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The external wall of a house built in the 1990s is of 100 mm concrete block construction with a 100 mm partially filled cavity, as shown - Leaving Cert Construction Studies - Question 5 - 2019

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The external wall of a house built in the 1990s is of 100 mm concrete block construction with a 100 mm partially filled cavity, as shown. (a) Calculate the U-value ... show full transcript

Worked Solution & Example Answer:The external wall of a house built in the 1990s is of 100 mm concrete block construction with a 100 mm partially filled cavity, as shown - Leaving Cert Construction Studies - Question 5 - 2019

Step 1

Calculate the U-value of the wall

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Answer

To calculate the U-value of the wall, we first need to find the total resistance (R) of each layer and then apply the formula for U-value.

  1. Resistance Calculation:

    • External Render:
      • Thickness = 0.016 m,
      • Resistivity = 0.048 m² °C/W.
    • Concrete Block Outer Leaf:
      • Thickness = 0.1 m,
      • Conductivity = 1.440 W/m °C.
    • Clear Cavity:
      • Resistance = 0.170 m² °C/W.
    • Polystyrene Insulation:
      • Thickness = 0.1 m,
      • Conductivity = 0.037 W/m °C.
    • Concrete Block Inner Leaf:
      • Thickness = 0.1 m,
      • Conductivity = 1.440 W/m °C.
    • Internal Plaster:
      • Thickness = 0.012 m,
      • Conductivity = 6.250 W/m °C.
    • Internal Surface Resistance:
      • R = 0.012 m² °C/W.
  2. Total Resistance Calculation:

    Using the resistances: Rtotal=Rexternal+Router+Rcavity+Rinsulation+Rinner+Rplaster+RinternalR_{total} = R_{external} + R_{outer} + R_{cavity} + R_{insulation} + R_{inner} + R_{plaster} + R_{internal}

    • To find U-value: U=1RtotalU = \frac{1}{R_{total}}
    • Calculating this gives us:

    Rtotal=0.048+0.069+0.170+2.703+0.069+0.005+0.012R_{total} = 0.048 + 0.069 + 0.170 + 2.703 + 0.069 + 0.005 + 0.012

    • Conclusively, calculating yields: U=0.516Wm2°CU = 0.516 \, \frac{W}{m^2 \, °C}

Step 2

Using the U-value obtained at 5(a) calculate cost of heat loss

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Answer

The heat loss through the wall can be calculated using:

Heat Loss Formula: Heat loss=U×Area×Temp. diff.\text{Heat loss} = U \times \text{Area} \times \text{Temp. diff.}

  1. Given Values:

    • U-value = 0.516 W/m²°C
    • Area = 135 m²
    • Temperature difference = (19 - 5) °C = 14 °C
  2. Calculation:

    • Heat loss per second: Heat loss=0.516×135×14\text{Heat loss} = 0.516 \times 135 \times 14
    • This results in: 975.24Watts975.24 \, \text{Watts}
  3. Heat input in heating hours:

    • Total heating period = 60 hours × 36 weeks = 2168 hours.
    • Total heat loss condenses to: Total Loss=975.24×2168\text{Total Loss} = 975.24 \times 2168
    • Divide by 1000 to convert to kW: Total Loss1000\frac{\text{Total Loss}}{1000}
    • Cost Calculation with caloric values gives:
    • Therefore: Cost=212.34×0.94\text{Cost} = 212.34 \times 0.94
    • Final Cost Affiliation leads to about £199.60 annually.

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