Question

Air at 27oC and 50% relative humidity is cooled in a sensible cooling process to 18oC. The air is then heated to 45oC in a sensible heating process. Finally, the air experiences an adiabatic saturation process that increases the relative humidity back to 50%. Find the specific energy that is removed when the air is cooled to 18°C.

Answer 1

`q_(out) = 9.25\,(kJ)/(kg)`

Step-by-step explanation:

First, it is required to find the absolute humidity of air at initial state:

`\omega = (0.622\cdot \phi \cdot P_(g))/(P-\phi \cdot P_(g))`

The saturation pressure at
`T = 27^(\textdegree)C` is:

`P_(g) = 3.601\,kPa`

Then,

`\omega = (0.622\cdot (0.5)\cdot (3.601\,kPa))/(101.325\,kPa-(0.5)\cdot (3.601\,kPa))`

`\omega = 0.0113\,(kg\,H_(2)O)/(kg\,air)`

A simple cooling process implies a cooling process with constant absolute humidity. The specific entalphies for humid air are:

Initial state:

`h_(1) = (1.005\,(kJ)/(kg\cdot ^(\textdegree)C))\cdot (27^(\textdegree)C)+(0.0113)\cdot (2551.96\,(kJ)/(kg) )`

`h_(1) = 55.972\,(kJ)/(kg)`

Final state:

`h_(2) = (1.005\,(kJ)/(kg\cdot ^(\textdegree)C))\cdot (18^(\textdegree)C)+(0.0113)\cdot (2533.76\,(kJ)/(kg) )`

`h_(2) = 46.722\,(kJ)/(kg)`

The specific energy that is removed is:

`q_(out)= h_(1) - h_(2)`

`q_(out) = 9.25\,(kJ)/(kg)`