00:01
In this question we have the value of a weight w which is equal to 0 .015 kg.
00:08
Now here this kg of the dry air.
00:12
Now we have the dry bulb, dry bulb temperature which is equal to 30 degrees celsius.
00:21
We can write it as tdb.
00:24
Then we can write the pressure of air which is equals to 1 .1.
00:30
0 .05 kilojoule per kg, kilo -kgy, and we can write pressure of water vapor, which is equal to 1 .88.
00:47
Now, using this value, let's first find out.
00:51
Here we have this pb which is equals to 85 khal okay so we can write the h which is equals to cp multiplied by this t d b plus w multiplied by h g at pv plus c pv multiplied by t d b minus t saturation so on simplification we can write w which is equals to 0 .622 multiplied by p v divided by p b minus pv therefore we can substitute all the values here so it becomes 0 .0 .15 which is equals to 0 .622 multiplied by pv divided by 85 minus pv.
01:59
So this becomes pv which is equals to 2 kilo pascal.
02:05
Now from the steam table, from the steam table we can write t t which is equals to t saturation which is equals to the 17 .51 degree celsius and hg which is equals to 253 .3 .0 .0.
02:27
Kilo temperature per kg.
02:32
And this both are at pv, which is equals to 2 kilo pascal.
02:40
Okay, so this is from the steam table.
02:44
Now using this value, let's substitute all the values in the equation number one...