U value calculation help required - Data sheet provided

[ruggers]

Can anyone help me with obtaining some material U values to input into the heat engineer software I've been using. I'm not smart enough to complete the calculations required when looking at the formulas 😚. I have attached two data sheets for the blocks I'm going to use. One for a dense block and one for an insulite block. Dense is showing as 1.17W/m2K and the insulite is 0.49W/m2K.

 

I would like to know what the U value would be for each type of block if used on their own for internal ground floor load bearing partition walls with a 12.5mm plasterboard (0.210W/m2K) dabbed onto each side. Am I correct in thinking that as a bare block with no plasterboard, the U value for internal use would just be as per data sheet being 1.17 and 0.49?

 

• For a little info. for anyone else - A member of the brick company provided me with some personalised data & an explanation for each when used in a cavity wall situation with PIR and strangely they both equal the same U-value. With both blocks used in the below layout, a U-value of 0.17W/m2K was obtained.
  
  12.5mm plasterboard (0.210W/m2K) - 15mm dab cavity - 100mm block - 100mm PIR (0.22W/m2K) - 50mm air gap - 100mm concrete brick - Total = 383mm
   
• Without the cavity air gap having the PIR sandwiched between the blocks made the insulite at 0.18W/m2K and using the dense block was 0.19W/m2K.

DENSE Datasheet - WEB July 17.pdf INSULITE Datasheet - WEB July 17 2.pdf

[Temp]

2 hours ago, ruggers said:

Dense is showing as 1.17W/m2K and the insulite is 0.49W/m2K.

 

2 hours ago, ruggers said:

Am I correct in thinking that as a bare block with no plasterboard, the U value for internal use would just be as per data sheet being 1.17 and 0.49?

 

Sorry but no.

 

It says the thermal conductivity is 1.17W/mK (eg not W/m^2k). Thermal Conductivity is the U value for a notional 1m thick wall. So 100mm thick wall would have a u-value 10 times worse at 11.7W/m^2k. That's consistent with the R-value quoted for 100mm blocks of 0.09m^2k/W....

 

U=1/R = 1/0.09 = 11.1W/m^2k.

 

For the insulite the U value of 100mm wall would be 4.9W/m^2k.

 

[Temp]

For a sandwich of materials you add up the R value...

 

R = R1 + R2 + R3

 

Or the reciprocal of the U values.. 

 

1/U = 1/U1 + 1/U2 + 1/U3

or..

U = 1 / {1/U1 + 1/U2 + 1/U3}

 

 

[ruggers]

1 hour ago, Temp said:

It says the thermal conductivity is 1.17W/mK (eg not W/m^2k). Thermal Conductivity is the U value for a notional 1m thick wall. So 100mm thick wall would have a u-value 10 times worse at 11.7W/m^2k. That's consistent with the R-value quoted for 100mm blocks of 0.09m^2k/W....

 

U=1/R = 1/0.09 = 11.1W/m^2k.

Thanks Temp for clearing this up. So to obtain U, I require the R value of the material? 

1 hour ago, Temp said:

 

For the insulite the U value of 100mm wall would be 4.9W/m^2k.

 

So on it's own, the insulite is is twice as good at preventing heat transfer compared to the dense? I'm not sure if this is of benefit for internal walls or not, I'm only losing heat to another room but the insulites sandwiching 100mm PIR are 0.01W/m2k better between my house and attached unheated garage if thats worth bothering about.
Insulite are around 20p per block more expensive. I was going to use them for my inner leaf of the cavity wall until i found out they are just the same as dense when combined with the PIR.

[ruggers]

1 hour ago, Temp said:

For a sandwich of materials you add up the R value...

 

R = R1 + R2 + R3

If i'm applying your methods correctly, looking at these two further examples attached & using the R values they've used, I'm getting 1.84W/m2K for plasterboard - air gap - dense block - air gap - plasterboard. And for the insulite it's 1.5W/m2K? 

I'm still baffled how both blocks equate to the same U value when combined with other materials.

Edited August 22, 2022 by ruggers

[Temp]

Just now, ruggers said:

Thanks Temp for clearing this up. So to obtain U, I require the R value of the material? 

 

The main property of the material is its Thermal Conductivity. Given that you can estimate the U or R values. The exact values depend on surface effects but its usually close enough.

 

To work out the U value of a sandwich you can calculate the R or U value of each layer and add them up. BUT the way you add them up is different as per my other post. 

[Temp]

6 minutes ago, ruggers said:

I'm still baffled how both blocks equate to the same U value when combined with other materials.

 

It's because the 100mm of PIR totally dominates the calculation. If you look at the R values it accounts for 4.545 out of the total 5.8 or 5.9.

 

[Temp]

Joe has 10p and Fred has 20p. So Fred is twice as rich as Joe.

