John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 80
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The heat flux leaving a surface may be known,if, say, it is an electrically powered radiant heater. In this case, the lefthand side of one of eqns. (10.34) can be replaced with the surface’s knownQnet , via eqn. (10.33b).For the adiabatic wall case just considered, if surface (1) had a specified heat flux, then eqn. (10.35) could be solved for eb1 and the unknowntemperature T1 .Example 10.9Two very long strips 1 m wide and 2.40 m apart face each other, asshown in Fig. 10.16. (a) Find Qnet1–2 (W/m) if the surroundings areblack and at 250 K.
(b) Find Qnet1–2 (W/m) if they are connected byan insulated diffuse reflector between the edges on both sides. Alsoevaluate the temperature of the reflector in part (b).Solution. From Table 10.2, case 1, we find F1–2 = 0.2 = F2–1 . Inaddition, F2–3 = 1 − F2–1 = 0.8, irrespective of whether surface (3)represents the surroundings or the insulated shield.In case (a), the two nodal equations (10.34a) and (10.34b) becomeB1 − B3B1 − B21451 − B1=+2.3331/0.21/0.8B2 − B3B2 − B1459.3 − B2=+11/0.21/0.8Equation (10.34c) cannot be used directly for black surroundings,since ε3 = 1 and the surface resistance in the left-hand side denominator would be zero. But the numerator is also zero in this case,since eb3 = B3 for black surroundings.
And since we now knowB3 = σ T34 = 221.5 W/m2 K, we can use it directly in the two equations above.Figure 10.16 Illustration forExample 10.9.557Radiative heat transfer558§10.4Thus,B1 − 0.14 B2 −0.56(221.5) = 435.6−B1 +10.00 B2 −4.00(221.5) = 2296.5orB1 − 0.14 B2 = 559.6−B1 + 10.00 B2 = 3182.515soB1 = 612.1 W/m2B2 = 379.5 W/m2Thus, the net flow from (1) to (2) is quite small:Qnet1–2 =B1 − B2= 46.53 W/m1 /(A1 F1–2 )Since each strip also loses heat to the surroundings, Qnet1 ≠ Qnet2 ≠Qnet1–2 .For case (b), with the adiabatic shield in place, eqn.
(10.34c) canbe combined with the other two nodal equations:0=B3 − B11/0.8+B3 − B21/0.8The three equations can be solved manually, by the use of determinants, or with a computerized matrix algebra package. The resultisB1 = 987.7 W/m2B2 = 657.4 W/m2B3 = 822.6 W/m2In this case, because surface (3) is adiabatic, all net heat transfer fromsurface (1) is to surface (2): Qnet1 = Qnet1–2 . Then, from eqn. (10.33a),we get987.7 − 657.4 987.7 − 822.6+= 198 W/mQnet1–2 =1/(1)(0.2)1/(1)(0.8)Of course, because node (3) is insulated, it is much easier to useeqn. (10.35) to get Qnet1–2 :Qnet1–2 = 5.67 × 10−8 4004 − 30040.70.3+111/0.8 + 1/0.8++ 0.20.50.5= 198 W/m§10.4Heat transfer among gray bodies559The result, of course, is the same. We note that the presence of thereflector increases the net heat flow from (1) to (2).The temperature of the reflector (3) is obtained from eqn.
(10.33b)with Qnet3 = 0:0 = eb3 − B3 = 5.67 × 10−8 T34 − 822.6soT3 = 347 KAlgebraic solution of multisurface enclosure problemsAn enclosure can consist of any number of surfaces that exchange radiation with one another. The evaluation of radiant heat transfer amongthese surfaces proceeds in essentially the same way as for three surfaces.For multisurface problems, however, the electrical circuit approach isless convenient than a formulation based on matrices. The matrix equations are usually solved on a computer.An enclosure formed by n surfaces is shown in Fig.
10.17. As before,we will assume that:• Each surface is diffuse, gray, and opaque, so that ε = α and ρ = 1−ε.• The temperature and net heat flux are uniform over each surface(more precisely, the radiosity must be uniform and the other properties are averages for each surface). Either temperature or fluxmust be specified on every surface.• The view factor, Fi−j , between any two surfaces i and j is known.• Conduction and convection within the enclosure can be neglected,and any fluid in the enclosure is transparent and nonradiating.We are interested in determining the heat fluxes at the surfaces wheretemperatures are specified, and vice versa.The rate of heat loss from the ith surface of the enclosure can conveniently be written in terms of the radiosity, Bi , and the irradiation, Hi ,from eqns. (10.19) and (10.21)qneti = Bi − Hi =εi σ Ti4 − Bi1 − εi(10.36)Radiative heat transfer560§10.4Figure 10.17 An enclosure composed of n diffuse, gray surfaces.whereBi = ρi Hi + εi ebi = (1 − εi ) Hi + εi σ Ti4(10.37)However, Ai Hi , the irradiating heat transfer incident on surface i, is thesum of energies reaching i from all other surfaces, including itselfAi Hi =n$Aj Bj Fj−i =j=1n$Bj Ai Fi−jj=1where we have used the reciprocity rule, Aj Fj−i = Ai Fi−j .
ThusHi =n$Bj Fi−j(10.38)j=1It follows from eqns. (10.37) and (10.38) thatBi = (1 − εi )n$Bj Fi−j + εi σ Ti4(10.39)j=1This equation applies to every surface, i = 1, . . . , n. When all the surface temperatures are specified, the result is a set of n linear equationsfor the n unknown radiosities.
