Question

فراشة مراسلات 31 AaBb AaBbce MaßbCal Aa Bbce AaBbc Tulous اعاده اتهامد 60 Part 2: Implement an analytical solution for the following problem: Use Liebmann's method (Gauss-Seidel) to solve for the temperature of the heated plate in the figure. Employ overrelaxation with a value of 1.5 for the weighting factor and iterate to e, - 10%. Then find the rate of heat flux (9) across the plate's surface at the points shown. Assume that the plate is 40 x 40 cm and is made out of aluminum with the coefficient of thermal conductivity K=0.5 cal/(s. cm. C) Use the last two digits of your ID for the missing temperature at the top of the plate. Constant Temperature, T-c T. Tg Constant Temperature, T-o'c Constant Temperature, T-80°C Constant Temperature, T- soc Prepared by Approved by: Reviewed by: Belen T. Lumeran Dr. Belen Luman Head of Mathematics & Verified by Wenn Dr. Ned. M. Nema Associate Dean Dr. Husham M. Ahmed Faculty Member Teachers Dr. Beda T. alte المتحدة

Implement an analytical solution for the following problem:
Use Liebmann’s method (Gauss-Seidel) to solve for the temperature of the heated plate in the figure. Employ overrelaxation with a value of 1.5 for the weighting factor and iterate to ε_s = 10%. Then find the rate of heat flux (q) across the plate’s surface at the points shown. Assume that the plate is 40 Χ 40 cm and is made out of aluminum with the coefficient of thermal conductivity K= 0.5 cal/(s. cm . ºC)
Use last two digits of your ID for the missing temperature at the top of the plate.

last two ID 14
please I need answers!!

Answer 1

Sol Given data :- ei Employ oveqqelazation 1:15 te per oror Weighing facto Es 10% Dimensions of the platé - 40 x 40 cm Coefficient of thermal conductivity K=05 cal The steady-state temperature of plate is given by 27 a aya Let I, is the figure be Ti 13 T2 > T2,3 T3 T3,3 T T1,2 57722 To T312 Tz Til To Tail Bil Discretized coun is t ao 2x2 AT Titl, jt TitT. i, jt, -uti, j =0 Tiij = Titujt Tinij & Ty, itt & Ti, ja 4

method cotob for gauss - seidel Assuming all integral model temperatuge guers is Zero, a 18t itigation 73, + T3 & T, 2 + 7,2 potent un 2 = T., desorgto rod worden NA 1714 wir 0+0+0+8094 sitio est, stuermon grorny broer tool T7 200 +T1,2 Ti = TiB = T2,3 + To, 3 + 7,4 4 0+0+93 +5 = 24.50 4 ototo+20 520 T = T12 - T212 +70,2 +7,3 tt,11 4 Tz = To +72,2507 13,3 +7,13 +72,4 13,3 - 0+245493 to 29.375 41 oll: to calenta 4 T3 - T3,3 s Tu, 347073 7'3,4+32. a 80129.375+9370 - 50-59€ ud ove borayot statut To = 3,2 Tu, 2472,2 +3,3 +73,1 4 32.658 80+0+50:59+0 4 4 a Tuil + T2,1 + T3, 2 + 730 Tg = T3,1 48.16 € 80+ 0+32.65780 4 q T3,1 + Tot +T+7210 Ts 48.46+20+0480 212 Tail 37.06c tie 40 Tg = T2,2 = 732 +T 12 +223 +T21 32.657542934 431-04 4, 1 4. 26.02°C

We found the above derived temperature are fa Eg - 100% for E = 10 present T., previous E, present present T, 1 Til previous 0.9 for &= 10% T., Previous 77 - Tel 20 09 = 22.22°C 0.9 24.5 0.9 = 27.22°C previay T = T13 09 previous T2 = 09 T23 29.375 32.64 € 0.9 T3 - 13,3 56.210 50.59 0.9 5. 55 c 4 Food 2 26.01 28. auc Tg 09 To 32.65 0.9 = 36.280 Tg 37.01 0.9 - 41. 12°C 48.16 53.515 Ta 09