Aluthermo Optima Attic Roof U-Value and Condensation Risk Calculations · 2020. 9. 29. · 6 BS EN...

Aluthermo Optima – Attic Roof

U-Value and Condensation Risk Calculations

For

LABC Consult

6th November 2014

Melin Consultants are accredited to provide a range of calculation and testing services. They are members of CIBSE Low Carbon Consultants.

Summary

1. Melin Consultants fully audit all work prior to completion and a robust audit trail exists

to demonstrate accountability.

2. All information within this document is based on evidence provided in the form of

drawings and specifications.

3. CPD (Continual Professional Development) records are kept and technicians are

required to complete a minimum 20 hours per year in training activities.

4. Low Carbon Consultants have the expertise and necessary qualifications to offer

advice in a professional capacity on matters relating to Part L of the Building

Regulations and sustainability within the construction sector.

This document contains the following information:

U-Value Calculation

Project Ref: 3336 Report Date: 6th November 2014 Report author: Kyle Jones Function: Technical Sustainability Consultant Authorised by: Darren Baker Function: Senior Consultant

Melin Consultants Ltd, The Beacon, Dafen Business Park, Llanelli, SA14 8LQ0845 094 1593; [email protected]

Documentation of the component 5. November 2014Thermal transmittance (U-value) according to BS EN ISO 6946Source: own catalogue - 3336 - AluthermoComponent: Slate roof 400ccs, MW quilt 0.13

Calculated with BuildDesk 3.4.5

OUTSIDE

INSIDE

This illustration of inhomogeneous layers is provided only to assist in visualising the arrangement.

On the basis of the given information about the inhomogeneous layers, it is not possible to estimate how and where bearing elements intersect each other. It was assumed that the layers intersect crosswise. The size of the areas was calculated corresponding to their percentage of the whole area.

Assignment: Pitched roof < 70°, with insulation between joists

Manufacturer Name Thickness[m],

number

Lambda[W/(mK)]

Q R[m²K/W]

Rse 0.04001 BS EN ISO 6946 Sheeted or tiled roof with felt/boards under tiles 0.5000 2.500 0.20002 Own catalogue Mineral wool quilt - Variable thickness 0.1700 0.040 4.2500

Air gaps Level 1: dU'' = 0.01 W/(m²K)3 Inhomogeneous material

layer consisting of: 0.1000 ø 0.051 1.9512

3a Own catalogue Mineral wool quilt - Variable thickness 87.50 % 0.040 -Air gaps Level 1: dU'' = 0.01 W/(m²K)

3b BS EN 12524 Softwood Timber [500 kg/m³] 12.50 % 0.130 -4 Aluthermo Aluthermo Optima 0.0250 0.038 0.6579



5a Own catalogue Low E cavity 94.20 % 0.074 -5b BS EN 12524 Softwood Timber [500 kg/m³] 05.80 % 0.130 -6 BS EN 12524 Gypsum plasterboard 0.0125 0.250 0.0500

Rsi 0.1000

0.8455

RT = (RT ' + RT '')/2 = 7.89 m²K/W

Correction to U-value for according to delta U[W/(m²K)]

Air gaps BS EN ISO 6946 Annex D 0.004Air gaps and fixings corrections need not be applied, as their total effect is less than 3% (Annex D BS 6946:1996).

0.000

U = 1/RT + U = 0.13 W/(m²K)

Q .. The physical values of the building materials has been graded by their level of quality. These 5 levels are the following.. A: Data is entered and validated by the manufacturer or supplier. Data is continuously tested by 3rd party... B: Data is entered and validated by the manufacturer or supplier. Data is certified by 3rd party.. C: Data is entered and validated by the manufacturer or supplier... D: Information is entered by BuildDesk without special agreement with the manufacturer, supplier or others... E: Information is entered by the user of the BuildDesk software without special agreement with the manufacturer, supplier or

others.

U = 0.13 W/(m²K) RT= 7.89 m²K/W

Page 1/7




Draft of the component (portion in %):

21.88 6.25 43.75 6.25 21.88

23.55

2.90

47.10

2.90

23.55

A

B

C

D

The intersection of the inhomogeneous layers results in 4Zones (A, B, C, D). Information given in %.

