Heat loss = HTC x DT. (There is no subtraction of PO in this equation, which I had wrongly included)

Power required for space heating = (HTC x DT) - PO, since the other sources of heat provide some of the power needed.

This was how I understood things, and is why you need to add PO to the primary source contribution (which I will call PP, Power Primary, as opposed to PO, Power Other), to get the total heat loss for IAT - OAT assessments. Given steady state:

Total heat loss = PP + PO

However, HTC x DT measures PP (the Y values that get used in the regression are from the primary source, they do not measure, and can't directly measure, PO), with PP remaining solely the primary heat source's energy contribution. It crosses the X axis at a positive intercept, the point at which PO is sufficient, and at that point, and at lower X axis values, no PP is needed. 

Since PP is only the primary source's contribution, we need to add PO to get the total heat loss, but for heat pump sizing we only need PP.

However, put another way:

If total heat loss = HTC x DT

then @rob-of-york's second equation 

Primary source power required for space heating (which is I think PP) = (HTC x DT) - PO

which after rearranging becomes

(HTC x DT) (which is the total heat loss) = PP + PO

And put yet another way, the energy vs IAT - OAT assessment is done after the PO has added its contribution, in effect the PO reduces the apparent heat loss the primary heat source has to deal with. To get back to the total heat loss, we need to add back the PO.   

It all depends on whether HTC x (IAT - OAT) represents the total heat loss (which includes the contribution from PO) or whether it represents the heat loss that the primary heat source provided after the PO has done its thing. My contention is that it is the latter, because the data used for the regression has no way of knowing what PO is, even if it can be derived indirectly, through the intercepts.

Or perhaps I am just hopelessly confused!

The DNO would only allow a maximum of 5.7kW export, probably because we live a long way from the nearest substation on a 1970's housing estate, and hence their wiring is likely not great. Occasionally, the voltage hits 250V+ on sunny days, so I think they were right to be concerned about lots of export raising the local voltage levels.

For the past couple of years, we have had the system configured to charge the battery from the grid on cheap rate overnight electricity, and discharge only when needed for consumption that is not covered directly by the solar PV output. We charge the batteries to 100% every 4 or 5 days to keep the BMS happy and able to track SOC correctly and then force discharge to 75% SOC the following morning. On other days we just charge to 70% overnight.

The reason we keep the batteries at a moderate SOC is two-fold. Firstly, for battery life, and secondly it allows the inverter to dump any excess power above 5kW that it cannot send to the grid or house directly into the batteries. This avoids any "clipping" losses. On cool spring days with lots of sunshine, this extra output can top the batteries back up to near 100%. At times we see over 7.2kW peak output from what are ostensibly 6.56kW panels. This is likely due to "cloud bounce" which happens when the panels receive the full effects of direct sunlight and reflected light from nearby clouds. On those occasions we are glad that the inverter can deal with up to 8kW.

We avoid any arbitrage, i.e. charging the batteries at overnight rates and then discharging them specifically to earn extra export as the economics don't work for us once the approx 6p/kWh cost in used up battery cycles is considered. We want the batteries to last as long as possible. We similarly don't use solar to charge the batteries except for recovering what would otherwise be clipping, the economics are better to just export the solar PV output and fill the batteries from the grid overnight. Our daily average electricity consumption is just under 7kWh and the batteries easily cover that (9.6kWh nominal and about 8.64kWh usable) along with direct electricity use from solar PV output.

We recently had the EPS backup power output of the inverter connected to some additional sockets and the downstairs lighting circuit, so we now have a 13A 240V uninterruptible power supply to those sockets and some lights that don't go out in a power cut. This involved having independent earths added, since the safety regulations assume that on a power cut the main earth may become disconnected.

We are fortunate that the orientation of our roof is near perfect for solar PV, close to south facing and a 30-degree slope. The best daily output so far was 48.7 kWh, and the total for 2025 was over 7500 kWh.

The system was fitted by a very good local installer who took pride in their work making sure that all of the little things were done well.

With insulation upgrades lowering the heat loss, two air-to-air heat pumps (AC system) that have an achieved efficiency of 5.2 on heating, a gas boiler that runs weather compensation, and a lot of kWhs export from the solar PV system, our annual energy bill is now at almost exactly the same level as the standing charges.

The degree day method of heat loss calculation is very useful because it only requires smart meter data on daily gas usage, which nearly everyone can obtain, and Met Office degree day data from a nearby weather station, which is again readily available.

The downside is that it is less accurate than using the difference between indoor and outdoor temperatures measured by a home weather station.

The question is what can be done to refine the degree day method to obtain better accuracy, without needing a lot more data. This post attempts to answer that question.

Degree Day method

Recall that the degree day method plots the measured power output of the heating source (PM) against degree day (DD) information typically obtained from local Met. Office data.

