No. COP relates to the efficiency of energy out of the heat pump vs energy in. Flow temperature is a key factor as less work is needed to raise the water temperature to 30C, for example, compared to 40C or 50C. Lots of big radiators at 30C can emit the same heat into the house as fewer radiators at higher temperature.
A further consideration is that while the heat loss from the building should be proportional to be difference between inside and outside temperature (although this may be influenced by other factors such as wind chill), the heat pump efficiency changes with the difference between outside temperature and the circulating water temperature. This is actually a compound effect. When it's colder outside the heat pump has to work harder but also the circulating water temperature has to rise in order emit more heat into the house which further increases the workload.
FWIW, here's a graph of heat out vs energy in for my heat pump since it started heating the house in mid-October. I have assumed that the energy in contributes to the total heat out although this may
I have assumed that the energy in contributes to the total heat out although this may not be completely correct.
It doesn't directly contribute.
The energy in basically powers the compressor (plus a bit for other things) and the energy out is the heat in the gas/liquid inside the compressor (which is raised to a useful temperature and liquified as a result of being compressed), minus losses in the transfer of this energy to the circulating water.
Of course all of the 'energy out' must, in fact, have gone in somewhere - its a heat pump not a perpetual motion machine. The excess of 'energy out' over 'energy in' is extracted from the outside ambient air, which is why you have a large fan that blows cold air from the front of the outdoor unit.
In terms of formulae
COP = energy out/energy in.
exactly how the figures are reported depends on your heat pump. Some heat pumps quote energy out and energy in. Some (eg Vaillant) quote energy in and 'environmental yield' which is energy out-energy in. So in the latter case the calculation becomes
COP = energy out/energy in = (energy in + environmental yield)/energy in.
This is unfortunately a bit confusing!
This post was modified 3 months ago 3 times by JamesPa
4kW peak of solar PV since 2011; EV and a 1930s house which has been partially renovated to improve its efficiency. 7kW Vaillant heat pump.
So, can someone please confirm that I’m calculating my COP correctly by using the graph below.
COP= approximately 2.85
Overall about 43/15 = 2.85
However this is made up of space heating COP and DHW COP, which normally are quoted separately, because they are typically very different and respond to different adjustments. As you have truncated the bottom plot its impossible to break them out. Can you post the rest of the plot please and what temperature do you heat your DHW to? It looks to me like you may be spending more on DHW than you need to, unless of course you actually use a lot.
4kW peak of solar PV since 2011; EV and a 1930s house which has been partially renovated to improve its efficiency. 7kW Vaillant heat pump.
day. My initial figures were from a day when I did use a lot more hot water than normal.
As I’m on an EV tariff, 6.7p midnight to 7am I heat my water then at a high of 60c. This way I feel I can make good use of the low rate. Also enables me to turn off legionnaires setting.
I have the heat pump set to 20c but at 18c during the day when no one is home.
As you imply sacrificing COP for low cost electricity saves money. Its possible that you would save money by cancelling the daytime setback, its the most efficient time to generate heat and would mean you might be able to drop the FT bit. The saving in loss from the house due to a 2C setback for say 8 hours is less than 5%. Only careful experimentation (which is difficult) will tell you.
Iv done a small amount of juggling with times monitoring my hour electricity bill throughout the day. Basically with the heating on during the daytime I’m paying around 30p an hour.
With a 2c setback it’s around 3p an hour.
So that 7hour period it’s approximately £1.75 a day extra to heat all day.
I really though should monitor how much less it cost in the evening after having the heating on throughout the day.
Following this topic, I am becoming less confident that I am adopting the correct policy for DHW heating. Summer is no problem. because I dont have to heat the house, but winter is problematic. I had assumed that it was best to use the ASHP to heat the water during the night when the tariff is reduced from €0.247 to €0.145, but I am now wondering if the higher daytime temperatures make it more logical to use the heat pump to do this at mid day when the OAT is at its highest. How can I best judge this?
A secondary question: On a Mitsubishi, is there any way to determine the energy used and delivered separately for heating and DHW so that SCOP can be evaluated for each?
If you are in the Algarve (as your forum name suggests) then the daily temperature range isn't very big. Heating the DHW to 50C with an outside temperature of 10C will, roughly, need 1/3rd more energy than at an OAT of 20C, but the electricity is only about 60% of the cost. Another consideration is when the DHW gets used. If it's primarily in the morning then the DHW tank won't have lost much heat after being warmed up a few hours earlier whereas it will have cooled more if heated in the middle of the day to be used the next morning, so you need to heat to a higher temperature for the latter condition to compensate for the heat loss.
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