Cold Weather Measurements of Heat Transfer Coefficient

Friends, it’s winter and the weather is reassuringly cold: average daily temperatures in Teddington are around 0 °C. And as I wrote the other week, that offers the opportunity to make measurements of the Heat Transfer Coefficient of a dwelling.

People with Gas Boilers

This is especially valuable for people with gas boilers who are thinking about getting a heat pump.

When the outside temperature is around 0 °C, the average heating power required to heat the majority of UK homes is typically in the range 5 kW to 10 kW.

Most gas boilers have a full power of 20 kW to 30 kW and so can heat a home easily. To keep the temperature just right, the boilers cycle on and off to reduce their average output to the required level. For most houses there is no possibility that a boiler will be undersized.

Heat pumps operate differently. They are typically less powerful than boilers and the maximum heat pump output must be chosen to match the maximum heat requirement of the house.

By measuring the daily use of gas (kWh) by a boiler on a cold day one can estimate the size of heat pump required to heat the dwelling to an equivalent temperature.

I wrote about this at great length here, but at its simplest one just takes the amount of gas used on a very cold day (say 150 kilowatt hours) and divides by 24 (hours) to give the required heat pump power in kilowatts (150/24 = 6.3 kW).

People with Heat Pumps

But the cold weather is not just for people with gas boilers: Heat pump custodians and people heating their house electrically can gain insights when it’s cold.

Starting on 1st December I looked up:

  • the average daily temperature;
  • the daily heat output from the heat pump (in kWh);
  • the daily electricity consumption of the heat pump (in kWh);

The internal temperature was a pretty stable 20 °C throughout this period. So I first worked out the so-called temperature demand: that’s the difference between the desired internal temperature and the actual external temperature.

I then plotted the daily heat output from the heat pump (in kWh) versus the average daily temperature demand (°C). The data fell on a plausible straight line as one might expect. Why? Because the colder it is outside, the faster heat flows out through the fabric of the dwelling, and the greater the rate at which one must supply heat to keep the temperature constant.

In the graph below I have re-plotted this  but instead of using the average daily heat output from the heat pump (in kWh) I have divided this by 24 to give the average daily heat pump power in kilowatts.

Click on graph for a larger version. Graph of average heating power (in kW) versus temperature demand (°C) for the first 10 days of December 2022. Notice that the line of best fit does not go through the origin.

The maximum heat output from the 5 kW Vaillant Arotherm plus heat pump varies with the external temperature, but for flow temperatures of up to 45 °C, it exceeds 5.6 kW.

The maximum daily average power for the first 10 days of December is just over 3 kW, so I think the heat pump will cope well in even colder weather. Indeed I could probably have got away the next model down. But it does seem to be a general rule of heat pumps that one ends up with the model one size above the size one actually needs.

Click on image for a larger version. Specifications for the Vaillant Arotherm plus heat pump. For the 5 kW model at an external temperature of -5 °C and heating water for radiators to between 40 °C and 45 °C, the maximum output is between 5.6 kW and 6 kW.

The slope and intercept of the graph

The slope of the graph is approximately 0.166 kW/°C or 166 W/°C. This figure is known as the Heat Transfer Coefficient for a dwelling. It is the figure which characterises the so-called fabric efficiency of a dwelling.

However, as I noted many years ago when I looked at this problem using gas boiler measurements, the straight line does not go through the origin. The best-fit line suggests zero power output when the external temperature is 2.8 °C below the internal temperature.

This would imply that the heat flow through the fabric of the building was not proportional to the difference between the inside and outside temperature.

The reason for this is that there are other sources of heating in the house, and not all the heat pump output goes into the house. Specifically:

  • People: each person heats the house with around 100 W, about 2.4 kWh/day.
  • Electrical Items: All the electricity consumed by items in the house ends up as heat. For my home this amounts to around 10 kWh/day.
  • Hot Water: Heat pump output that heats domestic hot water is mostly lost when the hot water is used. My guess for this house is that this amounts to around 3 kWh.day.

So to estimate the actual amount of heat dissipated in the house I should really take the heat pump output and:

  • Add 2.4 kWh/day for each person in the house
  • Add 10 kWh/day for all the electrical items
  • Subtract 3 kWh/day for the hot water lost.

