John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 78
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By inspection, we see that Fjet–room = 30/360 = 0.08333and Fjet–shield = 330/360 = 0.9167. Thus,44Qnetjet–room = Ajet Fjet–room σ Tjet− Troomπ (0.003) m2(0.08333)(5.67 × 10−8 ) 22734 − 3034=m length= 1, 188 W/mLikewise,44− TshieldQnetjet–shield = Ajet Fjet–shield σ Tjetπ (0.003) m2(0.9167)(5.67 × 10−8 ) 22734 − 9734=m length= 12, 637 W/mThe heat absorbed by the shield leaves it by radiation and convectionto the room. (A balance of these effects can be used to calculate theshield temperature given here.)To find the radiation from the inside of the shield to the room, weneed Fshield–room .
Since any radiation passing out of the slit goes to the539540Radiative heat transfer§10.3room, we can find this view factor equating view factors to the roomwith view factors to the slit. The slit’s area is Aslit = π (0.05)30/360 =0.01309 m2 /m length. Hence, using our reciprocity and summationrules, eqns. (10.12) and (10.15),Fslit–jet =Ajetπ (0.003)(0.0833) = 0.0600Fjet–room =Aslit0.01309Fslit–shield = 1 − Fslit–jet − Fslit–slit = 1 − 0.0600 − 0 = 0.940 0Fshield–roomAslit=Fslit–shieldAshield0.01309(0.940) = 0.08545=π (0.05)(330)/(360)Hence, for heat transfer from the inside of the shield only,44− TroomQnetshield–room = Ashield Fshield–room σ Tshieldπ (0.05)330=(0.08545)(5.67 × 10−8 ) 9734 − 3034360= 619 W/mBoth the jet and the inside of the shield have relatively small viewfactors to the room, so that comparatively little heat is lost throughthe slit.Calculation of the black-body view factor, F1–2 .
Consider two elements,dA1 and dA2 , of larger black bodies (1) and (2), as shown in Fig. 10.8.Body (1) and body (2) are each isothermal. Since element dA2 subtendsa solid angle dω1 , we use eqn. (10.6) to writedQ1 to 2 = (i1 dω1 )(cos β1 dA1 )But from eqn. (10.7b),i1 =σ T14πNote that because black bodies radiate diffusely, i1 does not vary withangle; and because these bodies are isothermal, it does not vary withposition. The element of solid angle is given bydω1 =cos β2 dA2s2Radiant heat exchange between two finite black bodies§10.3Figure 10.8 Radiant exchange between two black elementsthat are part of the bodies (1) and (2).where s is the distance from (1) to (2) and cos β2 enters because dA2 isnot necessarily normal to s.
Thus,σ T14 cos β1 cos β2 dA1 dA2dQ1 to 2 =πs2By the same token,dQ2 to 1ThenQnet1–2σ T24=π44= σ T1 − T2A1cos β2 cos β1 dA2 dA1s2A2cos β1 cos β2dA1 dA2π s2(10.16)The view factors F1–2 and F2–1 are immediately obtainable from eqn.(10.16). If we compare this result with Qnet1–2 = A1 F1–2 σ (T14 − T24 ), wegetF1–21=A1A1A2cos β1 cos β2dA1 dA2π s2(10.17a)541542Radiative heat transfer§10.3From the inherent symmetry of the problem, we can also writeF2–1 =1A2A2A1cos β2 cos β1dA2 dA1π s2(10.17b)You can easily see that eqns. (10.17a) and (10.17b) are consistent withthe reciprocity relation, eqn. (10.15).The direct evaluation of F1–2 from eqn. (10.17a) becomes fairly involved, even for the simplest configurations.
Siegel and Howell [10.4]provide a comprehensive discussion of such calculations and a large catalog of their results. Howell [10.5] gives an even more extensive tabulation of view factor equations, which is now available on the World WideWeb. At present, no other reference is as complete.We list some typical expressions for view factors in Tables 10.2 and10.3. Table 10.2 gives calculated values of F1–2 for two-dimensionalbodies—various configurations of cylinders and strips that approach infinite length.
Table 10.3 gives F1–2 for some three-dimensional configurations.Many view factors have been evaluated numerically and presentedin graphical form for easy reference. Figure 10.9, for example, includesgraphs for configurations 1, 2, and 3 from Table 10.3. The reader shouldstudy these results and be sure that the trends they show make sense.Is it clear, for example, that F1–2 → constant, which is < 1 in each case,as the abscissa becomes large? Can you locate the configuration on theright-hand side of Fig. 10.6 in Fig. 10.9? And so forth.Figure 10.10 shows view factors for another kind of configuration—one in which one area is very small in comparison with the other one.Many solutions like this exist because they are a bit less difficult to calculate, and they can often be very useful in practice.Example 10.2A heater (h) as shown in Fig.
10.11 radiates to the partially conicalshield (s) that surrounds it. If the heater and shield are black, calculate the net heat transfer from the heater to the shield.Solution. First imagine a plane (i) laid across the open top of theshield:Fh−s + Fh−i = 1But Fh−i can be obtained from Fig. 10.9 or case 3 of Table 10.3,Table 10.2 View factors for a variety of two-dimensional configurations (infinite in extent normal to the paper)ConfigurationEquation21.F1–2 = F2–1 =1+hw2−hw2.F1–2 = F2–1 = 1 − sin(α/2)3.F1–24.⎡⎤2 2h1⎣h ⎦1+− 1+=2wwF1–2 = (A1 + A2 − A3 ) 2A15.F1–26.r−1 b−1 atan− tan=b−accLet X = 1 + s/D.F1–2 = F2–17.F1–2 = 1,F2–2Then:1 3 21−X=X − 1 + sin−1πXr1, andr2r1= 1 − F2–1 = 1 −r2F2–1 =543Table 10.3 View factors for some three-dimensional configurationsConfiguration1.EquationLet X = a/c and Y = b/c.
