University:
Massachusetts Institute of Technology
Course:
2.58J | Radiative Transfer
Academic year:

2019
Views:

113

Pages:

9
Author:

camilobonilladz36

2.58 HW3 Solutions Prob 2.6 For this problem, we only consider normal incidence. (a) m. air m. Ngo ms Pt The reflection coefficient of a slab is givenby. hathse 42: 2mad λο 3 Where ris m-ma mitm The reflectivity is: R = |r| ^ 2 = 0 0.614 (b) For the average reflectivity, We can use eqn. (2.128) R = Pat 1- Pia Pase-zkad + rho_{12} = r_{12} ^ 2 , rho_{23} = r_{23} ^ 2 , R-P12 2k2 d=-Were k_{2} = (4pi*k_{2})/lambda_{6} +11-Wen R = 0.4 , d = 404.6mu*m For such a thick slab, the interference effects will rarely be observed. Prob 3.31 For a single slab of glass, egn. (3.89) gives R= PatPat, where Pu Pu -tau = e ^ (- (4pi*k_{a}*d)/lambda_{0}) Mair-Mglave Mair + Mglans Σ At lambda_{o} = 0.6mu*m , Pa=P23=0.0422, T = 0.9387 R= 0.0764 For the double-layer glass, apply eqns. 3.100 and 3.101 R= R+ TR 1-R 29 T = (T_{1} ^ 2)/(1 - R_{1} ^ 2) Where T. = (-) (1-4) 1-Pasで = 0.8625 R=0.o>64+ 8.86 i5 x s.o>64 i-0. s > 64 ^ 2 =0.1335 T = (8.86pi ^ 2)/(1 - 0.0764 ^ 2) = 0.7483 1 TVPOTREST 247 23(PORT) The effective absorptance of the solar collector is: Left = 20T+ (PBR)2T+ (PAR)2+……… 237 1-PR 0.90 x 0.7483 1-0.10x0.335 0.683 68.3% of the normally incident solar radiation is absorbed => Monte Carlo: Prob 5.34 h For this particular problem, it is relatively easy to set up the scheme of Monte Carlo Simulation. A (1) First of all, both surfaces eliminates the trouble the spectral dependence. are black which of taking into account (2) Secondly, the emission from both surfaces is diffuse Which allows us to use simple correlation to generate a bundle: Q= Sin Ro {५= 2TR9 Then we can determine the location where the bundle hits: X; = X + repAG cosy = Xit (3) We we Nij to represent the numble of bundles emitted from segment i that reach segment), We car write energy balance for Σ &i=lbi - 2+ 2 lbj N -each of the segment. froms. Where Li=1 and N; is the total bundle emitted-The next step is to solve the group of equations we obtain in the above, which can either be done by iteration or matrix elimination (Gaussian, LU, etc.)

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