In short
- Annual IT heat = IT MW × load factor × 8,760 h.
- COP ≈ 0.5 × Tsupply ÷ (Tsupply − Tsource), in kelvin.
- Heat out of the heat pump = recovered heat × COP ÷ (COP − 1), because its electricity becomes heat too.
- A 20 MW campus at 75 % load, 80 % capture, 40 → 70 °C and 70 % use delivers about 89 GWh a year, enough for about 8,900 homes.
Annual heat from the IT
Use the average load factor, not the design load: a new campus can take years to fill. Total site energy is EIT × PUE; at PUE 1.25 that is 164,250 MWh.
How much heat is captured
Only heat that reaches a water loop can be reused. Direct-to-chip racks put 70–90 % of their heat into liquid; air-cooled halls can recover heat from the chilled-water return at a much lower temperature. With 80 % capture: 131,400 × 0.8 = 105,120 MWh at about 40 °C.
The heat pump's COP
| Source | into 55 °C | into 70 °C | into 90 °C |
|---|---|---|---|
| 28 °C (air) | 6.1 | 4.1 | 2.9 |
| 40 °C (W32) | 10.9 | 5.7 | 3.6 |
| 55 °C (W45) | direct | 11.4 | 5.2 |
The heat pump's output is the recovered heat plus its own electricity: out = recovered × COP ÷ (COP − 1). Electricity = out ÷ COP. Try other lifts on the temperature lift slider.
Seasonal use
The network does not take all the heat all year. Multiply by the share it can use: 70 % is a reasonable first assumption for a city network with hot water demand, lower for a heating-only network without storage.
Homes, CO₂ and money
- Homes = heat delivered ÷ heat per home (about 10 MWh a year for a typical apartment in Central Europe).
- CO₂ avoided = heat delivered × 0.202 t/MWh ÷ 0.9 boiler efficiency − heat-pump electricity × grid factor.
- Net margin = heat delivered × heat price − heat-pump electricity × electricity price.
Energy reuse factor (ERF)
German targets for new sites are 10 %, 15 % and 20 %; see the rules guide.
A 20 MW liquid-cooled campus into a 70 °C network
| Quantity | Value |
|---|---|
| IT energy = 20 × 0.75 × 8,760 | 131,400 MWh |
| Recovered at 40 °C (80 %) | 105,120 MWh |
| COP, 40 → 70 °C | 5.7 |
| Heat-pump output = 105,120 × 5.72 ÷ 4.72 | 127,400 MWh |
| Heat delivered (70 % used) | ≈ 89,200 MWh · avg 10.2 MW |
| Heat-pump electricity used | ≈ 15,600 MWh |
| Homes at 10 MWh each | ≈ 8,900 |
| CO₂ avoided (grid 300 g/kWh) | ≈ 15,300 t/yr |
| Net margin at €40 heat, €120 power | ≈ €1.7M/yr |
| ERF = 73,600 ÷ 164,250 | ≈ 45 % |
Questions about the calculation
Why is the heat delivered more than the heat recovered?
The heat pump's electricity also ends up as heat. With a COP of 5, every 4 MWh recovered becomes 5 MWh delivered.
Why use 50 % of Carnot?
It is a typical efficiency for large industrial heat pumps. Manufacturers' data for the exact machine and temperatures replaces it in design.
How much heat does a home need?
Around 10 MWh a year for an apartment in Central Europe, more for older houses, less for new buildings. Use the network operator's figure.
Does a warmer cooling loop always help?
For heat reuse, yes: moving the source from 40 °C to 55 °C roughly doubles the COP into a 70 °C network. Check that the IT allows the warmer water.
Sources
- Calculator model on this site: COP from Carnot with η 0.5; gas 0.202 t CO₂/MWh at 90 % boiler efficiency.
- ERF definition as reported by EU sites under Delegated Regulation 2024/1364.
- Worked figures are screening estimates, not a design.