Question

АаВЬ АаВЫС АавьСcl АавьСcl AaBbc Tolo sc1 نوکیا العنوان bloul 2 الره 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 s = 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 TA Constant Temperature, T-o'c Constant Temperature, T-Bo'c I Constant Temperature, T-50°c Prepared by: Reviewed by Beden T. Luaran Dr. Belen Luna Head of Mathematiese Verified by: Monte Dr. NRM. Neaman Associate Deo Approved by D. Bet Dr. Beda T. Aleta Dr. Husham M. Ahmed Faculty Member Teachers IS

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

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 !!!

Answer 1

- State temperature of plate is given Q. Given Heat thee, Son K= 0.5 cal / (s.com.°c) the steady - DT + 87 ду» dze 2 - let Ti im thee figure be Ti, 3 Chere lipcolumn 3 is now) To T2,3 ; 73 733; Tu Tirits T2,2 To> 13,2 ; T7 Til i Tg > T2, 1; Ta> T3,1 Discere fized Tito, g + Tili+T; j-1 t.Ti, jt - 4T , j = 0 Tij Titi i to + Tij tToj jh + T5,5-1 Equation is 7,5-) 4 for Guess - Saidel metlice Assuming all interneel nodes temperature is Zelo constant temperature T-80°C 18+ texation : T3,1t + Tori +7,277,0 = Ty = Ti 4 - २००८ = 0+0+0+80 4 Ti = T1,3 = ( 723 + To, 3 + Ti,u+T2, 2) 4 Otot 93+5 24,5°C 4 = 24.5°C

14 = 1, = T2,2 + 10, +3,3+,,) 4 (0+0+0+20) -5° 1) - Таз (Т3, 3+ Tз+11,ч+T, 2. Ч "o+ 99-5€ 93+2 ач 54 93+9) - 2933г. *Тър.) 3 - Тъд -/Tч,3+ ta,2 + T3,4 +Tз, 4 80 +29 375 930 4 Бо. 59°a ) 6 = T3,2 =/T4, 2t 12, 2+ T3,34T3,1 2 ) 4 - 80 +0+ SO.S9+ь 4 Ta = T3,1 = Tullt Tai it T3, 2t 73,0 4. (1 80 +0+ 32-6+go - Ч8, 26 с **) - To = 3,1 = (Bitt, itt, 2t To 4 Ч. 18+ 20t Offo – 37 .0 °C 1 15-Тана - Т3,2 to+T2,3 + qa.5 +5+ 29.375+37.09 с

Ts - 26-02 °C the above desired temperceluxes are for & = 1067. от S = 16°/. prescuit Prolux 8, 1 T, present present T, Peurou8 This is from prefent Рекои - Ті, і (Till 0, 9 a boue o. Ti,, present п, precious 11 Present pregenet Т. О. 1 Ти) present Т. Preleious 0.9 T,,, 2 present Previous T, Т), 0.9 for &=10%. Т1 Т., 1 o'c T,, Рейну - а. 22°C - 0-9 prellious Ti Ті, Ті, з 0.4 29. S9 = 27. 22 0.9 24.37 PXPuous T, 2.3 32-63 (т 3 0.9 0.9 13,3 Previous Т3 - 12/3 Бо- S9°C 0.9 - 56. 2ја 0.9 polllidus Ty= T1,2 5° о. 9 S. 55°C ) 0.9 "J5 - 19,2 (19Очевоци * ) 26.02 09 - 28.91 °C 0.9

precious 139 3 Те: 2. 32-63 0.9 = 36-2-3са o 9 11 Ті, Tirl precious gooc 22. 22 °C - об.а 0-9 ТЕ. Т2 - T211 Precious - 37-69 °C о: - ч)-1 °C О. 9 Па = 131 1) 33, Росиоца ч -16°.53-5 ) 0.9 0.9