 

But if you give them both £5, Joe has £5.10 and Fred has £5.20 which is almost the same.

[ruggers]

I still thought it would affect the end result because the insulite R layer says 0.204 and the dense is 0.085 so once totalled and then divided the answer couldn't be the same considering as a bare block one is twice as good as the other.

Is my example ok to use? 12.5mm plasterboard R 0.060 + 15mm air gap R 0.170 + dense block R 0.085 + 15mm air gap R 0.170 + 12.5mm plasterboard R 0.060 = 0.545 Then U = 1/0.545 = 1.84W/m2K

[SteamyTea]

23 minutes ago, ruggers said:

I'm not sure if this is of benefit for internal walls or not,

It may be useful when doing a room by room heat loss calculation for sizing heating/cooling systems.

 

k = W.m^-1.K^-1 [thermal conductivity]

 

R = L / W.m^-1.K^-1 where L is thickness [thermal resistance]

 

U = 1 / R = W.m^-2.K^-1 [thermal transmittance]

 

 

 

[Temp]

5 hours ago, ruggers said:

would like to know what the U value would be for each type of block if used on their own for internal ground floor load bearing partition walls with a 12.5mm plasterboard (0.210W/m2K) dabbed onto each side

 

Using their R Layer values..

 

1. Surface effect/Rsi  0.130  

2. Plasterboard     0.06

3. 10mm cavity due to dot and dab approx 0.13

4. 100mm blocks  0.085 or 0.204

5. 10mm cavity due to dot and dab approx 0.13

6. Plasterboard     0.06

7. Surface effect /Rsi  0.130

 

Total R value 0.725 or 0.844 depending on the block.

 

U-values would be.. 1/0.725 = 1.38 or 1/0.844 = 1.18

 

 

 

 

 

 

[Temp]

12 minutes ago, ruggers said:

still thought it would affect the end result because the insulite R layer says 0.204 and the dense is 0.085 so once totalled and then divided the answer couldn't be the same considering as a bare block one is twice as good as the other.

 

See my Fred and Joe example.

 

I note they also round the answer to two decimal places one is 0.175 the other 0.170

 

Too late at night for me to spot any errors 🙂

Edited August 22, 2022 by Temp

[ruggers]

Thanks @SteamyTea I wasn't sure whats a figure worth caring about and what's insignificant and just looks better on paper. 0.17 vs 0.18 vs 0.19 

 

@Temp I missed the surface effect from the calculation both sides and see you have reduced the dab gap to 10mm from 15. Boffin of the night award for you tonight 🏆. I'll keep my dunce hat on, brains cooked for the night. Hopefully now I have enough figures to input everything as rockwool provided the partition walls, loft and intermediate floors values.

 

I did look at the examples again & used the formula you provided and now understand how close it makes it. 

 

Thanks again both of you.

[ruggers]

An after thought question on this....

16 hours ago, Temp said:

It says the thermal conductivity is 1.17W/mK (eg not W/m^2k). Thermal Conductivity is the U value for a notional 1m thick wall. So 100mm thick wall would have a u-value 10 times worse at 11.7W/m^2k. That's consistent with the R-value quoted for 100mm blocks of 0.09m^2k/W....

 

U=1/R = 1/0.09 = 11.1W/m^2k.

If W/mK is for a thickness of 1m, is W/m2K always based on 100mm?

[SteamyTea]

13 minutes ago, ruggers said:

If W/mK is for a thickness of 1m, is W/m2K always based on 100mm?

No.

It is based on the thickness you specify.

 

If the k is 1W.m^-1.K^-1 changing the m, thickness in metres, or the K, temperature difference, will change the 1W figure.

 

 

[Temp]

3 hours ago, ruggers said:

An after thought question on this....

If W/mK is for a thickness of 1m, is W/m2K always based on 100mm?

 

No. I just gave that as an example.

 

If you wanted to know the U-value of say 125mm insulation you multiply the Thermal Conductivity by 1000/125.

 

 

[JohnMo]

No.

Just divide the thickness (in metres) by thermal resistance.  Then you will have the R value, to get the U you need the inverse, so 1/R equals U.

 

0.022 is PIR thermal resistance

125mm thick.

 

0.125 / 0.022 = 5.68

 

1/5.68=0.17=U value.

 

 

 

 

[ruggers]

Thanks JohnMo

 

I have a ground floor SAP target of 0.10 W/m2K and the plans state, from the ground up.

 

1. Min 150mm sub base

2. 50mm sand blinding

3. DPM

4. 100 Concrete slab

5. 150mm PIR insulation

6. 50mm Liquid screed (With UFH pipes embedded) 

 

Someone said that I'd require 170mm of PIR to achieve 0.10 for the floor so I wanted to check if 150 was going to do it or not.

