Calculation of heat losses from temperature and power data:
| Date | Outside °C | Lounge W °C W/°C | Dining area W °C W/°C | Back room W °C W/°C | Study W °C W/°C |
|---|---|---|---|---|---|
| Jan 21 | 1.2 | 1173 17.0 | 1171 17.0 | 1132 18.6 | 373 16.3 |
| Jan 22 | -0.5 | 1190 16.8 | 1240 16.5 | 1279 18.4 | 447 16.3 |
| Jan 23 | 0.2 | 1089 16.7 | 1190 16.6 | 1090 17.7 | 432 16.1 |
| Jan 24 | 2.8 | 996 16.6 | 1019 16.1 | 1062 18.3 | 370 16.2 |
| Jan 25 | 3.3 | 1073 16.3 | 1129 16.1 | 1204 19.3 | 518 16.4 |
| Jan 26 | 4.7 | 1032 16.7 | 835 16.6 | 902 19.3 | 385 17.0 |
| Jan 27 | 4.4 | 989 17.0 | 710 17.2 | 848 18.6 | 307 16.6 |
| Jan 28 | 4.8 | 1069 16.7 | 813 16.5 | 1047 19.0 | 362 16.8 |
| Jan 30 | 6.4 | 1056 18.1 | 560 18.1 | 943 20.1 | 181 17.8 |
| Feb 1 | 6.5 | 835 17.3 | 454 17.2 | 1139 19.1 | 187 16.1 |
| Average | 3.38 | 1050 16.92 77.5 | 912 16.79 68.0 | 1065 18.84 68.9 | 356 16.56 27.0 |
For a design outside temperature of -2.2°C and design room temperatures of 21°C (lounge, dining area, back room, study, bathroom, hall) and 18°C (bedrooms, kitchen) we get the following losses, compared with predicted losses from my model and from Octopus. Note the use of 21°C as the hall design temperature - this is so that the hall radiator can assist with lounge and study heating.
| Room | Heat loss - measured W | Heat loss - Octopus W | Heat loss - model W | Heat loss - model W Main rooms heated |
|---|---|---|---|---|
| Lounge | 1798 | - | 1962 | 2872 |
| Dining area | 1578 | - | 631 | 687 |
| Lounge + dining area | 3376 | 1993 | 2593 | 3559 |
| Back room | 1598 (1262) | 1134 | 907 | 1178 |
| Study | 626 | 270 | 573 | 596 |
| Hall | - | 352 | 718 | - |
| Bathroom | - | 396 | 353 | - |
| Front bedroom | - | 687 | 228 | - |
| Back bedroom | - | 493 | 200 | - |
| Total | 5600 (5264) | 5325 | 5570 | 5332 |
I think the measured back room heat loss has been distorted by turning up the setpoint during the GO cheap rate period. I've attempted to compensate for this, producing the figures in parentheses.
There seem to be two approaches to calculating heat losses through a floor in contact with the ground.
The first approach assumes that the ground temperature is constant throughout the year, but otherwise treats the floor like any other external surface. MCS takes this approach (see document MIS-3005-D section 5.5.1):
When calculating the heat loss through a solid floor in contact with the ground, the temperature difference to be used is the internal design room temperature (Table 1) minus the local annual average external air temperature (see MGD 007 Section 5)."
For here, "South-eastern (Gatwick)" the relevant temperature is 10.2°C, or maybe "Southern (Hum)" at 10.4°C.
Concrete (as used for floors) has a conductivity of about 1.35 W/mK (see Tables of U-values and thermal conductivity Table 6.A.18). Our floor is probably 100mm thick (see Evolution of Building Elements - 3 Ground Floors), so it would have a U-value of 1.35/0.1 = 13.5 W/m²K.
For example, the lounge (excluding the dining area) has an area of 9.8m². The heat loss at the design temperatures would be 9.8*13.5*(21-10.4) = 1402W.
The lounge floor has 8mm thick wood flooring on top of the concrete. Wood has a conductivity of about 0.14 W/mK, so the wood flooring will have a U-value of about 0.18/0.008 = 22.5 W/m²K. Together, the concrete and wood will have a U-value of about 1/(1/13.5+1/22.5) = 8.4 W/m²K. So the wood flooring will reduce the heat loss to 9.8*8.4*(21-10.4) = 873W.
The second approach assumes that heat loss is primarily via the periphery of the floor and a short path through the ground. In this case, the temperature difference would be relative to the design air temperature of -2.2°C. A simplified formula for an uninsulated floor (see U-Values, building regulations and insulation) is:
U = 0.05 + 1.65(P/A) – 0.6(P/A)²
Where:
U = U-Value of the uninsulated floor (W/m²K).
P = Length of the exposed perimeter (m).
A = Area of the floor (m²)
For example, the lounge (excluding the dining area) has an area of 9.8m² and an exposed perimeter of 6.2m. P/A is 6.2/9.8 = 0.63, and U is 0.05 + 1.65*0.63 – 0.6*0.63² = 0.85. Heat loss through the lounge floor at the design temperatures would then be 9.8*0.85*(21+1.2) = 184W.
In my model I've taken the first approach, mainly because it's easier to apply (no need for exposed perimeter data), but also because MCS use.
However, the U-value as calculated above seems to produce losses that are too high, so I've experimented to get a U-value that produces heat losses similar to the measured ones.