Together this amounts to adding 9.4 kWh/day to each heating estimate. Pleasingly, plotting the same graph with these corrections, the graph now intercepts within 0.5 °C of the origin. To me this indicates that I am now accounting for all the significant sources of heat within the house reasonably well.

Click on graph for a larger version. Graph of average heating power (in kW) versus temperature demand (°C) for the first 10 days of December 2022.

I haven’t included any solar gain in these estimates because at this time of year solar gain is generally very low unless a home has large south facing windows. Previously I have noted that solar gain seemed to be much more important in spring and autumn with longer and generally sunnier days.

COP

I also took the opportunity to evaluate how the daily averaged coefficient of performance (COP) varies with external temperature. This is based on readings from a heat meter and an electricity meter which monitors the heat pump.

The COP values below 3 are a little bit lower than I would like, but still acceptable. On subsequent cold days I will be seeing if there are adjustments I can make to the heat pump operation which can improve this.

Click on graph for a larger version. Graph of daily average COP versus daily average outside temperature (°C) for the first 10 days of December 2022. Extrapolating the trend suggests that the COP would reach unity at a temperature of – 12 °C.

Out of curiosity, I also evaluated the heat output using the Vaillant SensoApp. The figures were massively in error. For example, on 10th December the app suggested the total heat delivered to the house by the heat pump was 58 kWh. In fact the correct answer was 73.9 kWh.

Conclusion

Cold weather offers an opportunity to assess the so-called Fabric Efficiency of a dwelling by direct measurements of its Heat Transfer Coefficient.

The cold weather will be with us for a few more days so there’s still a chance to make measurements in your dwelling.

Tags: , , ,

16 Responses to “Cold Weather Measurements of Heat Transfer Coefficient”

  1. Paul Rudman Says:

    150 kWh / day? I live in a flat with electric heating, and nowhere to put a heat pump. If I used that much energy it would cost me £1,575 a month!!! I dream of a gas boiler.

  2. Dan Grey Says:

    Two random thoughts; does the x-intercept being 2.8° suggest that your home has a heating degree day base temperature of 17.2°? And, your CoP only falling below 3 with average outside temperatures at freezing strikes me as excellent performance.

    • Dan Grey Says:

      heating degree day *base temperature*, that should’ve been!

    • protonsforbreakfast Says:

      Dan

      Yes, exactly that. It’s close enough to the standard ‘T – 3’ rule for most practical purposes.

      And yes, the COP is fine, but I wonder if it could be better. Many Twitter-Heat Pump colleagues have been achieving better results and I wonder if my system can be tweaked. I’ll get over it.

      All the best

      Michael

  3. protonsforbreakfast Says:

    Paul, Good afternoon. That’s a fairly typical daily winter consumption for a house when it’s very cold outside. It corresponds to around 6.3kW average power.

    Obviously flats are generally smaller and less exposed than houses. Often they are amenable to air-to-air heat pumps which have been designed for exactly your situation.

    Best wishes

    Michael

  4. Ioannis Palaiologos Says:

    Thank you for summarising your previous articles Michael.
    During this cold spell I have been recording daily readings of the gas and electricity meters and the size of the heat pump we need to replace our gas boiler has been determined.
    Your posts and videos have been a brilliant source of information.
    Thank you,
    Ioannis

  5. JoeW Says:

    hi
    I have noticed very similar things with our heat pump in the cold spell. I have independent electrical monitoring but have to work with Vaillant energy yield figures, Vaillant underestimate electrical consumption by 100 W when working hard.

    I don’t think this is just it ignoring the units circulation pump as the flow rate is very consistent. and in warmer weather the error is in the 50-60 W range. Perhaps it is related to the defrost cycles?

    My estimate have shown COPs of 2.5- 2.8 in these 0 degree days not ideal. But interesting that Vaillant underestimate the heat produced. I would think they would err on the side of generous. ~30% under is pretty off. I wonder where there error comes from? our flow rate is almost always 1200l/h perhaps this measurement has a large error margin, i did read it is problem if it is setting think it has glycol when it does not as the metering will have the wrong specific heat, and give junk figures. I suppose that as DT between flow and return is only ~3 degrees, temperature measuring error of +- 0.1 degrees is an potential error of .2 degrees giving 6.6% error not 30% perhaps their flow and return temps are worse than this. wish i had fitted proper monitoring.

