00:01
So let's start with the statement for our question.
00:03
It says that there is a small electric heating application that it uses wire of 2mm diameter diameter.
00:10
Right.
00:10
So let's write it down.
00:12
So the diameter of the wire is equal to 2mm .m.
00:21
And then it says that it has 0 .8 millimeter thickness, thick insulation, right? so the thickness of insulation that is equal to 0 .8 millimeter.
00:42
And then it says that k holds a value of 0 .15 watt measure per degree centigrade.
00:56
Right and it says that the heat transfer coefficient h .o.
01:03
That is equal to on the utility surface is equal to 31 watt per meter square degree centigrade, right? then it says that we need to determine the critical thickness of insulation, right? we need to determine the critical thickness.
01:24
So let's write it down as well.
01:26
So we need to determine critical thickness of insulation, right? and it says that r sub 1 is equal to, now what we need to do over here is that first of all, we need to determine the first radius, right, which is this one.
01:53
This r sub 1 radius.
01:55
Now, it would be basically 2mm divided by 2.
01:59
That would be equal to 1 millimeter and right and when we convert it into meters so that would be equal to 0 .001 meters right then similarly we need to calculate this r sub 2 right so this is basically the sum of this r sub 1 plus this point 8 so it's basically some of these two distances right so r sub 2 is r sub 1 is basically 1 millimeter right so 1 millimeter plus this 0 0 0 0 0 0 0 0 0 0 0 0 0 0000 right so that is equal to 1 .8 millimeter and again we need to convert it into meters a unit so r sub 2 is equal to 0 .018 meters right now that we have these two radius right this r sub 1 and r sub 2 right now what we need to do over here is that the next step in order to solve this question is to determine the r sub c, right? that is basically the critical radius of insulation.
03:11
So r sub c is equal to k divided by 8 sub o, right? it's sub o is the heat transfer coefficient, right? so what we need to do over here is that we simply need to substitute the values, right? so when we substitute the values, we get that's equal to 4 .383 .3 .3 .3.
03:33
838 into 10 raised to 1 negative 3 meters, right? or if we convert it into millimeter, so that would be equal to 4 .838 millimeter, right? now what we need to do is that now that we have r sub c, this critical radius, right? now what we are going to do over here is that now we are going to calculate the critical thickness.
03:57
And we know that the critical thickness, is equal to the difference of the critical radius, negative r sub 1, right? so now what you're going to do is that we are going to simply substitute the value.
04:18
So for my own easiness or whatever you can say, for my own benefit, i'm going to use the millimeter units while calculating it and then i'll convert it into meters, right? so 4 .838 negative 1 millimeters, right? so that is equal to 3 .838 meters, right? and this would be equal to this critical thickness would be equal to 3 .838 into 10 x .538 into 10 raised to bar negative 3 meters.
04:57
So this is the solution for the first.
05:01
And for the second part, it says that we need to calculate the percetage.
05:08
Change percentage change in the heat transfer rate right so how we are going to do that is that we're going to first of all solve two cases right now the first case would be that case one would be the heat flow through an insulated an insulated wire, right? and this heat flow through an insulated wire is given by q sub 1, right? and that is equal to 2 pi times a times the difference of the temperatures in the first and the second that is er.
06:02
That is divided by natural log, r sub 2 divided by r sub 1 divided by k, plus we have one divided by 8 sub o times r sub 2.
06:15
Now what we are going to do over here is that we know we do not have any of the variables or the parameters that are given in the numerator, but we do have all the values that are written over here in the denominator.
06:27
So what we are going to do is simply that we are going to plug in the values over here in the denominator, right? and then we are going to solve q in terms of the numerator that will remain the same.
06:38
And for the denominator, we would have some value.
06:41
So let's do it.
06:42
Right so u sub 1 is equal to 2 pi l t sub 1 negative t sub a right and that is divided by so this natural r sub 2 divided by r sub 1 is equal to 0 .5877 right and k holds the value of 0 .15 so that just comes as it is right and this h o times r sub 2 is equal to 0 .56 8, right? so let me calculate this denominator.
07:17
We would get that q sub 1 is equal to the numerator remains the same, right? and the denominator is equal to 21 .52.
07:29
Right.
07:29
So let's say that this is equation number one...