions are examples of the former category, while viscosity, naturalconvection, and density-difference-induced flow maldistributions are of the latter category. Flow maldistributions associated with two-phase and multiphase flow are too complex, with only limited information available in the literature.
The analysis methods and results for some of the above flowmaldistributions for single-phase flows are given by Shah (1985), Mueller and Chiou (1987), and Putnamand Rohsenow (1985).Fouling in Heat ExchangersFouling, Its Effect, and Mechanisms.Fouling refers to undesired accumulation of solid material (by-products of the heat transfer processes)on heat exchanger surfaces which results in additional thermal resistance to heat transfer, thus reducingexchanger performance. The fouling layer also blocks the flow passage/area and increases surfaceroughness, thus either reducing the flow rate in the exchanger or increasing the pressure drop or both.The foulant deposits may be loose such as magnetite particles or hard and tenacious such as calciumcarbonate scale; other deposits may be sediment, polymers, coking or corrosion products, inorganic salts,biological growth, etc.
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