Question

Room capacity is 250 m3, air temperature 18 ° C and relative humidity 45%.
How many kilograms of water in the room air and how much air can fit in the air
water as water vapor? The total pressure is 101.3 kPa.

Answer 1

Solve the first part by determining the absolute humility in $\text{g/m}^3$ using the psychrometric tables. For $T=18^\circ\text{C}$ find the saturation vapor density ($\rho_{SVD}$): it equals $15.4 \text{ g/m}^3$. Hence:

$$AH=\frac{RH}{100\%}\cdot \rho_{SVD}=\frac{45\%}{100\%}\cdot 15.4=6.93\text{ g/m}^3.$$

Since we know the total mass of water in grams per one cubic meter, it is easy to calculate the total mass of water in the room air:

$$m=\frac{AH}{1000}\cdot V=\frac{6.93}{1000}\cdot 250=1.73\text{ kg}.$$

How much vapor the air can contain in these conditions is determined by the saturation vapor density ($\rho_{SVD}$) which we have found in the psychrometric tables. It equals $15.4 \text{ g/m}^3$, so the air in the room can contain

$$m_{\text{max}}=V\cdot\frac{\rho_{SVD}}{1000}=250\cdot\frac{15.4}{1000}=3.85\text{ kg}$$

of water.