Question

The air at 35°C dry bulb temperature and 25°C Wet bulb temperature is passed
through the cooling coil at the rate of 280 m / min. The air leaves the cooling coil at
26.5°C dry bulb temperature and 50% relative humidity. Find 1. Capacity of the
cooling Tonnes of Refrigeration 2. Wet bulb temperature of the leaving air 3. Water
vapour removed per minute and 4. Sensible heat factor.​

Answer 1

Answer:

CHAPTER 7

DRYING (cont'd)

PSYCHROMETRY

Wet-bulb Temperatures

Psychrometric Charts

Measurement of Humidity

The capacity of air for moisture removal depends on its humidity and its temperature. The study of relationships between air and its associated water is called psychrometry.

Humidity (Y) is the measure of the water content of the air. The absolute humidity, sometimes called the humidity ratio, is the mass of water vapour per unit mass of dry air and the units are therefore kg kg-1, and this will be subsequently termed just the humidity.

Air is said to be saturated with water vapour at a given temperature and pressure if its humidity is a maximum under these conditions. If further water is added to saturated air, it must appear as liquid water in the form of a mist or droplets. Under conditions of saturation, the partial pressure of the water vapour in the air is equal to the saturation vapour pressure of water at that temperature.

The total pressure of a gaseous mixture, such as air and water vapour, is made up from the sum of the pressures of its constituents, which are called the partial pressures. Each partial pressure arises from the molecular concentration of the constituent and the pressure exerted is that which corresponds to the number of moles present and the total volume of the system. The partial pressures are added to obtain the total pressure.

EXAMPLE 7.5. Partial pressure of water vapour

If the total pressure of moist air is 100 kPa (approximately atmospheric) and the humidity is measured as 0.03 kg kg-1, calculate the partial pressure of the water vapour.

The molecular weight of air is 29, and of water 18

So the mole fraction of water = (0.03/18)/(1.00/29 + 0.03/18)

= 0.0017/(0.034 + 0.0017)

= 0.048

Therefore the water vapour pressure