John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 53
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(7.57) becomesε = 1 − exp (−NTU)(7.58)which we could have obtained directly, from either eqn. (3.20) or (3.21),by setting Cmin /Cmax = 0. A heat exchanger for which one stream isisothermal, so that Cmin /Cmax = 0, is sometimes called a single-streamheat exchanger.Equation (7.57) applies to ducts of any cross-sectional shape. We cancast it in terms of the hydraulic diameter, Dh = 4Ac /P , by substitutingHeat transfer surface viewed as a heat exchanger§7.4ṁ = ρuav Ac :Tbout − TbinhP L= 1 − exp −Tw − Tbinρuav cp Ach4L= 1 − exp −ρuav cp Dh(7.59a)(7.59b)For a circular tube, with Ac = π D 2 /4 and P = π D, Dh = 4(π D 2 /4) (π D)= D. To use eqn.
(7.59b) for a noncircular duct, of course, we will needthe value of h for its more complex geometry. We consider this issue inthe next section.Example 7.5Air at 20◦ C is hydrodynamically fully developed as it flows in a 1 cm I.D.pipe. The average velocity is 0.7 m/s. If it enters a section where thepipe wall is at 60◦ C, what is the temperature 0.25 m farther downstream?Solution.ReD =(0.7)(0.01)uav D== 422ν1.66 × 10−5The flow is therefore laminar. To account for the thermal entry region,we compute the Graetz number from eqn. (7.26)Gz =ReD Pr D(422)(0.709)(0.01)== 12.0x0.25Substituting this value into eqn. (7.29), we find NuD = 4.32.
Thus,h=3.657(0.0268)= 11.6 W/m2 K0.01Then, using eqn. (7.59b),Tbout − Tbin11.64(0.25)= 1 − exp −Tw − Tbin1.14(1007)(0.7) 0.01so thatTb − 20= 0.76460 − 20orTb = 50.6◦ C369370Forced convection in a variety of configurations7.5§7.5Heat transfer coefficients for noncircular ductsSo far, we have focused on flows within circular tubes, which are by far themost common configuration.
Nevertheless, other cross-sectional shapesoften occur. For example, the fins of a heat exchanger may form a rectangular passage through which air flows. Sometimes, the passage crosssection is very irregular, as might happen when fluid passes through aclearance between other objects. In situations like these, all the qualitative ideas that we developed in Sections 7.1–7.3 still apply, but theNusselt numbers for circular tubes cannot be used in calculating heattransfer rates.The hydraulic diameter, which was introduced in connection witheqn. (7.59b), provides a basis for approximating heat transfer coefficientsin noncircular ducts.
Recall that the hydraulic diameter is defined asDh ≡4 AcP(7.60)where Ac is the cross-sectional area and P is the passage’s wetted perimeter (Fig. 7.9). The hydraulic diameter measures the fluid area per unitlength of wall. In turbulent flow, where most of the convection resistance is in the sublayer on the wall, this ratio determines the heat transfer coefficient to within about ±20% across a broad range of duct shapes.In fully-developed laminar flow, where the thermal resistance extendsinto the core of the duct, the heat transfer coefficient depends on thedetails of the duct shape, and Dh alone cannot define the heat transfercoefficient. Nevertheless, the hydraulic diameter provides an appropriatecharacteristic length for cataloging laminar Nusselt numbers.Figure 7.9 Flow in a noncircular duct.Heat transfer coefficients for noncircular ducts§7.5The factor of four in the definition of Dh ensures that it gives theactual diameter of a circular tube.
We noted in the preceding sectionthat, for a circular tube of diameter D, Dh = D. Some other importantcases include:a rectangular duct ofwidth a and height ban annular duct ofinner diameter Di andouter diameter DoDh =Dh =2ab4 ab=2a + 2ba+b(7.61a) 4 π Do2 4 − π Di2 4π (Do + Di )= (Do − Di )(7.61b)and, for very wide parallel plates, eqn. (7.61a) with a b givestwo parallel platesa distance b apartDh = 2b(7.61c)Turbulent flow in noncircular ductsWith some caution, we may use Dh directly in place of the circular tubediameter when calculating turbulent heat transfer coefficients and bulktemperature changes.
Specifically, Dh replaces D in the Reynolds number, which is then used to calculate f and NuDh from the circular tubeformulas. The mass flow rate and the bulk velocity must be based on2/4 (seethe true cross-sectional area, which does not usually equal π DhProblem 7.46). The following example illustrates the procedure.Example 7.6An air duct carries chilled air at an inlet bulk temperature of Tbin =17◦ C and a speed of 1 m/s. The duct is made of thin galvanized steel,has a square cross-section of 0.3 m by 0.3 m, and is not insulated.A length of the duct 15 m long runs outdoors through warm air atT∞ = 37◦ C.
The heat transfer coefficient on the outside surface, dueto natural convection and thermal radiation, is 5 W/m2 K. Find thebulk temperature change of the air over this length.Solution. The hydraulic diameter, from eqn. (7.61a) with a = b, issimplyDh = a = 0.3 m371372Forced convection in a variety of configurations§7.5Using properties of air at the inlet temperature (290 K), the Reynoldsnumber isReDh =(1)(0.3)uav Dh== 19, 011ν(1.578 × 10−5 )The Reynolds number for turbulent transition in a noncircular ductis typically approximated by the circular tube value of about 2300, sothis flow is turbulent. The friction factor is obtained from eqn. (7.42)"−2!= 0.02646f = 1.82 log10 (19, 011) − 1.64and the Nusselt number is found with Gnielinski’s equation, (7.43)NuDh =(0.02646/8)(19, 011 − 1, 000)(0.713)3!" = 49.821 + 12.7 0.02646/8 (0.713)2/3 − 1The heat transfer coefficient ish = NuDh(49.82)(0.02623)k= 4.371 W/m2 K=Dh0.3The remaining problem is to find the bulk temperature change.The thin metal duct wall offers little thermal resistance, but convection resistance outside the duct must be considered.
