Moisture Remove Room Air Flow

Overview

Engineering reference data for Moisture Remove Room Air Flow in air psychrometrics.

Key Formulas

Humidity Ratio

$\omega = 0.622 \frac{P_v}{P_a}$

Mass of water vapor per mass of dry air.

Relative Humidity

$\phi = \frac{P_v}{P_{vs}} \times 100\%$

Ratio of actual to saturation vapor pressure.

Wet Bulb Temperature

$T_{wb} = T_{db} - \frac{P_{vs} - P_v}{\gamma}$

Temperature measured by wet-bulb thermometer.

Enthalpy of Moist Air

$h = c_p T + \omega h_g$

Sensible + latent heat per unit mass of dry air.

Variables

Example Calculation

The required air flow to remove moisture from a room can be calculated using the primary formula.

Given:

• Target room humidity ratio, $x_r = 0.010\ \text{kg}_{\text{H}_2\text{O}}/\text{kg}_{\text{dry air}}$
• Make-up air humidity ratio, $x_m = 0.001\ \text{kg}_{\text{H}_2\text{O}}/\text{kg}_{\text{dry air}}$
• Moisture production rate, $G = 100\ \text{kg/h}$

Calculation:

$$L = \frac{G}{x_r - x_m}$$ $$L = \frac{100\ \text{kg/h}}{0.010\ \text{kg}_{\text{H}_2\text{O}}/\text{kg}_{\text{dry air}} - 0.001\ \text{kg}_{\text{H}_2\text{O}}/\text{kg}_{\text{dry air}}}$$ $$L = \frac{100\ \text{kg/h}}{0.009\ \text{kg}_{\text{H}_2\text{O}}/\text{kg}_{\text{dry air}}}$$ $$L \approx 11,111\ \text{kg/h}$$

Conclusion: An air flow rate of approximately 11,111 kg/h of make-up air is required to maintain the desired humidity level in the room.

References

• Original Source