[Nick Laslett]

2 hours ago, ruggers said:

Thanks JohnMo

 

I have a ground floor SAP target of 0.10 W/m2K and the plans state, from the ground up.

 

1. Min 150mm sub base

2. 50mm sand blinding

3. DPM

4. 100 Concrete slab

5. 150mm PIR insulation

6. 50mm Liquid screed (With UFH pipes embedded) 

 

Someone said that I'd require 170mm of PIR to achieve 0.10 for the floor so I wanted to check if 150 was going to do it or not.

Not a recommendation, this was just the quickest PIR board U-value I could find. 

https://www.kingspan.com/gb/en/products/insulation-boards/floor-insulation-boards/

Table from Kingspan Thermafloor brochure.  Need 175mm for 0.10. 
 

 

That is their standard PIR board. They have two other products with better resistance to thermal conductivity, Kooltherm is a phenolic board and Optim-R which is a VIP board. These can be much thinner. Optim-R only needs to be 70mm thick to hit 0.10 u-value. 
 

Here is a calculator for another product, EcoTherm board, it only needed 140mm to hit 0.10. 
 

https://uvaluecalculator.ecotherm.co.uk/calculator/

 

Edited August 23, 2022 by Nick Laslett

[ruggers]

Thanks, I might give them a call tomorrow. If 150mm can get down to 0.11/0.12, it's possible that the 100mm slab under the insulation adds a small amount of value to the 150mm insulation and brigs it to 0.10W/m2K.

 

Eco therm and quinn therm were always cheaper than kingspan.

[JohnMo]

I would go into the link Nick gave you.  You need to know your floor area and perimeter and floor buildup.  Then just play with the calculation

[NIMAN]

On 22/08/2022 at 23:36, SteamyTea said:

It may be useful when doing a room by room heat loss calculation for sizing heating/cooling systems.

 

k = W.m^-1.K^-1 [thermal conductivity]

 

R = L / W.m^-1.K^-1 where L is thickness [thermal resistance]

 

U = 1 / R = W.m^-2.K^-1 [thermal transmittance]

 

 

I'd like to size a number of A2A minisplit heatpumps as zoned heat emitters in this 1958 Bugalow (targets being a mixture of 21C / 18C & 16C).

 

Internal walls have both sides finished with 15mm sand/cement plaster render & gypsum finish to 100mm dense concrete block

 

Estimated as:

                                                                                   thickness (mm)   λ (W/K.m)              R

15mm sand/cement plaster & gypsum                     15                                 0.5          0.030

100mm dense concrete block                                  100                               1.4           0.071

15mm sand/cement plaster & gypsum                     15                                 0.5          0.030

       

                                                                                                               Sum       0.131 K⋅m2/W

                                                                                                                 U          7.609 W/K⋅m2

 

For the external envelope total heat loss wall calculation I used rsi (intenal surface) = 0.14 K⋅m2/W & rso (external surface) = 0.04 without really investigating the underlying Physics involved.

 

For the internal walls the inclusion of rsi/rso would, obviously, substantially decrease the conductance calculated above.

 

If I should include them are there published values of what they should be and (if they are different) to which side of the wall (warm/cold) should the respective values be applied (not that I see this last point making any difference for emmitter sizing).

 

Depending on the modelling I will consider applying internal wall insulation to various regions of the internal walls.

Edited October 27 by NIMAN

[NIMAN]

17 hours ago, NIMAN said:

If I should include them are there published values of what they should be and (if they are different) to which side of the wall (warm/cold) should the respective values be applied (not that I see this last point making any difference for emmitter sizing).

 

 

 

For anyone following along and wondering the same thing - an AI query, initially suggested using an rsi of 0.14 K⋅m2/W for only one side of the wall but, afer I queried that, apologised (😮) and said that value should be used for both. In a relatively long conversation it supplied a lot of, obviously, technically accurate information and only made one other obvious mistake.

 

I will use 0.14 K⋅m2/W for both sides in future calculations. Without a deeper dive it, sort of, makes sense.

[JohnMo]

You really ignore internal wall U value - you have a a huge opening between rooms called a door, which could be either open or closed.

 

Be carefull using AI for anything - would generally be my last port of call, never the first

 

Rsi is the surface resistance of the internal face
Typical internal surface resistance: Roof / Ceiling 0.1, Wall 0.12, Floor 0.14.

Rso is the surface resistance of the external face.
Typical external surface resistance: Roof / Ceiling 0.04, Wall 0.06, Floor 0.04.

[Iceverge]

By the time you've started adding surface resistance's to the calculations you've lost the wood for the trees in my view. 

 

Things like uncontrolled airflow will be orders of magnitude more important. Even the best thermal models are only a guide in real life. 

 

If you want really accurate numbers for  heat loss a static test with an electric radiator and a thermometer would be far better.