For numerical purposes, it is sometimesconvenient to introduce the Kronecker delta,⎧⎨1 for i = j(10.40)δij =⎩0 for i ≠ jHeat transfer among gray bodies§10.4561and to rearrange eqn. (10.39) asn $δij − (1 − εi )Fi−j Bj = εi σ Ti4j=1 for i = 1, . . . , n(10.41)≡CijThe radiosities are then found by inverting the matrix Cij . The rate ofheat loss from the ith surface, Qneti = Ai qneti , can be obtained fromeqn. (10.36).For those surfaces where heat fluxes are prescribed, we can eliminatethe εi σ Ti4 term in eqn. (10.39) or (10.41) using eqn.
(10.36). We again obtain a matrix equation that can be solved for the Bi ’s. Finally, eqn. (10.36)is solved for the unknown temperature of surface in question.In many cases, the radiosities themselves are of no particular interest.The heat flows are what is really desired. With a bit more algebra (seeProblem 10.45), one can formulate a matrix equation for the n unknownvalues of Qneti :n$j=1δijεi−(1 − εj )εj AjAi Fi−j Qnetj =n$Ai Fi−j σ Ti4 − σ Tj4(10.42)j=1Example 10.10Two sides of a long triangular duct, as shown in Fig.
10.18, are madeof stainless steel (ε = 0.5) and are maintained at 500◦ C. The thirdside is of copper (ε = 0.15) and has a uniform temperature of 100◦ C.Calculate the rate of heat transfer to the copper base per meter oflength of the duct.Solution. Assume the duct walls to be gray and diffuse and thatconvection is negligible. The view factors can be calculated from configuration 4 of Table 10.2:F1–2 =A1 + A2 − A30.5 + 0.3 − 0.4= 0.4=2A11.0Similarly, F2–1 = 0.67, F1–3 = 0.6, F3–1 = 0.75, F2–3 = 0.33, and F3–2 =0.25. The surfaces cannot “see” themselves, so F1–1 = F2–2 = F3–3 =0.
Equation (10.39) leads to three algebraic equations for the three562Radiative heat transfer§10.4Figure 10.18 Illustration for Example 10.10.unknowns, B1 , B2 , and B3 .B1 = 1 − ε1 F1–1 B1 + F1–2 B2 + F1–3 B3 + ε1 σ T14 0.8500.40.60.150.50.6700.330.50.50.750.2500.5B2 = 1 − ε2 F2–1 B1 + F2–2 B2 + F2–3 B3 + ε2 σ T24 B3 = 1 − ε3 F3–1 B1 + F3–2 B2 + F3–3 B3 + ε3 σ T34 It would be easy to solve this system numerically using matrixmethods.
Alternatively, we can substitute the third equation into thefirst two to eliminate B3 , and then use the second equation to eliminate B2 from the first. The result isB1 = 0.232 σ T14 + 0.319 σ T24 + 0.447 σ T34Equation (10.36) gives the rate of heat loss by surface (1) asε1 σ T14 − B11 − ε1ε1σ T14 − 0.232 T14 − 0.319 T24 − 0.447 T34= A11 − ε1Qnet1 = A1Gaseous radiation§10.50.15(5.67 × 10−8 )= (0.5)0.85× (373)4 − 0.232(373)4 − 0.319(773)4 − 0.447(773)4 W/m= −1294 W/mThe negative sign indicates that the copper base is gaining heat.Enclosures with nonisothermal, nongray, or nondiffuse surfacesThe representation of enclosure heat exchange by eqn. (10.41) or (10.42)is actually quite powerful.
For example, if the primary surfaces in an enclosure are not isothermal, they may be subdivided into a larger numberof smaller surfaces, each of which is approximately isothermal. Then either equation may be used to calculate the heat exchange among the setof smaller surfaces.For those cases in which the gray surface approximation, eqn.
(10.8c),cannot be applied (owing to very different temperatures or strong wavelength dependence in ελ ), eqns. (10.41) and (10.42) may be applied ona monochromatic basis, since the monochromatic form of Kirchhoff’slaw, eqn. (10.8b), remains valid. The results must, of course, be integrated over wavelength to get the heat exchange. The calculation isusually simplified by breaking the wavelength spectrum into a few discrete bands within which radiative properties are approximately constant [10.2, Chpt.
7].When the surfaces are not diffuse — when emission or reflection varywith angle — a variety of other methods can be applied. Among them,the Monte Carlo technique is probably the most widely used. The MonteCarlo technique tracks emissions and reflections through various anglesamong the surfaces and estimates the probability of absorption or rereflection [10.4, 10.7]. This method allows complex situations to be numerically computed with relative ease, provided that one is careful toobtain statistical convergence.10.5Gaseous radiationWe have treated every radiation problem thus far as though radiant heatflow in the space separating the surfaces of interest were completelyunobstructed by any fluid in between.
However, all gases interact withphotons to some extent, by absorbing or deflecting them, and they can563564Radiative heat transfer§10.5even emit additional photons. The result is that fluids can play a role inthe thermal radiation to the the surfaces that surround them.We have ignored this effect so far because it is generally very small,especially in air and if the distance between the surfaces is on the order of meters or less. When other gases are involved, especially at hightemperatures, as in furnaces, or when long distances are involved, as inthe atmosphere, gas radiation can become an important part of the heatexchange process.How gases interact with photonsThe photons of radiant energy passing through a gaseous region can beimpeded in two ways.
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