= 82.43%5.15 + 10.30 + 5.15 + 10.30 + 20.61 + 10.30 + 5.15 + 10.30 + 5.15

consisting of material layers: 1, 2, 3a, 4, 5a, 6

= 11.78%1.47 + 2.94 + 1.47 + 1.47 + 2.94 + 1.47

consisting of material layers: 1, 2, 3b, 4, 5a, 6

= 5.08%0.63 + 1.27 + 0.63 + 0.63 + 1.27 + 0.63

consisting of material layers: 1, 2, 3a, 4, 5b, 6

= 0.73%0.18 + 0.18 + 0.18 + 0.18

consisting of material layers: 1, 2, 3b, 4, 5b, 6

Upper limit of the thermal transfer resistance R

UA [W/(m2K)] =( Ri,A ) + Rsi + Rse

1=

8.17 + 0.1 + 0.04

1= 0.12

UB [W/(m2K)] =( Ri,B ) + Rsi + Rse

1=

6.44 + 0.1 + 0.04

1= 0.15

UC [W/(m2K)] =( Ri,C ) + Rsi + Rse

1=

7.95 + 0.1 + 0.04

1= 0.12

UD [W/(m2K)] =( Ri,D ) + Rsi + Rse

1=

6.22 + 0.1 + 0.04

1= 0.16

RT ' = A * UA + B * UB + C * UC + D * UD

1= 8.03 m2K/W

Lower limit of the thermal transfer resistance R

Rse [m2K/W] = 0.04

R1 '' [m2K/W] = d 1/ 1= 0.5000 / 2.500 = 0.20

R2 '' [m2K/W] = d 2/ 2= 0.1700 / 0.040 = 4.25

R3 '' [m2K/W] = d 3/( 3a * (A + C) + 3b * (B + D)) = 0.1000 /( 0.040 * 87.50% + 0.130 * 12.50%) = 1.95

R4 '' [m2K/W] = d 4/ 4= 0.0250 / 0.038 = 0.66

R5 '' [m2K/W] = d 5/( 5a * (A + B) + 5b * (C + D)) = 0.0380 /( 0.074 * 94.20% + 0.130 * 5.80%) = 0.49

R6 '' [m2K/W] = d 6/ 6= 0.0125 / 0.250 = 0.05

Rsi [m2K/W] = 0.1

RT " = Ri " + Rsi + Rse = 7.74 m²K/W

Page 2/7


Documentation of the component 5. November 2014Calculation according BS EN ISO 13788Source: own catalogue - 3336 - AluthermoComponent: Slate roof 400ccs, MW quilt 0.13


OUTSIDE

INSIDE

The list of material layers shown below may differ from those in the U-value calculation printout. Only material layers which are used in the Condensation Risk Analysis are listed.

This calculation of the Condensation risk analysis according to BS EN ISO 13788:2002 has been performed on a construction containing inhomogeneous layers. This calculation is only valid through the selected section. It is advisable that you should also select the alternative position and recalculate the Condensation Risk Analysis for a more complete assessment of the construction.

The CRA calculation for pitched roofs can be very unreliable and caution should be used when interpreting these results. For further guidance the user is advised to follow the recommendation of BS 5250:202 (currently under review).


Name Thickn.[m]

lambda[W/(mK)]

Q µ[-]

Q sd[m]

R[m²K/W]

Sheeted or tiled roof with felt/boards under tiles 0.5000 2.500 1.00 0.50 0.2000Mineral wool quilt - Variable thickness 0.1700 0.040 1.00 0.17 4.2500Mineral wool quilt - Variable thickness 0.1000 0.040 1.00 0.10 2.5000Aluthermo Optima 0.0250 0.038 29231.00 730.78 0.6579Low E cavity 0.0380 0.074 1.00 0.04 0.5135Gypsum plasterboard 0.0125 0.250 4.00 0.05 0.0500


others.

Page 3/7




Condensation risk analysis - summary of main results Calculation according BS EN ISO 13788

Surface temperature to avoid critical surface moisture:

No danger of mould growth is expected.

Interstitial condensation:No condensation is predicted at any interface in any month.

