Example 1: 36k BTU, 1,200 hours, SEER 10 → SEER 16 at $0.14/kWh, $8,000 upgrade
- Cooling load = 36,000 BTU/hr; Old SEER = 10; New SEER = 16.
- kW_old = (36,000 ÷ 10) ÷ 1000 = 3.6 kW; Annual kWh_old = 3.6 × 1,200 = 4,320 kWh.
- kW_new = (36,000 ÷ 16) ÷ 1000 = 2.25 kW; Annual kWh_new = 2.25 × 1,200 = 2,700 kWh.
- Annual kWh savings = 4,320 − 2,700 = 1,620 kWh; Annual cost savings ≈ 1,620 × $0.14 ≈ $226.80.
- Payback ≈ $8,000 ÷ $226.80 ≈ 35.3 years.