00:01
Let's first determine the rate of heat transfer.
00:04
The rate of heat transfer is equal to our mass flow rate of cold water times our specific heat of the water, times the exit temperature of the cold water, minus the inlet temperature of the cold water.
00:19
So we plug in 0 .8 times 4180 times 70 minus 22, and we get 160 .5 kilowatts.
00:29
Now we would like to find the heat transfer surface area.
00:35
To find the heat transfer surface area, we can use this expression, q dot, is equal to f -u -a -d delta t.
00:44
F is our correction factor, u is our overall heat transfer coefficient, and delta t is our log -meant temperature difference.
00:53
We can find our correction factor first.
00:56
Our parameter p is t -c -out minus t -c -n, divided by t h in minus tc in.
01:03
So we plug in 70 minus 22 divided by 110 minus 22, and we get 0 .5454.
01:11
Now for our r factor is 10th -in minus t -h out divided by t -c -out minus t -c -n...