Compensating Husbanken’s requirements for Windows – a Practical Guide for Architects and Engineers

In a previous article we discussed how Husbanken introduced new strict requirements for windows on new buildings.
As we have already pointed out, Husbanken gave a way out to avoid people spending a fortune on windows: if you cannot reach the average U-value of  0.8 W/m2K, provided that you still are in average less than 1.0, you can compensate the Energy loss by adding additional insulation on other component of the house (i.e. the roof).

The question is: how much additional insulation shall you add?

Heads up!

Before cruncing the numbers we have to give a piece of advise.
Speaking from the experience with the Passivhaus Standard, we know that more the requirements become strict, the more Planning and Design become important.

In general Architecs and Engineers assume that performance of windows are automatically verified (meaning the average U-value is lower than the admissible limit). When the limit lowers we are not allowed anymore to take things for granted.

With limits as 0.8 W/m2K windows cannot be simply drawn on facades but they should be “designed”… meaning that their U-value should be pre-calculated in design phase based on their geometry. Missing this fundamental steps results in the possibility that the window simply cannot achieve an acceptable performance.
As a rule of thumb you want to have large glazed areas and less frame (because glass has a way far better U-value than the frame).

However, if your project is done already or if you care too much about aestethic to spoil your facade with “glassy” windows… here is the solution.

It is all about Energy

U-value is the measure of power irradiated from an area under the difference temperature of one degree Kelvin.
Fixing the U-value to some limit means fixing the maximum power that a building component can irradiate under a certanin difference of temperature. In this specific case we want windows to irradiate less than 0.8 W/m2K, in other words – considering a worst scenario of -30°C outside and working under a deltaT of 51°K – we want one square meter of window to irradiate not more than:

0.8W/m2K x 1.0m2 X 51°K = 40.8 W

This way, if our house has 30m2 of windows, we have a heat loss from windows of about 1.22 kW (30m2 x 40.8W/m2).

This is just an example and serves the purpose to clarify that U-value itself is just a number and what we really care about is the Energy (as power over time) that we irradiate from the building.

Let’s break it into Steps

Step1 – calculating average Energy loss

NOTE: each window has different U-value, depending on the ratio glass/frame.

First thing is to clarify how to calculate the average U-value:

– Multiply the area of each window for it’s own U-value, as declared by the manufacturer. The number you obtain is a quantity in W/K.
– Once you have this value for each window, sum the values all together. The sum is still W/K and represents the total power loss from windows (per deltaT = 1K).
– Divide the total power loss by the total area of the windows. The number you obtain is a measure in W/m2K and it represents the average U-value for your set of windows.

Make sure the average U-value it is less then 1.0W/m2 K. If the value exceeds 1.0, you really want to reconsider the geometry of the windows or to find a better supplier (more expensive windows).

Step2 – calculating excess Energy loss

Let’s start by figuring out what is the acceptable Energy loss and let’s remember what we speak about energy but we actually calculate Power.
Multiply the total window area by 0.8 W/m2K. This is the expected “acceptable Energy loss“, measured in W/K.
Now multiply the average U-value for windows (calculated before) by the total window area. This is the “real Energy loss“, measured in W/K.
The difference between “acceptable Energy loss” and “real Energy loss” gives the “excess Energy loss“, still in W/K.

Consider 30m2 of windowed area and an average U-value (calculated as per Step1) of 0.97 W/m2K.

Acceptable Energy loss: 0.8W/m2K x 30.0m2 = 24.0 W/K
Real Energy loss: 0.97W/m2K x 30.0m2 = 29.1 W/K
Excess Energy loss: 29.1W/K – 24.0W/K = 5.1 W/K

Good, now we have to distribute this Energy loss on other building components (i.e. the roof) in order to achieve an Energy performance better that the one we would have with windows meeting the 0.8W/m2K criteria.

Step3 – calculating the additional insulation

The roof is a perfect place to compensate for energy losses. In general it is quite easy to add insulation in a roof. We just have to calculate how much we should add.

For this exercise let’s suppose the warm part of our roof has a total area of 180m2.

All we have to do is to take the eccess Energy loss and divide it by the area of the roof:

Insulation improvement: 5.1W/K – 180m2 = 0.0283 W/m2K

This number is the quantity by which we have to improve the U-value of the roof.
Knowing this value does not bring us any immediate benefit… we are more interested to know which is the amount of insulation we should actually add in the roof construction.
In order to get that we have to perform a last couple of operations.

Assuming the roof has an insulation of 0.40m of the same insulating material, the current U-value of the roof is:

0.037 W/mK / 0.40 m = 0.093 W/m2

NOTE: this calculation does not take into account the effect of framing. For accurate results effect of framing shall be accounted for.

Now, we need to improve this U-value by the quantity calculated earlier:

0.093 W/m2K – 0.0283 W/m2K = 0.065 W/m2K

Almost done.
Now we just divide lambda by this improved U-value to obtain the total thickness of the improved insulation:

Insulation thickness: 0.037 W/mK / 0.065 W/m2K = 0.57 m

As the original thickness of insulation was 0.5m we have to add 70mm insulation in the roof to compensate for additional losses from windows.

In alternative we could just use insulation with better lambda. In this case, using a material with lambda=0.033 W/m2K, would be enough to compendate without the need to add additional insulation in the roof!

using better insulation –> 0.033 W/mK / 0.065 W/m2K = 0.508 m


The amount of additional insulation can be impractical therefore sometimes it is necessary to divide it further onto the walls as well.

The calculations are exactly the same.
It is up to the Architect/Engineer to find the best combination [Roof / Walls / Foundation] to minimize the costs and maximize the Energy performance.