The idea to use a reservoir of water as a heat pump’s heat source is not new. But now and then somebody dares to do it again in a more spectacular way. Provided governmental agencies give you permit, lakes or underground aquifers could be used.
Today a (German) press release about a European research project called Sinfonia caught my attention. The cities of Innsbruck (AT) and Bolzano (IT) plan to reduce energy demands by 40-50% and increase the percentage of renewable energies used for heating and electrical power by 30%. The results should serve as best practices applicable to other cities.
Diverse activities are planned, such as improving insulation, installing solar thermal collectors or photovoltaic panels, and developing ways to renovate even buildings that are subject to monument protection. The latter is quite a challenge in European cities as laws do typically not allow for installing anything that impacts the view of historical rooftops or the structure of facades.
In a smart grid infrastructure, energy demands and supply should be managed, for example by cooling down refrigerators to -30°C if too much electrical energy is available – effectively storing energy in the ice.
My favorite: Heat pumps should utilize water from a very special source – drain water from the tunnel underneath Brenner pass (called tunnel water or mountain water in German).
This source would provide water flowing at 200-300 liters per second at a temperature of 22°C, resulting in about 10 Megawatt of heating power.
I want to cross-check these numbers:
Assuming low-temperature floor heating loops, heat pumps would need to operate between 22°C ‘input’ temperature and about 40°C ‘output’ temperature.
The maximum theoretical efficiency is limited by principles of thermodynamics: This is Carnot’s Coefficient of Performance, which is:
Thot / (Thot – Tcold)
There are absolute temperatures in Kelvin, so 273 K needs to be added to the temperatures in °C.
Thus the COP is about:
COP = (40 + 273) / (40 + 273 – 22 – 273) = 313 / 18 ~ 17
Carnot’s perfect circular process does not include any phase change – as the evaporation and condensation in a real heat pump – and there are different sources of energy loss.
Thus a real-live heat pump shows a much lower COP. But there is a simple rule of thumb based on experience that is surprisingly accurate: Divide Carnot’s COP by 2 to calculate the realistic COP.
So these heat pumps would operate at a COP of 8,5 which is still very high. The temperature of the tunnel water is expected to be rather constant – as ground water – so COPs will also be high in winter.
Standard ‘geothermal’ brine-water heat pumps show COPs of about 4 – 4,5 when operating between the standard temperatures (values used in standardized tests) of 0°C brine temperature and 35° heating water temperature (B0/W35). As we discussed the meaning of ‘brine’ in the comments recently: In relation to heat pumps this term always refers to a solution of glycol-based frost protection in water.
Water-water heat pumps utilizing ground water with a temperature of about 10°C show a COP of 5,5 to 6 (W10/W35).
I have picked 40° rather than 35° in my estimate, accounting for losses in a district heating system attached to the gigantic heat pumps. Had I calculated with 35° my numbers for the tunnel water heat pumps would even be higher. So the whole exercise is more of an order-of-magnitude check.
The mass flow of about 300 kg (= 300 liters) per second can be converted into power retrieved by the heat pump: by multiplying it with the specific heat of water (4,19 kJ/kg) and with the temperature drop of the water caused by passing the heat pump’s evaporator unit.
The temperature difference in the brine circuit connecting the heat source to the heat pump is about 5 K for standard ‘small’ heat pumps. Water-water heat pumps might also use and additional brine circuit and thus an additional heat exchanger to transfer heat from the source water to the heat pump. I use those 5 K nonetheless as it is not a constant anyway – mentally insert error bars of several 10% here.
The drain water ‘carries’ approximately:
300 kg/s * 4,19 kJ/kgK * 5K ~ 6285 kJ/s = 6285 kW ~ 6,3 MW
The COP represents the factor the electrical energy feed into the heat pump is multiplied with to yield the heating power.
[Heating power] = COP * [Electrical power]
Given a COP of 8 an electrical power of 1 MW would result in 8 MW of heating power, delivered to floor heaters for example. However, the remaining 7 MW then need to be retrieved from the heat source, and the heat source really needs to be able to deliver them:
[Heating power] = [Power retrieved from source] + [Electrical power]
[Power to be retrieved from source] = [Heating power] * (COP – 1)/COP
Aiming at 10 MW heating power output using a heat pump with a COP of 8, the drain water would need to deliver:
[Required power from drain water] = 10 MW * 7/8 = 8,8 MW
This is higher than the calculated power of 6,3 MW but the temperature drop I used was just an estimate and 7 K would also as well be OK – let alone all the other assumptions for operating temperatures and COP. So the numbers from the press release are self-consistent.