## How to reduce energy consumption

#### by Adriaan Woonink

How can we reduce the energy needed to keep a building comfortable? What factors play a role? In this article we will explain how to reduce the energy consumption of buildings, whilst keeping the indoor temperature at a comfortable level.

**Insulation on walls, floors and roofs**

By applying insulation the need for heating is reduced. This is because insulation on a building is like wearing a thick winter coat. It reduces the amount of heat that passes through the construction over time. There are various types of insulation, all with different properties, but what they all have in common is that they have a low thermal transmittance value, or lambda (λ) value. This value stands for the amount of heat that passes from 1 space to another through this material. In general, a material with a lambda value lower than 0.05 is considered an insulation value. The lower the value, the less heat passes through the material, therefor the lower the heating loss.

**Thermal resistance**

However, the lambda value is not the only factor involved. The thickness of the material also plays an important role. The thicker the material, the better it insulates, even for materials that are not traditionally insulation materials. These factors are combined in the following formula:

R = d / λ in m^{2}K/W

R = R-value, the thermal resistance, or total insulation value of the material

d = thickness in meters of the material

λ (Lambda) = conductivity of the material, see a list of materials here: https://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

Example:

If we take the insulation material EPS with a thickness of 20cm and a lambda value of 0,035 we get the following formula:

R = 0,2 / 0,035 = 5,7 m^{2}K/W

The higher the R-value, the better the building is insulated.

**Thermal transmittance**

Once we know the amount of heat that is resisted we can calculate the amount of heat that passes through the construction. Remember that there will always be a flow of heat as long as there is a difference in temperature, no matter how high the R-value will be. The thermal transmittance is the inverse (opposite) of the thermal resistance and it is called the U-value, which is calculated by the following formula:

1 / R = U in W/m^{2}K

R = R-value, the thermal resistance, or total insulation value of the material

U = U-value, the thermal transmittance, or total amount of energy passing through the material

Example:

If we take the R-value of the previous example we get the following U-value:

1 / 5,7 = 0,175 W/m^{2}K

The lower the U-value, the better the building is insulated.

**Insulation for windows**

Every building has windows and doors, whose insulation value is calculated in the same way. However, because a window is a combination of a window frame, the glass and other materials we do not use the R-value for simplicity reasons. Instead we use the U-value as indicated by the supplier of the windows. Since windows can be made from different materials there is no single U-value for windows, but a certain range within different types of windows, as shown in the table below:

Type of window | U-value high in W/m^{2}K |
U-value low W/m^{2}K |

Single glazing | 5,4 | |

Double glazing | 3,2 | 2,4 |

Double glazing low-e | 2,3 | 1,2 |

Triple glazing | 1,5 | 0,6 |

**What does ‘W/m ^{2}K’ mean?**

W = Watts, the amount of energy that flows through a construction

m^{2} = square meter

K = Kelvin, a scientific denomination of temperature with the same interval as Celsius.

Together these denomination shows how much energy passes through every square meter of construction, per Kelvin/Celsius.

**Example for a house**

Here we demonstrate the amount of energy losses for a small fictional house. The house has a flat roof and 2 floors. 30% of the outside wall area is windows in all 4 orientations. We will do 2 calculations to demonstrate the effect of a higher and lower U-value.

Length | 5 | m^{1} |

Width | 8 | m^{1} |

Height | 6 | m^{1} |

Floors | 2 | |

Floor area | 80 | m^{2} |

Volume | 480 | m^{3} |

Wall area | 109 | m^{2} |

Window area (30%) | 47 | m^{2} |

Floor area | 40 | m^{2} |

Roof area | 40 | m^{2} |

We will assume the following U-values for the constructions:

Low U-value | High U-value | |

Wall | 0,15 W/m^{2}K |
0,50 W/m^{2}K |

Window | 1,1 W/m^{2}K |
2,50 W/m^{2}K |

Floor | 0,20 W/m^{2}K |
0,80 W/m^{2}K |

Roof | 0,10 W/m^{2}K |
0,80 W/m^{2}K |

The indoor temperature is set at 20˚C and the outdoor temperature is set at -10˚C. This results in a temperature difference of 30˚C. However, for the floor a different temperature difference must be calculated, because the ground has a higher temperature than the outside air in winter conditions. The ground is assumed to be 5˚C, resulting in a temperature difference of 15˚C

By using the U-value we can calculate the entire heat loss through every construction:

U-value * A * ΔT = Heat loss in W

The data as shown above here results in the following calculations for the well-insulated house with the low U-values:

**Walls:**

0,15 * 109 * 30 = 490,5W

**Windows:**

1,1 * 47 * 30 = 155,1W

**Roof:**

0,1 * 40 * 30 = 120W

**Floor:**

0,2 * 40 * 15 = 120W

It is interesting to note that the walls have the highest energy loss, even though the U-value of the windows is higher. If the windows would be larger this would quickly turn around however. The heat loss through the roof and the floor is the same, because of the temperature difference between the two. Applying the same insulation has more effect on the roof compared to the floor.

The total heat loss of the building under current conditions is: 490,5 + 155,1 + 120 + 120 = 885,6 W. This means that if these conditions remain the same for 1 hour the amount of heating that is required to keep the house at the same temperature is 0,89 kWh. If you would heat the building with a traditional gas heater with an efficiency the total amount of energy required would be: 0,89 / 0,9 = 0,99 kWh.

For comparison we will now calculate the badly insulated house with high U-values:

**Walls:**

0,5 * 109 * 30 = 1.635W

**Windows:**

2,5 * 47 * 30 = 3.525

**Roof:**

0,8 * 40 * 30 = 960W

**Floor:**

0,8 * 40 * 15 = 480W

It is interesting to note that the windows have a higher energy loss than the walls, even though the surface area is much lower. The amount of heat that is lost through the floor is way less than through the roof, even though the U-value is the same.

The total heat loss of the building under current conditions is: 1.635 + 3.525 + 960 + 480 = 6.600 W. This means that if these conditions remain the same for 1 hour the amount of heating that is required to keep the house at the same temperature is 6,6 kWh. If you would heat the building with a traditional gas heater with an efficiency the total amount of energy required would be: 6,6 / 0,9 = 7,33 kWh.

By analyzing the difference in energy consumption of 0,99 kWh and 7,33 kWh you can clearly see the effect of insulation on a building. Applying better insulating constructions can be a major step in reducing the need for energy consumption of buildings, whilst maintaining a comfortable indoor temperature.