00:01
Okay, so for this problem, we're told that in a cylindrical pipe where the area isn't constant, equation 16 .5 is going to look like this.
00:14
And we want to use this to show what the heat loss rate from a cylindrical pipe should be.
00:23
So in order to do this, i first want to substitute in the area for the surface area of a cylinder.
00:34
So h then is going to become equal to negative k times the surface area of a cylinder is going to be 2 pi r, l, where r is the radius of the cylinder and l is the length of the length of the.
00:54
The cylinder and then we're still multiplying by d t d r here so i'm going to get the r's on the left hand side leave d t on the right hand side so we'll end up with h over r d r is equal to negative k times two pi l d t now i'm going to integrate both sides of this equation so we'll just write in our integral sign there and we're left with h times the natural log of r.
01:36
And in this case, we're going to evaluate that between r1 and r2.
01:43
And that has to be equal to, again, minus k to pi -l times t, evaluated from t1 to t2.
01:59
So we're left with h times the natural log of r2 minus the natural log of r1 equals negative k times 2 p .l times t 2 minus t 1...