3.185 Problem Set 6. Radiation, Intro to Fluid Flow. Solutions

Transcription

1 3.85 Proble Set 6 Radiation, Intro to Fluid Flow Solution. Radiation in Zirconia Phyical Vapor Depoition (5 (a To calculate thi viewfactor, we ll let S be the liquid zicronia dic and S the inner urface of the heat hield. Then we ll create two additional fictitiou urface, S 3 and S : S 3 i the 0 cdiaeter dic at the top of the heat hield, and S i a 0 c-diaeter dic with a 5 c-diaeter hole in it between the zirconia and the bae of the hield. Since thee four urface for an encloure, we know that: F + F + F 3 + F = Since S, the zirconia, i not concave, F = 0. Since S and S are coplanar, none of the heat fro urface reache urface, o F = 0. Therefore, F + F 3 = e can get F 3 fro the attached graph, uing r =.5 c, r = 0 c and d = 0 c. Thi give u D = 0/.5 = and E = 0/0 =. Baed on thi, the graph tell u that F 3 = 0.5, o F = 0.5 (b Heat leave the zirconia urface with a flux given by: e = ɛσt The total heat leaving the zirconia i that flux tie it area. F give the fraction of that which reache S, and we divide that by A to etiate the flux at S : F A ɛσt q A 0.5 π(.5c K (00K = 500 = π 0c 0c (c Calculating F i a bit ore coplicated than F. Firt, we ll create a new S 5 which i the cobination of S and S, a 0 c-diaeter dik at the bae of the hield, which look a lot like S 3. Since S, S 3 and S 5 now for an encloure, we can ay: F + F 3 + F 5 = e can t get F 3 fro a graph, but we can get F 3 about the ae way we got F above: F 3 + F 33 + F 35 = Since S 3 i not concave, F 33 i zero, o F 3 = F 35. e can get F 35 fro the graph: r = r = d = 0 c, o D = E = and F 35 = 0.38, giving u F 3 = 0.6.

2 Now we can get F 3 fro: A F 3 = A 3 F 3 A i the inner area of the cylinder, πr = π(0. 0.; A 3 i the area of the top dik πr = π(0.. So A 3 i clearly half of A, and A 3 /A =, and F 3 = A 3 F 3 = A F 3 = 0.3 Since F 5 = F 3 = 0.3 by yetry, thi leave u with F = F 3 F 5 = = Radiation and Conduction in Metal Cating: Poirier & Geiger proble. (p. 3. et T be the teperature inide the old (500 K, T the outide old teperature (unknown, and T 3 the teperature of the large roo (300 K, and with no inforation given, aue the large roo i a black urface. e ll call the old outer urface S and the large roo S 3. The conductive flux through the old i T T q = k where i the old thickne given a 30. The radiative power tranferred fro the old to the roo i: Q 3 = A F 3 ɛ σt α 3 For a convex old, F = 0, o F 3 = ; alo if the roo i black, then α 3 =. The radiative power tranferred fro the roo to the old i: Q 3 = A 3 F 3 ɛ 3 σt 3 α Since α = ɛ for a grey body, α = ɛ and ɛ 3 = becaue the large roo i black. Alo, A 3 F 3 = A F 3, o with that and F 3 = we can rewrite Q 3 a: Q 3 = A ɛ σt 3 The net heat flux (not total power fro urface i therefore: q = Q 3 Q 3 = ɛ σ(t T 3 A Set the convective and radiative fluxe equal and olve for T : T T k = ɛ σ(t T 3 ɛ σt + k T ɛ σt 3 + kt = K T K T ( = 0 Call the left ide f (T and ue Newton ethod f (T i T i+ = T i f (T i Start with an initial gue halfway between T and T 3, and the iteration go (don t let the ubcript confue you, thi i till T : T 0 = 900K, T = K, T = K, T 3 = K, T = K

