Question

The dry bulb temperature and relative humidity of air inside a storage chamber are 37 °C and 50%, respectively. The saturation pressure of water vapour at 37 °C and barometric pressure are 6.28 kPa and 101.32 kPa, respectively. The humidity ratio of air inside the chamber is ________ kg water (kg dry air)⁻¹ (round off to three decimal places).
Given: Molecular weight of water vapour and dry air are 18.02 g mol⁻¹ and 28.97 g mol⁻¹, respectively.

Answer 1

Correct Answer: 0.017

We use the formula for the humidity ratio (\(W\)): \[ W = 0.622 \times \frac{P_w}{P - P_w} \] Where:
- \(P_w\) is the partial pressure of water vapor,
- \(P\) is the total atmospheric pressure.
Given: - Saturation pressure of water vapor \(P_{ws}\) at 37 °C = 6.28 kPa,
- Barometric pressure \(P\) = 101.32 kPa,
- Relative humidity (RH) = 50%.
The partial pressure of water vapor is: \[ P_w = \text{RH} \times P_{ws} = 0.50 \times 6.28 = 3.14 \, \text{kPa} \] Now, substituting into the formula: \[ W = 0.622 \times \frac{3.14}{101.32 - 3.14} = 0.622 \times \frac{3.14}{98.18} = 0.622 \times 0.032 = 0.020 \] Thus, the humidity ratio is \(0.020 \, \text{kg water per kg dry air}\).