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A house built to nZEB standards has a highly insulated solid concrete ground floor with a hardwood finish, as shown - Leaving Cert Construction Studies - Question 5 - 2020

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A house built to nZEB standards has a highly insulated solid concrete ground floor with a hardwood finish, as shown. (a) Calculate the U-value of the floor, given t... show full transcript

Worked Solution & Example Answer:A house built to nZEB standards has a highly insulated solid concrete ground floor with a hardwood finish, as shown - Leaving Cert Construction Studies - Question 5 - 2020

Step 1

Calculate the U-value of the floor

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Answer

To determine the U-value, we need to calculate the total resistance (R) of all the components and then apply the formula: U=1RtotalU = \frac{1}{R_{total}}

  1. Calculate individual resistances for each layer:

    • Hardwood flooring:

      R1=thicknessk=0.0206.666=0.003R_{1} = \frac{thickness}{k} = \frac{0.020}{6.666} = 0.003(m² °C/W)

    • Sand/cement fine screed:

      R2=0.0651.410=0.046R_{2} = \frac{0.065}{1.410} = 0.046

    • Floor insulation:

      R3=0.2000.022=9.091R_{3} = \frac{0.200}{0.022} = 9.091

    • Concrete floor slab:

      R4=0.1501.280=0.117R_{4} = \frac{0.150}{1.280} = 0.117

    • Radon Barrier:

      R5=0.000250.250=0.001R_{5} = \frac{0.00025}{0.250} = 0.001(m² °C/W)

    • Sand blinding:

      R6=0.0450.160=0.281R_{6} = \frac{0.045}{0.160} = 0.281(m² °C/W)

    • Hardcore:

      R7=0.2001.350=0.148R_{7} = \frac{0.200}{1.350} = 0.148(m² °C/W)

    • Subsoil:

      R8=0.3001.600=0.188R_{8} = \frac{0.300}{1.600} = 0.188(m² °C/W)

  2. Summing up all resistances,

    • Total R = R1 + R2 + R3 + R4 + R5 + R6 + R7 + R8 = 0.003 + 0.046 + 9.091 + 0.117 + 0.001 + 0.281 + 0.148 + 0.188 = 9.877 m² °C/W.

  3. Finally, calculate U-value:
    U=1Rtotal=19.993=0.100 m2/W°CU = \frac{1}{R_{total}} = \frac{1}{9.993} = 0.100 \ m²/W°C
    Hence, the U-value of the floor is 0.10 m²/W°C.

Step 2

Using the U-value of the floor obtained at 5(a), calculate the cost of heat lost annually through the floor

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Answer

We use the formula for heat loss: Heat lost=U×Area×(Tinternal−Texternal)\text{Heat lost} = U \times Area \times (T_{internal} - T_{external})(W) Where:

  • Area of floor slab = 6.5 m x 9.0 m = 58.5 m².
  • Temperature difference = 20°C - 6°C = 14°C.

  1. Calculate heat loss per hour: Q=0.10×58.5×14=81.9 WattsQ = 0.10 \times 58.5 \times 14 = 81.9 \text{ Watts}.

  2. Calculating for yearly heat loss:

    • Heating period = 39 weeks. Days = 7, Hours in a day = 24
    • Total hours = 39 weeks x 7 days/week x 24 hours/day = 6552 hours.
    • Total Heat loss = 81.9 Watts x 6552 hours = 536,653.8
    • Cost of energy: extCost=536,653.81000×0.96=€18.62 ext{Cost} = \frac{536,653.8}{1000} \times 0.96 = €18.62.

Step 3

Using notes and a freehand sketch, show best practice detailing that will prevent the formation of a thermal bridge

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Answer

To prevent thermal bridging at the junction of the concrete floor and external wall, consider the following best practices:

  1. Use an insulation layer that extends underneath the wall to eliminate direct contact between the cold wall and the warm concrete.
  2. Ensure that there is a continuous layer of insulation without gaps or thermal breaks.
  3. Incorporate a damp proof membrane to protect the insulation from moisture ingress.
  4. Additional measures may include using insulated foundation blocks and ensuring that thermal insulation wraps around the slab edge.

The sketch should visually demonstrate the layers of insulation, the positioning of the damp proof membrane, and any necessary junction details to achieve the recommended best practices.

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