a) Calculate the U-value of the uninsulated concrete ground floor, given the following data: Sand/cement fine screed thickness 50 mm Concrete floor slab thickness 100 mm Radon barrier thickness 0.25 mm Sand binding thickness 40 mm Hardcore thickness 200 mm Subsoil thickness 300 mm Thermal data of the concrete ground floor: Resistance of internal top surface of floor (R) 0.104 m² °C/W Resistivity of fine screed 0.720 m² °C/W Conductivity of concrete floor slab 1.280 W/m °C Conductivity of radon barrier 0.250 W/m °C Conductivity of sand binding 0.160 W/m °C Conductivity of hardcore 1.350 W/m °C Conductivity of subsoil 1.600 W/m °C b) Using the U-value of the concrete floor obtained at 5(a) above and the following data, calculate the cost of heat lost annually through the uninsulated concrete floor slab: - dimensions of floor slab 6.0 metres x 12.0 metres - average internal temperature 21 °C - average temperature of subsoil 5 °C - heating period 10 hours daily for 38 weeks per annum - cost of oil 96 cent per litre - calorific value of oil 37350 kj per litre - 1000 Watts c) It is proposed to redesign the above floor and upgrade its thermal properties to meet the Passive House standard by including expanded polystyrene in the design of the floor - Leaving Cert Construction Studies - Question 5 - 2018

1. Heat loss formula: $Q = U \cdot A \cdot \Delta T$

   • Where:
     • $U$ is the U-value
     • $A$ is the area
     • $\Delta T$ is the temperature difference.

2. Calculate the area: Area = 6.0 m x 12.0 m = 72 m²

3. Given:

   • Average internal temperature: 21 °C
   • Average temperature of subsoil: 5 °C
   • $\Delta T = 21 - 5 = 16 °C$

4. Substituting values: $Q = 1.363 \cdot 72 \cdot 16 = 1432.3968 \text{ Watts}$

5. Convert heat lost per day: $Q_{per ext{ day}} = 1432.3968 \times 10 \text{ hours} \times 38 \text{ weeks} \times 7 \text{ days} = 9,576,000 \text{ Joules}$

6. Convert to kilojoules: $9,576,000 \text{ Joules} = 9,576 \text{ kJ}$

7. Cost calculation:

   • Cost of oil = 96 cents per litre
   • Calorific value = 37350 kJ/litre

   $ext{Litres} = \frac{9,576}{37,350} \approx 0.256 \text{ litres}$

   • Cost per annum = 0.256 x 0.96 * 1000 = € 245.76 annually.

1. Given:

   • Current U-value: 1.2434 W/m²°C
   • Desired U-value: 0.15 W/m²°C.

2. Use the formula: $U = \frac{1}{R_{total}}$ $R_{total} = \frac{1}{U}$ $R_{total} = \frac{1}{0.15} = 6.6666 \ m^2 \degree C/W$

3. Calculate the total resistance of layers, including expanded polystyrene: $R_{total} = R_s + R_c + R_r + R_b + R_h + R_su + R_{polystyrene}$

4. Assuming all other resistances remain constant, solve for $R_{polystyrene}$: $R_{poly} = 6.6666 - (R_s + R_c + R_r + R_b + R_h + R_su)$

5. Substitute each layer's resistance: $R_{poly} = 6.6666 - 0.7341 \approx 5.9325 \ m^2 \degree C/W$

6. Calculate thickness: $R_{polystyrene} = \frac{t}{k} \Rightarrow t = R_{polystyrene} \times k$(k for polystyrene is 0.037) $t = 5.9325 \times 0.037 = 0.219 \text{ m} = 219 mm$

Thus, thickness required is approximately 216/217 mm.