Question

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 K= 0.5 cal/(s. cm . ºC)

Use the last two digits of your ID for the missing temperature at the top of the plate.

THE ID IS 58

Constant Temperature, T- °C T, T, T, T. Constant Temperature, T-0°c Constant Temperature, T-80°C T, T, - Constant Temperature, T-so'c

KINDLY AS I NEED IT BEFORE 21/7

Answer 1

So now looking back at the original point difference equation wel get, To + T3 tt Ta +2T Tot Tst 2T 4 4 for this st is mean that for isdated edge ure the alone modified difference earation to solve for to at each on that edge. Start with all grid point at Tao, This now include those additional 3-point on the power edge. Now I will show your solution By hand for one Interation, then write a matlas to solve For point T₂ +T+2T, a 62.5.0 8ot do +2786 T Now Apply over relaxation step to improve conna -genee we have given de 1.5 Now T = IT, + (1-d) T = 11TX $2.5+ (1-15) TZ 9350 xo 93.75.0 For point. Tz Tz + T + 2Ti 0493.75 +2x0 4 q Tz 23.43 T + T +2T, Tз do tot 2x0 z 40 4 z 4 CS Scanned with CamScanner