-1.00

-0.50

0.00

0.50

1.00

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Interstitial condensation andevaporation per month gc [g/m²]

Component, condensation range

CRA calculations according to BS EN ISO 13788:2002 are used as a guide in predicting interstitial condensation.

This methodology uses some simplifications of the dynamic processes involved and subsequently does have some

limitations. Further information can be found in Information Paper IP 2/05 'Modelling and controlling interstitial

condensation in buildings'' Feb 2005.

The CRA calculation for pitched roofs can be very unreliable and caution should be used when interpreting these

results. For further guidance the user is advised to follow the recommendation of BS 5250:202 (currently under

review).

Page 4/7




Surface temperature to avoid critical surface humidity Calculation according BS EN ISO 13788

Location: London (Heathrow); Humidity class according BS EN ISO 13788 annex A: Dwellings with high occupancy

1 2 3 4 5 6 7 8 9 10 11 12Month Te

[°C]phi_e

---Ti

[°C]phi_i

---pe

[Pa]delta p

[Pa]pi

[Pa]ps(Tsi)

[Pa]Tsi,min

[°C]fRsi

---Tsi

[°C]Tse[°C]

� January 4.9 0.840 20.0 0.695 727 897 1624 2030 17.7 0.851 19.6 5.0February 4.7 0.820 20.0 0.688 700 909 1609 2011 17.6 0.843 19.5 4.8March 6.9 0.770 20.0 0.661 766 778 1544 1930 16.9 0.767 19.6 7.0April 8.8 0.710 20.0 0.629 804 665 1469 1836 16.2 0.657 19.7 8.9May 12.6 0.690 20.0 0.619 1006 440 1446 1807 15.9 0.448 19.8 12.6June 15.7 0.690 20.0 0.636 1230 255 1486 1857 16.3 0.148 19.9 15.7July 17.9 0.680 20.0 0.650 1394 125 1519 1898 16.7 -0.579 19.9 17.9August 17.6 0.700 20.0 0.664 1408 143 1551 1938 17.0 -0.244 19.9 17.6September 14.9 0.750 20.0 0.673 1270 303 1573 1966 17.2 0.459 19.8 14.9October 11.2 0.810 20.0 0.685 1077 523 1600 2000 17.5 0.717 19.7 11.2November 7.6 0.840 20.0 0.690 876 737 1613 2016 17.6 0.809 19.6 7.7December 5.9 0.860 20.0 0.700 798 838 1636 2045 17.9 0.848 19.6 6.0

The critical month is January with fRsi,max = 0.851 fRsi = 0.970

fRsi > fRsi,max , the component complies.

Nr Explanation

1 External temperature

2 External rel. humidity

3 Internal temperature

4 Internal relative humidity

5 External partial pressure p e = e * psat (Te ); psat (Te ) according formula E.7 and E.8 of BS EN ISO 13788

6 Partial pressure difference. The security factor of 1.10 according to BS EN ISO 13788, ch.4.2.4 is already included.

7 Internal partial pressure p i = i * psat (Ti ); psat (Ti ) according formula E.7 and E.8 of BS EN ISO 13788

8 Minimum saturation pressure on the surface obtained by psat (Tsi ) = p i / si ,

where si = 0.8 ( critical surface humidity )

9 Minimum surface temperature as function of psat (Tsi ) , formula E.9 and E.10 of BS EN ISO 13788

10 Design temperature factor according 3.1.2 of BS EN ISO 13788

11 Internal surface temperature, obtained from Tsi = Ti - Rsi * U * (Ti - Te )

12 External surface temperature, obtained from Tse = Te + Rse * U * (Ti - Te )

Page 5/7




Interstitial condensation - main results Calculation according BS EN ISO 13788

No condensation is predicted at any interface in any month.

Climcatic conditions Location: London (Heathrow); Humidity class according BS EN ISO 13788 annex A: Dwellings with high occupancy

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov DecInternal temperature [°C] Ti 20.0 20.0 20.0 20.0 20.0 20.0 20.0 20.0 20.0 20.0 20.0 20.0Internal rel. humidity [%] phi_i 69.5 68.8 66.1 62.9 61.9 63.6 65.0 66.4 67.3 68.5 69.0 70.0External temperature [°C] Te 4.9 4.7 6.9 8.8 12.6 15.7 17.9 17.6 14.9 11.2 7.6 5.9External rel. humidity [%] phi_e 84.0 82.0 77.0 71.0 69.0 69.0 68.0 70.0 75.0 81.0 84.0 86.0

Page 6/7


Documentation of the component 5. November 2014Heat capacitySource: own catalogue - 3336 - AluthermoComponent: Slate roof 400ccs, MW quilt 0.13


The list of materials shown below may differ from those in the U-value calculation printout. Only material layers which are used in the heat capacity calculation are listed.

