Heat loss from the temperature difference

[Alexey_H]

TL;DR Summary

Is it possible to calculate heat loss improvements by knowing how the surface temperature changed?

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 as the walls and even doors are not made from a solid material. So in most cases it is not possible for me to get U-values.

So I thought if it is possible to calculate it based on temperature of surfaces?

Details:
* Ambient temp: -2°C
* Door outside temp 4°C
* Door inside temp 20°C
* Door height: 2.2m
* Door width: 0.8m

How many watts could be saved by decreasing heat loss if I insulate the door and its outside temperature drops to 3°C ?

I need this to reason the house insulation in many places to my landlord. As now he is not sure if it makes sense but I pay a lot on my electricity bills.

I would appreciate any help with this.

 

[russ_watters]

What is the room temp? Is the outside surface of the door in the sun? Is there any wind?

Typically insulation through walls and doors (but not windows) is good enough that the surface temperature comes near equalling ambient (adjacent air) temperature. Calculating convective heat loss from a small temperature difference is difficult/inaccurate. It's probably better to estimate the insulation value of the parts and calculate from there.

It gets harder to make changes to an existing building unless codes were not followed when it was built. E.g., if it's a hollow metal door and lacks the required insulation you can spray foam inside. But if the door is already insulated there's not much you can improve.

[edit, somewhat a reply to Chet..]
The numbers given imply substantially worse insulation in the door than the walls. Heat loss is proportional to temperature difference, at the wall surface outside you have a 6C delta-T vs the wall at 3C delta-T. So, twice the convective heat loss (per unit area).

 

[Chestermiller]

How does the heat loss through the door(s) compare with the heat loss through the walls and windows (in your judgment)?

 

[Alexey_H]

What is the room temp? Is the outside surface of the door in the sun? Is there any wind?

The room temp is 20°C. The door is not in the sun. No wind.

Calculating convective heat loss from a small temperature difference is difficult/inaccurate

The door is just an example. There are other surfaces like garage door, or some walls.

It's probably better to estimate the insulation value of the parts and calculate from there.

Unfortunately sometimes parts are not visible and usually consists of multiple materials.

It gets harder to make changes to an existing building unless codes were not followed when it was built. E.g., if it's a hollow metal door and lacks the required insulation you can spray foam inside. But if the door is already insulated there's not much you can improve.

That is true. But there are still options to improve a bit. Those improvements in multiple places multiplied by days could save some money in a month. This is what I need to prove.

Heat loss is proportional to temperature difference, at the wall surface outside you have a 6C delta-T vs the wall at 3C delta-T. So, twice the convective heat loss (per unit area).

Thank you. I wonder if I can convert it into absolute values and finally into watts.

To generalize the task I can say:
what is the conductivity coefficient of a complex material if we know temperature difference on two sides, the surface area and the ambient temperature.

 

[Alexey_H]

How does the heat loss through the door(s) compare with the heat loss through the walls and windows (in your judgment)?

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

 

[Alexey_H]

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

And I have no idea how can I measure the heat flux of a wall or garage door :)

Thank you for your answers.

 

[Lnewqban]

Please see:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatloss.html

https://www.hvaccomputer.com/start.asp

https://www.elitesoft.com/web/hvacr/chvacx.html

 

[hutchphd]

TL;DR Summary: Is it possible to calculate heat loss improvements by knowing how the surface temperature changed?

How many watts could be saved by decreasing heat loss if I insulate the door and its outside temperature drops to 3°C ?

With respect, these calculations are far too detailed. You need to identify the worst offenders for heat loss, ameliorate the ones that make sense and then repeat the process.
On a cold day, a thermal image of your house from the exterior and also one of the exterior walls from the interior will identify (quantitatively) the worst offenders. (Area) X (ΔT) is the figure of merit. From this data you can decide which problems are reasonable to attack and what level of expense makes sense. Repeat after fixing these areas.
There will be some heat leaks that are not really fixable. When the ones that are fixable reach this level you are finished. Demolition is the only more reasonable solution.

 

[Tom.G]

Here is a rough rule-of-thumb approach using the crazy units we use here in the USA.

R-value = BTU/Hr/sq.ft./degree F
BTU = British Thermal Unit = energy needed to raise 1 pound of water 1°F
Watts = BTU/3.41

R-value of an air film ≈ 0.5 ≈ 2BTU/Hr/sq.ft/degree F

Knowing the Ambient Temp of -2°C and door outer surface temperature 4°C, there is a 6°C temperature difference across an R-value of 0.5.

