How to calculate the thermal transmittance
Thermal insulation calculation and U-value -
Made understandable for the client
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If you want to carry out thermal insulation properly, you have to work your way through a mountain of terms and formulas. Without a degree in natural sciences, this can hardly be done. We therefore try to reduce everything to the essentials so that even the layman can carry out a thermal insulation calculation.
The terms that keep appearing in connection with thermal insulation are thermal conductivity, thermal resistance, heat transfer coefficient, heat transfer resistance and heat transfer coefficient. As you can see, you can work up a sweat! But don't panic, you don't have to memorize everything you have just read. For the most part, we spare ourselves precise definitions and formulas at this point, because these are at most irritating to the layperson.
The most important key figure in connection with the thermal protection of an individual component is U-value (heat transfer coefficient, heat permeability, U-value used to be the k-value). It indicates the amount of heat (in kWh) transported through a component area of 100 m² in one hour if there is a temperature difference of 10 degrees between inside and outside. The smaller the value, the better. (Please do not report physicists and engineers, we know that the exact definition is slightly different).
To give you a feeling for the difference between well and poorly insulated components, the following table lists reference values for the U-value of various components. The values listed under the heading "very good" are the values of a low-energy house and have therefore been the standard since the EnEV (Energy Saving Ordinance) came into force in February 2002.
|Component||very bad||bad||medium||Well||very good|
|top, roof||≥ 1,00||0,60||0,30||0,22||≤ 0,15|
maximum 4 cm
|6 to 10 cm|
|12 to 16 cm|
or 9 cm PUR
|18 to 20 cm|
or 12 cm PUR
|27 to 30 cm|
or 18 cm PUR
|Solid wall||≥1,50||0,80||0,40||0,30||≤ 0,20|
600 [kg / m³]
inside gypsum plaster
400 [kg / m³]
inside gypsum plaster
13 cm PUR
Walls of prefabricated houses were not mentioned in the above list, but they have very interesting U-values. Outer walls of the latest generation of prefabricated houses in timber frame or timber frame construction come with a thickness of only 25 cm - 27 cm and a U-value of 0.17. For comparison: A 36.5 cm solid wall made of brick masonry with an additional 13 cm PUR insulation comes to a comparable U-value, although the wall is about twice as thick.
The prefabricated house manufacturers argue quite correctly that with a floor space of 9 x 12 m for an average single-storey house, the living space with this wall structure is around 9 m² larger than with the solid masonry plus insulation variant.
This is what the Energy Saving Ordinance says about the U-value
In the case of a new building or changes to existing buildings, minimum requirements for the U-value of the component must be met. There is no general obligation for home owners to retrofit. The top floor slabs are an exception. Owners of residential buildings must insulate non-accessible, but accessible top floor ceilings above heated rooms by December 31, 2006 so that the U-value of the floor ceiling of 0.30 W / (m²K) is not exceeded. Otherwise the following values apply:
|Components||Measures (which are carried out on the components)||U-value|
|Exterior walls||Component replaced, installed for the first time; inside cladding / cladding (e.g. interior insulation) applied; Infills (e.g. half-timbering) renewed||0,45|
|External cladding / cladding as well as masonry cladding attached; Built-in insulation layers; Renewed external plaster (if the wall's U-value is greater than 0.9 W / m²K)||0,35|
|Component replaced, installed for the first time; additional front and inner windows built in||1,70|
|Curtain walls||Component replaced, installed for the first time; Replaced filling (glazing)||1,90|
|Ceilings, roofs, sloping ceilings over heated rooms||Component replaced, installed for the first time; Roof skin or external cladding / cladding replaced or rebuilt; inside cladding / insulation layers applied or renewed; Installed insulation layer||0,30|
|Flat roofs over heated rooms||Component replaced, installed for the first time; Roof skin or external cladding / cladding replaced or rebuilt; inside cladding / cladding applied or renewed; Insulation layers installed||0,25|
|Ceilings and walls against unheated rooms or the ground||External cladding / cladding, moisture barriers / drainage installed or renewed; Ceiling cladding (cold side e.g. on the basement ceiling) attached||0,40|
|Component replaced, installed for the first time; inside cladding / cladding attached; Floor structures (warm side) built up or renewed; Insulation layers installed||0,50|
The max.U-values of the EnEV specified above do not have to be complied with for:
1. Changes to the outer wall, windows, French doors and skylights of less than 20% of the component area with the same orientation,
2. Modification of all other external components (e.g. roof, floor) of less than 20% of the respective total area of the components.
