It may surprise heat pump owners just how much energy is required to reheat a house after any night-time setback, and what effect this has on heat demand from the HP.
To illustrate this, here’s a worked example, based on the following simplifying conditions (you can amend/refine the calculations below to suit your own situation):
- Your average required indoor temperature = 20degC (some rooms warmer, some rooms cooler)
- The outside temperature = 0degC (assumed constant in this example)
- Your estimated house heat loss at these temperatures = 5kW (this may be from a survey, or from your own operating experience)
- You have an 8kW heat pump
- Your emitters are capable of transferring 100% of the heat pump output at the weather compensated (WC) target LWT
First, you have to estimate the house thermal inertia (measured in kJ/K). This is the heat content of the entire house, including the boundaries (walls/floor/ceilings etc.), and furnishings and other house contents (including occupants, and even the air). You can do this with a once-off test in which you measure the fall in house temperature over a known time period without any heating, at a known (preferably constant) outdoor temperature. For this example let us assume that you establish one night that with the heating switched off, your house cools from 20degC average to 16degC average over 8 hours when the outside temperature is 0degC for the whole of the test.
You can estimate your thermal inertia from (m.Cp) = Q / DT where Q is the house heat loss (in kWh) and DT is the house temperature difference between the start and end of the test (in degC). Actually, the heat loss reduces proportionately as the house-to-ambient temperature difference falls. If this is 5kW when the house is at 20degC average, it will be 16 / 20 * 5 = 4kW when the house reaches 16degC at the end of the test. This gives an average heat loss of (5 + 4) / 2 = 4.5kW, and so the calculated (m.Cp) is 8h x 4.5kW / (20 – 16)degC x 3600s/h = 32400kJ/K. (Strictly, DT does not vary linearly with time as the house cools, but the assumption of linearity gives a good approximation.)
After setback finishes each morning, your heat pump will have to supply not only the ongoing heat loss, but also the “sensible heat” needed to re-warm the house fabric from your night-time setback – let’s assume that this is 2degC for this study (i.e. 18degC average house temperature) – back to 20degC in a reasonable time (occupant comfort considerations).
Assuming that your heat pump delivers exactly 8kW – not always true in practice of course – and that the outside temperature remains at 0degC, you can calculate how fast the house will warm up using the thermal inertia calculated above:
- With the house at 18degC, heat loss = 5 x (18 / 20) = 4.5kW, leaving 3.5kW available from the heat pump for house warming. At (m.Cp) = 32400kJ/K, the house temperature will rise at 3.5kW / 32400kJ/K x 3600s/h = 0.39degC/h.
- With the house at 20degC house, heat loss will have risen to 5kW, so only 3kW is available for sensible heat, and the house temperature will rise at 3 / 32400 x 3600 = 0.33degC/h.
- The average house temperature rise rate is thus (0.39 + 0.33) / 2 = 0.36degC/h (the above comments on non-linearity apply here too, but this figure would be a good approximation).
- So to recover the 2degC temperature setback it will take roughly 2 / 0.36 = 5.5h of heat pump operation at maximum output. If you start your heat pump at 06:00, your house won’t be back up to your required temperature until 11:30 if it is 0degC outside.
If there is less margin between heat pump duty and house heat loss than in the above example, the capacity for overcoming setback will be even less. For example, if your normal heat loss is 6kW (rather than the 5kW used above), but you still install an 8kW heat pump, it will take 7.7h to reheat your house from setback, again at full heat pump output.
Note that the above calculation assumed that your heat pump would be able to run continuously at full output during house reheating. If your weather compensator setup causes heat pump cycling, you will obviously get less total heat and thus even longer reheat times. (Cycling may occur if your WC target LWT is too low for your emitters to transfer the necessary heat, in which case LWT will reach WC setpoint and the compressor will slow down or even stop – not what you want if you are trying to extract maximum heat to reheat the house.)
To summarise, installing a heat pump where the nameplate capacity is well-matched to the maximum house heat loss may minimise the inefficiencies of cycling, but may limit your scope to operate a setback (timing and magnitude) at cold ambient conditions.
(The above notes do not address the economics of setback. Running a heat pump at maximum output for several hours per day to re-heat a house may or may not cost less than the savings from not operating it during the setback period, a matter further complicated by attractive off-peak TOU tariffs. You would need to assess the best operating strategy for your own situation and cost control strategy – some folk aim for a maximised CoP, others for a minimised electricity bill. These are not the same thing, of course.)
Any comments, please?