00:01
So the first thing that would want to do for this question is to determine the outer surface area of the tank.
00:06
We are doing this because the value of q representing the heat flow for a convection process is equal to h multiplied by a, multiplied by t infinity minus t at the surface.
00:21
So we would need to first of all determine this value of a and this is pi d squared.
00:27
So we are looking at a pi multiplied by 3 .02 raised to the power of 2, which gives us a 28 .6 in units of square meters.
00:39
So this gives us a qc that is equal to 30 multiplied by 28 .66 multiplied by 25 minus 0, giving us a heat transfer rate that is equal to 21 .5 in units of kilowatts.
00:55
Then the rate of heat transfer through radiation, q, rad, this is going to be ea sigma multiplied by t raised to the power of four of the surroundings minus t raise to the power of the surface.
01:12
So we are looking at 0 .75 multiplied by 2 .8 .66, multiplied by 5 .67 by 10 to the power negative 8, all of which is multiplied by 288, 8, all of which is multiplied by 288, raised to the power of 4 .6 ,000...