John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 52
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(7.41) or (7.43). For liquids, one then corrects by multiplying with a viscosity ratio. Over the interval 0.025 ≤ (µb /µw ) ≤ 12.5,⎧n⎨0.11 for Tw > TµbbNuD = NuDwhere n =(7.44)⎩0.25 for Tw < TbTb µwFor gases a ratio of temperatures in kelvins is used, with 0.27 ≤ (Tb /Tw ) ≤2.7,⎧⎨0.47 for Tw > T T nbbwhere n =(7.45)NuD = NuD⎩0Tb Twfor Tw < TbAfter eqn. (7.42) is used to calculate NuD , it should also be correctedfor the effect of variable viscosity. For liquids, with 0.5 ≤ (µb /µw ) ≤ 3⎧⎪⎨(7 − µb /µw )/6 for Tw > Tbwhere K =(7.46)f = f × K⎪Tb⎩(µb /µw )−0.24for Tw < TbFor gases, the data are much weaker [7.15, 7.16].
For 0.14 ≤ (Tb /Tw ) ≤3.3⎧ T m⎨0.23 for Tw > Tbbf = fwhere m ≈(7.47)⎩0.23 for Tw < TbTb Tw362Forced convection in a variety of configurations§7.3Example 7.3A 21.5 kg/s flow of water is dynamically and thermally developed ina 12 cm I.D. pipe. The pipe is held at 90◦ C and ε/D = 0. Find h andf where the bulk temperature of the fluid has reached 50◦ C.Solution.uav =ṁ21.5== 1.946 m/sρAc977π (0.06)2ReD =1.946(0.12)uav D== 573, 700ν4.07 × 10−7soandPr = 2.47,5.38 × 10−4µb== 1.74µw3.10 × 10−4From eqn. (7.42), f = 0.0128 at Tb , and since Tw > Tb , n = 0.11 ineqn.
(7.44). Thus, with eqn. (7.41) we haveNuD =(0.0128/8)(5.74 × 105 )(2.47)3 (1.74)0.11 = 16171.07 + 12.7 0.0128/8 2.472/3 − 1orh = NuD0.661k= 1617= 8, 907 W/m2 KD0.12The corrected friction factor, with eqn. (7.46), isf = (0.0128) (7 − 1.74)/6 = 0.0122Rough-walled pipes. Roughness on a pipe wall can disrupt the viscousand thermal sublayers if it is sufficiently large. Figure 7.6 shows the effectof increasing root-mean-square roughness height ε on the friction factor,f . As the Reynolds number increases, the viscous sublayer becomesthinner and smaller levels of roughness influence f .
Some typical piperoughnesses are given in Table 7.3.The importance of a given level of roughness on friction and heattransfer can determined by comparing ε to the sublayer thickness. Wesaw in Sect. 6.7 that the thickness of the sublayer is around 30 timesTurbulent pipe flow§7.3363Table 7.3 Typical wall roughness of commercially availablepipes when new.Pipeε (µm)GlassDrawn tubingSteel or wrought ironPipe0.311.546.Asphalted cast ironGalvanized ironCast ironε (µm)120.150.260.3ν/u∗ , where u∗ = τw /ρ was the friction velocity. We can define theratio of ε and ν/u∗ as the roughness Reynolds number, Reε2u∗ εεf= ReD(7.48)Reε ≡νD8where the second equality follows from the definitions of u∗ and f (anda little algebra). Experimental data then show that the smooth, transitional, and fully rough regions seen in Fig.
7.6 correspond to the followingranges of Reε :Reε < 5hydraulically smooth5 ≤ Reε ≤ 70transitionally rough70 < Reεfully roughIn the fully rough regime, Bhatti and Shah [7.8] provide the followingcorrelation for the local Nusselt numberNuD =(f /8) ReD Pr40.51 + f /8 4.5 Re0.2− 8.48ε Pr(7.49)which applies for the ranges104 ReD ,0.5 Pr 10,and 0.002 ε 0.05DThe corresponding friction factor may be computed from Haaland’s equation [7.17]:f =51.8 log101126.9ε/D 1.11+ReD3.7(7.50)364Forced convection in a variety of configurations§7.3The heat transfer coefficient on a rough wall can be several timesthat for a smooth wall at the same Reynolds number. The friction factor, and thus the pressure drop and pumping power, will also be higher.Nevertheless, designers sometimes deliberately roughen tube walls so asto raise h and reduce the surface area needed for heat transfer.
Several manufacturers offer tubing that has had some pattern of roughnessimpressed upon its interior surface. Periodic ribs are one common configuration. Specialized correlations have been developed for a numberof such configurations [7.18, 7.19].Example 7.4Repeat Example 7.3, now assuming the pipe to be cast iron with a wallroughness of ε = 260 µm.Solution. The Reynolds number and physical properties are unchanged. From eqn. (7.50)⎧⎡⎤⎫−2⎨−6 0.12 1.11 ⎬260×106.9⎦+f = 1.8 log10 ⎣⎭⎩573, 7003.7=0.02424The roughness Reynolds number is then2260 × 10−6 0.02424Reε = (573, 700)= 68.40.128This corresponds to fully rough flow. With eqn.