Single material layers shown in the U-value calculation printout may be separated to meet the exclusion criteria:

A .. The total thickness of the layers exceed 0.1 m.B .. The mid point in the construction is reached.

For insulation layers the following criteria applies:

C .. An insulating layer is reached (defined as lambda <= 0.08 W/(mK)).

OUTSIDE

INSIDE

Name Thickness[m]

lambda[W/(mK)]

Q Thermalcapacity

[kJ/(kgK)]

Q Density[kg/m³]

Q Thermalmass

kJ/(m²K)

CriteriaExclusion

End of calculation - Cold

1Sheeted or tiled roof with felt/boards under tiles

0.5000 2.500 1.01 1.2 0.6 A, -, C

2 Mineral wool quilt - Variable thickness 0.1700 0.040 1.03 12.0 0.0 A, -, C3 Inhomogeneous material layer consisting of: 0.0755 7.6 A, -, C3a Mineral wool quilt - Variable thickness 87.50% 0.040 1.03 12.0 0.8 A, -, C3b Softwood Timber [500 kg/m³] 12.50% 0.130 1.60 500.0 7.6 A, -, C3 Inhomogeneous material layer consisting of: 0.0245 2.5 -, -, C3a Mineral wool quilt - Variable thickness 87.50% 0.040 1.03 12.0 0.3 -, -, C3b Softwood Timber [500 kg/m³] 12.50% 0.130 1.60 500.0 2.5 -, -, C4 Aluthermo Optima 0.0250 0.038 0.88 57.6 0.0 -, -, C5 Inhomogeneous material layer consisting of: 0.0380 1.8 -, -, -5a Low E cavity 94.20% 0.074 1.01 1.3 0.0 -, -, C5b Softwood Timber [500 kg/m³] 05.80% 0.130 1.60 500.0 1.8 -, -, -6 Gypsum plasterboard 0.0125 0.250 1.00 900.0 11.3 -, -, -

Start of calculation - Warm

0.8455 13.0

Heat capacity = 13.0 kJ/(m²K)

The following exclusion criteria apply:A .. The total thickness of the layers exceed 0.1 m.C .. An insulating layer is reached (defined as lambda <= 0.08 W/(mK)).


others.

Page 7/7




OUTSIDE

INSIDE





number

Lambda[W/(mK)]

Q R[m²K/W]









Rsi 0.1000

0.7755

RT = (RT ' + RT '')/2 = 6.13 m²K/W



0.000

U = 1/RT + U = 0.16 W/(m²K)


others.

U = 0.16 W/(m²K) RT= 6.13 m²K/W

Page 1/7





21.88 6.25 43.75 6.25 21.88

23.55

2.90

47.10

2.90

23.55

A

B

C

D


= 82.43%5.15 + 10.30 + 5.15 + 10.30 + 20.61 + 10.30 + 5.15 + 10.30 + 5.15


= 11.78%1.47 + 2.94 + 1.47 + 1.47 + 2.94 + 1.47


= 5.08%0.63 + 1.27 + 0.63 + 0.63 + 1.27 + 0.63


= 0.73%0.18 + 0.18 + 0.18 + 0.18




1=

6.42 + 0.1 + 0.04

1= 0.15


1=

4.69 + 0.1 + 0.04

1= 0.21


1=

6.20 + 0.1 + 0.04

1= 0.16


1=

4.47 + 0.1 + 0.04

1= 0.22

RT ' = A * UA + B * UB + C * UC + D * UD

1= 6.27 m2K/W


Rse [m2K/W] = 0.04

R1 '' [m2K/W] = d 1/ 1= 0.5000 / 2.500 = 0.20

R2 '' [m2K/W] = d 2/ 2= 0.1000 / 0.040 = 2.50

R3 '' [m2K/W] = d 3/( 3a * (A + C) + 3b * (B + D)) = 0.1000 /( 0.040 * 87.50% + 0.130 * 12.50%) = 1.95