    • protonsforbreakfast Says:

      Joe,

      I just don’t understand why the Vaillant thermal readings are so poor. I agree with all your comments about sources of uncertainty, and I would not be surprised if the error were 5% or even 10%. But their numbers are just way off. Personally, I suspect they have good data internally but that this is a software problem. These great pumps really are let down by their software.

      In any case, good luck and keep warm!

      Michael

      • Joe Wentworth Says:

        In response to our earlier head scratching I have found in part why the energy yeild figures from the vaillant control unit are so much lower than your heat meter readings

        They are infact ‘environmental energy yeild’ values so need adding to the electrical consumption for the same period to get a better delivered heat value.

        See second last page of:

        22006-heat-pump-leaflet-v05-web-2-2343880.pdf

        Still annoying that you can only have today’s, this month and all times totals. But for me without independent heat metering it will improve my basic monitoring.

        Happy new year.

      • protonsforbreakfast Says:

        Joe, Good Morning and Happy New Year.

        The built-in monitoring may be better than nothing, but based on my results taken yesterday, I don’t think it’s very good.

        Mick Wall disagrees with me!

        Anyway – whatever: best wishes for a low energy new year. M

  6. Greg Says:

    Hi Michael
    I have read your article regarding calculation of the Heat Transfer Coefficient as I am interested in a heat pump for my radiant floor heating. I have a few questions:1)I am not clear on how you arrive on the average daily temperatures. Is that a value of each day that I can find for my specific location via a weather service? or Do I take the average of the low/high temps of that day?
    2)I have an electric boiler and I document daily kWh heating as well as daily low temps. I seem to recall that you mentioned to take the highest daily kWh usage for the heating season and divide that by a 24(hours/day?) to arrive on an estimated Heat Pump size- this being the minimum size for a heat pump that would run constantly.

    • Greg Says:

      I did not realize my weather station reports a daily temp. I will use these daily average numbers.

    • protonsforbreakfast Says:

      Greg, Good Morning.

      1. I thought long and hard about which temperature to use, but my conclusions was that – on balance – the average temperature was likely to be best estimate of heating demand. Why did I choose that? The minimum temperature generally occurs in the wee small hours and may not represent the demand to which occupants try to respond. Historically people estimated the average temperature as (T_max + T_min)/2 but in these days of 5 minute data-logging intervals, the true average is generally available on meteorological sites.

      2. So you are using an electric boiler for your under floor heating (UFH)? If so, then taking the heat input (kWh) – your boiler is 100% efficient ! – and dividing by 24 h will give you the average heating power – neglecting any domestic hot water heating.

      So I would recommend that you establish the HTC using the average internal and external temperatures and the average heat input to the house. Once you have the HTC then you can estimate the heat pump size from the heating demand at your design temperature – typically -3 °C in the southern half of the UK.

      Does that make sense?

      All the best

      Michael

  7. Greg Says:

    Hi Michael,
    Thank You for the info. Regarding HTC which is the slope in W/c: Now that I finally figured the x axis of heating demand, I am unclear how you arrive on the Y axis. With electric heat I would guess that I use the kW heat usage or should I use kW/hr (kW/24)?

    • protonsforbreakfast Says:

      Greg, Good Evening.

      It’s complicated.

      Simple Guess: If heating a dwelling gas, this will likely be the largest amount of heating. So you could just take the daily gas usage (kWh) and use this for the ‘y’ data.

      Better Guess: Assume the gas heating is only 90% efficient – 10% of the heat produced by the boiler does not enter your home but leaves through the flue.

      Best Guess: Include boiler inefficiency as above but also add in the heat produced all teh electricity used that day – typically another 10 to 20 kWh. And also the 2.4 kWh/day with which each adult human heats the house.

      In my experience, the simple guess will be not far off.

      Best wishes

      Michael

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s


%d bloggers like this: