Respuesta :
Answer:
(a) The proportion of dry air bypassing the unit is 14.3%.
(b) The mass of water removed is 1.2 kg per 100 kg of dry air.
Explanation:
We can express the proportion of air that goes trough the air conditioning unit as [tex]p_{d}[/tex] and the proportion of air that is by-passed as [tex]p_{bp}[/tex], being [tex]p_{d}+p_{bp}=1[/tex].
The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:
[tex]0.004 = 0.016*p_{bp}+ 0.002*p_{d}[/tex]
Replacing the first equation in the second one we have
[tex]0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143[/tex]
(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.
It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.
The water removed of every 100 kg of dry air is
[tex]100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW[/tex]
It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).
100 * 0.012 = 1.2 kgW
