Configuration of air-conditioning equipment
A central system has these parts:
- Heat source: a chiller makes cold water (about 7 C); a boiler or heat pump makes hot water for heating. A cooling tower or outdoor unit throws the heat outside.
- Pumps and pipes: carry the water.
- Air handling unit (AHU): filter, coil (cooling or heating), fan and sometimes a humidifier.
- Ducts and diffusers: carry and spread the air in the rooms.
- Controls: sensors, controller and valves.
The air path is: room air, filter, coil, fan, duct, diffuser, room.
Central and individual air-conditioning units
Central systems make cold water or air in one place and send it to many rooms. They suit large buildings: good control, high efficiency, one place for maintenance, but they need plant rooms and shafts.
Individual systems (window, split, cassette, VRF) serve one room or zone. They are cheap, easy to fit and each room is controlled separately, but many units can mean more noise outside and more upkeep. A split AC has an indoor unit and an outdoor unit joined by refrigerant pipes. VRF joins many indoor units to one outdoor unit.
Control of air-conditioning equipment
A control loop has a sensor (temperature or humidity), a controller and an actuator (valve, damper or fan speed). The controller compares the reading with the set value and acts. A two-way valve opens more when the room is warm. Other controls: fan speed, damper position for fresh air, and schedules that switch the plant off at night.
This is a feedback loop: measure, compare, act, repeat.
Energy saving in air conditioning
- Set the temperature at 24 to 26 C: each degree higher saves about 6%.
- Insulate walls and roof; shade windows; seal gaps.
- Use inverter compressors and variable-speed fans and pumps.
- VAV (variable air volume) lowers airflow when the load is small. Fan power falls roughly with the cube of speed.
- Recover heat from outgoing air; use free cooling when the outside air is cool.
- Clean filters and coils; fix leaks; switch off empty rooms.
COP = cooling output / electric input. A higher COP means less electricity for the same cooling.
Design of air-conditioning equipment
Design steps: (1) find the cooling and heating loads; (2) choose the plant capacity (kW or TR); (3) find the airflow V = Q / (rho x cp x ΔT); (4) size the ducts, area A = V / v (about 5 m/s in offices); (5) find the chilled-water flow m = Q / (cp x ΔT), with cp = 4.18 kJ/kg K; (6) choose pumps, fans and controls. Add a small safety margin but avoid oversizing.
Key formulas and definitions
- COP = cooling output / electric power
- Airflow V = Q / (rho x cp x deltaT), rho = 1.2 kg/m3, cp = 1.005 kJ/kg K
- Duct area A = V / v
- Water flow m = Q / (cp x deltaT), cp = 4.18 kJ/kg K
- Saving is about 6% per degree of higher set temperature
Worked examples
1. An AC gives 9 kW cooling for 3 kW of electricity. Find the COP.
COP = 9 / 3 = 3.
2. A room needs 6 kW sensible cooling with supply air 10 K cooler. Find the airflow.
V = 6 / (1.2 x 1.005 x 10) = 0.4975, about 0.5 m3/s.
3. Airflow 0.5 m3/s at 5 m/s. Find the duct area.
A = 0.5 / 5 = 0.1 m2.
4. A coil removes 84 kW; water warms by 5 K. Find the water flow.
m = 84 / (4.18 x 5) = 4.02 kg/s, about 4 L/s.
5. A unit uses 900 kWh a month. Set point up 2 degrees. Energy cost 8 per kWh. Find the monthly saving.
Saving 12% = 108 kWh. Money = 108 x 8 = 864.
Common mistakes
- Choosing a unit much bigger than the load, so it short-cycles.
- Placing the thermostat in the sun or near a heat source.
- Forgetting to clean filters, which makes the fan work harder.
- Mixing up COP (no unit) and power in kW.