00:01
Hello students, here given that thermal conductivity of slab a is 14 .5 watt per meter degree celsius and that of b is 210 watt per meter degree celsius and air is 0 .032 watt per meter degree celsius.
00:28
Area in contact is 30 percent of total area.
00:39
Temperature t1 is 220 degree celsius and t2 30 degree celsius that is nothing but t4 and temperature.
00:53
Then thickness of air gap is given 0 .025 millimeter.
01:01
So next by using resistance analogy we can consider total resistance is given by resistance of 1 2 plus 1 by resistance of air plus 1 by resistance of contact power minus 1 plus resistance of 3, 4.
01:36
Now from this assuming unit area we have to find resistance of earth 1, 2 we can consider length of a divided by thermal conductivity of a.
02:00
From this we obtain 8 .27 into 10 raised to minus 3.
02:07
Next resistance of air is given by thickness of the gap which is 0 .025 millimeter divided by thermal conductivity of air into 0 .7 into area is unit area.
02:29
From this we get 1 .11 into 10 power minus 3.
02:34
Similarly for contact again substitute for thickness of the gap thermal conductivity of b that is contact into 0 .3 into unit area...