John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 18
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In 1940, Bowman, Mueller and Nagle [3.2] organized suchcalculations for the common range of heat exchanger configurations. Ineach case they wrote⎞⎛⎟⎜⎜ Ttout − Ttin Tsin − Tsout ⎟⎟⎜,Q = U A(LMTD) · F ⎜Ttin Ttout − Ttin ⎟⎝Tsin − ⎠P(3.14)Rwhere Tt and Ts are temperatures of tube and shell flows, respectively.The factor F is an LMTD correction that varies from unity to zero, depending on conditions. The dimensionless groups P and R have the followingphysical significance:• P is the relative influence of the overall temperature difference(Tsin − Ttin ) on the tube flow temperature. It must obviously beless than unity.• R, according to eqn.
(3.10), equals the heat capacity ratio Ct /Cs .• If one flow remains at constant temperature (as, for example, inFig. 3.9), then either P or R will equal zero. In this case the simpleLMTD will be the correct ∆Tmean and F must go to unity.The factor F is defined in such a way that the LMTD should always becalculated for the equivalent counterflow single-pass exchanger with thesame hot and cold temperatures.
This is explained in Fig. 3.13.Bowman et al. [3.2] summarized all the equations for F , in various configurations, that had been dervied by 1940. They presented them graphically in not-very-accurate figures that have been widely copied. The TEMA[3.1] version of these curves has been recalculated for shell-and-tube heatexchangers, and it is more accurate. We include two of these curves inFig. 3.14(a) and Fig. 3.14(b). TEMA presents many additional curves formore complex shell-and-tube configurations. Figures 3.14(c) and 3.14(d)§3.2Evaluation of the mean temperature difference in a heat exchangerFigure 3.13 The basis of the LMTD in a multipass exchanger,prior to correction.are the Bowman et al. curves for the simplest cross-flow configurations.Gardner and Taborek [3.3] redeveloped Fig.
3.14(c) over a different rangeof parameters. They also showed how Fig. 3.14(a) and Fig. 3.14(b) mustbe modified if the number of baffles in a tube-in-shell heat exchanger islarge enough to make it behave like a series of cross-flow exchangers.We have simplified Figs. 3.14(a) through 3.14(d) by including curvesonly for R 1. Shamsundar [3.4] noted that for R > 1, one may obtain Fusing a simple reciprocal rule. He showed that so long as a heat exchanger has a uniform heat transfer coefficient and the fluid properties areconstant,F (P , R) = F (P R, 1/R)(3.15)Thus, if R is greater than unity, one need only evaluate F using P R inplace of P and 1/R in place of R.Example 3.45.795 kg/s of oil flows through the shell side of a two-shell pass, four-117a.
F for a one-shell-pass, four, six-, . . . tube-pass exchanger.b. F for a two-shell-pass, four or more tube-pass exchanger.Figure 3.14 LMTD correction factors, F , for multipass shelland-tube heat exchangers and one-pass cross-flow exchangers.118c. F for a one-pass cross-flow exchanger with both passes unmixed.d. F for a one-pass cross-flow exchanger with one pass mixed.Figure 3.14 LMTD correction factors, F , for multipass shelland-tube heat exchangers and one-pass cross-flow exchangers.119Heat exchanger design120§3.3tube-pass oil cooler. The oil enters at 181◦ C and leaves at 38◦ C. Waterflows in the tubes, entering at 32◦ C and leaving at 49◦ C.
In addition,cpoil = 2282 J/kg·K and U = 416 W/m2 K. Find how much area theheat exchanger must have.Solution.LMTD ==R=(Thin − Tcout ) − (Thout − Tcin )Thin − TcoutlnThout − Tcin(181 − 49) − (38 − 32)= 40.76 K181 − 49ln38 − 32181 − 38= 8.41249 − 32P=49 − 32= 0.114181 − 32Since R > 1, we enter Fig. 3.14(b) using P = 8.412(0.114) = 0.959 andR = 1/8.412 = 0.119 and obtain F = 0.92.2 It follows that:Q = U AF (LMTD)5.795(2282)(181 − 38) = 416(A)(0.92)(40.76)A = 121.2 m23.3Heat exchanger effectivenessWe are now in a position to predict the performance of an exchanger oncewe know its configuration and the imposed differences. Unfortunately,we do not often know that much about a system before the design iscomplete.Often we begin with information such as is shown in Fig.
3.15. Ifwe sought to calculate Q in such a case, we would have to do so byguessing an exit temperature such as to make Qh = Qc = Ch ∆Th =Cc ∆Tc . Then we could calculate Q from U A(LMTD) or UAF (LMTD) andcheck it against Qh . The answers would differ, so we would have to guessnew exit temperatures and try again.Such problems can be greatly simplified with the help of the so-calledeffectiveness-NTU method.
