Approximation Problem

Academic year

2023
Author

Jake Tomson
Views

0

write down the fisite diferexe approximation to dT - fax the following derivative dr ST Q. Consider a Square plate bring 4 nodes 500 500 as shown in the given dingram 1 1 lee 100 4 100 100 100 - 500c the The tamp on the upper basiany temp o the loft and Righ Randy are 100°C and tap on the tower bandaves are looce Tetcomine the temperature at the point 1 2 3 and 4 (steady state conduction) for a 1-D steady state heat Conduction equator (with internal a Tm aT + V -O. ox2 K 112 mH for a point m,n we approximal the or ax first derivatives at point m-1/2m and AX mty2 A2 as aT or 1. Troin - Train = on 1/2,2 Ar aT on then : Imis, Troin Ax

T1= (500+13+100+T) -1 T==1 (500+74+100+7) - -2 4-1-10 600 1 + 0 1 6 00 T3=1/4 -3 10-41200 T4. 1/4 (100172+1001T3) - 4 01/4200 4T1-72-T3=600 exays T1-422+74=600 from 1 71-473+74-200 T2- 471-53600 in 2 T2+T3-474-200 +Tu=600 it's the THE to in 4 +91-75*600173-474 =200 ATI-474-800 J31600 11-T4=200 TIE 459253 Ti= gottly 14 T3= 3750 +14T4 4. 15000 50 +T4 - 4x3750 4x14XTU +T4 -200 4 soo 4ST4 door 21.00 44874-2100

T1-T4-200 T1. 2001 T4. II 4T3.144-200 18th -700 2TO 105° 493-114 T4 34800-1050 also .50 3.000 -4T3 1980 XEG 1960 473-14T4=00 - to -4731274-210 3 the - 4000 36 40 2016, 1-1274==000 1ito Tu= 210 the -12 T4=-1800 80 - T4==1800 = 150 - 12 1078 T1= 200. + 150 = 350 T1=350 T4=150 the T3= -1800 + 14x180 300 121/10 It X 900 4 300 tooc 4 1 look 3 4002 500c

5.487.5% 362.5° t4=14 (700+(2++100th) b:437.5 4,412.50 t-= 4 t3 13-14 ) 800 the 3200 = = 4/3 G+122+900 4(-800-t21444) - t3-t4 1800 4+12-444 = - $ 800 4th ++++++ = 1100 =1300 = 600 = 4300- 2 -900 4t2-t3-t4 = 600-3 4t3-4t4 = 4 463-454-100 402-43-44 = 600 49X used t3= 429.1 - 4t3-4t4=100 = 20400 II 204 = 200 455 4sty = 20600 2575 t3.415 tu 20600 ur 126 t4-429.1°