John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 12
Текст из файла (страница 12)
2.12.Solution.Step 1. T = T (r )Step 2.1 ∂r ∂rr∂T∂r+1 ∂2T∂2Tq̇++=222r ∂φ∂zk =0, since T ≠ T (φ, z)Step 3. Integrate once: r=01 ∂Tα ∂t=0, since steady∂T= C1 ; integrate again: T = C1 ln r + C2∂rStep 4. T (r = ri ) = Ti and T (r = ro ) = To68Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.3Figure 2.12 Heat transfer through a cylinder with a fixed walltemperature (Example 2.5).Step 5.Ti = C1 ln ri + C2To = C1 ln ro + C2Step 6. T = Ti −⎧∆TTi − To⎪⎪⎪=−⎨ C1 =ln(ri /ro )ln(ro /ri )⇒⎪∆T⎪ C =T +⎪ln ri⎩ 2iln(ro /ri )∆T(ln r − ln ri ) orln(ro /ri )ln(r /ri )T − Ti=To − Tiln(ro /ri )(2.20)Step 7.
The solution is plotted in Fig. 2.12. We see that the temperature profile is logarithmic and that it satisfies both boundaryconditions. Furthermore, it is instructive to see what happenswhen the wall of the cylinder is very thin, or when ri /ro is closeto 1. In this case:ln(r /ri ) rr − ri−1=ririThermal resistance and the electrical analogy§2.3andln(ro /ri ) ro − ririThus eqn. (2.20) becomesT − Tir − ri=To − Tiro − riwhich is a simple linear profile. This is the same solution thatwe would get in a plane wall.Step 8. At any station, r :qradial = −kk∆T 1∂T=+∂rln(ro /ri ) rSo the heat flux falls off inversely with radius.
That is reasonable, since the same heat flow must pass through an increasinglylarge surface as the radius increases. Let us see if this is the casefor a cylinder of length l:Q (W) = (2π r l) q =2π kl∆T≠ f (r )ln(ro /ri )(2.21)Finally, we again recognize Ohm’s law in this result and writethe thermal resistance for a cylinder: ln(ro /ri ) KRtcyl =(2.22)2π lkWThis can be compared with the resistance of a plane wall: KLRtwall =kA WBoth resistances are inversely proportional to k, but each reflects a different geometry.In the preceding examples, the boundary conditions were all the same—a temperature specified at an outer edge.
Next let us suppose that thetemperature is specified in the environment away from a body, with aheat transfer coefficient between the environment and the body.6970Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.3Figure 2.13 Heat transfer through a cylinder with a convectiveboundary condition (Example 2.6).Example 2.6A Convective Boundary ConditionA convective heat transfer coefficient around the outside of the cylinder in Example 2.5 provides thermal resistance between the cylinderand an environment at T = T∞ , as shown in Fig. 2.13.
Find the temperature distribution and heat flux in this case.Solution.Step 1 through 3. These are the same as in Example 2.5.Step 4. The first boundary condition is T (r = ri ) = Ti . The secondboundary condition must be expressed as an energy balance atthe outer wall (recall Section 1.3).qconvection = qconductionat the wallorh(T − T∞ )r =ro∂T = −k∂r r =roStep 5. From the first boundary condition we obtain Ti = C1 ln ri +C2 . It is easy to make mistakes when we substitute the generalsolution into the second boundary condition, so we will do it inThermal resistance and the electrical analogy§2.3detail:h (C1 ln r + C2 ) − T∞r =ro= −k∂(C1 ln r + C2 )∂rr =ro(2.23)A common error is to substitute T = To on the lefthand sideinstead of substituting the entire general solution.
