 Energy in buildings

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# 2.4.2 Degree days

We can also use this heat loss coefficient together with the number of degree days to understand how much space heating energy a building might use in different locations. Over a long period, such as a day or so, the heat loss from a building will be proportional to the average temperature difference between the interior and the outside air. If on a given day the average internal temperature was 20°C and the average external temperature was 10°C, then the difference would be 10°C. We would describe that particular day as having ‘10 degree days’. If on another day the average internal temperature was the same and the external temperature was zero, 0°C, i.e. an average difference of 20°C, we would describe that day as having 20 degree days and expect the building to lose twice as much heat as on the first day.

However if the average external temperature was higher than the interior, then there would not be any heating requirement, and the number of degree days would be zero (rather than a negative number). The total heating requirement over a month will be proportional to the sum of all the degree days of the individual days.

Table 8 gives some long-term averages for sample UK locations. Given their long history of use, it is not surprising that they are normally produced in the UK with a standard indoor base temperature of 60°F, equivalent to 15.5°C.

## Table 8 20-year averages of degree days (1985–2004) to base 15.5°C for sample UK areas

South WesternLondon (Thames Valley)MidlandsNorthern IrelandBordersNorth-East Scotland
January281319356343345367
February257282314305304327
March239242278286295313
April193180220227248255
May11297136150180183
June58447285106110
July251836465762
August231937535566
September5048769295112
October111120167174171200
November193227264258260285
December252293334323327358
Total179418892290234224432638
(Source: EST, 2005)

Table 8 gives an annual total of 1889 degree days for the London area. A first estimate of an annual heating energy consumption of our house in watt-hours would be the heat loss coefficient, 127.3 W K–1, multiplied by the number of degree days multiplied by 24 (to convert from days to hours). Dividing by 1000 then gives the result in kilowatt-hours (kWh).

• Annual consumption = 127.3 × 1889 × 24/1000 = 5771 kWh

If the house had been located in Berlin, instead, which has 2600 degree days, then the heating load would have been much higher:

• Annual consumption = 127.3 × 2600 × 24/1000 = 7944 kWh

Put another way, it would have to be better insulated to achieve the same heating demand.

We can go further and say that if we managed to trim 1 W K–1 off the heat loss coefficient by better insulation or airtightness, then the marginal saving in space heating demand would be 1889 × 24 = 45.3 kWh in London or 2600 × 24 = 62.4 kWh in Berlin. This could then be used to analyse the relative cost effectiveness of further energy-saving investments.

## Activity 10

Based on the degree day data, is our sample house likely to have a higher heating demand in Berlin or north-east Scotland?

The more degree-days, the higher the likely heating demand. North-east Scotland has slightly more degree days, 2638, than Berlin with 2600, so our sample house is likely to have the highest heating demand in Scotland.

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