Part 1: Answers

In this chapter, we will provide you with the answers to the question at the end of each chapter of this first part.

Question 1.1

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I have an ASHP with a measured efficiency of 3.9, and a GSHP with a measured efficiency of 3.8. When both buildings had the identical demand for heating and domestic hot water, and no cooling was required, which one is the cheapest to operate during winter?

Having a higher efficiency means that the same amount of heat (or cold) can be given to the building with a lower electricity consumption. In this case, since the ASHP has a higher efficiency its electrical usage will be lower on a yearly level, however, that does not necessarily mean that it would be cheaper to operate.

When the electricity price is a constant over the year, consuming less electricity is linearly proportional to a lower electricity bill, however with dynamic prices, one can have a slightly higher electricity consumption on a yearly level (for the GSHP in this case) and still have a cheaper electricity bill, since it is probably more efficient at times when the electricity cost is higher.

Hence, without further information related to the electricity prices, it is not possible  to give a definitive answer to this question.

Question 3.1

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Given a surface ground temperature of 10°C, a geothermal heat flux of 0.06 W/m² and a ground thermal conductivity of 3 W/(m·K), what is the ground temperature at a depth of 100 m? Assume a linear temperature gradient.

With the geothermal heat flux $\dot{q}$ and the ground thermal conductivity $\lambda$ given, the geothermal gradient $\Delta T$ can be calculated as follows: $$\Delta T=\frac{\dot{q}}{\lambda}=\frac{0.06}{3}=0.02K/m$$

This means that when we go 100 m deep, our ground temperature is 2 K (or 2°C, which is identical) higher than our surface temperature, when we assume a linear gradient, meaning that at a depth of 100m, the expected ground temperature is 12°C.

Question 3.2

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Using the same data as above, calculate the undisturbed ground temperature if the borehole starts 10 metres below the surface.

Using the same gradient of 0.02 K/m, the ground temperature at 10m deep is (theoretically) 10.2°C. The average ground temperature between 10 m and 100 m is hence: $$\frac{10.2+12}{2}=11.1°C$$

Question 3.3

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I have a very shallow borefield, only 40 m deep. The initial measurement of ground conductivity was taken at the end of the rainy season. What effect could this have on the accuracy of the test, and what are the consequences if my borefield has a strong imbalance?

The saturation degree of the ground (i.e. percentage of voids in the ground that are filled with water) is a very important factor when it comes down to the determination of the ground thermal conductivity. The higher the moisture content/saturation, the better this conductivity will be. The first layers of the ground are the most sensitive to weather impact (like rain or snow) and since our borefield is rather shallow, the impact of this will be greater than when you have boreholes of 100-200 m deep.

Since the measurements were taken at the end of the rainy season, chances are that you measured the most optimistic ground thermal conductivity and that on average, its value is lower.

When there is a large imbalance, it is important to have a good conductivity to cope with the long-term temperature drift of the ground. Since we probably overestimated our ground thermal conductivity, chances are that we underestimate this temperature drift. Therefore, it is better to make another measurement at a more representative time of the year or to take some extra safety factor and decrease the conductivity with which you calculate.

Question 4.1

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I have a residential building of 120 m², with floor heating. I know from the installer that it can deliver 35 W/m² in heating and about half that in cooling. The heat pump has a capacity of 6 kW and is modulating. What values would you estimate for the peak heating, peak cooling and yearly heating and cooling?

The emission system can deliver in total $120\cdot35=4200W=4.2kW$ of power in heating and 2.1 kW in cooling. Our heat pump however, has a capacity which is greater than the emission power of the floor, which means that it will never deliver more than 4.2 kW. This gives us a final maximum peak heating demand of 4.2 kW for our geothermal calculation.

From the tables, we know that a residential building has around 1200-1500 full load hours in heating, giving us approximately 5670 kWh/year of heating demand and 700 full load hours in cooling (for Belgium) gives us a cooling demand of 1470 kWh/year.

Question 5.1

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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?

Given the SCOP values, we can calculate the energy extracted from the ground $Q_l$ as follows:$$Q_l=Q_h-E=Q_h-\frac{Q_h}{SCOP}=Q_h\left(1-\frac{1}{SCOP}\right)$$

where $E$ is the electricity used by the compressor and $Q_h$ is the energy given to the building. This means that for our heating demand, 3.2 MWh/year is extracted from the ground and for our domestic hot water 0.714 MWh/year. The total energy extracted is hence 3.914 MWh/year.

Question 5.2

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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?

The efficiency of a heat pump is driven by two temperatures: the inlet or source temperature and the outlet or supply temperature. In our case above both change (from 0°C to 5°C for the source and 35°C to 40°C for the supply), but the temperature lift stays the same in both cases (35°C).

Given the Carnot efficiency, for the regime B0/W35, we would expect an efficiency of:$$COP_c=\frac{35+273.15}{(35+273.15)-(0+273.15)}=\frac{308.15}{35}=8.804$$

For the regime of B5/W40, we would expect an efficiency of: $$COP_c=\frac{40+273.15}{(40+273.15)-(5+273.15)}=\frac{313.15}{35}=8.947$$

Therefore, we would expect the efficiency to be higher for the B5/W40 case than for the B0/W35 situation. From the Carnot efficiency, the increase is about 1.6%, giving us an estimated COP of 4.67.

Question 5.3

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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?

Every heat pump creates simultaneously heat and cold. Typically, in the case of a ground source heat pump, this cold is dumped into the ground and the heat is used for the building. However, the operation can simply be reversed to dump the heat into the ground and cool the building. This is exactly what is happening here.

It is important to note the convention of the heating and cooling mode of a heat pump. The value Bx/Wy means that x is the primary temperature entering the heat pump and y is the secondary temperature exiting the heat pump. Hence, in heating mode, B15/W35 means that 15°C enters the heat pump from the borefield (and leaves at 10°C, due to our 5°C temperature difference) whilst 35°C leaves the heat pump (and returns at 30°C).

If we now switch to active cooling, keeping our temperature regime the same, the entering primary fluid temperature will be 30°C and it will exit the borefield at 35°C. Similarly, the secondary temperature out of the heat pump will be now 10°C and it returns from the building to the heat pump at 15°C. Given our definition, this regime is given as B30/W10 (and not, as you would perhaps expect B35/W15).

Given the definitions of the COP and EER being respectively $\dot{Q}_h/\dot{E}$ and $\dot{Q}_l/\dot{E}$ we know that:$$\dot{Q}_h=\dot{Q}_l+\dot{E} \Rightarrow \dot{E}\cdot COP = \dot{Q}_l+\dot{E} \Rightarrow \dot{Q}_l = \dot{E}(COP-1) \Rightarrow \frac{\dot{Q}_l}{\dot{E}}=EER=COP-1$$