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Required inputs: Heat pump efficiency

The final piece of the input puzzle for geothermal simulations is the efficiency of your heat pump and heat exchanger. This information is required to convert the building load (which we discussed in our previous chapter) to the final ground extraction and injection load. But what is this efficiency, and how do you account for seasonal variation and modulating heat pumps?

What is efficiency?

The efficiency of a heat pump can be both expressed in terms of its Coefficient of Performance (COP) or Seasonal Coefficient of Performance (SCOP). Both are explained below.

Coefficient of Performance (COP)

A heat pump uses electricity to transfer heat from a low-temperature source (in our case, the ground) to our building. This electricity is used to power a compressor within the heat pump, which determines the system’s final efficiency. The efficiency can be defined as follows: $$COP=\frac{\dot{Q}_h}{\dot{E}}$$where $\dot{Q}_h$ is the heat emitted to the building in (kW) and $\dot{E}$ in (kW) is the required electricity for this. The power extracted from the borefield $\dot{Q}_c$ can hence be defined as follows: $$\dot{Q}_h=\dot{E}+\dot{Q}_c$$

When talking about power and energy, it is important to work with certain conventions. When units like $m$ (mass (kg)) or $Q$ (energy (kWh)) are written with a dot on top ($\dot{m}$ or $\dot{Q}$), this means that the unit is expressed per second (so, for example, $\dot{m}$ is the mass flow rate in (kg/s) and $\dot{Q}$ is the power in (kW)). Furthermore, in the case of energy/power, when the unit is written with a small ($\dot{q}$) instead of a capital letter ($\dot{Q}$), the unit is expressed per length unit of the borehole, so $\dot{q}$ is the power per meter borehole in (kW/m).

This COP value varies over time and is a function of both the source (i.e. the fluid temperature of the borefield) and the emission temperature (35°C, for example, in the case of floor heating, or 55°C for domestic hot water). Although a detailed derivation is out of the scope of this course, the theoretical and maximal COP (called the Carnot efficiency) is defined as follows: $$COP=\frac{T_h}{T_h-T_c}$$where $T_h$ is the temperature of the emission in (°K), and $T_c$ the temperature of the source in (°K).

Please be cautious of this theoretical formula, since the temperatures here are in degrees Kelvin, relative to the absolute zero. Given that 0°C equals 273.15K, 10°C is, for example, 283.15K.

In this formula one key principle of heat pumps becomes clear: the larger the difference between the required emission temperature $T_h$ and the source temperature $T_c$ the lower your heat pump efficiency, leading to higher electricity consumption. This is the reason why, as we discussed in Chapter 1 of this part, a ground-source heat pump at the coldest time of the year has a better efficiency than the air-source counterpart. The $T_c$ of the latter is namely the environmental temperature, which can be way below 0°C, whereas for the GSHP it is the fluid temperature in the borefield.

Working principle of a heat pump. (Source: IEA, https://www.iea.org/reports/the-future-of-heat-pumps/how-a-heat-pump-works)
Working principle of a heat pump. (Source: IEA, https://www.iea.org/reports/the-future-of-heat-pumps/how-a-heat-pump-works)
For completeness, the $T_h$ and $T_c$ in the formulas above refer to the temperatures at the heat pump side of the emission and the source, not the emission temperature and the source itself. Imagine, for example, that for our floor heating, water comes out of the heat pump at 35°C ($T_H$). Therefore, in the refrigeration cycle, the temperature ($T_h$) has to be higher than 35°C, in order to transfer the heat to the water. Similarly, if the fluid from the borefield is 10°C ($T_C$), the coldest temperature in the heat pump ($T_c$) has to be lower than 10°C in order for the heat to flow into the evaporator: $T_c \le T_C$. Since the efficiency of the heat pump is determined by the temperature difference between $T_h$ and $T_c$, the highest efficiency can be obtained when these are equal to $T_H$ and $T_C$. This, of course, is only possible in theory.

