Leaving Cert Construction Studies Heat & U-Value Calculations: Using the following data, calcul

Using the following data, calculate the U-value of the insulated concrete ground floor: Concrete floor slab thickness 125 mm Extruded polystyrene insulation thickness 200 mm Damp proof membrane (DPM) thickness 0.3 mm Sand binding thickness 30 mm Hardcore thickness 125 mm Thermal data of floor: Resistance of top surface of floor (R) 0.104 m² °C/W Conductivity of concrete (k) 1.600 W/m °C Conductivity of polystyrene insulation (k) 0.031 W/m °C Conductivity of damp proof membrane (DPM) (k) 0.450 W/m °C Conductivity of sand binding (k) 0.160 W/m °C Conductivity of hardcore (k) 1.260 W/m °C (b) Using the U-value of the concrete ground floor obtained at 5(a) above and the following data, calculate the cost of heat lost per annum through the concrete ground floor: Dimensions of floor Area: 13.0 metres x 7.0 metres Average internal temperature 21°C Average external temperature 7°C Heating period P/A: 12 hr / day for 39 weeks per annum Cost of oil 95 cent per litre Calorific value of oil 37350 kJ per litre 1000 Watts (c) A solid concrete ground floor abuts a 350 mm concrete block external wall with an insulated cavity - Leaving Cert Construction Studies - Question 5 - 2014

Question 5

Using the following data, calculate the U-value of the insulated concrete ground floor: Concrete floor slab thickness 125 mm Extruded polystyrene insulation thickn... show full transcript

Worked Solution & Example Answer:Using the following data, calculate the U-value of the insulated concrete ground floor: Concrete floor slab thickness 125 mm Extruded polystyrene insulation thickness 200 mm Damp proof membrane (DPM) thickness 0.3 mm Sand binding thickness 30 mm Hardcore thickness 125 mm Thermal data of floor: Resistance of top surface of floor (R) 0.104 m² °C/W Conductivity of concrete (k) 1.600 W/m °C Conductivity of polystyrene insulation (k) 0.031 W/m °C Conductivity of damp proof membrane (DPM) (k) 0.450 W/m °C Conductivity of sand binding (k) 0.160 W/m °C Conductivity of hardcore (k) 1.260 W/m °C (b) Using the U-value of the concrete ground floor obtained at 5(a) above and the following data, calculate the cost of heat lost per annum through the concrete ground floor: Dimensions of floor Area: 13.0 metres x 7.0 metres Average internal temperature 21°C Average external temperature 7°C Heating period P/A: 12 hr / day for 39 weeks per annum Cost of oil 95 cent per litre Calorific value of oil 37350 kJ per litre 1000 Watts (c) A solid concrete ground floor abuts a 350 mm concrete block external wall with an insulated cavity - Leaving Cert Construction Studies - Question 5 - 2014

Step 1

Using the following data, calculate the U-value of the insulated concrete ground floor:

96%

114 rated

Answer

To calculate the U-value, we first need to determine the total thermal resistance (R) of the floor:

Calculate the Resistance (R) for each layer:
- For the concrete floor slab: $R_{1} = \frac{T_{1}}{k_{1}} = \frac{0.125}{1.600} = 0.078125 \ m^{2} °C/W$
- For the extruded polystyrene insulation: $R_{2} = \frac{T_{2}}{k_{2}} = \frac{0.200}{0.031} = 6.451612903 \ m^{2} °C/W$
- For the damp proof membrane: $R_{3} = \frac{T_{3}}{k_{3}} = \frac{0.0003}{0.450} = 0.00066666667 \ m^{2} °C/W$
- For sand binding: $R_{4} = \frac{T_{4}}{k_{4}} = \frac{0.030}{0.160} = 0.1875 \ m^{2} °C/W$
- For hardcore: $R_{5} = \frac{T_{5}}{k_{5}} = \frac{0.125}{1.260} = 0.09920634921 \ m^{2} °C/W$
Total Resistance (R_total): Now add all the resistances: $R_{total} = R_{1} + R_{2} + R_{3} + R_{4} + R_{5} = 0.078125 + 6.451612903 + 0.00066666667 + 0.1875 + 0.09920634921 = 7.71611191988 \ m^{2} °C/W$
Calculate U-value: The U-value is the reciprocal of total thermal resistance: $U = \frac{1}{R_{total}} = \frac{1}{7.71611191988} = 0.12919 \ W/m^{2} °C$

Step 2

Using the U-value of the concrete ground floor obtained at 5(a) above and the following data, calculate the cost of heat lost per annum through the concrete ground floor:

99%

104 rated

Answer

Calculate Area: $Area = 13.0 \times 7.0 = 91 \ m^{2}$
Calculate Temperature Difference: $Temp\_diff = 21°C - 7°C = 14°C$
Calculate Heat Loss Using the U-value: $Heat\_loss = U \times Area \times Temp\_diff\times 24 \times 39/1000$$$$ Substitute values:$ Heat_loss = 0.12919 \times 91 \times 14 \times 24 \times 39 / 1000 = 19.50186969 \ kJ/s $For annual heat loss:$ Annual_loss = 19.50186969 \times 3600 \times 24 \times 39$$ Simplifying gives total annual heat loss in kJ.
Calculate the Cost of Heat Loss: $Cost = Annual\_loss \times (cost\_per\_litre/Calorific\_value)$ Use provided values to find the cost.

Step 3

A solid concrete ground floor abuts a 350 mm concrete block external wall with an insulated cavity. Using notes and freehand sketches, show best practice design detailing which will prevent the formation of a cold bridge at the junction of the floor and the wall. Show the typical design detailing from the bottom of the hardcore to the top of the skirting board.

96%

101 rated

Answer

Sketch Description: Provide a detailed sketch showing layers of insulation and how they interact to prevent cold bridging.
Detailing: Illustrate:
- The insulation layer extending from the hardcore up to the level of the wall insulation.
- Use of sheathing or a thermal break at the junction.
- Ensure no direct thermal path exists from slab to wall.
Notes: Additional notes emphasizing the materials used and their effectiveness in mitigating thermal bridging should accompany the sketch.

Using the following data, calculate the U-value of the insulated concrete ground floor:

Using the U-value of the concrete ground floor obtained at 5(a) above and the following data, calculate the cost of heat lost per annum through the concrete ground floor:

Join the Leaving Cert students using SimpleStudy...