00:01
Hi, everybody.
00:01
So, we need to find the temperature of t2, the mass of liquid per kilogram of dry air, and the overall heat transfer rate.
00:13
So to do this, we just need to find the pv1, which is 5 times pg1 equals 0 .9 times 5 .6.
00:30
628 equals 5 .0625k pascels.
00:39
Okay.
00:41
And then we can find the absolute community of 1, 0 .622 times 5 .0625, 100 minus 5 .0625, which is 0 .03318 from the absolute community equation.
01:12
So then the property of water and salt water be three and state three for that one is going to be 0 .3 times 2 .3.
01:38
Equals 0 .7017 k pascels.
01:45
And at absolutely many of the third one is 0 .622 times 0 .7117, 100 minus 0 .717.
02:03
Equals 0 .0439.
02:07
This is also going to be absumany of two.
02:11
So now we need find the mass flow weight of air.
02:17
So from that we have 100 minus 0 .7017 times 0 .01 divided by 0 .287 times 0 .08, divided by 0 .287 times.
02:35
293 .15 and we get 0 .118 .02 kilograms per second.
02:49
Okay.
02:50
And now we can find the mass of the liquid, which is mv1 minus mv2, which equals ma of m1 of 1 minus w2, which equals ma of 1 minus w2.
03:08
Equals ma here of 0 .0333118, minus 0 .00439.
03:23
And so we put m and lma together, massive error, and we get 0 .02879.
03:34
And this is before we continue on to add.
03:38
So we'll just leave it like that, so don't worry.
03:41
So we need to attain temperature 2, and it's going to be 0 .01 plus 5 minus 0 .01 times 0 .7017 minus 0 .6113, divided by 0 .8721 minus 0 .6113...