Maths for Air Source Heat Pumps
Air source heat pump maths answers one simple question:
How much heat will a heat pump deliver, how much electricity will it use, and what will it cost?
We’ll build up the maths slowly, using only what’s needed.
1. Heat, Power, and Units (The Starting Point)
Before any heat pump maths, you need the units:
- kW (kilowatts) = power (how fast heat or electricity flows)
- kWh (kilowatt-hours) = energy (how much you used over time)
Example:
- A 2 kW heater running for 3 hours uses:
- Energy = 2 kW × 3 h = 6 kWh
That “kW × hours = kWh” idea comes up everywhere.
2. The Core Number: COP
The main heat pump number is COP (Coefficient of Performance).
COP tells you how much heat you get for each unit of electricity you put in.
Formula:
COP = Heat output (kW) ÷ Electricity input (kW)
Worked example
If your heat pump gives 6 kW of heat while using 2 kW of electricity:
COP = 6 ÷ 2 = 3
So you get 3 units of heat for every 1 unit of electricity.
COP changes with outdoor temperature (more on that soon).
3. Seasonal COP (SCOP): The “Year Round” Version
COP is a snapshot at a moment in time. SCOP looks at performance across a season (or year).
SCOP = Total heat output (kWh) ÷ Total electricity input (kWh)
Worked example
If you need 12,000 kWh of heat over a year, and the heat pump uses 4,000 kWh of electricity:
SCOP = 12,000 ÷ 4,000 = 3.0
That means 3:1 efficiency on average across the year.
4. Why COP Drops in Cold Weather (Simple Numbers)
As the outside air gets colder, the heat pump has to “work harder” to lift that heat up to your heating temperature. So COP usually drops.
A simple (typical) way to show this is with a small table.
Example COP vs outdoor temperature
(These are illustrative “typical” values, not a promise for your system.)
| Outdoor temp | Typical COP idea |
|---|---|
| +7°C | around 4–5 |
| 0°C | around 3 (often) |
| -7°C | around 2–3 |
Key point:
Heat pumps still work below zero — they just run less efficiently.
5. Heat Loss: The Number That Drives Everything
Heat pumps don’t “guess” how much heat you need. Your home’s heat loss sets the target.
You can split heat loss into two parts:
- Fabric heat loss (through walls, roof, floor, windows)
- Ventilation heat loss (air leaving and entering the home)
Fabric heat loss (simple form)
For each building part:
Heat loss (W) = U-value × Area × Temperature difference
The MCS heat loss guidance uses this idea: U-value × area × ΔT.
6. A Worked Heat Loss Example (Small and Practical)
Let’s do a simplified example so you can see the shape of the maths.
Assumptions
- Indoor target: 21°C (a common design indoor temperature)
- Cold outdoor design temp: -3°C (varies by location; we’ll just use an example)
- So temperature difference:
ΔT = 21 - (-3) = 24°C
Example fabric losses (made-up but realistic-ish inputs)
| Part | U-value | Area (m²) | Loss = U × A × ΔT |
|---|---|---|---|
| Walls | 0.30 | 80 | 0.30×80×24 = 576 W |
| Roof | 0.20 | 50 | 0.20×50×24 = 240 W |
| Floor | 0.25 | 50 | 0.25×50×24 = 300 W |
| Windows | 1.40 | 15 | 1.40×15×24 = 504 W |
Total fabric heat loss:
576 + 240 + 300 + 504 = 1,620 W ≈ 1.62 kW
Now add ventilation (we’ll keep it simple):
Let’s say ventilation adds 0.8 kW (example).
Total heat loss:
1.62 kW + 0.8 kW = 2.42 kW
So, on a cold design day, this home needs roughly 2.4 kW of heat to hold temperature.
7. Heat Pump Size (A Simple “Match the Loss” Rule)
At steady conditions:
Heat pump output (kW) should roughly match heat loss (kW) at design conditions.
So if your calculated heat loss is 2.4 kW, you’d look at something like a 3 kW class system (exact choices depend on hot water needs, defrost, margins, etc.).
MCS design guidance and tools focus heavily on heat loss as the starting point for system sizing.
8. Running Cost Maths (The One People Actually Want)
Once you know your heat demand and SCOP, you can estimate electricity use:
Electricity used (kWh) = Heat needed (kWh) ÷ SCOP
Worked example
Heat needed for the year: 12,000 kWh
SCOP: 3.0
Electricity = 12,000 ÷ 3.0 = 4,000 kWh
Now apply your electricity price:
Cost (£) = Electricity (kWh) × Price (£/kWh)
If electricity costs £0.25/kWh (example):
Cost = 4,000 × 0.25 = £1,000/year
That’s the cleanest “back-of-envelope” method.
9. Why Flow Temperature Matters (A Quick Rule)
Heat pumps run more efficiently when they supply lower-temperature heating water (often with larger radiators or underfloor heating).
Simple idea:
Lower flow temperature → higher COP.
Higher flow temperature → lower COP.
This is why good system design focuses on emitters and temperatures, not just the box outside.
10. A Simple COP “Reality Check” Example
Let’s say on a cold day your home needs 6 kW of heat.
Case A: COP = 3
Electricity = 6 ÷ 3 = 2 kW
Case B: COP = 2 (colder weather)
Electricity = 6 ÷ 2 = 3 kW
Same heat demand — but electricity use jumps by 50% when COP falls from 3 to 2.
That’s why temperature and system setup matter so much.
11. That’s All the Maths You Need
Every air source heat pump calculation comes back to:
- Heat loss (kW)
- COP / SCOP (ratio)
- Heat demand over time (kWh)
- Electricity use (kWh)
- Cost (£)
No advanced physics required — just:
- Multiplication
- Division
- Temperature differences
- Percentages (if you want)