Recent content by Alexey_H

I managed to calculate the R-value and the heat loss of the door. I was using the up to date temperature values. Measurements Door inside 19 °C Room temp: 20.8 °C Outside temp: 1 °C Calculations: Room and the inside door surface temp difference is 20.8 - 19 = 1.8 °C , Ambient and the inside...

It seems I found what I was looking for. Matthias Wandel solves this by using walls/room temp difference and the outside temperature. He explained well the approach in his article Measuring your house walls R-value with an infrared thermometer and also made a video showing his measurements.

Does it take into account an R-value of the material itself? Thank you for the support, will do.

That is a reasonable approach. I'd do it if it were my house. But in my current situation I need to count roughly how many kW are lost. Thanks for the idea anyway. 😂

It looks like my task is not possible. I found out that it requires the heat flux (Φ) besides the temperature gradient (ΔT). The heat flux is defined as the amount of watts transferred per square meter. Knowing the heat flux I can get the R-value of the material using this formula R = ΔT / Φ...

I can say from just observing temperature on those surfaces. The walls are significantly better and have 1°C while the door has 4°C. Windows temperature is 5°C

The room temp is 20°C. The door is not in the sun. No wind. The door is just an example. There are other surfaces like garage door, or some walls. Unfortunately sometimes parts are not visible and usually consists of multiple materials. That is true. But there are still options to improve a...

Hello, My house has some heat loss. As an example I know the outside door temperature. How can I calculate the heat loss delta if I reduce the door temperature by 1 degree? I know that there is a formula to calculate a heat loss based on the U-value of a fabric. But I can't get those values...