R4 '' [m2K/W] = d 4/ 4= 0.0250 / 0.038 = 0.66

R5 '' [m2K/W] = d 5/( 5a * (A + B) + 5b * (C + D)) = 0.0380 /( 0.074 * 94.20% + 0.130 * 5.80%) = 0.49

R6 '' [m2K/W] = d 6/ 6= 0.0125 / 0.250 = 0.05

Rsi [m2K/W] = 0.1

RT " = Ri " + Rsi + Rse = 5.99 m²K/W

Page 2/7




OUTSIDE

INSIDE





Name Thickn.[m]

lambda[W/(mK)]

Q µ[-]

Q sd[m]

R[m²K/W]



others.

Page 3/7








-1.00

-0.50

0.00

0.50

1.00










review).

Page 4/7






1 2 3 4 5 6 7 8 9 10 11 12Month Te

[°C]phi_e

---Ti

[°C]phi_i

---pe

[Pa]delta p

[Pa]pi

[Pa]ps(Tsi)

[Pa]Tsi,min

[°C]fRsi

---Tsi

[°C]Tse[°C]




Nr Explanation














Page 5/7








Page 6/7









OUTSIDE

INSIDE

Name Thickness[m]

lambda[W/(mK)]

Q Thermalcapacity

[kJ/(kgK)]

Q Density[kg/m³]

Q Thermalmass

kJ/(m²K)

CriteriaExclusion



0.5000 2.500 1.01 1.2 0.6 A, -, C



0.7755 13.0




others.

Page 7/7




OUTSIDE

INSIDE





number

Lambda[W/(mK)]

Q R[m²K/W]









Rsi 0.1000

0.8255

RT = (RT ' + RT '')/2 = 7.59 m²K/W



0.000

U = 1/RT + U = 0.13 W/(m²K)


others.

U = 0.13 W/(m²K) RT= 7.59 m²K/W

Page 1/7





23.55 2.90 47.10 2.90 23.55

23.55

2.90

47.10

2.90

23.55

A

B

C

D


= 88.74%5.55 + 11.09 + 5.55 + 11.09 + 22.18 + 11.09 + 5.55 + 11.09 + 5.55


= 5.46%0.68 + 1.37 + 0.68 + 0.68 + 1.37 + 0.68


= 5.46%0.68 + 1.37 + 0.68 + 0.68 + 1.37 + 0.68


= 0.34%0.08 + 0.08 + 0.08 + 0.08




1=

7.67 + 0.1 + 0.04

1= 0.13


1=

5.94 + 0.1 + 0.04

1= 0.16


1=

7.45 + 0.1 + 0.04

1= 0.13


1=

5.72 + 0.1 + 0.04

1= 0.17

RT ' = A * UA + B * UB + C * UC + D * UD

1= 7.67 m2K/W


Rse [m2K/W] = 0.04

R1 '' [m2K/W] = d 1/ 1= 0.5000 / 2.500 = 0.20

R2 '' [m2K/W] = d 2/ 2= 0.1500 / 0.040 = 3.75

R3 '' [m2K/W] = d 3/( 3a * (A + C) + 3b * (B + D)) = 0.1000 /( 0.040 * 94.20% + 0.130 * 5.80%) = 2.21

R4 '' [m2K/W] = d 4/ 4= 0.0250 / 0.038 = 0.66

R5 '' [m2K/W] = d 5/( 5a * (A + B) + 5b * (C + D)) = 0.0380 /( 0.074 * 94.20% + 0.130 * 5.80%) = 0.49

R6 '' [m2K/W] = d 6/ 6= 0.0125 / 0.250 = 0.05

Rsi [m2K/W] = 0.1

RT " = Ri " + Rsi + Rse = 7.50 m²K/W

Page 2/7




OUTSIDE

INSIDE





Name Thickn.[m]

lambda[W/(mK)]

Q µ[-]

Q sd[m]

R[m²K/W]



others.

Page 3/7








-1.00

-0.50

0.00

0.50

1.00










review).