If you don't get tangled up in the conversions, you should be able to find the energy lost thru the door.

This site may help with the conversion:
https://www.unitsconverters.com
(and don't forget the conversions between °F and °C)

Please keep us updated on what you find... and if you make any progress with your landlord!

Cheers,
Tom

 

[Alexey_H]

On a cold day, a thermal image of your house from the exterior and also one of the exterior walls from the interior will identify (quantitatively) the worst offenders. (Area) X (ΔT) is the figure of merit.

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.

Demolition is the only more reasonable solution.

 

[Alexey_H]

R-value of an air film ≈ 0.5 ≈ 2BTU/Hr/sq.ft/degree F

Knowing the Ambient Temp of -2°C and door outer surface temperature 4°C, there is a 6°C temperature difference across an R-value of 0.5.

Does it take into account an R-value of the material itself?

Please keep us updated on what you find... and if you make any progress with your landlord!

Thank you for the support, will do.

 

[Alexey_H]

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.

 

[hutchphd]

So you needneed to report your numbers for your door (please). My guess: Doors are notoriuosly hard to estimate. But the bad one will be r=~1.5 and a really really good door wil be R=10 (in godawful USA units (F/sqrft/BtU/hr if I recall)

 

[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 door surface temp difference is 19 - 1 = 18 °C

Checked the spreadsheet.
Found my inside temp difference in the first column. Then found the column with the outside difference and the R value:
R-value (imperial) = 7.3 ft²·°F·h/BTU
Calculated the metric RSI value
RSI (metric) is 7.3 / 5.678 = 1.29 m²·K/W

The heat loss would be
Q = ∆T / RSI = (19 - 1) / 1.29 ≈ 14 W/m²

My only question is what temperature difference should I use when calculating heat loss? ambient - room or inside surface - ambient or surface inside - surface outside ?

 

[hutchphd]

For the door itself I would use (surface inside - surface outside) but the different methods should change the (imperial) R by less than one. Your overall errors will be that large anyway, so don't worry.

 

[S P]

TL;DR Summary: Is it possible to calculate heat loss improvements by knowing how the surface temperature changed?

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 as the walls and even doors are not made from a solid material. So in most cases it is not possible for me to get U-values.

So I thought if it is possible to calculate it based on temperature of surfaces?

Details:
* Ambient temp: -2°C
* Door outside temp 4°C
* Door inside temp 20°C
* Door height: 2.2m
* Door width: 0.8m

How many watts could be saved by decreasing heat loss if I insulate the door and its outside temperature drops to 3°C ?

View attachment 320300

I need this to reason the house insulation in many places to my landlord. As now he is not sure if it makes sense but I pay a lot on my electricity bills.

I would appreciate any help with this.

Your idea is good, but... impractical. General rule of thumb for most of the houses built before the era of consciousness (literally almost every building until year ~1990) is that heat loss ratio of building due to thermal conductivity of walls, windows, doors, roof - all the external surfaces AND natural ventilation is 60:40, for better insulated houses the ratio is 50:50 for even better insulated is 40:60 and so on. That is assuming that they are all equally hermetic and there are no ventilation systems with heat recovery.

I am assuming, that you live in pretty old building, so from practical standpoint the amount of effort and money (!!!) needed to reduce your heat loss has to go to reduce unwanted ventilation. In simple terms - hunt for all the tiny holes in windows, doors, walls and so on. It will cost you pennies to seal them, but the effect will be tremendous. The hermetic house is an energy efficient house. Most of the gaps can be felt easily with hand. To insulate the external surfaces will cost a fortune...

From construction standpoint I would suggest to solve your exact task by gluing even thinnest XPS type foam on the OUTSIDE of the doors. Gluing is very important: you have to avoid air gap between door and the foam. It is critical. It will be hard to source something thinner than 2 cm /1 inch and the visual and practical result will be unsatisfactory... Also you can do the same on the inside, but there is a catch: any, even smallest area on the surface, where you leave foam unglued will be much colder than ambient air, and condensate may form. Basically on the inside you can use any foam type you want. As a bonus - you will get huge additional soundproofing effect.

If your entire door (external skin) is made from heat conductive material, like metal, which I doubt, then if you insulate the sash, you have not let it to touch directly with the frame with some special foamy hermetic tape. Hermetic type of tape is very recommended in any case.

p.s. In your case 2 degree difference of door surface is nothing. Your effort will pay itself back only in many many years...

Those are the practical advice... No formulas from me, and apologies for poor English :)

 

[S P]

p.s. Some creativity + leather + foam rubber - and the result from the inside can look decent