John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 17
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A 1980 Chevette radiator. Cross-flow exchanger with neither flow mixed. Edges of flat vertical tubes can be seen.c. The basic 1 ft. × 1 ft.× 2 ft. module for a waste heat recuperator. It isa plate-fin, gas-to-air cross-flow heatexchanger with neither flow mixed.b. A section of an automotive air conditioning condenser. The flow through the horizontal wavy fins is allowed to mix with itselfwhile the two-pass flow through the U-tubesremains unmixed.Figure 3.6 Several commercial cross-flow heat exchangers.(Photographs courtesy of Harrison Radiator Division, GeneralMotors Corporation.)105Figure 3.7 Four typical heat exchanger configurations (continued on next page). (Drawings courtesy of the Tubular Exchanger Manufacturers’ Association.)106§3.2Evaluation of the mean temperature difference in a heat exchangerFigure 3.7 ContinuedOur problem then reduces to finding the appropriate mean temperaturedifference that will make this equation true.
Let us do this for the simpleparallel and counterflow configurations, as sketched in Fig. 3.8.The temperature of both streams is plotted in Fig. 3.8 for both singlepass arrangements—the parallel and counterflow configurations—as afunction of the length of travel (or area passed over). Notice that, in theparallel-flow configuration, temperatures tend to change more rapidlywith position and less length is required. But the counterflow arrangement achieves generally more complete heat exchange from one flow tothe other.Figure 3.9 shows another variation on the single-pass configuration.This is a condenser in which one stream flows through with its tempera-107108Heat exchanger design§3.2Figure 3.8 The temperature variation through single-passheat exchangers.ture changing, but the other simply condenses at uniform temperature.This arrangement has some special characteristics, which we point outshortly.The determination of ∆Tmean for such arrangements proceeds as follows: the differential heat transfer within either arrangement (see Fig.
3.8)isdQ = U ∆T dA = −(ṁcp )h dTh = ±(ṁcp )c dTc(3.2)where the subscripts h and c denote the hot and cold streams, respectively; the upper and lower signs are for the parallel and counterflowcases, respectively; and dT denotes a change from left to right in theexchanger. We give symbols to the total heat capacities of the hot andcold streams:Ch ≡ (ṁcp )h W/KandCc ≡ (ṁcp )c W/K(3.3)Thus, for either heat exchanger, ∓Ch dTh = Cc dTc .
This equation canbe integrated from the lefthand side, where Th = Thin and Tc = Tcin for§3.2Evaluation of the mean temperature difference in a heat exchangerFigure 3.9 The temperature distribution through a condenser.parallel flow or Th = Thin and Tc = Tcout for counterflow, to some arbitrarypoint inside the exchanger. The temperatures inside are thus:CcQ(Tc − Tcin ) = Thin −ChChCcQ−(Tcout − Tc ) = Thin −ChChparallel flow:Th = Thin −counterflow:Th = Thin(3.4)where Q is the total heat transfer from the entrance to the point of interest.
Equations (3.4) can be solved for the local temperature differences:CcCc∆Tparallel = Th − Tc = Thin − 1 +Tc +TcChCh inCcCcTc −∆Tcounter = Th − Tc = Thin − 1 −TcChCh out(3.5)109110Heat exchanger designSubstitution of these in dQ = Cc dTc = U ∆T dA yieldsU dA dTc= CcCcCc parallel− 1+Tc +Tc + ThinChCh indTcU dA = CCccCc counter− 1−Tc −Tc + ThinChCh outEquations (3.6) can be integrated across the exchanger: Tc outAUdTcdA =Tc in [− − −]0 CcIf U and Cc can be treated as constant, this integration gives⎡ ⎤CcCcTcout +Tc + Thin ⎥⎢− 1 +UAChCh in⎢⎥⎥ =−parallel: ln ⎢ 1+CcCc⎣⎦CcTcin +− 1+Tc + ThinChCh in⎤⎡ CcCcT−T+T−1−coutchin ⎥⎢UAChCh out⎥⎢ ⎥ =−1−counter: ln ⎢CcCc⎦⎣CcTcin −− 1−Tcout + ThinChCh§3.2(3.6)(3.7)CcChCcCh(3.8)If U were variable, the integration leading from eqn. (3.7) to eqns.
(3.8)is where its variability would have to be considered. Any such variabilityof U can complicate eqns. (3.8) terribly. Presuming that eqns. (3.8) arevalid, we can simplify them with the help of the definitions of ∆Ta and∆Tb , given in Fig. 3.8:(1 + Cc /Ch )(Tcin − Tcout ) + ∆Tb11= −U A+parallel: ln∆TbCcCh∆Ta11counter: ln= −U A−(−1 + Cc /Ch )(Tcin − Tcout ) + ∆TaCcCh(3.9)Conservation of energy (Qc = Qh ) requires thatTh − ThinCc= − outChTcout − Tcin(3.10)§3.2Evaluation of the mean temperature difference in a heat exchangerThen eqn. (3.9) and eqn.
