How much will my heat pump cost to run?

One of the most common questions clients ask when considering a new heating system is, "How much will it cost to run?" While it’s a perfectly reasonable question, the answer depends on several interconnected factors, including heat loss, heat pump efficiency, and degree days, as well as how much hot water is needed. Each of these variables affects the energy consumption and, in turn, the running cost of your system.

In this post, we’ll explain how these factors interact and explain the process of estimating seasonal energy demand (kWh) from heat loss (kW). By incorporating degree days, we'll show how to calculate running costs with real-world seasonal data.

We’ll then get into the details of how much hot water production can cost with a well engineered hot water system for a heat pump.

Central Heating

The Three Key Variables for heating the building (we’ll cover hot water separately)

1. Heat Loss from the Building

The heat loss rate (measured in kW) represents the amount of heat energy your building loses to the outside environment at the design outside temperature. It depends on:

- Insulation quality: Better insulation reduces heat loss.

- Windows and doors: Double or triple glazing helps minimise heat loss.

- Building size and shape: Larger homes lose more heat.

- Thermal bridging: Poorly insulated areas (like gaps or junctions) increase heat loss.

At Alpine Heating, early on in the design process we will come to your building or look at your plans and build a very detailed thermal model to fully understand how much heat your building will lose at your chosen indoor temperature, at the design outside temperature for where your building is situated.

For example:

- A home might have a heat loss of 10 kW at the Chamonix design temperature of -18°C, assuming an internal temperature of 21°C.

2. Heat Pump Efficiency (Coefficient of Performance - COP)

The COP of a heat pump measures its efficiency: the ratio of heat energy delivered to the electricity consumed. However, the COP varies based on:

- Outdoor temperature: Higher outdoor temperatures improve efficiency, while extreme cold reduces it.

- Flow temperature: Lower flow temperatures (e.g., for underfloor heating) improve efficiency.

The outside temperature can’t be controlled, and we live in a cold place. What is within our control is the design flow temperature of the heating system, and the efficiency of the hydraulic system. At Alpine Heating we are in a league of our own when it comes to hydraulic design for exceptionally low flow temperatures, routinely designing for 35 degrees flow temperature at -18 outside. Weather compensation on heat pumps means that when it’s not -18 (most of the time) the flow temperature is even lower than 35 degrees, ensuring the highest possible levels of efficiency.

For example:

- A heat pump and associated heating system might have a COP of 5 in mild conditions and drop to 2.5 on very cold days.

3. Degree Days

Degree days are a measure of how much heating (or cooling) a building needs over a specific period. They are calculated using the difference between the outdoor temperature and a base indoor temperature (commonly 15-17°C).

The Base Temperature is the point at which no heating is needed (even if you want your home at 21 degrees inside). You may have observed this before where you don’t need the heating on to keep the house warm during mild weather, humans, cooking, showers etc all generate heat which is enough to keep the building warm, when it is above the Base Temperature outside.

- Heating degree days (HDD): The colder it is, the more degree days accumulate.

- Degree days are location-specific and vary annually based on weather patterns.

For example:

- A cold region might experience 2,500 HDD per year, while a milder region may experience 1,500 HDD. This difference reflects the total heating required for the year.

Our nearest weather station which tracks degree days and makes this data available is Bourg St Maurice at 815m altitude. We will use this as a base data set, and correct it for the 1000m altitude we have in Chamonix.

Bourg St Maurice had 2402.6 degree days Feb 24-Feb25, with a base temperature of 16 degrees.

Correcting this for altitude and the different design temperature we get 2658.3 degree days.

From Heat Loss (kW) to Seasonal Energy Demand (kWh)

To estimate the total energy demand (kWh) for the heating season, we use the heat loss rate (kW) and apply degree days. Here’s how to calculate it:

Step 1: Heat Loss Rate (kW)

Start with the calculated heat loss of the building at the design outside temperature. For example:

- A home with a heat loss of 10 kW at -18°C.

