00:01
Hi, everybody.
00:02
So what are the required liquid water input and heat transfer rates for this purpose? so we have the pressure v1 equals 5 pg1 equals 0 .5 times 1 .276 equals 0 .6 .1 .38 kilopascals.
00:30
Okay.
00:31
And we have absolute humidity as 0 .622, and it's pv1 over p minus pv1.
00:42
So this is equal to 0 .622, 0 .6138 divided by 100 minus 0 .6138.
00:56
Okay equals 0 .0384.
01:04
And let me 84.
01:10
Okay.
01:12
And so we're going to at table b, what was it, table b 12? and so for the numbers.
01:24
So we got pb2 equals 52 pg2 equals 0 .5 times 3 .169 equals 1 .584 .5k pascal.
01:44
And again, we're going to find absolute humidity.
01:47
And i'm just going to write it quickly here as 1 .584 .5100 minus 1 .5845.
02:05
10014.
02:09
And now what we can do is find the mass of a2, and that's going to be p -a -2, the r -a -t -2, and it's going to be 100 minus 1 .5845 times 1, divided by 0 .257 times 298 .15 equals 1 .152 kilograms per second.
02:52
And your ma2 is equal to the ma1...