00:01
Let's find out the heat flow rate for this composite mass.
00:05
So the given data is the conductivity for slab a is 152 watt meter degrees celsius.
00:16
For b, this is 31 and for c, this is 68.
00:25
And for d, this is 53 watt meter degrees celsius.
00:32
So if you find out the heat flow rate, you know that the heat transfer is temperature difference divided by resistance so the temperature difference is you can see 400 degrees celsius and this is 60 so minus 60 so this will become 3 40 degrees celsius is the temperature difference now we are finding out the resistance so you can see this diagram so b and c are in parallel this is b, c, a, and d.
01:09
So let's find out the resistance.
01:12
So r a, the resistance, this is r a.
01:16
So if you find out this, this is thickness of the wall a, conductivity, and the area.
01:26
So the thickness is 3 centimeters.
01:30
So we can write in meters.
01:32
The conductivity for a is 152.
01:36
And if you find out the area this is 10 centimeters so we can write point one and this is point one so the area is this cross -sectional area we are considering so the resistance for slab a is 0 .01973 similarly for b this is tb kb and this is ab so if you find out this 0 .0 8 this is 8 this is 8 centimeter thickness of the wall b.
02:14
Conductivity is 31.
02:16
The cross -sectional area is 0 .10 .03.
02:19
All the units are in meters.
02:21
So we can easily get the resistance after solving this...