Question

Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air conditioning unit which reduces humidity to 0.002 kg per kg. However, the required humidity at the drier inlet is 0.004 kg per kg and part of the air therefore must by- pass the conditioning unit. Determine: (a) the percentage dry air by passing the unit and (b) the mass of water removed per 100 kg dry air feed.

Answer 1

(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.

Step-by-step explanation:

We can express the proportion of air that goes trough the air conditioning unit as
`p_(d)` and the proportion of air that is by-passed as
`p_(bp)`, being
`p_(d)+p_(bp)=1`.

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

`0.004 = 0.016*p_(bp)+ 0.002*p_(d)`

Replacing the first equation in the second one we have

`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)=(-0.012)/(-0.014)=0.857\\\\p_(bp)=1-p_(d)=1-0.857=0.143`

(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

`100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW`

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