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The external wall of a house built in the 1970s is of single-leaf, hollow-block construction with an external render - Leaving Cert Construction Studies - Question 5 - 2017

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The external wall of a house built in the 1970s is of single-leaf, hollow-block construction with an external render. Plasterboard, with bonded expanded polystyrene,... show full transcript

Worked Solution & Example Answer:The external wall of a house built in the 1970s is of single-leaf, hollow-block construction with an external render - Leaving Cert Construction Studies - Question 5 - 2017

Step 1

Calculate the U-value of the external wall

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Answer

To calculate the U-value of the external wall, we need to find the total resistance of the wall components. The total resistance (R) is calculated as the sum of all resistances:

Rtotal=Rext+Rexternal+Rconcrete+Rpolystyrene+Rplasterboard+RintR_{total} = R_{ext} + R_{external} + R_{concrete} + R_{polystyrene} + R_{plasterboard} + R_{int}

Using the given resistances:

  • External surface resistance: 0.048 m²C/W
  • External render resistance: 0.034272.170=0.0158 m2C/W\frac{0.03427}{2.170} = 0.0158 \ m^2C/W
  • Concrete block resistance: 0.210 m²C/W
  • Expanded polystyrene resistance: 0.0370.037=1.000 m2C/W\frac{0.037}{0.037} = 1.000 \ m^2C/W
  • Plasterboard resistance: 0.01250.160=0.078125 m2C/W\frac{0.0125}{0.160} = 0.078125 \ m^2C/W
  • Internal surface resistance: 0.104 m²C/W

Calculating: Rtotal=0.048+0.0158+0.210+1.0+0.078125+0.104=1.455925 m2C/WR_{total} = 0.048 + 0.0158 + 0.210 + 1.0 + 0.078125 + 0.104 = 1.455925 \ m^2C/W

U-value is the reciprocal of total resistance: U=1Rtotal=11.455925=0.686W/m2CU = \frac{1}{R_{total}} = \frac{1}{1.455925} = 0.686 W/m²C

Step 2

Calculate the thickness of expanded polystyrene required

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Answer

To meet the U-value of 0.15 W/m²C, we need to calculate the resistance required:

Rrequired=10.15=6.6666 m2C/WR_{required} = \frac{1}{0.15} = 6.6666 \ m^2C/W

Using the formula for total resistance to solve for the required thickness of expanded polystyrene: Rtotal=Rexisting+RpolystyreneR_{total} = R_{existing} + R_{polystyrene}

Where:

  • Rexisting=1.151 m2C/WR_{existing} = 1.151 \ m^2C/W
  • Difference in resistance is: 6.6666−1.151=5.5156 m2C/W6.6666 - 1.151 = 5.5156 \ m^2C/W

The thickness of expanded polystyrene: T=Rpolystyrene∗Resistivitypolystyrene=5.5156∗0.037=204.08mT = R_{polystyrene} * Resistivity_{polystyrene} = 5.5156 * 0.037 = 204.08 m (approximately 205 mm accepted)

Step 3

Calculate the cost of heat lost annually

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Answer

Using the formula for heat loss: Heat Loss=U×A×(Tinside−Toutside)\text{Heat Loss} = U \times A \times (T_{inside} - T_{outside})

Where:

  • Area (A) = 140 m²
  • Temperature difference = 18 °C - 6 °C = 12 °C
  • U-value = 0.15 W/m²C

Calculating heat loss: Heat Loss=0.15×140×12=252KJ/s\text{Heat Loss} = 0.15 \times 140 \times 12 = 252 KJ/s

Total heating period:

  • 60 hours/week for 36 weeks: 60×36=216060 \times 36 = 2160 hours
  • Converted to seconds: 2160×3600=7,776,0002160 \times 3600 = 7,776,000 seconds

Total heat loss = 252KJ/s×7,776,000s=1,962,672,000KJ252 KJ/s \times 7,776,000 s = 1,962,672,000 KJ

Cost Calculation: Cost of oil (98 cent/litre) with calorific value at 37350 KJ/litre:

  • Total litres used: rac{1,962,672,000}{37350} \approx 52455.8 litres
  • Total cost = 52,455.8×0.98€≈51,406.29€52,455.8 × 0.98€ ≈ 51,406.29€

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