Book - School of Science and Technology

hot water radiators. In consequence, the supply air leaving the plant would be at the roomtemperature, i.e. 20 C.Hall empty, preconditioning for occupationThe following calculation assumes that the plant will handle 100% outside air when theambient temperature is 4 C, below which level recirculation would be used. In thisexample, it is assumed that humidification will be by steam using a packaged injectiontypesteam humidifier.Dry bulb temperature and other conditionsin roomˆ 20 Cmass flow of air ˆ 15:52 m 3 /sspecific heat of airˆ 1:02 kJ/kgTemperature of air entering plantˆ 4 Cand hence, after heater duty(15:52 1:02)(20 4) ˆ 253 kWFrom Figure 18.1, moisture content of:air at room conditionˆ 0:00885 kg/kgoutside airˆ 0:00505 kg/kgThus, moisture added by humidifier:(0:00885 0:00505) 15:52 ˆ 0:059 kg/s ˆ 214 litre/hand energy absorbed in humidification:(0:059 2450) ˆ 146 kWHence, total energy absorbed excluding all losses:(253 ‡ 146) ˆ 399 kWAs an alternative, if permissible, the cooling coil would be pump sprayed and thus wouldprovide adiabatic saturation to the air flow. For the purpose of the example, the figuringassumes 100% coil efficiency, i.e. saturating the air to room dew point but a level of, say,80±90% would probably arise in practice.From Figure 18.1, at room condition of 20 C:60% saturation, dew point ˆ 12:1 Centhalpy at dew pointˆ 34:18 kJ/kgThus, duty of preheater(34:18 16:70) 15:52 ˆ 271 kWand that of after-heater(42:58 34:18) 15:52 ˆ 130 kWThus, total energy capacity absorbed excluding all losses:(271 ‡ 130) ˆ 401 kWMoisture added at sprayed coil (as before)0:059 kg/s ˆ 214 litre/hApplication 525