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

`\omega_(out) = 0.867\,(kg\,H_(2)O)/(kg\,DA)`

Step-by-step 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) = (\dot m_(in)\cdot \omega_(in)+\dot m_(w))/(\dot m_(out))`

`\omega_(out) = \omega_(in) + (\dot m_(w))/(\dot m_(out))`

`\omega_(out) = 0.6\,(kg\,H_(2)O)/(kg\,DA) + (0.4\,(kg\,H_(2)O)/(s) )/(1.5\,(kg\,DA)/(s) )`

`\omega_(out) = 0.867\,(kg\,H_(2)O)/(kg\,DA)`