How to Calculate Space Heating Requirements

To enable you to understand the heatceiling and floor areas.
requirements of a space you need to understandWith the above carried out we can start to
how heat is lost from a space and how quicklycalculate the loss of heat through the material of
this happens. This is to enable you to calculate thethe space e.g. Walls, floor, windows and roof or
heat output you require and how quickly youceiling. In addition, calculate the heat requirements
need to heat a space to maintain the desired orof the infiltration ventilation, from areas of the
design temperature.space that allow air into the space from outside
Quick Rule of Thumb Routefrom under doors or through windows with no
Step Onedraft proofing, etc...
Calculate the area of the space to be heated inThis we use for air change rates within the space.
m2The way heat loss is calculated is by measuring
Measure and record the external or internal lengththe area and multiplying it by the known U value
in metres.for the material. This is the thermal resistance of
Measure and record the external or internal widththe material multiplied by the thickness of the
in metres.material. There are tables available giving U values
Example: Length 10m x Width 5m = 50m2of materials e.g. an unfilled cavity wall has a U
Step Twovalue of 1.6.
Calculate the heat requirement of the space byThese values are calculated from the K value of
using rules of thumb for different spaces.the material multiplied by the thickness of the
Below are guides to heat requirements:material.
Living Room 60W/m2 21 Deg CCalculation example:
Bathroom 70W/m2 21 Deg CSpace Temp 21/Outside -4/Air Changes 3
Kitchen 60W/m2 16 Deg CDifferential Temp 25
Bedroom 60W/m2 18 Deg CRoom: Living Dimensions x U Values = specific
Hall and Landing 60W/m2 16 Deg Cheat loss
This is based on solid brick or block dwellings withAir 10 x 5 = 50 x 3 = 150m3 x 0.33 = 50
cavity thermal insulation and double glazedGlass 1 x 2 = 2m2 x 3.0 = 9.5
windows.Walls (external) 10 + 5 x 3 = 45-2=43m2 x 0.45
Example:= 19
Length 10m x Width 5mFloor 10 x 5 = 50m2 x 0.10 = 5
= 50m2 x 60W/m2Roof/Ceiling 10 x 5 = 50m2 x 0.54 = 27
= 3,000WSpecific heat loss total = 110 x 25
= 3 kWTotal Heat Loss = 2,750W HLoss Total
This example can also be used to calculate theIf we round up the result to 3,000w or 3kW you
output required for a new boiler installationcan see that we have lost 250W from our original
Specific Heat Loss Calculation Routerule of thumb calculation. This will not always be
To calculate the specific heat loss in a space youthe case and if we had more window area, more
have to consider the temperatures of the outsideoutside walls or more air changes we would have
space and the desired inside space. There is anmore heat loss. In addition, it should be
accepted temperature criteria, which is - 4 Deg Cremembered that the rule of thumb is an
minimum outside temperature and the desiredapproximation.
temperatures as shown above in the Rule ofThis can now be applied to each space and to
Thumb step two.radiator sizing calculations and schedules.
To carry out the specific heat loss calculation youHistorically a margin was always added to the
measure the space dimensions internally i.e. height,calculation when radiator sizing of 11%. So, if we
width and length in metres. Calculate the area oftake our 3kW and add 11% it becomes 3.33kW
the windows in m2 and subtract this from thesay 3.5kW.
wall area and measure and calculate in m2 the