Calculation of the U-value
The most important term related to thermal insulation is thermal conductivity. This is a thick, independent material property and should not be confused with the U-value. The thermal conductivity of a material indicates what amount of heat (in kWh) is transported through a building material surface of 100 m² and 1 m thick in one hour if there is a temperature difference of 10 degrees between inside and outside. As with the U-value, the smaller the value, the better.
In Germany, insulation materials are classified in a "thermal conductivity group" (thermal conductivity group, WLG) depending on their thermal conductivity. This group corresponds to the decimal places of the thermal conductivity. Example: Insulation with a thermal conductivity of 0.030 has the thermal conductivity group WLG 030.
In order to be able to calculate the thermal permeability (U-value) of a component, you also need the insulation value of the individual layers. In technical jargon, this thermal insulation value is called thermal resistance. This indicates the resistance of a layer to the flow of heat. To determine it, the thickness of the relevant layer (in meters) must be divided by the material-related thermal conductivity. In the case of multi-layer components, the individual value must be determined for each layer using this calculation method. The sum of all individual values then gives the thermal transmittance or thermal insulation value for the entire component. The greater the resistance, the better the thermal insulation.
Even calm air layers (no flowing rear ventilation) have a certain insulation value depending on the thickness and incline. This resistance is 0.16 for static layers of air up to an incline of 60 degrees. If the layer is inclined more than 60 degrees 0.18. These values must be taken into account when calculating the total resistance.
The calculation of the U-value becomes more complicated if the thermal insulation of components is interrupted. This is the case with roofs, e.g. with insulation between the rafters. Two areas need to be examined here. On the one hand the compartment where the thermal insulation is located and on the other hand the rib that interrupts the thermal insulation. The ribs (i.e. the rafters or the beams) are only taken into account in the thickness of the thermal insulation lying on the side. Vapor barriers and waterproofing membranes are not taken into account in the thermal insulation calculation.
The following formulas for calculating the thermal protection of a component result from the above explanations:
The insulation value of a layer, called resistance (thermal resistance), is calculated using the following formula:
If the component has several layers, all layers (including static air layers, see above) must be added to the total resistance (thermal resistance):
The thermal permeability (the U-value) of a component is calculated from the reciprocal of the total resistance:
If the compartment and rib are added, the following formula results for the heat permeability:
You should now be able to carry out a thermal insulation calculation yourself.
The thermal conductivity of some important component layers can be found in the table below. If you are missing values, take a look at the relevant construction tables or ask the manufacturer or supplier of the building material.
|Claddings / panels|
|Wooden formwork (softwood)||0,13|
|Chipboard (flat press)||0,13|
|Gypsum plaster without surcharge||0,35|
|(Standard interior plaster)||0,90|
|Sand-lime brick masonry, 1600 [kg / m³]||0,79|
|Light vertical perforated brick masonry, 900 [kg / m³]||0,42|
|Aerated concrete 350 [kg / m³] |
600 [kg / m³]
|Rafters / ribs made of softwood||0,13|
|Solid brick masonry, 1600 [kg / m³]||0,68|
|> Glass fleece bitumen roofing membranes V 13||0,17|
|Wood fiber insulation boards, WLG 050||0,050|
|Mineral wool WLG 035 ("glass or stone wool")||0,035|
|Mineral wool WLG 040 ("glass or stone wool")||0,040|
|Polystyrene (PS 20 SE) ("Styrofoam") WLG 040||0,040|
|PUR rigid foam, WLG 030||0,030|
|FOAMGLAS® sheets T4-040||0,040|
|isofloc cellulose insulation material WLG 040||0,040|
|Sheep wool insulation mats DWS 8/90||0,044|
|(Standard thermal insulation)||0,040|
|Synthetic resin plaster||0,70|
|Standard exterior plaster)||0,90|
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