Respuesta :
Explanation:
The given data is as follows.
Temperature of dry bulb of air = [tex]25^{o}C[/tex]
Dew point = [tex]10^{o}C[/tex] = (10 + 273) K = 283 K
At the dew point temperature, the first drop of water condenses out of air and then,
Partial pressure of water vapor [tex](P_{a})[/tex] = vapor pressure of water at a given temperature [tex](P^{s}_{a})[/tex]
Using Antoine's equation we get the following.
[tex]ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}[/tex]
[tex]ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}[/tex]
= 0.17079
[tex]P^{s}_{a}[/tex] = 1.18624 kPa
As total pressure [tex](P_{t})[/tex] = atmospheric pressure = 760 mm Hg
= 101..325 kPa
The absolute humidity of inlet air = [tex]\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}} [/tex]
[tex]\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}[/tex]
= 0.00735 kg [tex]H_{2}O[/tex]/ kg dry air
Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg [tex]H_{2}O[/tex]/ kg dry air.
Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.
0.07 kg - 0.00735 kg
= 0.06265 kg [tex]H_{2}O[/tex] for every 1 kg dry air
Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.
[tex]0.06265 kg \times 100[/tex]
= 6.265 kg
Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.
