Book - School of Science and Technology

526 Calculations for air-conditioning designHall occupiedThe following calculation assumes that the air entering the plant is a mixture of outsideand recirculated air as for summer operation, outside air conditions being as above. Inthis example, again it is assumed that humidification will be by steam using a packagedinjection-type steam humidifier. The internal heat gains, 44 kW sensible and 15 kW latent,are those due to occupancy and lighting only.Dry bulb temperature and other conditions:in roomˆ 20:0 Cmass flow of air ˆ 15:52 m 3 /sspecific heat of airˆ 1:02 kJ/kgTemperature of air entering plant[(10:7 20) ‡ (4:82 4)]/15:52 ˆ 15:0 CAir temperature rise in room from supply due to 64 kW sensible gain :64/(15:52 1:02) ˆ 4:0 CThus, air temperature leaving plant:(20 4:0) ˆ 16:0 Cand main heater duty(15:52 1:02)(16:0 15:0) ˆ 15:8kWFrom Figure 18.1, moisture content of:air at room conditionˆ 0:00885 kg/kgair entering plant[(10:7 0:00855) ‡ (4:82 0:00505)/15:52 ˆ 0:00746 kg/kgand rise in moisture content in room due to 15 kW latent gain15/(15:52 2450) ˆ 0:00039 kg/kgThus, moisture content of supply air:(0:00885 0:00039) ˆ 0:00846 kg/kgand moisture added by humidifier(0:00846 0:00746) 15:52 ˆ 0:016 kg/s ˆ 57 litre/hEnergy absorbed in humidification(0:016 2450) ˆ 39:2kWHence, total energy absorbed:(15:8 ‡ 39:2) ˆ 55 kWAs an alternative, once again if permissible, the cooling coil would be pump sprayed andthus would provide adiabatic saturation to the air flow.Temperature of air entering plant (as before)Moisture content of air mixture (as before)ˆ 15:0 Cˆ 0:00746 kg/kg