Confused about U-Values

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Macdesigner

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Hello, I'm new to this forum and I appreciate your help !

I have a timber frame building, and below is what I've calculated so far for the u-value. I understand that by UK regulations you need the value to be 0.3 (?)

Cladding – Softwood

Thickness m Density Kg/m3 Conductivity W/mk

0.020 500 0.130
0.20/0.180 = 0.015

Plywood Sheathing

0.020 650 0.140

0.20/0.140 = 0.142

Mineral Wool

0.145 12 0.042

0.145/0.042 = 0.345

Timber studs ( 145*50 )

0.050 700 0.180
0.050/0.180 = 0.277

Plasterboard
0.015 600 0.160
0.015/0.160 = 0.093


Total 0.872

Thats the total of 0.872, which is way to high, where have I gone wrong with my calculations ? What you see is exactly what I have done, I didn't take into account the inside and outside thermal resistance.

Thank You
 
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Or just get on the technical advice line of the manufacturer of your chosen insulation to bang it out in about 1 minute on the phone! :rolleyes:
 
I've looked at that website and here's what I've got s far ( see attached ) I understand that the U value needs to be 0.3, but how is this possible when the insulation alone is 3.8? I haven't divided or taken away the internal or external surface resistance because I don't know how to. By looking at the screenshot, can I achieve a U value of 0.3 ?

Thank you so much, I'm sure it's simple once you know how to do it.
Untitled

Untitled

 
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Retyped due bad example.

Don't confuse the U and R values

U = 1/Rt = 1/R1+1/R2+1/R3...

where R1,R2 R3 are the Resistance of the different materials and Rt is the Total The above is oversimplified but should start you thinking correctly
 
Thermal Conductivity = U value
Thermal Resistance = R value

R x U = 1

100 mm mineral wool = R2.5 = U 0.40

The better something is insulated, the smaller the U value ( as it is a measure of how many Watts you are losing ).

For this reason you cannot add U values as you have tried to do and get an answer .

What you have to do is add up the R values and then convert to a final U value
 
mointainwalker said:
Thermal Conductivity = U value
Thermal Resistance = R value

R x U = 1

100 mm mineral wool = R2.5 = U 0.40

The better something is insulated, the smaller the U value ( as it is a measure of how many Watts you are losing ).

For this reason you cannot add U values as you have tried to do and get an answer .

What you have to do is add up the R values and then convert to a final U value
Click to expand...

So when I add up all the R values ( which is about 3.9 ) what do I do ? By looking at my screenshot, what would be the next steps ?

Thank you all very much.
 
They improve your R values so you don't have to do anything with them if you are just looking to meet 0.30.

I haven't been involved with this but I seriously doubt whether most BCO's would bother with that anyaway.

You should realise , that although the numbers in themselves give a reassuring appearance of accuracy, once you start building and there are gaps, extra battens, 2 cm clearance etc etc nobody is likely to be calculating to the last (theoretical )2.5 %.
 
Hi

Just to confirm a couple of points

Thermal conductivity = k

Thickness of material (metre's)/k = R1 ; R2 etc.

Sum of R's = Total Resistance and 1/SoR = U

Example:-
R
Cladding - 0.020m / 0.13 = 0.154
Sheathing - 0.020m/0.14 = 0.143
Insulation - 0.145m/0.042 = 3.452
Plasterboard - 0.015/0.16 = 0.938
Rsi + Rso = 0.18

Sum of Rs = 4.867

U = 1/SoR = 1/4.867 = 0.21W/m2K

As a note, you would only need around 80mm thickness of PIR insulation to provide an equivalent R value of 3.45

Regards
 
Macdesigner said:
So when I add up all the R values ( which is about 3.9 ) what do I do ? By looking at my screenshot, what would be the next steps ?
Click to expand...

Try U = 1/(0.04 + (T1/C1) + (T2/C2) + (T3/C3) + 0.13)

where C1 is the conductivity of the first material and T1 is its thickness etc. As you add more layers and/or thickness, the overall U value must drop, not go up.
 
Macdesigner said:
So when I add up all the R values ( which is about 3.9 ) what do I do ? By looking at my screenshot, what would be the next steps ?
Click to expand...

Simply take its reciprocal: U = 1/R = 0.26

Try U = 1/(0.04 + (T1/C1) + (T2/C2) + (T3/C3) + 0.13)

where C1 is the conductivity of the first material and T1 is its thickness etc. As you add more layers and/or thickness, the overall U value must drop, not go up.
 
I think the above is a good example of totally needless complication just for the sake of it.

Alittlerespect is quoting R values to 3 - yes 3 - decimal places ! AJRobb limits himself to a modest 2.

I'm sorry but that is simply demented.

Mathematically you get to three places, but that completely ignores the fact that the original numbers are not usually quoted to any decimal place at all i.e. they are rounded off. Then you ought to reflect on the fact that there is undoubtedly a tolerance variance for the material , both in thickness and insualtion properties that we are unaware of

Then you have to take into account the environment and accepted practices: this is not an experiment in a lab dealing with nano-particles but something prctical for a building-site.

You wouldn't try to give instructions for mixing concrete to the nearest 100 g , would you ?

Everything has to relate to its surroundings, so here the simplest method of calculating/explaining is the best.
 
mointainwalker said:
I think the above is a good example of totally needless complication just for the sake of it.

Alittlerespect is quoting R values to 3 - yes 3 - decimal places ! AJRobb limits himself to a modest 2.
Click to expand...
There were 2 questions here:
1. Is the U value low enough?
2. How the f*** do you calculate it?

I was only trying to help with number 2.

As far as the U value is concerned, start with the insulation. That alone has a low enough U value for the whole wall. Then ask yourself, are you wasting money with that insulation thickness? Could you use cheaper materials or construction?

Only if the building inspector is unhappy and thinks your windows and doors are too big, say, would you need to scrape the barrel for every last contributor to the insulation calculation (including those tiny surface effects). That is not common sense, it is bureaucracy, but a game that must be played. If you need to drop a couple of digits in the 2nd significant figure, then you should use 3 significant figures in your calculation.
 
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