**Questions and Solutions from SEC**

The external wall of a house of timber frame construction has the following specification:

External render thickness 19 mm

Concrete block outer leaf thickness 100 mm

Oriented Strand Board (OSB) sheeting thickness 18 mm

Timber stud inner leaf thickness 120 mm

Mineral wool insulation between studs thickness 120 mm

Plasterboard thickness 12.5 mm

Thermal data of outer leaf and cavity:

Resistance of the external surface (R) 0.048 m2 °C/W

Resistivity of the external plaster (r) 2.170 m °C/W

Conductivity of concrete block (k) 1.320 W/m °C

Resistance of the cavity (R) 0.170 m2 °C/W

Thermal data of inner leaf:

Conductivity of OSB sheeting (k) 0.130 W/m °C

Conductivity of mineral wool (k) 0.040 W/m °C

Conductivity of plasterboard (k) 0.160 W/m °C

Resistance of the internal surface (R) 0.104 m2 °C/W

Note: The timber studs of the inner leaf need not be considered in your calculations.

(a) Calculate the U-value of the above external wall.

Calculate the cost of heat lost annually through this wall, using the following data:

Area of external wall 150 m2

Average internal temperature 18 °C

Average external temperature 6 °C

U-value of wall as calculated above

Heating period 8 hours per day, every day for 38 weeks per annum

Cost of oil 96 cent per litre

Calorific value of oil 37350 kJ per litre

1000 Watts 1 kJ per second.

It is proposed to upgrade the thermal properties of the above wall to meet the Passive House standard by fixing expanded polystyrene to the external surface. Given the thermal conductivity (k) of expanded polystyrene as 0.037 W/m °C, calculate the thickness of expanded polystyrene required to achieve a U-value of 0.15 W/m2 °C.

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