Thermal bridges

Barrage hydraulique

Let's go back to our hydraulic analogies.
Let's watch a calm river flowing. The flow is uniform, laminar, the twigs and the dead leaves (we are studying a theoric case without any plastic bottles and fatty papers) stream regularly. A bit further, some rocks limit the section of its bancks. We can see that, 2 meters way from the rocks, the floating objects speed up to rush into the trickle of current which dive between the rocks.
The section being reduced, in order to maintain the flow of the river constant, the current, that is to say the speed of the particles of water, increase : D = V×S = constant.
Just like the dam must resist water pressure, the wall of the house must resist the thermal pressure. As soon as there is a gap, the water in the first case, the flux of thermal energy in the other case, rush with speed and strength through the gap. Put your hand at the end of watering pipe. Of course, the strenght of the thermal flux, you do not feel it, it is the furnace which feels it.
Hot volumes under the screed are in common with cold volums, thus there is a direct transfer of energy hence the name of "thermal bridge", it could also be called "thermal rift".


Pont thermique

As the sketch opposite shows, the leak lines of the calories withhold thermal energy at ground level and at the level of the ceiling of the house and spread this energy in the wall which plays the part of a radiator, like the radiators which are stuck on the microprocessors and other power components, completed most of the time by a fan, which in the case of the wall will be efficiently replaced by the wind.
The bottleneck, or ejection pipe, is the part of the floor tile which is inside the thickness of the insulating wall.

We will do the calculations for a slice of mur of L=1 meter, with a conductivity for the concrete of λ=1, e=10 cm d'isolation et h=15 cm d'épaisseur de dalle, 20 cm de distance de dispersion sur 50 cm de hauteur, 50 cm de zone de captage, 30°C between the inside and the outside, this gives


conductivity of the dispersion zone = 1×0,50/2/0,20 =1,25

conductivity of the bottleneck = 1×0,15/0,10 =1.5

conductivity of the reception zone = 1×0,50/0,25 =2

total resistance = 1/1,25+1/1,5+1/2 = 1,97,


thus : global conductivity = 0,5
and Qpsol(Watt) = 0,5×30 = 15 Watt per linear meter of facade.
For the ceiling tile, because the air at the ceiling is hotter by about 5°C, we have Qppla = 35/30×Qpsol = 17,5 Watt
That is to say a total of 32,5 Watt.
The losses by the wall of 2,50 meter of height on 1 meter of length in breeze blocks of 20 cm of thickness, insulated with 10 cm of polystyrene give in the same conditions
Qm = 0,4×2/(0,4+2)×2,5×1×30 = 25 Watt. (The placing in chain of the conductivity gives : Geq = G1×G2/(G1+G2).)
Now you know where the liters of oil or the m3 of gas go.

This particularly harmful case is not utopian : sometimes, there is not even any insulation under the tile on the earth platform. Moreover, we did not take into account the partition walls if there is any, and we did not take into account the wind when there is any.
And if a balcony lenghten the tiling, imagine the radiator that this represents, it is the tongue of the dog which hangs outside its mouth to evacuate the calories that its coat can not evacuate.


Curves edited by the journal "Revue technique du bâtiment" N° 92 october 82 show that the coefficient of inside insulation is twice as bad as outside insulation, for the same thickness of insulating. The article of this magazine present so convicing arguments in favor of outside insulation that it is incomprehensible that some still build outside-insulated house ! It is the reading of this journal that drove me into build my house which is used as an example in this article.


Fortunatly, recent norms about house construction are stricter at the insulation level. We will analyze what it's all about. Pont thermique

The opposite sketch show that the bottom-floor tile, which is normally insulated from the ground either by a thickness of insulating, either by an enclosed crawl space (do not forget that the air is the best insulating after vacuum, with the condition of being stationary), bears a screed cast on a thickness of 4 to 5 cm of insulating, called floating tile.

The ground of the house is thus not in direct contact anymore with the thermal bridge of the floor tile. But the surface is big. Let's suppos that the house has a surface S=100 m2, this gives a conductivity of

G = 0,04/e×S = 0,04/0,04×100 = 100 Watt/°C, that is to say a loss of 3 kWatt for a cold of -7°C, only through the ground, if the tile was not insulated.

But the sketch show that only few leak lines are attracted by the thermal bridge, the others go toward the ground where they meet the insulating of the tile.

Let's call "reception zone" this strip of ground that border the outside walls and be 2x its width, x being then the distance to the thermal baricenter at the entrance of the thermal bridge. We make the calculations always considering a width of 1 meter of wall.


The conductivity of the reception strip is : Gc = 0,04/0,04×1×2×x = 2x soit Rc = 1/2x

The conductivity of the thermal bridge is: Gth = 1/(x+0,1)×1×h = 0,15/(x+0,1) soit Rpth = (x+0,1)/0,15

The conductivity of the dispersion sone is : Gd = 1/0,2×0,50/2 = 1,25 soit Rp = 0,8
thus Rtt = Rc + Rpth + Rp = 1/2x + (x+0,1)/0,15 + 0,8 = [1,5 + 2x(10x+1)]/3x + 0,8

Rtt = (20×x2+2×x+1,5)/3x + 0,8

By virtue of the principle of Fermat-Maupertuis, let's look for the point of maximum enthalpy, that is to say of maximum conductivity, by looking fo the zero of the derived of Rtt :

R' = [3x(40x+2)-3(20x2+2x+1,5)]9x2 = (60x2-4,5)/9x2

R' = 0 donne x2 = 1/13,33 et x = 0,27 , that is to say a reception strip of 54 cm.

et Rtt = (1,5 + 0,54 + 1,5)/0,81 + 0,8 = 5,17 et G = 0,193 ≈0,2 that is to say Qpth = 0,2×30 = 6 Watt per linear meter of wall.

The thermal bridge of the floor is divide by 2,5, we passed from 15 W to 6 W, but in general the thermal bridge of the ceiling remains, which still makes 23,5 W of thermal bridge with regards to the 25 W of the 2,5 mètres of insulated wall. We are still on the CSTB's curves of 1982 which show that the inside insulation is twice less efficient than outside insulation.


Then some builder, with the laudable goal of improving the situation, will partly neutralize, in the same way, the thermal bridge of the ceiling by sticking on its wall surface a thickness of alveolar placo.


But then goodbye to the vault effect, goodbye to the comfort of the cavemen!
Without forgetting the partition walls, and the walls of the garage ! Are the contiguous walls of the garage insulated, the ceiling when there is a room above, the gable wall, the junk romom : everywhere small walls either uselessly covered with insulation, either unfortunately not insulated ? It is impossible to eliminate all the thermal bridges of every nook in an house inside-insulated. To get convinced, go to the CSTB website and read the non-exhaustive list of thermal bridges.


Why should you outside-insulate ? Because you can non inside-insulate.