00:01
Hi there, so for this problem we are told that a radioactive waste is packing a long thin wallet cylindrical container.
00:09
The waste generates thermal energy not uniformly according to the following relationship.
00:15
That is that the heat generated is equal to the initial heat q -0 and this times 1 minus the radius r.
00:33
Divided by the radius r sub 0 and that to the square, where q sub 0 is a constant and row sub 0 is the radius of the container.
00:47
So, steady state, conditions are maintained by submerging the content in a liquid at a temperature, that we're going to call t the temperature at infinity, and provides an uniform convention coefficient.
01:08
In h.
01:09
So we need to obtain an expression for the temperature, ts.
01:14
So we need to find that expression for ts of the container wall under steady conditions.
01:27
So with that said, do solve this problem.
01:36
We start by first find the rate at which the thermal energy is generated inside the s sphere.
01:44
So as specified in this problem.
01:47
The heat generated per unit volume is only a function of the radius.
01:53
So the total heat generated inside a sphere can hence be taken to be the following.
01:59
So the total heat generated is equal to the integral.
02:14
So this is from the radius equals to zero to the radius equal to r sub 0, and then this times the heat generated, which is the expression that we are given, and this times the shell volume, which is 4 times pi, times the radius squared times the differential in the radius.
02:40
Now, since the expression of the heat generated is given, we substitute that in heel, so the total heat generated is equal to the integral, the integral, from the radius equal to 0 to the radius equal to r sub 0 is equal to q sub 0 times 1 minus the radius divided by the radius r sub 0 and that to the square.
03:08
And all of this times 4 times pi times the radius to the square.
03:17
So first, what we are going to do is to multiply this by each term in here.
03:24
So we will have to follow in the integral from the radius.
03:29
Equals to 0 to the radius equal to r sub 0 of the heat q sub 0 times 4 times pi in here we're gonna have the radius to the square minus the radius to the 4 divided by the radius of 0 to the square is integrated over the radius so we just now do the integral in here so we will have q sub zero we can take we can take out q sub zero out of the integral four times pi so though the solution is the radius to the three divided by two this is the integral of the first term in the second term we are going to have the radius to the five divided by five times the radius of zero to the square of course we need to evaluate this from zero to the radius r sub zero so so at the radius 0, we know that it's going to be 0, but at the radius r sub 0, we're going to obtain the following q sub 0 times 4 times pi, the radius r sub 0 to the 3 divided by 2.
04:53
Oh, sorry, in here it is divided by 3.
04:55
Sorry, there is divided by 3.
04:59
And this minus, in here, if we substitute the radius r sub 0, we can cancel 2 with 2 in 2.
05:08
The denominator.
05:10
So we're going to call the radius of zero to the three divided by phi...