Air conditioner size
Cooling load per room, sensible and latent, kW and BTU/h
How the calculation works
Choose a quick estimate or a room-by-room calculation. Enter the location, latitude, design month, outdoor temperature with daily range and humidity, and indoor conditions. For each room enter the dimensions, orientation, boundaries of walls, ceiling and floor with U-values, windows with g-value and shading, people, lighting, equipment, ventilation and infiltration. The calculator computes the load for every hour of the design day and shows the peak, the contribution of items and the comparison with a unit.
Formula and assumptions
Wall: Q = U·A·(t_e + α·I/h_e − t_i). Window: Q = U·A·(t_e − t_i) + A·g·F_c·I. Air: Q_sens = m·c_p·(t_e − t_i), Q_lat = m·(W_e − W_i)·2501 kJ/kg. Peak = largest sum in the same hour; 1 kW = 3412 BTU/h.
Worked example
Room 5 × 4 × 2.6 m, south wall U = 0.3 with a 2 × 1.5 m window (U = 1.1, g = 0.5), 2 people, 60 W lighting, 150 W equipment, infiltration 0.3 h⁻¹; 45.8° N, July, 33 °C with an 11 K daily range and 40 % humidity, indoors 26 °C and 50 %. Peak at 12 h: 1.26 kW sensible + 0.12 kW latent = 1.38 kW (4695 BTU/h); of which sun through the window 803 W.
More worked examples
- Wall 10 m², U = 0.3, no sun, 33 °C outdoors, 26 °C indoors: 0.3 · 10 · 7 = 21 W.
- Ventilation 100 m³/h at sea level, outdoors 33 °C and 40 % (12.6 g/kg), indoors 26 °C and 50 % (10.5 g/kg): sensible ≈ 0.23 kW, latent ≈ 0.17 kW.
The same room with a 2 × 1.5 m window on different sides (July, 45.8° N)
| Window facing | Peak hour | Total (W) | BTU/h |
|---|---|---|---|
| N | 18 | 812 | 2772 |
| E | 9 | 1561 | 5327 |
| S | 12 | 1376 | 4695 |
| W | 16 | 1752 | 5977 |
What affects real consumption
- Orientation, size and shading of windows.
- Outdoor design temperature, humidity and altitude.
- Number of people, equipment and outdoor air flow.
Common mistakes
- Adding up the solar maxima of all façades as if they occurred at the same time.
- The same outdoor air counted both as ventilation and as infiltration.
- Comparing the load with the unit's rated capacity regardless of the conditions at which the capacity is stated.
Frequently asked questions
How many kW of air conditioning does a room need?
It depends mostly on windows and sun, people and equipment. The quick estimate multiplies the area by a rule-of-thumb value; the room-by-room calculation adds up the items hour by hour and gives the peak in kW and BTU/h.
What is the latent load?
The heat needed to remove moisture from the air – from people and more humid outdoor air. It is calculated from the difference in humidity ratio outdoors and indoors.
Why is the building peak not the sum of room peaks?
East rooms peak in the morning and west rooms in the afternoon. The coincident peak is the largest sum in the same hour and is smaller than the sum of the separate peaks.