This method was first developed in full detail2Notice that, for a 1 shell-pass exchanger, these R and P lines do not quite intersect[see Fig. 3.14(a)]. Therefore, one could not obtain these temperatures with any singleshell exchanger.Heat exchanger effectiveness§3.3121Figure 3.15 A design problem in which the LMTD cannot becalculated a priori.by Kays and London [3.5] in 1955, in a book titled Compact Heat Exchangers.
We should take particular note of the title. It is with compact heatexchangers that the present method can reasonably be used, since theoverall heat transfer coefficient is far more likely to remain fairly uniform.The heat exchanger effectiveness is defined asε≡Ch (Thin − Thout )Cc (Tcout − Tcin )=Cmin (Thin − Tcin )Cmin (Thin − Tcin )(3.16)where Cmin is the smaller of Cc and Ch . The effectiveness can be interpreted asε=actual heat transferredmaximum heat that could possibly betransferred from one stream to the otherIt follows thatQ = εCmin (Thin − Tcin )(3.17)A second definition that we will need was originally made by E.K.W.Nusselt, whom we meet again in Part III.
This is the number of transferunits (NTU):NTU ≡UACmin(3.18)122Heat exchanger design§3.3This dimensionless group can be viewed as a comparison of the heatcapacity of the heat exchanger, expressed in W/K, with the heat capacityof the flow.We can immediately reduce the parallel-flow result from eqn. (3.9) tothe following equation, based on these definitions: CcCminCminCminNTU = ln − 1 +ε++1(3.19)−CcChChCcWe solve this for ε and, regardless of whether Cmin is associated with thehot or cold flow, obtain for the parallel single-pass heat exchanger:1 − exp [−(1 + Cmin /Cmax )NTU]Cminε≡= fn, NTU only(3.20)1 + Cmin /CmaxCmaxThe corresponding expression for the counterflow case isε=1 − exp [−(1 − Cmin /Cmax )NTU]1 − (Cmin /Cmax ) exp[−(1 − Cmin /Cmax )NTU](3.21)Equations (3.20) and (3.21) are given in graphical form in Fig.
3.16.Similar calculations give the effectiveness for the other heat exchangerconfigurations (see [3.5] and Problem 3.38), and we include some of theresulting effectiveness plots in Fig. 3.17. To see how the effectivenesscan conveniently be used to complete a design, consider the followingtwo examples.Example 3.5Consider the following parallel-flow heat exchanger specification:cold flow enters at 40◦ C:◦hot flow enters at 150 C:2A = 30 mCc = 20, 000 W/KCh = 10, 000 W/KU = 500 W/m2 K.Determine the heat transfer and the exit temperatures.Solution. In this case we do not know the exit temperatures, so itis not possible to calculate the LMTD. Instead, we can go either to theparallel-flow effectiveness chart in Fig.
3.16 or to eqn. (3.20), usingNTU =500(30)UA= 1.5=Cmin10, 000Cmin= 0.5CmaxHeat exchanger effectiveness§3.3Figure 3.16 The effectiveness of parallel and counterflow heatexchangers. (Data provided by A.D. Kraus.)and we obtain ε = 0.596. Now from eqn. (3.17), we find thatQ = ε Cmin (Thin − Tcin ) = 0.596(10, 000)(110)= 655, 600 W = 655.6 kWFinally, from energy balances such as are expressed in eqn. (3.4), wegetQ655, 600= 84.44◦ C= 150 −Ch10, 000Q655, 600+= 40 += 72.78◦ CCc20, 000Thout = Thin −Tcout = TcinExample 3.6Suppose that we had the same kind of exchanger as we consideredin Example 3.5, but that the area remained unspecified as a designvariable.
Then calculate the area that would bring the hot flow out at90◦ C.Solution. Once the exit cold fluid temperature is known, the problem can be solved with equal ease by either the LMTD or the effective-123Figure 3.17 The effectiveness of some other heat exchangerconfigurations. (Data provided by A.D. Kraus.)124Heat exchanger effectiveness§3.3125ness approach.Tcout = Tcin +Ch1(Thin − Thout ) = 40 + (150 − 90) = 70◦ CCc2Then, using the effectiveness method,ε=Ch (Thin − Thout )10, 000(150 − 90)== 0.5455Cmin (Thin − Tcin )10, 000(150 − 40)so from Fig.
3.16 we read NTU 1.15 = U A/Cmin . ThusA=10, 000(1.15)= 23.00 m2500We could also have calculated the LMTD:LMTD =(150 − 40) − (90 − 70)= 52.79 Kln(110/20)so from Q = U A(LMTD), we obtainA=10, 000(150 − 90)= 22.73 m2500(52.79)The answers differ by 1%, which reflects graph reading inaccuracy.When the temperature of either fluid in a heat exchanger is uniform,the problem of analyzing heat transfer is greatly simplified.
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