3 Thi i converging pretty quickly to T = 856. K. To check thi, the conductive flux through the old i The radiative flux fro the old urface i 500K 856.K q = 0.7 =.50 0 K K q = K [(856.K (300K ] =.50 0 K The two are equal (to ten digit uing the lat iteration above, which i a it hould be. Thi i a nice exaple of a proble with coupled radiation and conduction. 3. Radiation in Facing Target Sputtering (a There are two way to ue the diagra to calculate F, one i by the paralell rectangle ethod: F A F B F = For F A, x = b/c = 0c/0c =, and y = a/c = 0c/0c =, o the graph give F A = 0.. For F B, x = b/c = 0c/0c = 0.5, and y = a/c = 0c/0c = 0.5, o the graph give F B = Thee and the equation above reult in F = The other way i by the coon-edge rectangle ethod, uing φ = 90 : 3

4 F = (F C F D For F C, N = a/b = 0c/0c = and = c/b = 0c/0c =, o the graph give F C = 0.. For F D, N = a/b = 0c/0c =, and = c/b = 0c/0c =, o the graph give F D = 0.0. Thee and the equation above reult in F = 0.080, which i, well, oewhat cloe to the parallel rectangle viewfactor. Now ue the reciprocity rule to calculate F : A F = A F A F F = S include two 0 0 quare, o A = 00c = 0.0 ; S i one 0 0 quare, o A = 00c = 0.0. Thee and the F calculated by the two ethod above give F = 0.03 to (b Thi i the total power eitted by the target ultiplied by the viewfactor and by the aborptivity of the ubtrate: Q = α F A ɛ σt Uing α = ɛ and F give u: Q = K (00K = 3 Not a whole lot of power, but for a all ubtrate, enough to get it decently hot. (c Thi i a iilar ituation; uing F = 0.07 give u: Q = K (900K = 7.3 Note that the ratio Q /Q i iply T /T. Thi ethod ie one thing, which i reflection of a urface radiation off another urface and back onto itelf. ith viewfactor a all a thee, thi econdary aborption of reflected radiation will be negligible. But in general, thi i the otivation for the total exchange viewfactor decribed in the text, uing circuit diagra.. Poirier and Geiger proble. (p. 36. (5 Thi wa a traightforward Arrheniu extrapolation, like you ve done a zillion tie before in Coure 3, but with a warning at the very end: ΔGvi η = A exp RT Thi can be ade into a -point linear fit by taking the log of both ide: A ln η = ln A + ΔG vi R T 500K : ln 00 N = ln A + ΔG vi R 500K 700K : ln 0 N = ln A + ΔG vi R 700K Subtracting get rid of ln A: ln 00 N ln 0 N = ΔG vi R 500K 700K

5 ΔG vi ln 00 0 = = 050K R 500K 700K Now olve the 500K equation for ln A: ln A = ln 00 N ΔG ( vi N = ln R 500K N So, A =. 0. Now jut plug 50K into the original equation: ΔGvi η = A exp =. 0 N ( 050K exp = 60 N RT 50K Now the warning. Thi kind of extrapolation downward i very dangerou to do with olten glae, a you ight hit the gla tranition teperature, which would end the vicoity upward by order of agnitude! So, okay for thi proble et, but pleae avoid doing thi in real life. 5. Shear tre and couette flow Calculate the velocity of the lower plate for: 3 N (a ater, with vicoity of 0. Chooing x a the direction of plate otion and z noral to the plate, we et up an x-oentu balance: accuulation = in out + generation There i no gravity in the x-direction, and no preure drop wa entioned, o there i no generation. Alo, becaue thi ak for jut one velocity and not the velocity a a function of tie, we ll aue teady-tate o there i no accuulation. On each plane with contant z, we have a contant velocity, o we can do our hell balance for thin horizontal heet of thickne Δz: 0 = [A τ zx ] z [A τ zx ] z+δz Ue the fact that A i contant, and divide by V (= AΔz: 0 = τ zx z τ zx z+δz Δz et Δz go to zero, then ubtitute the Newtonian contitutive equation tau zx = η ( du x with dz contant η, and factor out η: Now integrate twice: 0 = dτ zx dz d 0 = η du x dz dz 0 = η d u x dz u z = C x + C The boundary condition are zero velocity at the top plate which we ll call z = u x = 0 and contant velocity V at the botto plate o z = 0 u x = 0. Thi give u V u x = z + V The hear force on the botto plate i the integral of hear tre over the area, ince the hear tre there i contant thi i jut the product: du x V F d = τ zx A = η A = η A dz 5