Page 4/7






1 2 3 4 5 6 7 8 9 10 11 12Month Te

[°C]phi_e

---Ti

[°C]phi_i

---pe

[Pa]delta p

[Pa]pi

[Pa]ps(Tsi)

[Pa]Tsi,min

[°C]fRsi

---Tsi

[°C]Tse[°C]




Nr Explanation














Page 5/7








Page 6/7









OUTSIDE

INSIDE

Name Thickness[m]

lambda[W/(mK)]

Q Thermalcapacity

[kJ/(kgK)]

Q Density[kg/m³]

Q Thermalmass

kJ/(m²K)

CriteriaExclusion



0.5000 2.500 1.01 1.2 0.6 A, -, C



0.8255 13.0




others.

Page 7/7




OUTSIDE

INSIDE





number

Lambda[W/(mK)]

Q R[m²K/W]









Rsi 0.1000

0.7755

RT = (RT ' + RT '')/2 = 6.33 m²K/W



0.000

U = 1/RT + U = 0.16 W/(m²K)


others.

U = 0.16 W/(m²K) RT= 6.33 m²K/W

Page 1/7





23.55 2.90 47.10 2.90 23.55

23.55

2.90

47.10

2.90

23.55

A

B

C

D


= 88.74%5.55 + 11.09 + 5.55 + 11.09 + 22.18 + 11.09 + 5.55 + 11.09 + 5.55


= 5.46%0.68 + 1.37 + 0.68 + 0.68 + 1.37 + 0.68


= 5.46%0.68 + 1.37 + 0.68 + 0.68 + 1.37 + 0.68


= 0.34%0.08 + 0.08 + 0.08 + 0.08




1=

6.42 + 0.1 + 0.04

1= 0.15


1=

4.69 + 0.1 + 0.04

1= 0.21


1=

6.20 + 0.1 + 0.04

1= 0.16


1=

4.47 + 0.1 + 0.04

1= 0.22

RT ' = A * UA + B * UB + C * UC + D * UD

1= 6.41 m2K/W


Rse [m2K/W] = 0.04

R1 '' [m2K/W] = d 1/ 1= 0.5000 / 2.500 = 0.20

R2 '' [m2K/W] = d 2/ 2= 0.1000 / 0.040 = 2.50

R3 '' [m2K/W] = d 3/( 3a * (A + C) + 3b * (B + D)) = 0.1000 /( 0.040 * 94.20% + 0.130 * 5.80%) = 2.21

R4 '' [m2K/W] = d 4/ 4= 0.0250 / 0.038 = 0.66

R5 '' [m2K/W] = d 5/( 5a * (A + B) + 5b * (C + D)) = 0.0380 /( 0.074 * 94.20% + 0.130 * 5.80%) = 0.49

R6 '' [m2K/W] = d 6/ 6= 0.0125 / 0.250 = 0.05

Rsi [m2K/W] = 0.1

RT " = Ri " + Rsi + Rse = 6.25 m²K/W

Page 2/7




OUTSIDE

INSIDE





Name Thickn.[m]

lambda[W/(mK)]

Q µ[-]

Q sd[m]

R[m²K/W]



others.

Page 3/7








-1.00

-0.50

0.00

0.50

1.00










review).

Page 4/7






1 2 3 4 5 6 7 8 9 10 11 12Month Te

[°C]phi_e

---Ti

[°C]phi_i

---pe

[Pa]delta p

[Pa]pi

[Pa]ps(Tsi)

[Pa]Tsi,min

[°C]fRsi

---Tsi

[°C]Tse[°C]




Nr Explanation














Page 5/7








Page 6/7









OUTSIDE

INSIDE

Name Thickness[m]

lambda[W/(mK)]

Q Thermalcapacity

[kJ/(kgK)]

Q Density[kg/m³]

Q Thermalmass

kJ/(m²K)

CriteriaExclusion



0.5000 2.500 1.01 1.2 0.6 A, -, C



0.7755 13.0




others.

Page 7/7

Date post:	25-Aug-2021
Category:	Documents
Upload:	others
View:	2 times
Download:	0 times

Aluthermo Optima Attic Roof U-Value and Condensation Risk Calculations · 2020. 9. 29. · 6 BS EN...

Documents