Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 19
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This is completely analogous to electrical resistance, which,according to Ohm’s law, is defined as the ratio of the voltage difference to the currentflow. With this definition, the thermal resistance of the plane wall, the hollow cylinder,and the hollow sphere are, respectively,Rcond =LkA(3.77)Rcond =ln(r2 /r1 )2πkL(3.78)Rcond =1/r1 − 1/r24πk(3.79)BOOKCOMP, Inc. — John Wiley & Sons / Page 182 / 2nd Proofs / Heat Transfer Handbook / BejanSTEADY ONE-DIMENSIONAL CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445183When convection occurs at the boundaries of a solid, it is convenient to define theconvection resistance from Newton’s law of cooling:q = hA(Ts − T∞ )(3.80)where h is the convection heat transfer coefficient and T∞ is the convecting fluidtemperature.
It follows from eq. (3.80) thatRconv =3.4.5Ts − T∞1=qhA(3.81)Composite SystemsThe idea of thermal resistance is a useful tool for analyzing conduction throughcomposite members.[183], (23)Composite Plane Wall For the series composite plane wall and the associatedthermal network shown in Fig.
3.9, the rate of heat transfer q is given byLines: 1188 to 1259q=T∞,1 − T∞,21/ h1 A + L1 /k1 A + L2 /k2 A + 1/ h2 A———(3.82)Once q has been determined, the surface and interface temperatures can be found:1h1 AL11T2 = T∞,1 − q+h1 A k 1 A1L1L2Ts,2 = T∞,1 − q++h1 A k 1 A k 2 ATs,1 = T∞,1 − qTs,1 − Ts,2L1 /k1 H1 + L2 L3 /(k2 H2 L3 + k3 H3 L2 ) + L4 /k4 H4[183], (23)(3.84)(3.85)(3.86)where H1 = H2 +H3 = H4 and the rule for combining resistances in parallel has beenemployed. Once q has been determined, the interface temperatures may be computed:T1 = Ts,1 − qL1k1 H1(3.87)T2 = Ts,2 + qL4k4 H4(3.88)BOOKCOMP, Inc.
— John Wiley & Sons / Page 183 / 2nd Proofs / Heat Transfer Handbook / Bejan———Long Page* PgEnds: Eject(3.83)Figure 3.10 illustrates a series–parallel composite wall. If materials 2 and 3 havecomparable thermal conductivities, the heat flow may be assumed to be one-dimensional. The network shown in Fig. 3.10 assumes that surfaces normal to the directionof heat flow are isothermal. For a wall of unit depth, the heat transfer rate isq=0.37024pt PgVar184123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFER[184], (24)Lines: 1259 to 1259———*Figure 3.9 Series composite wall and its thermal network.T1Ts,1H1 k1L1H2L2H3k2Ts,2L4L2k2 H2L1qTs,1L4k4 H4k1 H1T1T2Ts,2L3k3 H3Figure 3.10 Series–parallel composite wall and its thermal network.BOOKCOMP, Inc. — John Wiley & Sons / Page 184 / 2nd Proofs / Heat Transfer Handbook / Bejan———Normal PagePgEnds: TEX[184], (24)H4k3L3k4T235.45401pt PgVar185STEADY ONE-DIMENSIONAL CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445Cold fluidT⬁2, h2qT2Ts,2Ts,1k2r2r1k1[185], (25)LLines: 1259 to 1286———r31.82707pt PgVarHot fluidT⬁1, h1———Normal PagePgEnds: TEXT⬁11ln(r2 /r1)ln(r3 /r2)1h1 2r1 L2k1 L2k2 Lh 22r3 LTs,1T2Ts,2[185], (25)T⬁2Figure 3.11 Series composite hollow cylinder and its thermal network.Composite Hollow Cylinder A typical composite hollow cylinder with bothinside and outside experiencing convection is shown in Fig.
3.11. The figure includesthe thermal network that represents the system. The rate of heat transfer q is given byq=T∞,1 − T∞,21/2πh1 r1 L + ln(r2 /r1 )/2πk1 L + ln(r3 /r2 )/2πk2 L + 1/2πh2 r3 L(3.89)Once q has been determined, the inside surface Ts,1 , the interface temperature T2 , andthe outside surface temperature Ts,2 can be found:Ts,1 = T∞,1 − q12πh1 r1 LBOOKCOMP, Inc.
