00:01
Okay, this is our question.
00:03
We have a sphere at 120 degrees celsius and in the second part of the question we have an insulation and this is insulation and let's say the insulation is of thickness t.
00:16
Okay, so we can straight away write that t is equal to r0 minus r i and our i is given to be 0 .5 meter because the diameter is given to be 1 meter see let's first analyze the situation when we do not have an insulation.
00:35
So straight away we can write by the newton's law of cooling, q1 equals delta t into a into h.
00:45
And we might write a minus because we have to make the heat transfer positive.
00:52
So we'll be getting, substituting the values.
00:57
We have dts as 15 minus 120 multiplied by the area.
01:07
Since it says sphere it is 4 pi r square and multiplied by the h the coefficient so h in this case is given to be 25 so we get an answer 8246 .68 watts so this is our first part of the problem without insulation and now let's do the problem with the insulation so before that we have to derive an expression for a sphere which is having r i as initial radius and r o as the final radius so we can write q by a equals q by a can be written as minus d t by d r into k this is the basic heat conduction equation so area can be written in this form 4 pi r square and taking d r to this side we'll be getting an expression like this and let's integrate both sides and finally after integrating it and applying the limits the limits is r i to r o t i to t o if we're finally getting an expression like q equals k into t o minus t i into four pi and divided by one by r o minus 1 by r i it is simple integration nothing else and we can also do this in the form of the resistance method whereby we can mark the resistance as can take this k and 4 pi down here and mark this as the resistance we can also do the problem in that manner okay so now this is the expression that we get for the heat transfer and let's supply the values okay this is actually the heat that is conducted heat that is conducted from the surface of the sphere to the outer surface so in the steady state this heat must also be the heat that is given away to the fluid okay that is the concept that we are using here and let's apply the values and solving it we get an expression like minus 20 .106 divided by 1 by r o minus 2 and we're substituting every values because we know what is ri we know t t -i -t -o -k.
03:55
In this case, t -o is the outer temperature of the sphere.
04:00
It's given to be 40 degrees celsius.
04:02
And t -i, we know it is an inner temperature sphere.
04:04
Okay.
04:05
So to clarify it, it's given that the sphere is having an outer temperature of 40 degrees celsius and inner temperature of 120 degrees celsius.
04:13
We're directly substituting the values.
04:16
Okay...