The degree-day measure is 15.5C minus the daily average outdoor temperature, i.e. DD = 15.5 – OT, where OT is the outdoor temperature.

In an ideal world, the indoor temperature would be held perfectly constant at its set point, and hence the degree day value would reflect indoor temperature (IT) minus outdoor temperature (OT).

Recall that the balance of heat energy into and out of the house is:

PM = HTC (IT – OT) – PO

Where PO is the power output from other sources.

If the indoor temperature is constant, then we have DD = (15.5 – OT) and IT = K, the constant set point value. Since DD = (15.5 – OT) the above equation becomes:

PM = HTC(DD + K – 15.5) – PO

or

PM = HTC(K-15.5) + HTC(DD) – PO.

Note that the term HTC(K-15.5) is a constant and represents an offset of (K-15.5) degrees on the x-axis of the graph, e.g. 3-degrees in the example above. The slope of the regression line is HTC, and the heat loss is HTC x DT, where DT is the design temperature difference.

This means that all the data points and the slope of the regression line on a graph of PM versus (IT-OT) and a graph of PM versus (DD) are the same, it is just the x-axis scale that is shifted by (15.5 – K) degrees. For example, if the constant indoor temperature is 18.5C, then the point at which the indoor and outdoor temperatures match and the heat loss is zero corresponds to a DD value of -3C, i.e. the vertical line at x=-3C. The negative of the intercept of a regression line with this line gives PO, the average power output from other sources.

The issue

The equation PM = HTC (IT – OT) – PO states that the power output needed to keep the house at some indoor temperature depends on the temperature difference.

Using DD instead of (IT – OT) implies that all that temperature difference is due to changes in outdoor temperature; however, this is not strictly the case. A lower average outdoor temperature can often result in a lower average indoor temperature. This will happen if the heating system is turned off over-night and the house is allowed to cool.

Below is a graph of daily average indoor temperature plotted against daily average outdoor temperature for my house for the period Oct-24 to Mar-25. For these six months, the heating was off between 10pm and 6am every day.

Aside from some outliers, there is a clear trend of lower average outdoor temperatures resulting in lower average indoor temperatures.

Refinement of the Degree Day method

Instead of assuming that the indoor temperature is fixed, we could instead take the view that a theoretical degree-day from the Met Office data of say 20 results in an actual 19-degree temperature difference due to a 1-degree lowering of the average indoor temperature. In other words, we multiply the degree day value by 0.95 and use that value in the plot. This would mean that a degree day value of 20 results in an adjusted value of 19, a DD of 10 would give an adjusted value of 9.5 and so on. This makes sense as a larger degree day value (temperature difference) has a proportionately larger effect on the indoor temperature.

In our house, the heating is off for 8 hours over-night. If it is subzero outside, then the house may cool by as much as 6 degrees in total, i.e. an average of about 3 degrees lower indoor temperature for 8 hours overnight. Taken over a 24-hour period, this equates to a 1 degree decrease in average indoor temperature.

So, what happens if we plot gas power output against 0.95 DD instead of just DD.

The plot of power (gas only) vs. adjusted degree days gives HTC = 263.3W/C and power from other sources of 556W (negative of x = -3C line intercept, given my average indoor temperature of 18.5C, and degree days definition of 15.5C). Hence the heat loss at a 24-degree temperature difference is 6320W less 180W for hot water gives 6140W, which is less than 150W below the results from previous calculations based on more accurate temperature difference data. The power from other sources of 556W is probably still underestimated here. The power required for space heating is 5580W and 180W for hot water, which is within 100W of the more accurate estimates obtained using temperature differences.

In conclusion, adjusting the degree days by a factor of 0,95 provides a way of accounting for the fact that lower outdoor temperatures tend to feed through into lower indoor temperatures, and are certain to do so if the heating is turned off overnight.

Why pick 0.95 as an adjustment? The logic for that comes from a simple estimate of the indoor temperature loss overnight on a very cold day when the heating is turned off. If a heating system is running 24/7 and is very good at holding the indoor temperature at the set point, then degree days can be used directly. Otherwise, a well-reasoned adjustment to the degree days value can improve the accuracy of heat loss estimation. Certainly, this is the case when the heating system is turned off overnight, as is common practice with gas central heating.

Most of the time our hot water heating runs every 48 hours rather than every 24, since with relatively low use it takes that long for the tank temperature to drop 8-degrees (the hysteresis of the thermostat) and trigger the hot water heating again.

I really appreciate that, particularly given your reservations before reading it.

"Compulsory reading" is a pretty extraordinary endorsement and exactly the sort of response I hoped the book might earn once people got past the title and into what it is actually saying.

Thanks again for buying it, reading it properly and backing it publicly!