Heat travelsfirst from the air at T∞ through the outside heat transfer coefficientto the duct wall, through the duct wall, and then through the insideheat transfer coefficient to the flowing air — effectively through threeresistances in series from the fixed temperature T∞ to the rising temperature Tb . We have seen in Section 2.4 that an overall heat transfercoefficient may be used to describe such series resistances.
Here, withAinside Aoutside , we find U based on inside area to be−1111+ Rt wall +U=Ainside (hA)inside(hA)outside =11+4.371 5−1neglect= 2.332 W/m2 KWe then adapt eqn. (7.59b) by replacing h by U and Tw by T∞ :Tbout − TbinU4L= 1 − exp −T∞ − Tbinρuav cp Dh4(15)2.332= 0.3165= 1 − exp −(1.217)(1)(1007) 0.3The outlet bulk temperature is thereforeTbout = [17 + (37 − 17)(0.3165)] ◦ C = 23.3 ◦ C§7.5Heat transfer coefficients for noncircular ductsThe results obtained by substituting Dh for D in turbulent circulartube formulæ are generally accurate to within ±20% and are often within±10%.
Worse results are obtained for duct cross-sections having sharpcorners, such as an acute triangle. Specialized equations for “effective”hydraulic diameters have been developed for specific geometries and canimprove the accuracy to 5 or 10% [7.8].When only a portion of the duct cross-section is heated — one wall ofa rectangle, for example — the procedure for finding h is the same. Thehydraulic diameter is based upon the entire wetted perimeter, not simply the heated part. However, in eqn. (7.59a) P is the heated perimeter:eqn. (7.59b) does not apply for nonuniform heating.One situation in which one-sided or unequal heating often occurs isan annular duct, for which the inner tube might be a heating element.The hydraulic diameter procedure will typically predict the heat transfercoefficient on the outer tube to within ±10%, irrespective of the heatingconfiguration.
The heat transfer coefficient on the inner surface, however, is sensitive to both the diameter ratio and the heating configuration.For that surface, the hydraulic diameter approach is not very accurate,especially if Di Do ; other methods have been developed to accuratelypredict heat transfer in annular ducts (see [7.3] or [7.8]).Laminar flow in noncircular ductsLaminar velocity profiles in noncircular ducts develop in essentially thesame way as for circular tubes, and the fully developed velocity profilesare generally paraboloidal in shape.
For example, for fully developedflow between parallel plates located at y = b/2 and y = −b/2, 2 y3u1−4=(7.62)uav2bfor uav the bulk velocity. This should be compared to eqn. (7.15) for acircular tube. The constants and coordinates differ, but the equationsare otherwise identical. Likewise, an analysis of the temperature profilesbetween parallel plates leads to constant Nusselt numbers, which maybe expressed in terms of the hydraulic diameter for various boundaryconditions:⎧⎪⎪⎨7.541 for fixed plate temperatureshDh= 8.235 for fixed flux at both plates(7.63)NuDh =⎪k⎪⎩5.385 one plate fixed flux, one adiabaticSome other cases are summarized in Table 7.4.
Many more have beenconsidered in the literature (see, especially, [7.5]). The latter include373374Forced convection in a variety of configurations§7.6Table 7.4 Laminar, fully developed Nusselt numbers based onhydraulic diameters given in eqn. (7.61)Cross-sectionTw fixedqw fixedCircularSquareRectangulara = 2ba = 4ba = 8bParallel plates3.6572.9764.3643.6083.3914.4395.5977.5414.1235.3316.4908.235different wall boundary conditions and a wide variety cross-sectionalshapes, both practical and ridiculous: triangles, circular sectors, trapezoids, rhomboids, hexagons, limaçons, and even crescent moons! Theboundary conditions, in particular, should be considered when the ductis small (so that h will be large): if the conduction resistance of the tubewall is comparable to the convective resistance within the duct, then temperature or flux variations around the tube perimeter must be expected.This will significantly affect the laminar Nusselt number.
The rectangular duct values in Table 7.4 for fixed wall flux, for example, assume auniform temperature around the perimeter of the tube, as if the wall hasno conduction resistance around its perimeter. This might be true for acopper duct heated at a fixed rate in watts per meter of duct length.Laminar entry length formulæ for noncircular ducts are also given byShah and London [7.5].7.6Heat transfer during cross flow over cylindersFluid flow patternIt will help us to understand the complexity of heat transfer from bodiesin a cross flow if we first look in detail at the fluid flow patterns that occurin one cross-flow configuration—a cylinder with fluid flowing normal toit. Figure 7.10 shows how the flow develops as Re ≡ u∞ D/ν is increasedfrom below 5 to near 107 . An interesting feature of this evolving flowpattern is the fairly continuous way in which one flow transition follows§7.6Heat transfer during cross flow over cylindersFigure 7.10 Regimes of fluid flow across circular cylinders [7.24].375376Forced convection in a variety of configurations§7.6Figure 7.11 The Strouhal–Reynolds number relationship forcircular cylinders, as defined by existing data [7.24].another.
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