Step 2: Convert Degree Days to Heating Hours

Degree days represent the cumulative temperature difference over a season. To estimate the total heating hours, we use the following formula:

Heating Hours = Degree Days x 24 / Temperature Difference (ΔT)

Where:

- Degree Days = total HDD for the season (e.g., 2658.3 HDD).

- Temperature Difference (ΔT) = difference between the indoor base temperature (e.g., 21°C) and the design outside temperature (e.g., -18°C). For this example:

Delta T = 21 - (-18) = 39°C

Heating hours for 2658.3 HDD would be:

Heating Hours = 2658.3 x 24 / 39 = 1635.88 hours.

Step 3: Calculate Seasonal Energy Demand

Multiply the heat loss rate (kW) by the total heating hours (hours):

Energy Demand (kWh) = Heat Loss Rate (kW) x Heating Hours (hours).

For a heat loss rate of 10 kW and 1635.88 heating hours:

Energy Demand = 10 x 1635.88 = 16358.8 kWh.

This is the total heat energy required to maintain a comfortable indoor temperature over the heating season.

Adjusting for Heat Pump Efficiency

Once we know the seasonal energy demand (kWh), we can adjust for the heat pump’s efficiency (COP) or Seasonal COP (SCOP). The electricity consumption is calculated as:

Electricity Consumption (kWh) = Energy Demand (kWh) / SCOP

If the SCOP over the season is 3.5, the electricity required is:

Electricity Consumption = 16358.8 / 3.5 = 4673.9 kWh

Estimating Running Costs

Finally, multiply the electricity consumption by the cost per kWh of electricity. For example:

- If electricity costs €0.22 per kWh, the running cost is:

Running Cost = 4673.9 x 0.22 = €1028.27 annually to heat the building.

Conclusion (heating)

Estimating the running cost of a heat pump involves moving from the heat loss rate (kW) to the seasonal energy demand (kWh) using real-world data like degree days. This approach accounts for the building’s specific heat loss, local climate, and the efficiency of the heat pump.

At Alpine Heating we design our systems for the highest possible efficiency, within the constraints of the overall project. The SCOP in the example above is an average figure, we aim high, design well and yet have to recognise we live in a cold place, hence the conservative 3.5 for this post.

By using degree days, we avoid making oversimplified assumptions and base our calculations on real seasonal conditions. While these calculations involve some variables (like SCOP and electricity prices), they provide a practical framework for giving a realistic estimate.

What about hot water?

As we increase insulation levels and improve windows and draft proofing, hot water production represents a significant proportion of the electricity that a heat pump will use. Let’s take a look at that:

Assumptions: our “10 kW” house above has a family of 4 people who each take a good shower each day and do a typical amount of cleaning and washing up consuming a typical 50 litres per person of hot water each day.

The family take a total of 4 weeks of holidays per year, yet host extended family members at other times of year so the hot water demand averages out at 200 litres of water at 45 degrees per day. The water arrives from the mains at 10 degrees.

As hot water is required to be hotter than our signature low temperature central heating, the heat pump needs to work a little harder to get a cylinder of water hot.

At Alpine Heating we work with our clients to understand their needs and optimise the efficiency and space requirements for the hot water system. By increasing the cylinder coil size and storing a larger volume of slightly less hot water the efficiency improvements are there to be attained, contributing to significantly higher SCOPs even when generating hot water.

Like with central heating, the COP changes with outdoor temperature, and we know the heat pump has to work a little harder to get the water to 45 degrees, so we assume a SCOP of 3.0

The maths:

45-10 =35 degree increase in temperature in the cylinder

Daily energy requirement = 200 x 35 x 0.00116 = 8.12 kWh per day

(0.00116 is the specific heat capacity of a litre of water in kWh)

Annual energy requirement = 365 x 8.12 = 2963.8 kWh

Adjusted for heat pump SCOP = 2963.8 / 3.0 = 987.93 kWh

Annual hot water cost = 987.93 x 0.22 = €217.34

Clearly if you add an Airbnb annex and or teenagers into the mix, then this number could easily double or triple.