Then:F1–2⎧ 1/2(1 + X 2 )(1 + Y 2 )2 ⎨ln=π XY ⎩1 + X2 + Y 2− X tan−1 X − Y tan−1 Y3+ X 1 + Y 2 tan−12.⎫⎬3XY√+ Y 1 + X 2 tan−1 √1 + Y21 + X2 ⎭Let H = h/ and W = w/. Then:⎧⎪−1/231 ⎨1W tan−1− H 2 + W 2 tan−1 H 2 + W 2F1–2 =πW ⎪W⎩⎧⎨ (1 + W 2 )(1 + H 2 )11−1+ H tan+ lnH4 ⎩1 + W 2 + H2⎫W 2 H 2 ⎫⎪⎬⎬222222W (1 + W + H )H (1 + H + W )×⎭⎪(1 + W 2 )(W 2 + H 2 )(1 + H 2 )(H 2 + W 2 )⎭3.BLet R1 = r1 /h, R2 = r2 /h, and X = 1 + 1 + R22R12 .
Then:41X − X 2 − 4(R2 /R1 )2F1–2 =24.Concentric spheres:F1–2 = 1,544F2–1 = (r1 /r2 )2 ,F2–2 = 1 − (r1 /r2 )2545Figure 10.9 The view factors for configurations shown in Table 10.3Figure 10.10 The view factor for three very small surfaces“looking at” three large surfaces (A1 A2 ).546Radiant heat exchange between two finite black bodies§10.3Figure 10.11 Heat transfer from a disc heater to its radiation shield.for R1 = r1 /h = 5/20 = 0.25 and R2 = r2 /h = 10/20 = 0.5.
Theresult is Fh−i = 0.192. ThenFh−s = 1 − 0.192 = 0.808Thus,Qneth−s = Ah Fh−s σ Th4 − Ts4π= (0.1)2 (0.808)(5.67 × 10−8 ) (1200 + 273)4 − 37344= 1687 WExample 10.3Suppose that the shield in Example 10.2 were heating the region wherethe heater is presently located. What would Fs−h be?Solution.
From eqn. (10.15) we haveAs Fs−h = Ah Fh−sBut the frustrum-shaped shield has an area of4As = π (r1 + r2 ) h2 + (r2 − r1 )23= π (0.05 + 0.1) 0.22 + 0.052 = 0.09715 m2547Radiative heat transfer548§10.3andAh =π(0.1)2 = 0.007854 m24soFs−h =0.007854(0.808) = 0.06530.09715Example 10.4Find F1–2 for the configuration of two offset squares of area A, asshown in Fig. 10.12.Solution.
Consider two fictitious areas 3 and 4 as indicated by thedotted lines. The view factor between the combined areas, (1+3) and(2+4), can be obtained from Fig. 10.9. In addition, we can write thatview factor in terms of the unknown F1–2 and other known view factors:(2A)F(1+3)–(4+2) = AF1–4 + AF1–2 + AF3–4 + AF3–22F(1+3)–(4+2) = 2F1–4 + 2F1–2F1–2 = F(1+3)–(4+2) − F1–4And F(1+3)–(4+2) can be read from Fig. 10.9 (at φ = 90, w/ = 1/2,and h/ = 1/2) as 0.245 and F1–4 as 0.20. Thus,F1–2 = (0.245 − 0.20) = 0.045Figure 10.12 Radiation between twooffset perpendicular squares.Heat transfer among gray bodies§10.410.4549Heat transfer among gray bodiesElectrical analogy for gray body heat exchangeAn electric circuit analogy for heat exchange among diffuse gray bodieswas developed by Oppenheim [10.6] in 1956. It begins with the definitionof two new quantities:flux of energy that irradiates the2H (W/m ) ≡ irradiance = surfaceandtotal flux of radiative energy2B (W/m ) ≡ radiosity = away from the surfaceThe radiosity can be expressed as the sum of the irradiated energy thatis reflected by the surface and the radiation emitted by it.
Thus,B = ρH + εeb(10.18)We can immediately write the net heat flux leaving any particular surface as the difference between B and H for that surface. Then, with thehelp of eqn. (10.18), we getqnet = B − H = B −B − εebρ(10.19)This can be rearranged asqnet =ε1−ρeb −Bρρ(10.20)If the surface is opaque (τ = 0), 1 − ρ = α, and if it is gray, α = ε. Then,eqn.
(10.20) givesqnet A = Qnet =eb − Beb − B=ρ/εA(1 − ε) εA(10.21)Equation (10.21) is a form of Ohm’s law, which tells us that (eb − B) canbe viewed as a driving potential for transferring heat away from a surfacethrough an effective surface resistance, (1 − ε)/εA.Now consider heat transfer from one infinite gray plate to anotherparallel to it. Radiant energy flows past an imaginary surface, parallelto the first infinite plate and quite close to it, as shown as a dotted line550Radiative heat transfer§10.4Figure 10.13 The electrical circuit analogy for radiation between two gray infinite plates.in Fig.
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