Heating a Single Room

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The 3 considerations when thinking about space heating for a room or building are:
  • The desired room temperature
  • The outside temperature (at its worst case)
  • The Ventilation rate of the room
The “space heating” requirements should match the “heat loss” from the room. Once you know the heat loss of a room (or the sum of the heat loss calculations for all rooms in a house), you can then use this figure to decide what boiler size you need. There are two types of heat loss from a room:
1. Fabric Heat Loss: Where a temperature difference occurs between two separated areas, heat will naturally flow from the hotter area to the cooler area. These heat flow losses will occur naturally through floors, walls, windows, roofs and even between two rooms that have differing temperatures. For ease of calculation, these heat flow losses are assumed to be at a uniform rate through each surface, and are calculated by:
Fabric Heat Loss = Surface Area (m2) * Temperature Difference (Celsius) * U Value
The U value is the heat transfer co-efficient of a specific material and is measured in watts per square meter. It basically measures how quickly heat escapes through a specific material. The calculations of the U value are quite complex, and outside of the scope of this paper. Most U values are given on manufacturers websites or tables from the Energy Savings Trust.
2. Ventilation Heat Loss: this is heat loss caused by air flowing through the building. They are quoted in “air changes per hour”, i.e. the volume of air flowing through the room in one hour, divided by the volume of the room itself and the Ventilation heat loss is calculated by:
Ventilation Heat Loss = Room Volume (m3) * Air Change Rate * Temperature Diff * Ventilation Factor (w/m3 oC)
The Ventilation Factor is taken as the specific heat factor for air, and is a constant factor of 0.33. It can be calculated as:
Vent Factor = [Specific Heat Capacity of Air * 1000 (to convert from kJ to Joules) * Air density]/3600 (to convert from hours to seconds)
Vent Factor = (1.01 * 1000 * 1.2)/(3600)
Vent Factor = 0.34
The minimum design temperatures and air change rates required by BS 5449 are given for each room as being:
Room Temperature Air Change
(Celsius)
Lounge Room 21 1.50
Living Room 21 1.50
Dining Room 21 1.50
Kitchen 18 2.00
Breakfast Room 21 2.00
Kitchen/Breakfast Room 21 2.00
Hall 18 2.00
Cloakroom 18 2.00
Toilet 18 2.00
Utility Room 18 1.50
Study 21 1.50
Games Room 21 1.50
Bedroom 18 1.00
Bedroom/en Suite 18 2.00
Bedsitting 21 1.50
Bedroom/Study 21 1.50
Landing 18 2.00
Bathroom 22 2.00
Dressing Room 21 1.50
Storeroom 16 1.00
Here, you can see that rooms are generally required to be heated to a level between 18 and 21 degrees Celsius, depending on their usage. The Air Flow figures are for modern buildings and when designing for older properties should be adjusted to take into account poor insulation of windows, floors and walls etc. Similarly, if there is an extractor fan in any of the rooms then (such as in a bathroom), then the air flow should be adjusted higher to compensate for this.
Now, let’s take a look at a working example for a bedroom in a house:
bedroom
We are also assuming that the temperature of the room above (i.e. the roof) is -1 degree Celsius (as assumed for the outside temperature, and the temperature of the room below is 21 degrees Celsius.
We can now get the heat loss for this room as being:
Fabric Heat Loss Area Temp Diff U Value Heat Loss
m2 oC w/m2 oC Watts
External Wall 5.50 * 19.00 * 0.92 = 96.14
Window 2.00 * 19.00 * 5.00 = 190.00
Neighbours Wall 10.00 * 8.00 * 2.10 = 168.00
Internal Wall 1 7.50 * 0.00 * 1.70 = 0.00
Internal Wall 2 10.00 * -2.00 * 1.70 = -34.00
Floor 12.00 * -3.00 * 1.36 = -48.96
Ceiling 12.00 * 19.00 * 0.34 = 77.52
448.70
Ventilation Heat Loss
Air Changes Room Volume Temp Diff Vent Factor Heat Loss
(m3) (oC) (W/m3 oC) (Watts)
2.00 * 30.00 * 19.00 * 0.34 = 387.60
Thus, the total heat loss for the room would be 448.7 + 387.6 = 836.3 Watts. However, additional heating capacity will be needed in both the heating system and the boiler system to allow for extra demands for heating and to allow for heating up periods. This is usually taken as a 10% adjustment (upwards), although can be as much as 30% should a shorter “heat-up” period be required. Thus for this room, the heating system would be required to be a minimum of 919.93 Watts.
If we were to repeat this process for the rest of the building, obtaining heat loss values for every single room of the house, we would get:
Room Heat Adjustment Required Heating
Loss 10% System Size per room
Lounge Room 2015 2217 2217
Living Room 2150 2365 2365
Dining Room 1050 1155 1155
Kitchen 400 440 440
Hall 1150 1265 1265
Cloakroom 150 165 165
Toilet 200 220 220
Utility Room 250 275 275
Study 400 440 440
Bedroom 1 919 1011 1011
Bedroom 2 873 960 960
Bedroom 3 745 820 820
Bedroom/en Suite 1346 1481 1481
Landing 950 1045 1045
Bathroom 278 306 306
Here we get a total heating requirement of 14,164 Watts and thus for this house we would need to choose the next largest boiler size up from this for the space heating requirements. Assuming there were no Hot Water Requirements for the house, this would be a 15kW boiler size. If there was Hot Water Requirements, then we would need to add this to the space heating requirements. Note, that no further adjustments should be made for pipe lagging etc, as over sizing the boiler actually reduces its efficiency, thereby wasting energy.
For systems that have to provide Domestic Hot Water, we also need to add in the amount of energy required to heat the water in the hot water tank whilst not disrupting the hot water for the central heating system (note that this does not apply for combi-boilers which have no hot water tank needs). The first consideration is the size of the hot water tank. Obviously larger houses will have larger tanks. Let us assume you have a 120 litre hot water tank.
We then calculate how much energy it would take to heat the water in the tank as if it had been emptied and needed a complete heating. We know that UK mains water is provided at 4 degrees Celsius and that we have to heat the water to at least 60 degrees Celsius. Thus we need to heat the water by at least 56 degrees in the tank. We also know that 1 litre of water requires 1.16 watts of energy to heat it by 1 degree Celsius in an hour. Thus, the total amount of energy required to heat the tank would be:
(120 * 1.16 * 56) = 7,600 watts hours = 7.6 kWh
Thus, for our house above we would need to add 7.6kWh to the system size, making the boiler size at least 23kW.
Other areas of interest in this section are:

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