Seasonal Coefficient of Performance (SCOP)

The COP expresses the efficiency of the heat pump at one unique moment, however, since the ground temperature as well as the required emission temperature vary over time, the COP changes significantly (as will be discussed below). Therefore, the measure of the Seasonal Coefficient of Performance (SCOP) was introduced. This can be defined as: $$SCOP=\frac{Q_h}{E}$$where now the $Q_h$ is the energy emitted to the building and the $E$ the required electricity used for that. This formula is hence the energy equivalent of the COP, which can be seen as an average value for a whole season. Therefore, typically, an SCOP is higher than a COP value.

Besides SCOP, there is also the term Seasonal Performance Factor. Both are essentially the same, although SCOP is often used when we talk about theoretical, on-paper efficiencies, whereas SPF values are typically measured.

There are also heat pumps that do not require any electricity to run, like (gas) absorption heat pumps. Here, the compressor in our traditional heat pump is replaced by an adsorption and desorption cycle driven by a heat source at higher temperature, as illustrated by the dotted rectangle in the image below. The efficiency here is defined as the delivered thermal energy $Q_h$ divided by the required energy (at high temperature) $Q_s$ going into the desorber.

Absorption heat pump cycle. (<a href="https://commons.wikimedia.org/wiki/File:Absorption_heat_pump_configuration_(refrigeration).jpg">Fuliyehuanshi</a>, <a href="https://creativecommons.org/licenses/by-sa/4.0">CC BY-SA 4.0</a>, via Wikimedia Commons)
Absorption heat pump cycle. (Fuliyehuanshi, CC BY-SA 4.0, via Wikimedia Commons)

Whereas a regular heat pump has only two thermal connections (one on the hot side of the heat pump, the condenser and one on the cold side, the evaporator), in the case of an absorption heat pump, there is a third connection at the absorber. In order for the absorption-desorption-cycle to work, the absorber needs to be cooled. This power, $\dot{Q}_a$, can be given, together with the energy from the condenser, to the building.

These heat pumps are usually quite large, ranging from 200 kW to 10 MW, and are used in district heating and cooling networks, as well as in buildings with high energy demands, such as hospitals.

Energy Efficiency Ratio (EER) and Seasonal Energy Efficiency Ratio (SEER)

The paragraphs above discussed the efficiency of a heat pump during heating, but a similar concept can be made for the cooling mode. Here, the terminology is Energy Efficiency Ratio (EER) defined as follows: $$EER=\frac{\dot{Q}_c}{\dot{E}}$$where $\dot{Q}_c$ is now the energy extracted from the building. Similarly, the SEER is defined as $$SEER=\frac{Q_c}{E}$$The energy injected into the borefield $Q_h$ is hence $$Q_h=E+Q_c$$

Similar to the heating mode, also a theoretical EER can be calculated as follows:$$EER=\frac{T_c}{T_h-T_c}$$where $T_c$ is the temperature of the building and $T_h$ is the temperature of the ground.

The terms $Q_h$, $T_h$, $Q_c$ and $T_c$ relate to the hot and cold sides of the heat pump. Therefore, when the heat pump is heating the building, the $Q_c$ is the energy extracted from the borefield and $Q_h$ is the energy emitted to the building at a higher temperature. During cooling, the building serves as the cold side, with the energy $Q_c$ extracted from it and the energy $Q_h$ injected into the ground, at higher temperature.

Modulating heat pumps

Although heat pumps may appear similar on the outside, there can be significant differences on the inside. One key difference is between on-off and modulating heat pumps.

With an on-off heat pump, your system is always either fully on or fully off, as the name suggests. This means that when your building has only a small heating requirement, your heat pump will switch on at full power, run for a short time, and then shut down. This cycling behaviour puts a lot of strain on the compressor, which is why more and more manufacturers are switching to modulating heat pumps.

