Question

Radiation from a heater is used to heat the still and produce vapor steam as shown here. The heater is maintained at 1000C. The surrounding temperature is 27C. By employing a hypothetical surface for the side surface between Al and A2 to form an enclosure, a radiative circuit can be plotted as shown. Take D. = 200mm, D2 = 250mm, H = 100mm.&z = 0.8. Water latent heat of vaporization is 2.26 x 106J/kg. vapa 18 RE J E J 1. Ry RE R Coaxial Parallel Disks (Figure 13.5) R-T/LR-/L 1 +R S 1 + R J & Ro Water, 100°c - sunfua , & -Heater, semfan Az, Eu Jan FX = 115 - 15° - 45,/10943 Q1(5pts): Determine R4, R3, E61, Eby, and Eb3 in the circuit. Q2(15 pts): Determine view factors F12, F13, and F32. Q3(5 pts): Determine space resistances R12, R13, R23.

Please help and solve. I am just learning this and do not understand it. Thanks.

Answer 1

Assuming Steady State conditions, surrounding room is large and therefore behaves as a black body. and Ez = 1 09 QB) R, and R3 - surface desistances; R = 1- E, Eg An 1 - 0.8 = 7.961 0,8 (40,22) 0 Ebb Kelvin R3 1-E3 1 - 1 Ez A3 Ez A3 Emissive power of black body Ebg = ST4= = 5.67X 10-8 (Lot 273) 1097,535 W/m2 Ebg = 872 ² = 5.67x108 (1000+ 273) * = 148 900.677 W ma Eb3 = 8 T3 t = 8,678108 ( 27 +273) 4 459.27 W/m2

Q2 for coarial parallel disks from the Fij formula given in question ¿ 2 སེབ་ 1 Dr/2 250 Ri = R2 = 2 2 H - 1.25 1o Rj = Rs = de es D/2 20% = = 1 H To It R2 1 + 12 S = 1 + 1 + = 2.28 R 2² 1.25² = 1 Careers 2 29²2 250/ Fij = F2 s= 1 {s - [524 ( 22 23 ²4 1/2 [2.28 - [2282 4/200/2 F21 = 0.327 from reciprocity relation A, FI2 = A2 F21 faz = A2 for = 250² XO. 327 = 0.512 I 2002 40 from summation rule & fij = 1 jet F12 t Fiz + Fr. = 1 the 0.512 + f 13 to = 1 Faz= 0.487

3 R = 125mi F32 = A 2 F23 = I x 2.50² lateral surface area of surfare 3 shape of frustum of cone Which is in ♡ Azi Ala+R) SR-r) +h² de somm -1 (50+125)S(125-50)+1002 halomo = 68722, 34 mm? Now E fij = 1 i for surfare 2; F22 + f21 + 723 = 1 o +0.327 +823 = 1 F2 3 = 0; 673 Reciprocity melation A 2 F23 = A₃F32 A3 XO.673 68722.34 f32 = 0.480 Q.3) space resistances: 1 1 R12= Az F12 - 248,805 4 x 0.12x0,512 heter met R13 11 = 261,578 A, F13 7x0,1² 0,487 223 = 23 7 x 0.25² x 0,673 ܚܙ = 30.285

Q.4) for Node J2 Jy - Ja J3 - J2 R23 t Ebz-J 2 R2 + Raz node J3 J, - J3 30 t E63 - J3 J2-J3 R23 + 0 R13 R3

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