Maths for Ground Source Heat Pumps

Ground source heat pump maths answers one clear question:

How much heat will a ground source heat pump deliver, how much electricity will it use, and what will it cost?

The maths is built from a few simple ideas: power, energy, heat loss, and efficiency.

1. Heat, Power, and Units

Two units appear everywhere:

kW (kilowatts) = power (how fast energy flows)
kWh (kilowatt-hours) = energy (how much is used over time)

Example:
A 3 kW heater running for 4 hours uses:

Energy = 3 kW × 4 h = 12 kWh

That simple rule — kW × hours = kWh — underpins all running-cost maths.

2. The Core Number: COP

The key performance number for any heat pump is COP (Coefficient of Performance).

COP = Heat output (kW) ÷ Electricity input (kW)

Worked example

If a ground source heat pump delivers 8 kW of heat while using 2 kW of electricity:

COP = 8 ÷ 2 = 4

So you receive 4 units of heat for every 1 unit of electricity.

Ground source heat pumps usually achieve higher and steadier COP values than air source systems because the ground temperature stays fairly constant all year.

3. Seasonal Performance (SCOP)

COP is a snapshot. SCOP measures performance over a whole season or year.

SCOP = Total heat output (kWh) ÷ Total electricity input (kWh)

Worked example

Annual heat needed: 14,000 kWh
Electricity used: 3,500 kWh

SCOP = 14,000 ÷ 3,500 = 4.0

That means the system averages 4:1 efficiency across the year.

Because the ground stays at a stable temperature, ground source systems often achieve SCOP values around 3.5–4.5 in real homes.

4. Why Ground Systems Stay Efficient

Air temperature swings widely. Ground temperature does not.

At a few metres below the surface, soil often sits near 8–12°C all year. The heat pump therefore works against a much smaller temperature gap than an air source system in winter.

Simple effect:

Smaller temperature lift
Less compressor work
Higher COP
More heat per unit of electricity

That stability shows up directly in the maths.

5. Heat Loss: The Number That Sets Everything

A heat pump must replace the heat a building loses. That loss sets the target.

A simple building-part formula looks like this:

Heat loss (W) = U-value × Area × Temperature difference

You repeat this for walls, roof, floor, and windows, then add ventilation loss.

6. A Small Worked Heat Loss Example

Assume:

Indoor target: 21°C
Cold outdoor design: -2°C

ΔT = 21 - (-2) = 23°C

Part	U-value	Area (m²)	Loss (W)
Walls	0.28	90	0.28×90×23 = 579
Roof	0.18	60	0.18×60×23 = 248
Floor	0.25	60	0.25×60×23 = 345
Windows	1.40	18	1.40×18×23 = 580

Fabric loss:

579 + 248 + 345 + 580 = 1,752 W ≈ 1.75 kW

Add ventilation (say 0.7 kW):

Total heat loss = 1.75 + 0.7 = 2.45 kW

So this home needs about 2.5 kW of heat on a cold design day.

7. Sizing the Heat Pump

At steady conditions:

Heat pump output (kW) ≈ heat loss (kW)

So a home with a 2.5 kW design heat loss might suit a 3 kW ground source system, with margin for hot water and defrost cycles.

Ground systems often run continuously at low power, matching demand smoothly.

8. Running Cost Maths

Once you know heat demand and SCOP:

Electricity used (kWh) = Heat needed (kWh) ÷ SCOP

Worked example

Annual heat demand: 14,000 kWh
SCOP: 4.0

Electricity = 14,000 ÷ 4.0 = 3,500 kWh

Now apply price:

Cost = Electricity × Price per kWh

At £0.25/kWh:

Cost = 3,500 × 0.25 = £875/year

That single division gives a clear, quick estimate.

9. Flow Temperature and Efficiency

Ground source systems work best with low heating temperatures.

Simple rule:

Lower flow temperature → higher COP
Higher flow temperature → lower COP

That’s why underfloor heating and large radiators pair so well with ground source heat pumps.

10. A Quick Reality Check

If your home needs 6 kW of heat:

With COP = 4:

Electricity = 6 ÷ 4 = 1.5 kW

With COP = 3:

Electricity = 6 ÷ 3 = 2.0 kW

A small drop in COP raises electricity use by 33%.
That is why design, temperatures, and ground conditions matter.

In Short

Ground source heat pump maths rests on five numbers:

Heat loss (kW) – what your building needs
Heat demand (kWh) – energy over time
COP / SCOP – how efficiently heat is delivered
Electricity use (kWh) – heat ÷ SCOP
Cost (£) – electricity × price

With just multiplication and division, you can:

Size a system
Compare options
Estimate annual running costs

The ground does the hard work. The maths simply tells you how well you use it.