Question 13.11: 0.08 m³/s of air at 305 K and 60% humidity is to be cooled t...

At 305 K and 60% humidity, from Fig. 13.4, the wet-bulb temperature is 299 K and \mathscr{H} = 0.018 kg/kg. Thus, as the air is cooled, the per cent humidity will increase until saturation occurs at 299 K and the problem is then one of cooling saturated vapour from 299 K to 275 K.
Considering the cooling in 10 deg K increments, the following data are obtained from Fig. 13.4:

At 305 K: the specific volume of dry air = 0.861 m³/kg

the saturated volume = 0.908 m³/kg
and hence the specific volume at 60% humidity = [0.861 + (0.908 – 0.861)60/100]
= 0.889 m³/kg

Thus: mass flow of moist air = (0.08/0.889) = 0.090 kg/s

Thus the flowrate of dry air = 0.090/(1 + 0.018) = 0.0884 kg/s.

From Fig. 13.4, specific heat of dry air (at \mathscr{H} = 0) = 0.995 kJ/kg K.

∴ enthalpy of moist air = (0.0884 × 0.995)(299 – 273) + (0.018 × 0.0884)
× [4.18(299 – 273) + 2435] + 0.090 × 1.032(305 – 299) = 6.89 kW

At 295 K: Enthalpy of moist air = (0.0884 × 0.995)(295 – 273) + (0.017 × 0.0884)
× [4.18(295 – 273) + 2445] = 5.75 kW

At 285 K: Enthalpy of moist air = (0.0884 × 0.995)(285 – 273) + (0.009 × 0.0884)
× [4.18(285 – 273) + 2468] = 3.06 kW

At 275 K: Enthalpy of moist air = (0.0884 × 0.995)(275 – 273) + (0.0045 × 0.0884)
× [4.18(275 – 273) + 2491] = 1.17 kW

and hence in cooling from 305 to 295 K, heat to be removed = (6.89 – 5.75) = 1.14 kW

in cooling from 295 to 285 K, heat to be removed = (5.75 – 3.06) = 2.69 kW

in cooling from 285 to 275 K, heat to be removed = (3.06 – 1.17) = 1.89 kW

The mass of water condensed = 0.0884(0.018 – 0.0045) = 0.0012 kg/s.

The humid heats at the beginning and end of the process are:

1.082 and 1.001 kJ/kg K respectively.