01What this calculator estimates
This heat pump sizing calculator estimates the capacity in kilowatts (kW) a home needs to stay warm on the coldest design day. It works from the building fabric: it multiplies the heated volume (floor area × ceiling height) by a heat-loss factor that reflects how well insulated the property is, and by the design temperature difference between inside and outside. The result is the design heat loss in kW, which is the figure a heat pump must be able to meet.
The approach mirrors the method used by heating engineers and echoed in U.S. Department of Energy building guidance: a heat pump should be matched to the design-day heat load, not chosen from floor area or bedroom count alone. For context on efficient equipment and incentives, see the ENERGY STAR air-source heat pump criteria and the DOE Building Technologies Office. For related tools, try our energy efficiency upgrade calculator and embodied carbon calculator.
02Capacity by property type
As a sanity check, the calculator maps the recommended output onto typical property bands. These ranges are widely quoted by installers and energy suppliers and match the People Also Ask guidance for common home types. They are a cross-check, not a substitute for a heat-loss calculation — a well-insulated large home can need less than a small, leaky one.
03What changes the result
The estimate is a robust early figure, but several factors move the final kW:
- Insulation and airtightness. Fabric is the single biggest driver. Upgrading loft, wall and floor insulation and sealing draughts — see ENERGY STAR guidance on sealing and insulating — can move a home a whole band and let you fit a smaller, cheaper heat pump.
- Design temperature. A colder design outdoor temperature raises ΔT and therefore the heat loss. A home in a mild climate is sized for a smaller load than the same home in a cold one.
- Ceiling height and volume. Rooms with high or vaulted ceilings hold more air to heat, so volume — not just floor area — matters.
- Ventilation and infiltration. Air changes carry heat out; this simplified factor bundles them into the fabric figure, but a leaky old house loses far more.
- Flow temperature and emitters. Radiators sized for a low flow temperature let the heat pump run efficiently; undersized emitters can force a higher flow temperature and lower the COP.
- Incentives. Efficient systems may qualify for support — check the federal tax credits listed by ENERGY STAR.
- Enter the total heated floor area of the home in square metres (add up all storeys).
- Set the average ceiling height (2.4 m is typical; raise it for tall or vaulted rooms).
- Set the design temperature difference — indoor design temp minus the coldest outdoor design temp (e.g. 18 − −3 = 21°C).
- Choose the insulation / construction quality that best matches the property.
- Adjust the assumed COP if you have a manufacturer figure, then press Calculate.
Fitting out the wider project too? Our conduit fill calculator helps size electrical containment for the new circuits a heat pump needs.
This is an early-stage estimate for planning and comparison, not a certified heat-loss survey. It uses a single whole-house heat-loss factor rather than a room-by-room calculation, and does not account for:
- Individual room orientation, glazing area, exposure and thermal bridging
- Detailed ventilation and infiltration rates or mechanical ventilation with heat recovery
- Hot-water demand, which can add to the required output on some systems
- Emitter sizing, flow temperature and the effect of the weather compensation curve on real COP
01The sizing formula
Heat pump sizing comes down to one relationship: how fast the home loses heat on the coldest design day. Multiply the heated volume by a fabric heat-loss factor and the design temperature difference to get the design heat loss, then match the heat pump output to it.
Where:
- floor area= total heated floor area of the home in m² (all storeys).
- ceiling height= average floor-to-ceiling height in m (default 2.4).
- factor= whole-house heat-loss factor in W per m³ per K, set by insulation quality.
- ΔT= design temperature difference: indoor design temp minus outdoor design temp, in K (°C).
- COP= coefficient of performance — heat delivered per unit of electricity used.
02Worked example
Take an average 3-bed semi of 90 m² with 2.4 m ceilings, a design ΔT of 21°C and an assumed COP of 3.2. Here is the calculation carried through to a recommended capacity:
A 5.5 kW heat pump lands squarely in the 5–7 kW band for a 3-bed semi, confirming the estimate is plausible. That is about 60 W/m² of floor area, typical for an average existing home. Switching the insulation setting to “well insulated” (factor 0.80) drops the heat loss to about 3.6 kW and a 4 kW unit — a good illustration of why sealing and insulating first can shrink the heat pump you need before you ever pick a model.