If your heat pump is modulating, it can work at 100 percent of its capacity, but also at 70 percent or sometimes even 30 percent. This means that if your building has a low demand, the heat pump can switch on at a much lower power and follow the building’s demand more accurately. Therefore, modulating heat pumps have fewer start-stops, less wear on the compressor, and are generally quieter.

When a modulating heat pump is working in a part-load regime (i.e. at a lower than maximum capacity), there is another great benefit. Strictly speaking, the internal components of the heat pump are now oversized relative to the heating demand, so its efficiency is higher. Therefore, modulating heat pumps tend to be more efficient and have a higher SCOP than on-off ones, since both types depend on source and emission temperatures (as discussed above), but the modulating heat pump has the additional advantage of part-load operation.

For completeness, on-off heat pumps can also have a form of modulation. For example, heat pumps with a higher capacity (for example, 64 kW) typically have multiple compressors in parallel. If you have, for example, 2 compressors which together can deliver the full 64 kW, one compressor can deliver 32 kW, which can be interpreted as being 50% part-load.

Efficiencies in borefield design

When you are designing borefields with a building load, you need to convert this one way or another to a ground load (that is, extraction and injection of heat). This is traditionally done by using the SCOP and SEER values that can be found in the technical data sheets, for a certain temperature regime. For example, in the case of floor heating, the SCOP value given at B0/W35 (meaning 0°C entering from the borefield and 35°C exiting to the building) can be used. For domestic hot water or radiators, B0/W55 values are available.

In the case of passive, or free cooling, when no heat pump compression is used, the electricity consumption for the circulation pump is used to calculate the SEER. Since this is typically small when compared to the required compression energy during heating (or active cooling), the efficiency of passive cooling is between 15-30.

Using the SCOP and SEER values is rather straightforward, but it also comes with a couple of challenges, which are explained below.

  1. By using the SCOP to convert the peak power heating to an extraction peak power, you overestimate the peak power, since the COP during peak conditions is typically lower than the SCOP. This can lead to an oversized borefield.

  2. Using an SCOP at B0/W35 to convert heating and domestic hot water demand into ground load assumes that the temperature going into our heat pump is 0°C. However, in most designs, this only occurs after a couple of years at the earliest, meaning that the average temperature is higher. This gives a higher SCOP, so using a B0/W35 value is an underestimation of the real efficiency and therefore of the imbalance, which can result in an undersized borefield.

  3. The efficiency of a heat pump depends on the fluid temperature coming out of the borefield, so it will change depending on the design. However, since the SCOP is typically an input rather than an output of a borefield design, the SCOP does not vary when the design changes. This is rather counterintuitive.

It should be clear that there are quite a few challenges and uncertainties when using only an SCOP for borefield design. That is why, when we discuss advanced borefield design insights, we will focus on the part-load efficiency module in GHEtool for more accurate simulations.

Conclusion

In this last chapter, we discussed the different efficiency concepts (COP, SCOP, EER, SEER, etc.) for heat pumps and how they relate to one another. With this, the final piece of required input data is known and we have all the background knowledge needed to start talking about the physics of geothermal borefields in Part 2.

Questions

My building has an annual heating demand of 4 MWh and an annual domestic hot water demand of 1 MWh. If my heat pump has a SCOP of 5 for heating (B0/W35) and 3.5 for domestic hot water (B0/W55), what is the annual energy extracted from the borefield?
The COP of my heat pump is 4.6 at B0/W35. Based on the Carnot efficiency, would you expect the COP to be higher or lower at B5/W40?
I want to use my heat pump for active cooling, but I only know the COP value at B15/W35. How can I find or calculate the EER value at B30/W10 given a temperature difference across the evaporator and condenser of 5°C?

References

  • https://en.wikipedia.org/wiki/Absorption_heat_pump [last visited: 23/01/2026]
  • Peere, W. (2025). Integrating Temperature and Part-Load Dependent COP in Shallow Geothermal Borefield Design. In Proceedings of German Geothermal Congress DGK 2025. Frankfurt (Germany), 18-20 November 2025.

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