That will dono good, because To is not an accessible piece of information.Equation (2.23) reduces to:h(T∞ − C1 ln ro − C2 ) =kC1roWhen we combine this with the result of the first boundary condition to eliminate C2 :T∞ − TiTi − T∞C1 = − =1/Bi + ln(ro /ri )k (hro ) + ln(ro /ri )ThenC2 = Ti −T∞ − Tiln ri1/Bi + ln(ro /ri )Step 6.T =T∞ − Tiln(r /ri ) + Ti1/Bi + ln(ro /ri )This can be rearranged in fully dimensionless form:T − Tiln(r /ri )=T∞ − Ti1/Bi + ln(ro /ri )(2.24)Step 7. Let us fix a value of ro /ri —say, 2—and plot eqn.
(2.24) forseveral values of the Biot number. The results are includedin Fig. 2.13. Some very important things show up in this plot.When Bi 1, the solution reduces to the solution given in Example 2.5. It is as though the convective resistance to heat flowwere not there. That is exactly what we anticipated in Section1.3for large Bi. When Bi 1, the opposite is true: (T −Ti ) (T∞ −Ti )7172Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.3Figure 2.14 Thermal circuit with tworesistances.remains on the order of Bi, and internal conduction can be neglected.
How big is big and how small is small? We do notreally have to specify exactly. But in this case Bi < 0.1 signalsconstancy of temperature inside the cylinder with about ±3%.Bi > 20 means that we can neglect convection with about 5%error.Ti − T∞1∂T=k∂r1/Bi + ln(ro /ri ) rThis can be written in terms of Q (W) = qradial (2π r l) for a cylinder of length l:Step 8. qradial = −kQ=Ti − T∞Ti − T∞=ln(ro /ri )Rtconv + Rtcond+2π klh 2π ro l1(2.25)Equation (2.25) is once again analogous to Ohm’s law. But this timethe denominator is the sum of two thermal resistances, as would bethe case in a series circuit. We accordingly present the analogouselectrical circuit in Fig.
2.14.The presence of convection on the outside surface of the cylindercauses a new thermal resistance of the formRtconv =1hA(2.26)where A is the surface area over which convection occurs.Example 2.7Critical Radius of InsulationAn interesting consequence of the preceding result can be brought outwith a specific example. Suppose that we insulate a 0.5 cm O.D.
coppersteam line with 85% magnesia to prevent the steam from condensingThermal resistance and the electrical analogy§2.3Figure 2.15 Thermal circuit for aninsulated tube.too rapidly. The steam is under pressure and stays at 150◦ C. Thecopper is thin and highly conductive—obviously a tiny resistance inseries with the convective and insulation resistances, as we see inFig. 2.15. The condensation of steam inside the tube also offers verylittle resistance.3 But on the outside, a heat transfer coefficient of h= 20 W/m2 K offers fairly high resistance. It turns out that insulationcan actually improve heat transfer in this case.The two significant resistances, for a cylinder of unit length (l =1 m), areln(ro /ri )ln(ro /ri )=K/W2π kl2π (0.074)11=K/W=2π (20)ro2π ro hRtcond =RtconvFigure 2.16 is a plot of these resistances and their sum. A very interesting thing occurs here.
Rtconv falls off rapidly when ro is increased,because the outside area is increasing. Accordingly, the total resistance passes through a minimum in this case. Will it always do so?To find out, we differentiate eqn. (2.25), again setting l = 1 m:1(Ti − T∞ )1dQ=0=+2 −dro2π ro2 h 2π kroln(ro /ri )1+2π k2π ro hWhen we solve this for the value of ro = rcrit at which Q is maximumand the total resistance is minimum, we obtainBi = 1 =hrcritk(2.27)In the present example, adding insulation will increase heat loss in3Condensation heat transfer is discussed in Chapter 8. It turns out that h is generallyenormous during condensation so that Rtcondensation is tiny.73Heat conduction, thermal resistance, and the overall heat transfer coefficientrcrit = 1.48 ri4Thermal resistance, Rt (K/W)74§2.3Rtcond + RtconvRtconv2Rtcond01.01.52.02.52.32Radius ratio, ro/riFigure 2.16 The critical radius of insulation (Example 2.7),written for a cylinder of unit length (l = 1 m).stead of reducing it, until rcrit = k h = 0.0037 m or rcrit /ri = 1.48.Indeed, insulation will not even start to do any good until ro /ri = 2.32or ro = 0.0058 m.