(3.10) give⎤⎡∆Ta −∆Tb⎥⎢⎢ (Tcin − Tcout ) + (Thout − Thin ) +∆Tb ⎥⎥parallel: ln ⎢⎥⎢∆Tb⎦⎣= lncounter:ln∆Ta∆Tb − ∆Ta + ∆Ta= ln∆Ta∆Tb∆Ta∆Tb= −U A= −U A11+CcCh11−CcCh(3.11)Finally, we write 1/Cc = (Tcout − Tcin )/Q and 1/Ch = (Thin − Thout )/Q onthe right-hand side of either of eqns. (3.11) and get for either parallel orcounterflow,Q = UA∆Ta − ∆Tbln(∆Ta /∆Tb )(3.12)The appropriate ∆Tmean for use in eqn. (3.11) is thus the logarithmic meantemperature difference (LMTD):∆Tmean = LMTD ≡∆Ta − ∆Tb∆Taln∆Tb(3.13)Example 3.1The idea of a logarithmic mean difference is not new to us. We havealready encountered it in Chapter 2. Suppose that we had asked,“What mean radius of pipe would have allowed us to compute theconduction through the wall of a pipe as though it were a slab ofthickness L = ro − ri ?” (see Fig.
3.10). To answer this, we comparermean∆T= 2π kl∆TQ = kALro − riwith eqn. (2.21):Q = 2π kl∆T1ln(ro /ri )111112Heat exchanger design§3.2Figure 3.10 Calculation of the mean radius for heat conduction through a pipe.It follows thatrmean =ro − ri= logarithmic mean radiusln(ro /ri )Example 3.2Suppose that the temperature difference on either end of a heat exchanger, ∆Ta , and ∆Tb , are equal. Clearly, the effective ∆T must equal∆Ta and ∆Tb in this case. Does the LMTD reduce to this value?Solution. If we substitute ∆Ta = ∆Tb in eqn. (3.13), we getLMTD =∆Tb − ∆Tb0= = indeterminateln(∆Tb /∆Tb )0Therefore it is necessary to use L’Hospital’s rule:limit∆Ta →∆Tb∂(∆Ta − ∆Tb )∂∆Ta∆Ta − ∆Tb∆Ta =∆Tb=ln(∆Ta /∆Tb )∂∆Ta ln∂∆Ta∆Tb ∆T =∆Tab1== ∆Ta = ∆Tb1/∆Ta ∆Ta =∆TbEvaluation of the mean temperature difference in a heat exchanger§3.2It follows that the LMTD reduces to the intuitively obvious result inthe limit.Example 3.3Water enters the tubes of a small single-pass heat exchanger at 20◦ Cand leaves at 40◦ C.
On the shell side, 25 kg/min of steam condenses at60◦ C. Calculate the overall heat transfer coefficient and the requiredflow rate of water if the area of the exchanger is 12 m2 . (The latentheat, hfg , is 2358.7 kJ/kg at 60◦ C.)Solution.Q = ṁcondensate · hfg60◦ C=25(2358.7)= 983 kJ/s60and with reference to Fig. 3.9, we can calculate the LMTD withoutnaming the exchanger “parallel” or “counterflow”, since the condensate temperature is constant.LMTD =(60 − 20) − (60 − 40)= 28.85 K60 − 20ln60 − 40ThenQA(LMTD)983(1000)= 2839 W/m2 K=12(28.85)U=andṁH2 O =983, 000Q== 11.78 kg/scp ∆T4174(20)Extended use of the LMTDLimitations.
There are two basic limitations on the use of an LMTD.The first is that it is restricted to the single-pass parallel and counterflow configurations. This restriction can be overcome by adjusting theLMTD for other configurations—a matter that we take up in the followingsubsection.113Heat exchanger design114§3.2Figure 3.11 A typical case of a heat exchanger in which Uvaries dramatically.The second limitation—our use of a constant value of U — is moreserious.
The value of U must be negligibly dependent on T to completethe integration of eqn. (3.7). Even if U ≠ fn(T ), the changing flow configuration and the variation of temperature can still give rise to seriousvariations of U within a given heat exchanger. Figure 3.11 shows a typical situation in which the variation of U within a heat exchanger mightbe great. In this case, the mechanism of heat exchange on the water sideis completely altered when the liquid is finally boiled away. If U wereuniform in each portion of the heat exchanger, then we could treat it astwo different exchangers in series.However, the more common difficulty that we face is that of designing heat exchangers in which U varies continuously with position withinit.
This problem is most severe in large industrial shell-and-tube configurations1 (see, e.g., Fig. 3.5 or Fig. 3.12) and less serious in compact heatexchangers with less surface area. If U depends on the location, analysessuch as we have just completed A[eqn. (3.1) to eqn. (3.13)] must be doneusing an average U defined as 0 U dA/A.1Actual heat exchangers can have areas well in excess of 10,000 m2 . Large powerplant condensers and other large exchangers are often remarkably big pieces of equipment.Figure 3.12 The heat exchange surface for a steam generator.
This PFT-type integral-furnace boiler, with a surface areaof 4560 m2 , is not particularly large. About 88% of the areais in the furnace tubing and 12% is in the boiler (Photographcourtesy of Babcock and Wilcox Co.)115116Heat exchanger design§3.2LMTD correction factor, F. Suppose that we have a heat exchanger inwhich U can reasonably be taken constant, but one that involves suchconfigurational complications as multiple passes and/or cross-flow. Insuch cases it is necessary to rederive the appropriate mean temperaturedifference in the same way as we derived the LMTD. Each configurationmust be analyzed separately and the results are generally more complicated than eqn. (3.13).This task was undertaken on an ad hoc basis during the early twentieth century.
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