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A house built in the 1970s has an un-insulated solid concrete ground floor with a sand/cement fine screed finish - Leaving Cert Construction Studies - Question 5 - 2012
Question 5
A house built in the 1970s has an un-insulated solid concrete ground floor with a sand/cement fine screed finish. (a) Calculate the U-value of the concrete ground f... show full transcript
Worked Solution & Example Answer:A house built in the 1970s has an un-insulated solid concrete ground floor with a sand/cement fine screed finish - Leaving Cert Construction Studies - Question 5 - 2012
Step 1
Calculate the U-value of the concrete ground floor
Answer
To determine the U-value of the concrete ground floor, we will first calculate the thermal resistances for each component involved:
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Sand/cement fine screed:
- Thickness: 60 mm = 0.060 m
- Conductivity (k): 0.710 W/m°C
- Resistance (R) = Thickness / Conductivity = 0.060 / 0.710 = 0.0845 m²°C/W
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Concrete floor slab:
- Thickness: 100 mm = 0.100 m
- Conductivity (k): 0.160 W/m°C
- Resistance (R) = 0.100 / 0.160 = 0.625 m²°C/W
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Damp proof membrane (DPM):
- Thickness: 0.1 mm = 0.0001 m (approx.)
- Conductivity (k): 0.250 W/m°C
- Resistance (R) = 0.0001 / 0.250 = 0.0004 m²°C/W
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Sand binding:
- Thickness: 100 mm = 0.100 m
- Conductivity (k): 0.160 W/m°C
- Resistance (R) = 0.100 / 0.160 = 0.625 m²°C/W
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Hardcore:
- Thickness: 225 mm = 0.225 m
- Conductivity (k): 1.330 W/m°C
- Resistance (R) = 0.225 / 1.330 = 0.1695 m²°C/W
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Subsoil:
- Thickness: Considered very thick so we don’t calculate additional resistance here.
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Residual resistance of internal surface:
- R = 0.104 m²°C/W (given)
Now we add up all the resistances:
Total Resistance (R_total) = R_screed + R_concrete + R_DPM + R_sand_binding + R_hardcore + R_internal
= 0.0845 + 0.625 + 0.0004 + 0.625 + 0.1695 + 0.104
= 1.6094 m²°C/W
Finally, to compute the U-value, we use the formula:
[ U = \frac{1}{R_{total}} = \frac{1}{1.6094} = 0.620 W/m²K ]
Step 2
Calculate the cost of heat lost per year
Answer
Using the U-value calculated in part (a), we can now determine the cost of heat loss:
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Heat loss formula:
[ \text{Heat loss (Q)} = U \times \text{Area} \times \Delta T ]
Where:- ΔT = T_internal - T_subsoil = 20°C - 5°C = 15°C
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Area of room:
- Area = 9 m x 7 m = 63 m²
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Calculate heat loss:
- Q = 0.620 W/m²K x 63 m² x 15°C
- Q = 0.620 x 63 x 15 = 588.6 W
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Convert to kJ per year:
- Heating period: 12 hours per day for 40 weeks = 12 x 7 x 40 = 3360 hours = 12 x 3600 seconds
- Total seconds = 3360 x 3600 = 12,096,000 seconds
- Total energy loss (in kJ) = 588.6 W x 12,096,000 s = 7,121,424,000 J = 7,121,424 kJ
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Cost calculation:
- Cost of heating oils = 37350 kJ/litre.
- Litres required = Total Energy / Cost per litre = 7,121,424 kJ / 37350 kJ/litre = 190.6 litres
- At 85 pence per litre:
- Total Cost = 190.6 x 0.85 = £161.01
Step 3
Design of ground floor - Passive standard
Answer
For a passive house standard, the following elements must be included in the design of the ground floor:
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Floor Construction:
- 25mm fine screed on top
- 75mm concrete floor slab
- Use of expanded polystyrene insulation with a minimum thickness of 250mm under the slab.
- The hardcore should not be less than 100 mm in thickness.
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Radon Barrier:
- Installation of a radon-proof barrier beneath the insulation layer.
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Thermal Mass:
- Use materials that absorb and retain heat efficiently.
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Links to HVAC system:
- Ensure all systems account for ground floor insulation requirements.
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Breathability:
- The assembly must allow for moisture control to avoid condensation issues.
Include sketches or diagrams to show these components clearly.
The general goal is to ensure that the floor is built to minimize heat loss and meet any necessary ventilation guidelines.