The proposed external wall design detail for a new house is a 215 mm single leaf wall of solid block construction with external insulation, as shown - Leaving Cert Construction Studies - Question 5 - 2022

Using the U-value from part (a), we calculate the heat loss:

Heat loss through wall = U-value x Area x Temp. diff
Heat loss = 0.191 x 144 x (20 - 5) Heat loss = 0.191 x 144 x 15 = 413.068 Watts.

Determine the total heating hours per year:

Heating period:

• 10 hours daily for 37 weeks:
  $10\ hours/day \times 7 \ days/week \times 37 \ weeks = 2,590 \ hours.$

Total heat loss in kJ per year:

$9,324,000 \ seconds \times 413.068\ J/s \div 1000 = 3,851,446.032 \ kJ/sec$

Now we find the cost of oil:

• Calorific value of oil = 37,350 kJ per litre
  Let’s convert total heat loss into litres:

$\frac{3,851,446.032}{37350} = 103.118 \ litres.$

Finally, the cost:

• Cost of oil = 0.97 per litre Total cost = 103.118 x 0.97 = €100.02 (to two decimal points).

To find the additional thickness, we start with the target U-value:

Required U-value = 0.12 W/m²°C. Using the formula: $R_{total} = \frac{1}{U} = \frac{1}{0.12} = 8.333 \ m^2\ °C/W$

Knowing the total resistance R is: $R_{total} = R_{existing} + R_{additional}$

We already calculated the existing: $R_{existing} = 5.229 \ m^2 \ °C/W$

Thus: $R_{additional} = 8.333 - 5.229 = 3.104 \ m^2 \ °C/W$

Now we convert this back to additional thickness: $t = R \times k$ Using k for external insulation (0.031): $t = 3.104 \ m^2 \ °C/W \times 0.031 \ W/m °C = 0.096 \ m$ Therefore, thickness = 96 mm.