Air heat pump vs geothermal energy, new build 400m2

I still consider the 18kw to be too high. If you can provide some information about the aerated concrete block, the window areas / geometry, I can roughly estimate whether that might be accurate.

from [unterwegs]

Me too... even with a house made entirely of glass (uw 0.75), it would mean an envelope area of 700 sqm...(just as a meaningless number game for now)

I still consider the 18kw to be too high. If you can provide some information about the aerated concrete block, the window areas/geometry, I can roughly calculate whether this could be correct.

Thanks for the offer.

I can now provide the following more detailed information:

- All exterior walls are built with Ytong at 36.5cm, also in the basement (black tank)
- The geometry is mostly cubic, except for the ground floor, where the house has an L-shape (approx. 28 sqm attached to a cube of approx. 130 sqm)
- The house is now fully basemented, so also L-shaped
- On the upper floor, at the L position, there is a roof terrace with about 28 sqm accordingly
- Attic purely cubic with a hipped roof

I have now learned the corresponding U-values and recalculated the envelope areas myself (which almost coincide with the details in the heating load calculation, so that is at least okay):

- 154.40 sqm with Uw 0.21: basement floor slab
- 150.80 sqm with Uw 0.18: exterior wall in the basement (net, i.e. without windows and doors)
- 227.40 sqm with Uw 0.21: exterior wall ground floor - upper floor (net)
- 174.66 sqm with Uw 0.17: attic / roof (net)
- 118.30 sqm with Uw 0.86: window areas (mostly safety glass)
- 10.30 sqm with Uw 0.7: exterior doors (Uw estimated here, as I have no data)
- 28.02 sqm with Uw 0.24: roof terrace areas upper floor

In total, about 860 sqm envelope area and naively calculated an average U-value of about 0.29 (which is given as 0.34 in the heating load calculation) — thus differing by about 15% (in contrast to the envelope area). Unfortunately, the heating load calculation does not specify individual U-values, only the average U-value, so I currently do not know how the difference arises.

Do you think the above U-values correspond to the standard, or does one of the values deviate from the usual range? Unfortunately, I have no experience with this so far...

However, I do not know how to get from the envelope area and the U-value to the heating load. As far as I know, one also needs the building air exchange, which is assumed to be 0.6/h in the heating load calculation, as well as the minimum outdoor temperature, which is assumed to be -14°C. The room interior temperature is 20°C and the heated net volume 1230m3, gross volume 1620m3.

Many numbers to keep track of ...

The U-values are basically okay. For me, the windows and walls are better, the walls/floor in the basement are less well insulated. If you take these data, the transmission heat losses amount to 7.5 kW heat loss at design temperature (plus a little for thermal bridges, depending on execution quality, so about 8 - 8.5 kW maximum). The calculation is simply surface area * u-value * temperature difference (the ground remains above the design temperature). At a building air exchange rate of 0.6/h, this results in a ventilation heat loss of 8.5 kW without heat recovery at the mentioned volume and design temperature. However, I consider this value unrealistic since ventilation is not based on volume but on actual air consumption, i.e. the number of people. If 5-6 people live there, I would estimate the ventilation heat loss rather at a max. of about 4 kW (think about how much hot air a 4 kW hot air device produces... that's how much you actually have to ventilate - and when it's cold, you can quickly reduce the humidity with just a little air exchange). For hot water, 1-2 kW power can be assumed. The design temperature will not apply continuously, and with the sum of then 14 kW, a lot of water is heated quickly (and if more heat is needed/if it is colder, an electric heating element in the heat pump covers the difference for a few hours). The calculation you have is, in my opinion, already reasonable, but for the ground heat pump I would choose one in the range of 14-15 kW, for the air-to-water heat pump I would not be so sure, I would try to get one that offers somewhat more power but can modulate (so that efficiency/performance is right at low temperatures, but there is no excess power at high temperatures and the heating does not operate effectively).

