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Q5- U-Value 2010 HL 5. (a) Calculate the U-value of an uninsulated external solid concrete wall of a dwelling house built in the 1950s given the following data: External render thickness 16 mm Solid concrete wall thickness 225 mm Internal plaster thickness 13 mm Thermal data of external wall of house: Resistivity of the solid concrete wall (r) 1.190 m °C/W Resistivity of external render (r) 2.170 m °C/W Resistivity of internal plaster (r) 6.250 m °C/W Resistance of external surface (R) 0.048 m2 °C/W Resistance of internal surface (R) 0.122 m2 °C/W (b) Phenolic foam insulation is to be fitted to the external surface of the solid concrete wall. Given the conductivity (k) of phenolic foam as 0.025 W/m °C, calculate the thickness of phenolic foam required to achieve a U-value of 0.27 W/m2 °C. (c) Discuss in detail, using notes and freehand sketches, the importance of thermal mass in improving the thermal performance of a dwelling house.

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MaterialResistivity (r)Thickness (m)Resistance (R) Formulae: R=T x r U= 1/Rt U Value: U = 1 / 0.55372 = 1.805 W/m2 C External Render Wall Internal plaster Internal Surface External surface 2.17 1.19 6.25 0.016 0.225 0.013 Rt = 0.03472 0.26775 0.08125 0.122 0.048 0.55372 W/m2 °C.

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Q.5 (b) To find the Resistance for a U-value of 0.27 W/m2 /K Use formula U= 1/Rt. & Solve for R. R = 1/ U-value R = 1/ 0.27 = 3.703 m2 C/W (this is the resistance of the wall after the insulation has been added) Total resistance of existing wall(before the insulation was added) = 0.55372 m2 C/W Therefore, the resistance of the insulation on its own is 3.703 – 0.55372 = 3.15 m2 C/W

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To find the thickness of insulation required Thickness X Resistivity = Resistance Therefore, Thickness = Resistance / Resistivity Resistivity = 1 / conductivity (k) Resistivity= 1 / 0.025 = 40 Therefore, Thickness = 3.15 / 40 = 0.07875 m To change to mm, 0.07875 x 1000 = 78.75 mm This is the thickness of phenolic foam insulation that needs to be added to achieve the required resistance.

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Thermal mass or fabric energy storage Thermal mass improves thermal performance and enhances thermal comfort, thus requiring less use of fossil fuels. Dense materials such as brick, stone, concrete, glass, marble are most effective in storing and releasing substantial quantities of thermal energy In Ireland’s mild climate solar gain can make a substantial contribution ( 20%) to space heating requirements. Thermal mass may best be utilised as a passive heating strategy in residential buildings by o A: Rigid insulation; B: Concrete o C: Ceramic tile; D: Concrete internal walls

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Passive Solar Collection Orientation – South +/- 15 degrees due South best Large south facing glazed areas Tall windows allow increased sunlight into rooms Design for a mass to glass ration of 6:1 Windows – double or triple glazed with soft low e- coatings improve solar heat gain Passive Solar Storage Locate thermal mass elements in south facing spaces that receive large amounts of sunlight Thermal mass elements positioned in direct sunlight are 30% more efficient

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