Question

A stream of moist air flows into an air conditioner with an initial humidity ratio of 0.6 kg(vapor)/kg(dry air), and a dry air flow rate of 1.5 kg/s. If the moist air stream mixes with a separate stream of water vapor at 0.4 kg/s, what will the humidity ratio be at the exit in kg(vapor)/kg(dry air)?

Answer 1

Answer:

$\omega_{out} = 0.867\,\frac{kg\,H_{2}O}{kg\,DA}$

Explanation:

The final humidity ratio is computed by the Principle of Mass Conservation:

Dry Air

$\dot m_{in} = \dot m_{out}$

Moist

$\dot m_{in} \cdot \omega_{in} + \dot m_{w} = \dot m_{out}\cdot \omega_{out}$

Then, the final humidity ratio is:

$\omega_{out} = \frac{\dot m_{in}\cdot \omega_{in}+\dot m_{w}}{\dot m_{out}}$

$\omega_{out} = \omega_{in} + \frac{\dot m_{w}}{\dot m_{out}}$

$\omega_{out} = 0.6\,\frac{kg\,H_{2}O}{kg\,DA} + \frac{0.4\,\frac{kg\,H_{2}O}{s} }{1.5\,\frac{kg\,DA}{s} }$

$\omega_{out} = 0.867\,\frac{kg\,H_{2}O}{kg\,DA}$