— John Wiley & Sons / Page 185 / 2nd Proofs / Heat Transfer Handbook / Bejan(3.90)186123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERT2 = T∞,1 − qTs,2 = T∞,2 + q1ln(r2 /r1 )+2πh1 r1 L2πk1 L(3.91)12πh2 r3 L(3.92)Composite Hollow Sphere Figure 3.12 depicts a composite hollow spheremade of two layers and experiencing convective heating on the inside surface andconvective cooling on the outside surface.
From the thermal network, also shown inFig. 3.12, the rate of heat transfer q can be determined asq=T∞,1 − T∞,21/4πh1 r12 + (r2 − r1 )/4πk1 r1 r2 + (r3 − r2 )/4πk2 r3 r2 + 1/4πh2 r32(3.93)[186], (26)Once q has been determined, temperatures Ts,1 , T2 , and Ts,2 can be found:Lines: 1286 to 1307———3.39914pt PgVar———Normal Page* PgEnds: EjectCold fluidT⬁2, h2qk2[186], (26)k1Ts,1T2Ts,1r1r2r3Hot fluidT⬁1, h1qT⬁111/r1 ⫺ 1/r21/r2 ⫺ 1/r31h1 4r 214k14k2h2 4r 23Ts,1T2Ts,2T⬁2Figure 3.12 Series composite hollow sphere and its thermal network.BOOKCOMP, Inc. — John Wiley & Sons / Page 186 / 2nd Proofs / Heat Transfer Handbook / BejanSTEADY ONE-DIMENSIONAL CONDUCTION12345678910111213141516171819202122232425262728293031323334353637383940414243444514πh1 r121111+−T2 = T∞,1 − q4πk1 r1r24πh1 r12Ts,1 = T∞,1 − qTs,2 = T∞,2 + q3.4.614πh2 r32187(3.94)(3.95)(3.96)Contact ConductanceThe heat transfer analyses for the composite systems described in Section 3.4.5 assumed perfect contact at the interface between the two materials.
In reality, however, the mating surfaces are rough and the actual contact occurs at discrete points(asperities or peaks), as indicated in Fig. 3.13. The gaps of voids are usually filledwith air, and the heat transfer at the interface is the sum of solid conduction acrossthe contact points and fluid conduction through the gaps. Because of the imperfectcontact, there is a temperature drop across the gap or interface, ∆Tc . The contactconductance hc (W/m 2 · K) is defined as the ratio of the heat flux q/A through theinterface to the interface temperature drop:hc =q/A1= ∆TcRc(3.97)where q/A is the heat flux through the interface and Rc (m 2 · K/W), the inverse of hc ,is the contact resistance.The topic of contact conductance is of considerable contemporary interest, asreflected by a review paper of Fletcher (1988), a book by Madhusudana (1996),Figure 3.13 Contact interface for actual surfaces.BOOKCOMP, Inc.
— John Wiley & Sons / Page 187 / 2nd Proofs / Heat Transfer Handbook / Bejan[187], (27)Lines: 1307 to 1334———0.55927pt PgVar———Normal PagePgEnds: TEX[187], (27)188123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERand numerous contributions from several research groups.
Chapter 4 of this bookis devoted exclusively to this subject.3.4.7Critical Thickness of InsulationWhen a plane surface is covered with insulation, the rate of heat transfer alwaysdecreases. However, the addition of insulation to a cylindrical or spherical surfaceincreases the conduction resistance but reduces the convection resistance becauseof the increased surface area.
The critical thickness of insulation corresponds to thecondition when the sum of conduction and convection resistances is a minimum. Fora given temperature difference, this results in a maximum heat transfer rate, and thecritical radius rc is given byk(cylinder)(3.98)hrc =2k(sphere)(3.99)hwhere k is the thermal conductivity of the insulation and h is the convective heattransfer coefficient for the outside surface.The basic analysis for obtaining the critical radius expression has been modifiedto allow for:• The variation of h with outside radius• The variation of h with outside radius, including the effect of temperaturedependent fluid properties• Circumferential variation of h• Pure radiation cooling• Combined natural convection and radiation coolingThe analysis for a circular pipe has also been extended to include insulationboundaries that form equilateral polygons, rectangles, and concentric circles.
Suchconfigurations require a two-dimensional conduction analysis and have led to theconclusion that the concept of critical perimeter of insulation, Pc = 2π(k/ h), ismore general than that of critical radius. The comprehensive review article by Aziz(1997) should be consulted for further details.3.4.8Effect of Uniform Internal Energy GenerationIn some engineering systems, it becomes necessary to analyze one-dimensional steadyconduction with internal energy generation. Common sources of energy generationare the passage of an electric current through a wire or busbar or a rear windowdefroster in an automobile. In the fuel element of a nuclear reactor, the energy isgenerated due to neutron absorption.
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