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Calculate the U-value of an uninsulated external solid concrete wall of a dwelling house built in the 1950s given the following data: External render thickness 16 mm Solid concrete wall thickness 225 mm Internal plaster thickness 13 mm Thermal data of external wall of house: - Resistivity of the solid concrete wall: (r) 1.190 m * °C / W - Resistivity of external render: (r) 2.170 m * °C / W - Resistivity of internal plaster: (r) 6.250 m * °C / W - Resistance of external surface: (R) 0.048 m² * °C / W - Resistance of internal surface: (R) 0.122 m² * °C / W (b) Phenolic foam insulation is to be fitted to the external surface of the solid concrete wall - Leaving Cert Construction Studies - Question 5 - 2010

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Question 5

Calculate-the-U-value-of-an-uninsulated-external-solid-concrete-wall-of-a-dwelling-house-built-in-the-1950s-given-the-following-data:--External-render----------------thickness-16-mm-Solid-concrete-wall------thickness-225-mm-Internal-plaster----------------thickness-13-mm--Thermal-data-of-external-wall-of-house:---Resistivity-of-the-solid-concrete-wall:-(r)-1.190-m-*-°C-/-W---Resistivity-of-external-render:-(r)-2.170-m-*-°C-/-W---Resistivity-of-internal-plaster:-(r)-6.250-m-*-°C-/-W---Resistance-of-external-surface:-(R)-0.048-m²-*-°C-/-W---Resistance-of-internal-surface:-(R)-0.122-m²-*-°C-/-W--(b)-Phenolic-foam-insulation-is-to-be-fitted-to-the-external-surface-of-the-solid-concrete-wall-Leaving Cert Construction Studies-Question 5-2010.png

Calculate the U-value of an uninsulated external solid concrete wall of a dwelling house built in the 1950s given the following data: External render ... show full transcript

Worked Solution & Example Answer:Calculate the U-value of an uninsulated external solid concrete wall of a dwelling house built in the 1950s given the following data: External render thickness 16 mm Solid concrete wall thickness 225 mm Internal plaster thickness 13 mm Thermal data of external wall of house: - Resistivity of the solid concrete wall: (r) 1.190 m * °C / W - Resistivity of external render: (r) 2.170 m * °C / W - Resistivity of internal plaster: (r) 6.250 m * °C / W - Resistance of external surface: (R) 0.048 m² * °C / W - Resistance of internal surface: (R) 0.122 m² * °C / W (b) Phenolic foam insulation is to be fitted to the external surface of the solid concrete wall - Leaving Cert Construction Studies - Question 5 - 2010

Step 1

Calculate the U-value of the wall

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Answer

To determine the U-value, we first need to calculate the total resistance (R_total) of the wall system:

Rtotal=Rexternalrender+Rsolidconcretewall+Rinternalplaster+Rexternalsurface+RinternalsurfaceR_{total} = R_{external\,render} + R_{solid\,concrete\,wall} + R_{internal\,plaster} + R_{external\,surface} + R_{internal\,surface}

Where:

  • Thickness of external render = 0.016 m, Resistivity = 2.170 m * °C / W, hence Rexternalrender=0.0162.170=0.00737m2°C/WR_{external\,render} = \frac{0.016}{2.170} = 0.00737\, m²\, °C/W

  • Thickness of solid concrete wall = 0.225 m, Resistivity = 1.190 m * °C / W, thus Rsolidconcretewall=0.2251.190=0.18908m2°C/WR_{solid\,concrete\,wall} = \frac{0.225}{1.190} = 0.18908\, m²\, °C/W

  • Thickness of internal plaster = 0.013 m, Resistivity = 6.250 m * °C / W, so Rinternalplaster=0.0136.250=0.00208m2°C/WR_{internal\,plaster} = \frac{0.013}{6.250} = 0.00208\, m²\, °C/W

  • Resistance of external surface = 0.048 m² * °C/W and internal surface = 0.122 m² * °C/W are provided.

Now substitute to find R_total:

Rtotal=0.00737+0.18908+0.00208+0.048+0.122=0.36853m2°C/WR_{total} = 0.00737 + 0.18908 + 0.00208 + 0.048 + 0.122 = 0.36853\, m²\, °C/W

Finally, the U-value (U) can be calculated as:

U=1Rtotal=10.368532.71W/m2°CU = \frac{1}{R_{total}} = \frac{1}{0.36853} \approx 2.71\, W/m²\, °C

Step 2

Calculate the thickness of phenolic foam insulation

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Answer

Given the desired U-value of 0.27 W/m² °C:

To find the required resistance (R_required) for this U-value:

Rrequired=1U=10.273.70m2°C/WR_{required} = \frac{1}{U} = \frac{1}{0.27} \approx 3.70\, m²\, °C/W

Now calculate the difference in resistance needed:

DifferenceinResistance=RrequiredRexisting=3.700.36853=3.33m2°C/WDifference\,in\,Resistance = R_{required} - R_{existing} = 3.70 - 0.36853 = 3.33\, m²\, °C/W

Using the formula for the resistance of insulation:

R=Tk,R = \frac{T}{k}, where k = 0.025 W/m °C.

Let T be the thickness of the phenolic foam:

R=T0.025R = \frac{T}{0.025}

Setting it equal to the Required Resistance:

3.33=T0.0253.33 = \frac{T}{0.025}

Thus,

T=3.33×0.0250.08325m=83.25mmT = 3.33 \times 0.025 \approx 0.08325\, m = 83.25\, mm

So, the thickness of phenolic foam insulation required is approximately 83.25 mm.

Step 3

Discuss the importance of thermal mass

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Answer

Thermal mass is crucial in improving the thermal performance of a dwelling house. It refers to the ability of a material to absorb, store, and later release heat.

Key Points:

  • Energy Efficiency: Materials with high thermal mass, like concrete and brick, can reduce energy consumption by moderating indoor temperatures. They absorb heat during the day and release it at night, minimizing the need for heating or cooling.
  • Thermal Comfort: A well-designed thermal mass structure enhances comfort by maintaining stable indoor temperatures, avoiding peaks of heat or cold.
  • Passive Solar Gain: Thermal mass can store excess heat from sunlight during the day, contributing positively to the thermal performance, especially in cooler climates.

Freehand Sketches:

  1. Example of a building with high thermal mass showing sun exposure and heat storage.
  2. Illustration of heat flow in different seasons, demonstrating how thermal mass works in regulating indoor climate.

In summary, optimizing the use of thermal mass in residential buildings leads to energy efficiency and improved comfort, thus enhancing overall sustainability.

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