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

Answer:

$q_{out} = 9.25\,\frac{kJ}{kg}$

Explanation:

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

$\omega = \frac{0.622\cdot \phi \cdot P_{g}}{P-\phi \cdot P_{g}}$

The saturation pressure at $T = 27^{\textdegree}C$ is:

$P_{g} = 3.601\,kPa$

Then,

$\omega = \frac{0.622\cdot (0.5)\cdot (3.601\,kPa)}{101.325\,kPa-(0.5)\cdot (3.601\,kPa)}$

$\omega = 0.0113\,\frac{kg\,H_{2}O}{kg\,air}$

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

Initial state:

$h_{1} = (1.005\,\frac{kJ}{kg\cdot ^{\textdegree}C})\cdot (27^{\textdegree}C)+(0.0113)\cdot (2551.96\,\frac{kJ}{kg} )$

$h_{1} = 55.972\,\frac{kJ}{kg}$

Final state:

$h_{2} = (1.005\,\frac{kJ}{kg\cdot ^{\textdegree}C})\cdot (18^{\textdegree}C)+(0.0113)\cdot (2533.76\,\frac{kJ}{kg} )$

$h_{2} = 46.722\,\frac{kJ}{kg}$

The specific energy that is removed is:

$q_{out}= h_{1} - h_{2}$

$q_{out} = 9.25\,\frac{kJ}{kg}$