Subtopic: Psychrometry – properties of air-water vapour mixture
Question Statement:
At standard atmospheric pressure of $P = 101.325\text{ kPa}$, moist air has a water vapor partial pressure of $p_v = 2.45\text{ kPa}$. The humidity ratio (specific humidity) of the moist air is ________ $\text{g water / kg dry air}$ (round off to two decimal places).
Answer Key:
15.41
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Molecular Weight Ratio Formula**
Humidity ratio ($W$) is given by:
$W = \frac{M_w}{M_a} \cdot \frac{p_v}{P - p_v} = 0.622 \cdot \frac{p_v}{P - p_v}$
Given:
- $P = 101.325\text{ kPa}$
- $p_v = 2.45\text{ kPa}$
- Partial pressure of dry air: $p_a = P - p_v = 101.325 - 2.45 = 98.875\text{ kPa}$
$W = 0.622 \times \frac{2.45}{98.875} = \frac{1.5239}{98.875} \approx 0.0154124\text{ kg water / kg dry air}$
In grams per kg of dry air:
$W = 0.0154124 \times 1000 = 15.41\text{ g water / kg dry air}$
**Method 2: Ideal Gas Mass Ratio**
$W = \frac{m_v}{m_a} = \frac{p_v V / (R_v T)}{p_a V / (R_a T)} = \frac{R_a}{R_v} \frac{p_v}{p_a} = \frac{287.05}{461.5} \frac{2.45}{98.875} = 0.6220 \times 0.024778 = 0.01541\text{ kg/kg} = 15.41\text{ g/kg}$