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Adventures in heat loss calculations

Rob of York · 2026-08-29T15:49:23Z

I thought I’d post my recent adventures in heat loss calculations, in the hope that they may be of interest to others embarking on the same. Over the last few days, I’ve done some heat loss calculation for my house, a 1970's 4-bed detached (146 sqm), which has as much insulation as we could reasonably add without it turning into a full retrofit. As a stubborn and pragmatic scientist, I decided I’d work out the heat loss as far as possible myself from the basics. I used 24 degrees as the design temperature difference (i.e. -3C outside and 21C inside). I calculated the heat loss in two ways and also tried out a simple estimate: Theoretical: based on the attributes of the house (room sizes, construction materials and their U-values etc.). Practical: based on the energy going into the house that kept it significantly warmer than outside during December 2024 and January 2025. (I used those months because our only heating system then was a gas boiler). Quick estimate: using the Michael De Podesta Formula 1. Heat loss (theoretical): 1a. Thermal conduction: I worked out the areas of the various building components (floors, ceilings, walls, windows etc) on a room-by-room basis and either looked up their U-values from the rdSAP 10 manual or worked them out from the U-values for the individual materials used and their thicknesses. (The open university web pages were useful in that respect, seeand the next few pages). From the various U-values, areas, and the temperature difference, I worked out the heat loss. The overall thermal conduction losses came to about 3720 W. 1b. Thermal Bridging: This takes account of the joins between walls and roof, walls and floor, windows and walls etc. which tend to be areas of high heat loss. rdSAP 10 accounts for this by increasing the U-values of everything in 1a above by a fixed amount depending on the age of the property, e.g.by 0.15 for my 1970's house. The overall thermal bridging losses came to about 1180 W 1c. Ventilation Losses: These are given by a simple formula: 0.33 x temp difference x house internal volume x ACH. Where ACH is the number of air changes per hour. Appropriate values for ACH are an area of some debate as they can vary widely for two ostensibly similar houses. As I’ve taken care that my house is well sealed, I used a value of 0.5 for ACH, however, I currently have no hard evidence for that value. The overall ventilation losses came to about 1340 W Total theoretically determined heat loss 6240 W Thermal conduction was 60% of the total, thermal bridging 19% and ventilation losses 21%. 2. Heat loss (practical): I have a semi-professional weather station in my back garden. The outdoor and indoor units record temperature every 30 minutes, hence I could easily get average indoor and outdoor temperatures for Dec 2024 and Jan 2025 and the average difference in temperature over each of those months. I then worked out the total heat energy going into the house from the following: Gas consumption figures, assuming 0.9 boiler efficiency (condensing except when heating hot water). Subtracted gas consumption for hot water heating, since the water is still hot when it drains out of the house. Electricity consumption figures, since all use of electricity within the house eventually turns into heat. Two people in the house at 100W each for an average of 16 hours per day. Solar gains from 10 kWh per sqm incident energy on east and west facing windows over each whole month. (Double glazed unit g-value of 0.49, glass to frame ratio of 0.7, overall window area measured). Correcting from the actual difference in average indoor and outdoor temperatures gave an input energy of 147.8 kWh per day to achieve a 24-degree temperature difference. Total practically determined heat loss 6160W 3. Michael De Podesta Formula The Michael De Podesta Formula is simply annual gas usage divided by 57.3 multiplied by the 24-degree temperature difference. Total heat loss based on Michael De Podesta Formula 6031W Thoughts: The first thing that struck me was that the figures from the two completely different heat loss calculations are surprisingly close. As they are worked out completely differently, this gives me some confidence that they are about right. Further, the simple Michael De Podesta Formula also gives a value that is within 5%. The ACH value has the most uncertainty and varying that has a big effect. I don't know if there is anything I can do to reduce that uncertainty short of paying for an air tightness test. There is quite some heat that goes into the house from sources other than the heating system. This amounts to about 570W from non-heating electricity use, people burning calories, and solar gain through the windows even at the dullest time of year. If I just want the maximum demand on the central heating system, then it looks like this is about 5500 to 5700W, not including hot water heating. Comments welcome!

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JamesPa

(@jamespa)

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10/11/2026 4:21 pm

Posted by: @susannasaunders

↑

Posted by: @toodles

↑

@rob-of-york In my experience, energy used to supply DHW will be higher in the colder months - have you estimated and made any allowance for this factor? Toodles.

Agreed because the ground water is colder and will require more heating to provide showers etc... In summer the ground temperature is around +10C whereas in winter its much closer to +3C. Maybe nearer zero in some instances. But as the central heating is also running, I don't see how to estimate this increased power draw needed to warm the water an additional 7 to 10C degrees during winter? Unless I guess you just double it during the winter months and maybe 1.5 times for the shoulder months? Any thoughts on this?

I think you may be overthinking this. It really is most unlikely matter, unless its for pure interest or you have really high usage.

But if you do want to work out the power needed then its easy from basic physics:

energy required = specific heat capacity X mass X temperature rise.

Since 1l of water weights 1kg, all you need to do is estimate your daily volume of (hot) water used and multiply it by 4200 (the specific heat capacity of water in J/kgC) and then again by the (extra) temperature rise needed. This gives energy required in joules.

1kWh is 3,600,000 J, so you can get energy in the more familiar unit kWh by dividing by this number. Then divide by 24 to get average power required in kW.

So for example if you use 150l of DHW a day and the winter time additional temp uplift needed is 7C then the additional energy is 150*4200*7= 4, 410,000 J which is 1.225 kWh or an average (over a day) of a little over 0.05kW.

In other words (with these figures) you can ignore it completely when sizing your heat pump!

This post was modified 58 minutes ago 2 times by JamesPa

This post was modified 27 minutes ago 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.

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SusannaSaunders

(@susannasaunders)

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Joined: 3 weeks ago

Posts: 34

10/11/2026 4:48 pm

@toodles thanks! So as suspected, around a doubling of the cost for DHW during winter! 👍🫶

Installed solar [7.44 kWp] and batteries [Sigenstor 3 x 8kWh] and 12 kW inverter in July 2025.

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