6 The hear force i alo the weight of the a, g, which i 9.8N. Now olve for V : F d V = = 3kg = 9 ηa 0 3 N 0.5 (b Molten teel at 600 C (ue the vicoity graph on p. 8, be a precie a you can Uing the graph on p. 8, it not poible to be very precie, o the bet we can do i about 3 N 5 0. Then we jut repeat the calculation above: F d 9.8N V = = = 60 ηa N 0.5 (c The hear tre due to thi weight i 3 kg tie 9.8/ divided by the area 0.5, which i about 60 N/. Since thi inu the yield tre i 0 N/, we ue that for F d /A which we can call τ net = τ τ 0, the net hear tre: 6. Poirier and Geiger proble. (τ τ 0 (60 N 0 N V = = = 0.5 η P 0. N e re given the general olution and a bunch of contant, which we can plug in to each other, firt for the gla on top: g co β δ x δ δ + + (ρ g ρ δ (δ δ u z =, 0 x δ ν g ν η And in the etal on the botto: g co β δ x + (ρ g ρ δ (δ x u z =, δ x δ ν η hen we take the appropriate derivative, both of the equation and all four of the boundary condition are atified. The velocitie are decreaing with repect to x, o the axiu velocitie are at the top of each layer; firt for the gla at x = 0 (note both denitie given in the book are three order of agnitude off, like titaniu on tet : g co β δ δ δ + (ρ g ρ δ (δ δ u ax,gla = + νg 9.8 ( (0 3 + ( 0 3 (0 3 kg 3800 u 3 ax,gla = kg u ax,gla =.57 ( = 0. And for the etal at x = δ : g co β δ δ + (ρ g ρ δ (δ δ u ax,etal = ν η ( 0 3 ( kg (0 3 u 3 ax,etal = kg ν η u ax,etal =.57 (7.565 =

7 A for the average velocitie, thoe will be given by the flow rate divided by the area, in thi cae: ū = u z dx δ where δ i the thickne of the appropriate layer. For the top gla layer, thi give u: δ g co β ( δ x + δ δ + (ρ g ρ δ (δ δ u gla = dx δ 0 ν g ν η [ x x 3 /3 δ ] δ g co β δ δ x + (ρ g ρ δ (δ δ ū gla = + x δ ν g ν η 0 g co β ( δ δ δ + (ρ g ρ δ (δ δ ū gla = + 3νg ν 9.8 ( (0 3 + ( 0 3 (0 3 kg 3800 ū 3 gla = kg u gla =.57 ( = 0.7 No big uprie here: the axiu and iniu velocitie were very cloe, thi i between the. The etal i a bit trickier: δ g co β δ x + (ρ g ρ δ (δ x u etal = dx δ δ δ ν η [ g co β δ ] δ x x 3 /3 (ρ g ρ δ (δ x x / u etal = + (δ δ ν η δ g co β δ 3 3δ δ + δ 3 δ δ + δ + (ρ g ρ δ (δ u etal = (δ δ 3ν η g co β ( δ δ δ δ + (ρ g ρ δ (δ δ u etal = 3ν η ( 9.8 ( ( 0 3 ( 0 3 (0 3 ( kg u = 3 etal kg u etal =.57 ( =.9 Thi i jut above half of the axiu velocity in the etal, o the etal velocity i roughly linear. hen we plot the velocitie in the gla and etal layer together, it look oething like: η 7