Air conditioner capacity converter
Convert between air-conditioner horsepower, cooling capacity in watts and KFR model numbers, then estimate the suitable room area, input power and hourly consumption.
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Result
Capacity, horsepower, model number, suitable area and power draw, plus the KFR model decoded.
Enter a value to convert
Cooling W: 0
What this tool does
- Size a unit before buying: a 15 m² bedroom at 175 W/m² needs about 2625 W, so 1 hp or 1.5 hp covers it.
- Read KFR model numbers: KFR-35GW means 3500 W of nominal cooling, the “1.5 hp” class.
- Estimate running cost: 3500 W of cooling at an EER of 3.5 draws about 1000 W, roughly one kWh per hour at full load.
Example
Input
3500
Output
3500 W 制冷量 ≈ 1.4 匹（KFR-35），适用面积约 23.333333–17.5 m² 制冷量 = 3500 W 制冷量（kW） = 3.5 kW 匹数（按 1 匹 ≈ 2500 W） = 1.4 匹 型号数字（KFR-xx） = KFR-35 输入功率（能效比 3.5） = 1000 W 适用面积（按 175 W/m²） = 20 m² 适用面积区间（150–200 W/m²） = 23.333333–17.5 m² 每小时耗电（额定制冷） = 1 度 KFR-35：额定制冷量约 3500 W（1.4 匹），G 表示挂壁式室内机，W 表示室外机 「匹」是行业俗称：1 匹 ≈ 2500 W 制冷量来自定频时代的经验值，并不是严格物理量；变频机、新一级能效与不同品牌的实际额定制冷量会有 5%–10% 偏差，选购时以铭牌上的「额定制冷量 W」和「KFR 型号数字」为准。 面积估算只考虑冷负荷指标（150–200 W/m² 为普通住宅区间，顶层、西晒、大面积玻璃或人员密集场所要上调到 200–250 W/m²，层高超过 3 m 还需按体积另算）。
1.5 hp ≈ 3750 W of cooling (trade rule 1 hp ≈ 2500 W), i.e. KFR-37.5, suitable for about 18.75–25 m².
Frequently asked questions
How many watts is one horsepower?
The trade rule of 1 hp ≈ 2500 W of cooling is an approximation, not a physical definition: it comes from the fixed-speed era rule of 735.49875 W input times a typical EER. Inverter models and efficiency classes vary by 5–10%, so trust the nameplate and the KFR number.
Why does 1.5 hp compute to 3750 W while shops sell KFR-35?
The industry labels the 1.5 hp class at 3500 W of nominal cooling (KFR-35), while strict multiplication gives 3750 W — a 250 W naming convention. Use the nameplate value in practice.
How is the suitable area estimated?
From a cooling-load index: 150–200 W/m² for ordinary homes, 175 W/m² by default, so the range is capacity ÷ 200 to capacity ÷ 150. Top floors, west-facing rooms and large glazing need 200–250 W/m².
How is power consumption calculated?
Input power = cooling capacity ÷ EER (3.5 by default). 3500 W of cooling draws about 1000 W, roughly one kWh per hour at full load; inverters settle at 30–60% of that.
Keywords：空调匹数匹数换算制冷量KFR 型号空调选型适用面积空调耗电air conditioner size