Thank you for the calculation. So I can understand it:

Area*u-value*temperature difference

So in my case:

560 * 0.29 * 37 = 6kW (above ground, temperature difference of 37 resulting from 20°C to -14°C)
300 * 0.29 * 20 = 1.7kW (below ground, temperature difference of 20 resulting from 20°C to 0°C ground?)

In total, about the 7.5kW transmission losses you mentioned. Is that the calculation?

Building air exchange rate of 0.6/h results in a ventilation heat loss of 8.5kW at the mentioned volume and design temperature without heat recovery. But I consider this value unrealistic, since ventilation is based not on volume but on actual air consumption, i.e. number of people. If 5-6 people live there, I would estimate the ventilation heat loss at max about 4 kW

Here too, so I can follow:

Are you assuming the heat capacity of air as 1.2 kJ/(m3 K)? (So about 0.28 W/(m3 K)). If you assume the net volume of 1230m3 and again the temperature difference of 37°, you get:

0.28 * 1230 * 37 = ~12.8 kW energy to heat the air.

Multiplied by the building air exchange rate of 0.6, about 7.5 kW per hour. You get 8.5 kW, where is my mistake? Is the heat capacity different?

And these two numbers are roughly just added (including buffer for thermal bridges), add a bit for hot water, and that is the "simplified method"?

Then overall little can be done about the transmission heat loss, but there is a large margin of maneuver with the loss due to the air exchange rate (as you also say).

I will ask the energy consultant where the 0.6/h comes from.

The calculation you have is, in my opinion, reasonable, but for the brine heat pump I would choose one in the 14-15 kW range, for the air heat pump I wouldn’t be so sure, I would try to get one that offers somewhat more power but can modulate

If it really goes towards 14-15 kW for the brine, it could possibly save one of the five boreholes and the price difference to the air heat pump would be smaller. That would be something and would make the decision much easier.

The offer we have for the air heat pump also includes a fully modulating device – so that is at least already planned.

I calculated the heat losses as follows:
qm U heat loss
Exterior wall ground floor upper floor 227.4 0.21 1623.636
Exterior wall basement 150.8 0.18 434.304
Base plate kg 154.4 0.21 518.784
Roof surfaces 174.6 0.17 1009.188
Windows 118.3 0.86 3459.092
Doors 10.3 0.7 245.14
Terrace area 28.02 0.24 228.6432
each A * U * dT. According to the standard, there would be an additional surcharge of 0.05 to 0.1 for thermal bridges, which would result in a total of 8.7 kW. But the standard is not always the same as reality, and the larger the building, the smaller the area proportion of thermal bridges, I think (where, for example, does a large window have a thermal bridge?). From -14°C it is in my case only +34° to 20°. However, you can gladly calculate with 23°C as room temperature (but we are quite satisfied with about 20°, 23 is too warm for me).

For the air I calculated as follows (actually rather transformed, and thought about the units... you can remember it as a simple rule of thumb):
1 m^3/h = 1/3600 m^3/s
Density: 1/3600 m^3/s * 1.21 kg/m^3 = 0.0003361 kg/s
Specific heat capacity: 0.0003361 kg/s * 1010 J/(kg K) = 0.3394 J/(Ks) = 0.34 W/K
thus 0.34 W ventilation capacity per degree Kelvin for 1 m^3/h ventilation loss.

1230 m^3 volume * 0.6 = 738 m^3 exchange per hour, with 34° as dT gives
34K * 738 m^3 * 0.34 W/(K m^3) = 8531 W = 8.5 kW.

The 0.6 comes from a standard. The whole procedure is somewhat more complicated overall (especially the basement area; depending on depth, you can calculate the soil temperature and specific special thermal conductivity coefficients for the basement slab via heat conduction, because generally less heat goes into the ground there or the heat partly “remains” in the soil close to the basement, roughly speaking — as a physicist you also know nonlinear differential equations with boundary conditions, which could make it more exact). The estimate with the temperature I am doing here is still “poor”, actually less heat should be lost.
Otherwise, you repeated it correctly in principle.