We call rcrit the critical radius of insulation.There is an interesting catch here. For most cylinders, rcrit < ri andthe critical radius idiosyncrasy is of no concern. If our steam line had a 1cm outside diameter, the critical radius difficulty would not have arisen.When cooling smaller diameter cylinders, such as electrical wiring, thecritical radius must be considered, but one need not worry about it inthe design of most large process equipment.Resistance for thermal radiationWe saw in Chapter 1 that the net radiation exchanged by two objects isgiven by eqn. (1.34):Qnet = A1 F1–2 σ T14 − T24(1.34)When T1 and T2 are close, we can approximate this equation using aradiation heat transfer coefficient, hrad .
Specifically, suppose that thetemperature difference, ∆T = T1 − T2 , is small compared to the meantemperature, Tm = (T1 + T2 ) 2. Then we can make the following expan-Thermal resistance and the electrical analogy§2.3sion and approximation:Qnet = A1 F1–2 σ T14 − T24= A1 F1–2 σ (T12 + T22 )(T12 − T22 )= A1 F1–2 σ(T12 + T22 ) (T1 + T2 ) (T1 − T2 )2 + (∆T )2 /2= 2Tm=2Tm=∆T3 A1 4σ TmF1–2 ∆T(2.28)≡hrad2 or (∆T /T )2 /4 1.where the last step assumes that (∆T )2 /2 2TmmThus, we have identified the radiation heat transfer coefficient⎫Qnet = A1 hrad ∆T ⎬3F1–2hrad = 4σ Tm⎭for ∆T Tm241(2.29)This leads us immediately to the introduction of a radiation thermal resistance, analogous to that for convection:Rtrad =1A1 hrad(2.30)For the special case of a small object (1) in a much larger environment(2), the transfer factor is given by eqn.
(1.35) as F1–2 = ε1 , so that3hrad = 4σ Tmε1(2.31)If the small object is black, its emittance is ε1 = 1 and hrad is maximized.For a black object radiating near room temperature, say Tm = 300 K,hrad = 4(5.67 × 10−8 )(300)3 6 W/m2 KThis value is of approximately the same size as h for natural convectioninto a gas at such temperatures. Thus, the heat transfer by thermal radiation and natural convection into gases are similar. Both effects must betaken into account. In forced convection in gases, on the other hand, hmight well be larger than hrad by an order of magnitude or more, so thatthermal radiation can be neglected.7576Heat conduction, thermal resistance, and the overall heat transfer coefficient§2.3Example 2.8An electrical resistor dissipating 0.1 W has been mounted well awayfrom other components in an electronical cabinet.
It is cylindricalwith a 3.6 mm O.D. and a length of 10 mm. If the air in the cabinetis at 35◦ C and at rest, and the resistor has h = 13 W/m2 K for naturalconvection and ε = 0.9, what is the resistor’s temperature? Assumethat the electrical leads are configured so that little heat is conductedinto them.Solution.
The resistor may be treated as a small object in a largeisothermal environment. To compute hrad , let us estimate the resistor’s temperature as 50◦ C. ThenTm = (35 + 50)/2 43◦ C = 316 Kso3hrad = 4σ Tmε = 4(5.67 × 10−8 )(316)3 (0.9) = 6.44 W/m2 KHeat is lost by natural convection and thermal radiation acting inparallel. To find the equivalent thermal resistance, we combine thetwo parallel resistances as follows:1Rtequiv=1Rtrad+1Rtconv = Ahrad + Ah = A hrad + hThus,Rtequiv =1 A hrad + hA calculation shows A = 133 mm2 = 1.33 × 10−4 m2 for the resistorsurface.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
