Question 13.4: Air containing 0.005 kg water vapour per kg of dry air is he......

For each of the four sets of shelves, the condition of the air is changed to 60 per cent humidity along an adiabatic cooling line.

Initial condition of air: θ = 325 K, \mathscr{H} = 0.005 kg/kg

On humidifying to 60 per cent humidity:
θ = 301 K, \mathscr{H} = 0.015 kg/kg and \theta _{w} = 296 K
At the end of the second pass: θ = 308 K, \mathscr{H} = 0.022 kg/kg and \theta _{w} = 301 K
At the end of the third pass: θ = 312 K, \mathscr{H} = 0.027 kg/kg and \theta _{w} = 305 K
At the end of the fourth pass: θ = 315 K, \mathscr{H} = 0.032 kg/kg and \theta _{w} = 307 K
Thus the temperatures of the material on each of the trays are:

Total increase in humidity = (0.032 – 0.005) = 0.027 kg/kg
The air leaving the system is at 315 K and 60 per cent humidity.
From Figure 13.4, specific volume of dry air = 0.893 m³/kg
Specific volume of saturated air (saturatedvolume) = 0.968 m³/kg
Therefore, by interpolation, the humid volume of air of 60 per cent humidity = 0.937 m³/kg
Mass of air passing through the dryer = \left({\frac{5}{0.937}}\right)=5.34\ {\mathrm{kg/s}}

Mass of water evaporated = (5.34 x 0.027) = \underline{\underline{0.144\ kg/s}}

If the material is to be dried by air in a single pass, the air must be heated before entering the dryer such that its wet-bulb temperature is 307 K.
For air with a humidity of 0.005 kg/kg, this corresponds to a dry bulb temperature of \underline{\underline{370\ K}}.

The various steps in this calculation are shown in Figure 13.6.