Project Design
Heat pump selection
Example
Example
Note In a new building, the building fabric generally dries out in the rst two
years after occupancy, whereby moisture from the construction phase
escapes from the building fabric; in this phase, the heat requirement is
higher than after the building has dryed. This increase in heat require-
ment should be offset by the additional electric heater.
• For a residential house in London, UK a required heating load of
9.6 kW and an outside temperature θe = -1.5 °C
• Water heating for four people with a normal comfort level
(45 litres per person and day at 45 °C tap temperature or 1.8 kWh):
4 Hence 1.8 = 7.2 kWh per day. A heat pump with a heating power
of 9.6 kW would require an operating time of 7.2 kWh / 9.6 kW = 0.75 h.
Thus, if rounded up:
Tank charging = 1 h
• The pipe correction factor, owing to a pipe length of 15 m
(one-way length) with means of 1.0 and 0.83 results in
line correction factor = 0.92
Total heating capacity ≥
9.6 • 24 h
=
230.4
= 10.89 kW
(24 h – 1 h) • 0.92 21.16
Factoring in power supplier outages of 2 hours per day:
Total heating capacity ≥
9.6 • 24 h
=
230.4
= 11.93 kW
(24 h – 1 h – 2 h) • 0.92 19.32
The calculated overall heating capacity must be calculated using a
continuous water supply temperature of 35 °C for a underoor heating
system.
Heat pump capacity ≥
standard heating load
•
24 h
(24 h – tank charging time – power supplier outage time)
•
pipe correction factor
Note The illustrated calculation of the overall heating load may differ slightly
from the detailed calculation of the Aquarea Designer, but can still be
used as a rule of thumb for fast calculation without the need of a calcu-
lation program.
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