AC Running Cost Calculator 💡

Estimate how much electricity an air conditioner uses and what it costs per day, month and year, from its capacity, efficiency (EER, SEER or COP) and running hours, and compare it with a more efficient unit.

kW = Btu/h ÷ EER ÷ 1000  ·  kWh = kW × hours × load  ·  cost = kWh × rate

🧮 Inputs

Average load is the share of rated power the compressor uses over the running hours: 60–80% for fixed-speed units on a hot day, lower for inverter units.
Running cost per month
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AC Running Cost Formula

Power input (kW) = cooling capacity (Btu/h) ÷ EER ÷ 1000   (or cooling kW ÷ COP)
Energy (kWh/day) = power input × hours × average load
Cost = energy × electricity rate
1 TR = 12,000 Btu/h = 3.517 kW  ·  COP = EER ÷ 3.412

AC Power Consumption Chart (kW at Full Load)

Electrical input at rated conditions. Multiply by hours and average load to get kWh.

AC sizeEER 8EER 10EER 12EER 14
0.75 TR (9,000 Btu/h)1.120.900.750.64
1 TR (12,000 Btu/h)1.501.201.000.86
1.5 TR (18,000 Btu/h)2.251.801.501.29
2 TR (24,000 Btu/h)3.002.402.001.71
2.5 TR (30,000 Btu/h)3.753.002.502.14
3 TR (36,000 Btu/h)4.503.603.002.57
4 TR (48,000 Btu/h)6.004.804.003.43
5 TR (60,000 Btu/h)7.506.005.004.29

Worked Example: 1.5 Ton Split AC

A 1.5 TR (18,000 Btu/h) split AC with EER 10 draws 18,000 ÷ 10 = 1.8 kW. Running 8 hours a day at 70% average load uses 1.8 × 8 × 0.7 = 10.1 kWh per day, or 302 kWh in a 30-day month. At 0.15 per kWh that is about 45 per month. An inverter unit with EER 13 uses 1.38 kW and about 233 kWh a month, a saving of about 23%.

Check the AC size first with the AC Tonnage Calculator or the AC Tonnage Chart. For the types of system and their efficiency, see Types of HVAC Systems.

FAQ

How much electricity does a 1.5 ton AC use?

A 1.5 ton AC with an EER of 10 draws about 1.8 kW at full load. Over 8 hours at 70% average load it uses about 10 kWh per day, or 300 kWh a month. A higher-efficiency inverter unit uses roughly 20–30% less.

How do I calculate AC running cost?

Divide the capacity in Btu/h by the EER to get watts, multiply by running hours and the average load, then by the electricity rate. 18,000 Btu/h ÷ 10 = 1,800 W; × 8 h × 0.7 = 10.1 kWh; × rate = daily cost.

What is the difference between EER, SEER and COP?

EER is cooling output in Btu/h per watt of input at one rating condition. SEER is a seasonal average over a range of outdoor temperatures, so it is higher for inverter units. COP is the same ratio in watts per watt: COP = EER ÷ 3.412.

Does an inverter AC really save electricity?

Usually yes, because it slows the compressor at part load instead of cycling on and off. Savings of 20–40% are common in climates where the AC runs many hours at part load, less where it runs flat out all day.

How can I reduce AC running cost?

Set the thermostat 1–2°C higher, clean filters monthly, shade windows, seal gaps around doors, keep the outdoor unit in shade with clear airflow, and replace old units with higher-EER inverter models.

📚 References

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