(7.49) we haveNuD =(0.02424/8)(5.74 × 105 )(2.47)3!"1 + 0.02424/8 4.5(68.4)0.2 (2.47)0.5 − 8.48= 2, 985soh = 29850.661= 16.4 kW/m2 K0.12In this case, wall roughness causes a factor of 1.8 increase in h and afactor of 2.0 increase in f and the pumping power. We have omittedthe variable properties corrections here because they were developedfor smooth-walled pipes.§7.3Turbulent pipe flow365Figure 7.7 Velocity and temperature profiles during fully developed turbulent flow in a pipe.Heat transfer to fully developed liquid-metal flows in tubesA dimensional analysis of the forced convection flow of a liquid metalover a flat surface [recall eqn. (6.60) et seq.] showed thatNu = fn(Pe)(7.51)because viscous influences were confined to a region very close to thewall. Thus, the thermal b.l., which extends far beyond δ, is hardly influenced by the dynamic b.l.
or by viscosity. During heat transfer to liquidmetals in pipes, the same thing occurs as is illustrated in Fig. 7.7. The region of thermal influence extends far beyond the laminar sublayer, whenPr 1, and the temperature profile is not influenced by the sublayer.Conversely, if Pr 1, the temperature profile is largely shaped withinthe laminar sublayer.
At high or even moderate Pr’s, ν is therefore veryimportant, but at low Pr’s it vanishes from the functional equation. Equation (7.51) thus applies to pipe flows as well as to flow over a flat surface.Numerous measured values of NuD for liquid metals flowing in pipeswith a constant wall heat flux, qw , were assembled by Lubarsky and Kaufman [7.20].
They are included in Fig. 7.8. It is clear that while most of thedata correlate fairly well on NuD vs. Pe coordinates, certain sets of dataare badly scattered. This occurs in part because liquid metal experimentsare hard to carry out. Temperature differences are small and must oftenbe measured at high temperatures.
Some of the very low data might possibly result from a failure of the metals to wet the inner surface of thepipe.Another problem that besets liquid metal heat transfer measurementsis the very great difficulty involved in keeping such liquids pure. Most366Forced convection in a variety of configurations§7.3Figure 7.8 Comparison of measured and predicted Nusseltnumbers for liquid metals heated in long tubes with uniformwall heat flux, qw . (See NACA TN 336, 1955, for details anddata source references.)impurities tend to result in lower values of h. Thus, most of the Nusselt numbers in Fig. 7.8 have probably been lowered by impurities in theliquids; the few high values are probably the more correct ones for pureliquids.There is a body of theory for turbulent liquid metal heat transfer thatyields a prediction of the formNuD = C1 + C2 Pe0.8D(7.52)where the Péclét number is defined as PeD = uav D/α.
The constants arenormally in the ranges 2 C1 7 and 0.0185 C2 0.386 accordingto the test circumstances. Using the few reliable data sets available foruniform wall temperature conditions, Reed [7.21] recommendsNuD = 3.3 + 0.02 Pe0.8D(7.53)(Earlier work by Seban and Shimazaki [7.22] had suggested C1 = 4.8 andC2 = 0.025.) For uniform wall heat flux, many more data are available,Heat transfer surface viewed as a heat exchanger§7.4and Lyon [7.23] recommends the following equation, shown in Fig. 7.8:NuD = 7 + 0.025 Pe0.8D(7.54)In both these equations, properties should be evaluated at the averageof the inlet and outlet bulk temperatures and the pipe flow should haveL/D > 60 and PeD > 100.
For lower PeD , axial heat conduction in theliquid metal may become significant.Although eqns. (7.53) and (7.54) are probably correct for pure liquids,we cannot overlook the fact that the liquid metals in actual use are seldompure. Lubarsky and Kaufman [7.20] put the following line through thebulk of the data in Fig. 7.8:NuD = 0.625 Pe0.4D(7.55)The use of eqn. (7.55) for qw = constant is far less optimistic than theuse of eqn. (7.54).
It should probably be used if it is safer to err on thelow side.7.4Heat transfer surface viewed as a heat exchangerLet us reconsider the problem of a fluid flowing through a pipe with auniform wall temperature. By now we can predict h for a pretty widerange of conditions. Suppose that we need to know the net heat transferto a pipe of known length once h is known. This problem is complicatedby the fact that the bulk temperature, Tb , is varying along its length.However, we need only recognize that such a section of pipe is a heatexchanger whose overall heat transfer coefficient, U (between the walland the bulk), is just h.
Thus, if we wish to know how much pipe surfacearea is needed to raise the bulk temperature from Tbin to Tbout , we cancalculate it as follows: Q = (ṁcp)b Tbout − Tbin = hA(LMTD)orA= (ṁcp)b Tbout − TbinhTbout − TwlnTbin − Tw Tbout − Tw − Tbin − Tw(7.56)By the same token, heat transfer in a duct can be analyzed with the effectiveness method (Sect. 3.3) if the exiting fluid temperature is unknown.367368Forced convection in a variety of configurations§7.4Suppose that we do not know Tbout in the example above. Then we canwrite an energy balance at any cross section, as we did in eqn.
(7.8):dQ = qw P dx = hP (Tw − Tb ) dx = ṁcP dTbIntegration can be done from Tb (x = 0) = Tbin to Tb (x = L) = TboutL Tbd(Tw − Tb )(Tw − Tb )Tbin0LTw − TboutPh dx = − lnṁcp 0Tw − TbinhPdx = −ṁcpoutWe recognize in this the definition of h from eqn. (7.27). Hence,hP L= − lnṁcpTw − TboutTw − Tbinwhich can be rearranged asTbout − TbinhP L= 1 − exp −ṁcpTw − Tbin(7.57)This equation can be used in either laminar or turbulent flow to compute the variation of bulk temperature if Tbout is replaced by Tb (x), L isreplaced by x, and h is adjusted accordingly.The left-hand side of eqn. (7.57) is the heat exchanger effectiveness.On the right-hand side we replace U with h; we note that P L = A, theexchanger surface area; and we write Cmin = ṁcp . Since Tw is uniform,the stream that it represents must have a very large capacity rate, so thatCmin /Cmax = 0. Under these substitutions, we identify the argument ofthe exponential as NTU = U A/Cmin , and eqn.
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