Not quite. The max temperature difference tells you the total size of heating system you require. To figure out the total heating demand for the year in kWh you need to know the rate your building loses heat in W K<sup>-1<sup>. This will be the sum of your fabric heat loss and ventilation heat loss rates. You then multiply that by 24 times your HDD to get the heat demand in kWh. Add your DHW, subtract any free gains and multiply by the system's efficiency to find out how much fuel you'll use.

I think that you need to empirically derive your base temperature.

i.e. plot the data - say in imeasure and look at which gives you the best fit.

The base temperature effects the degree days number and takes care of the deta T i.e. HDD15

Technically the 26C in your equation should be the difference between your degree day base temperature and the lower temperature your max heat loss is derived at.

so for HDD15 the heat required to keep your house at 21 is:

E = (5.39/20) x 1937 x 24 = 4624 kWh

then as seret says add your DHW.

I think that (fixed) free gains are included in your hdd base temperature if work it out from real data.

Posted By: SeretYou don't use the delta T at all to work out the demand.

Yes, starting from scratch this is right.

However, if you have somebody else's calculation of the heat loss rate at maximum temperature difference (e.g., done by the plumber sizing the heating system who's taking the fabric and ventilation heat losses into account but not told you their numbers) then dividing the maximum heat-loss rate by the maximum temperature difference to work back to the heat-loss coefficient (the sum of the conductive and ventilation heat losses for each degree Celsius) seems reasonable.

Also, doing it experimentally this sort of calculation would make sense. If you pick a chilly time and find the average outside temperature for a period (say a day or two) and the total energy use of the house you could divide energy use by the time period to get the average power and divide that by the temperature difference to get the heat-loss coefficient.

As long as you're keeping the inside of the house at a steady temperature (so there's not more or less energy stored in the internal building fabric to take into account) and the outside temperature doesn't vary too much (so there's not a great change in the energy in the more external parts of the building fabric) this should give you a reasonable estimate. A run of chilly, cloudy (so it doesn't get too cold at night), dry and not too windy days would probably be best.

Posted By: JT101Heat loss coefficient. I've searched through books and online, and can't seem to find any decent info.

Depends what you mean by heat loss coefficient I think.

Right

Watts per Kelvin or W/K or W.K^-1

All it is is the heat flux at each temperature difference.

So say your wall looses 10 W when the difference between inside and outside (the deltaT) is 1 K (or °C but not °K) and it looses 20 W when it is 2 K different, 30 W when it is 3 K different and so forth.

HDDs are a number that describes the K bit but on a day basis (you can if you want reduce it all to SI units so a day is 86400 seconds but will not help you any here, just remember it is per day)

So if your HDD is 10, that means it has a delta T of 10 K for 1 day (sometimes you get monthly figures so it may be something like March 270 @12HDD. All that means is that it is the sum of the daily figures for that month when the base is 12).

So if you are looking to find the kWh (which are just kJ really) you need to do a bit of equation rearranging.

So if your wall looses 10 W/(m^2.K) for 1 day, that will be 10W.Day, that will be 240 Wh or 0.24 kWh for every square meter and for every 1 K temperature difference.

So if your HDD is 270 for the month and your losses are 0.24 kWh you just multiply the two together for the months losses 64.8 kWh.

Not sure if that will help or hinder.

Posted By: JT101I've calculated all my heat loss in Watts, using Q = ∆T * U * Area. Not Watts / Deg K.

So how do I calculate heat loss coefficient from scratch. Do I just omit ∆T from the above calculation?

Yes, for the definition of heat-loss coefficient I was using.

(I haven't come across another definition but wouldn't be at all surprised if somebody uses the term differently.)

Posted By: JT101Just to finalise my calcs, can I just verify if specific heat demand (e.g. for passive house 15kWh/m2.K) is based on Treated Floor Area?

Yes, but for passivhaus the heat demand is whatever PHPP tells you it is, not some other number that you work out in some other way.