The external wall of a house of timber frame construction has the following specification: External layer Thickness 120 mm Concrete block outer leaf Thickness 100 mm Timber Strand Board (OSB) sheeting Thickness 18 mm Mineral wool insulation between studs Thickness 120 mm Plasterboard Thickness 12.5 mm Thermal data of outer leaf and cavity: Resistance of the external surface (R) = 0.0483 m²°C/W Resistivity of the concrete block Thickness (1): Conductivity of OSB sheeting (k) = 0.130 W/m°C Conductivity of mineral wool (k) = 0.040 W/m°C Conductivity of plasterboard (k) = 0.160 W/m°C Resistance of inner surface (R) = 0.104 m²°C/W Thermal data of inner leaf: Conductivity of OSB sheeting Conductivity of mineral wool Conductivity of plasterboard Resistance of the interior surface. Note: The inner studs of the inner leaf need not be considered in your calculations. (a) Calculate the U-value of the above external wall. (b) Calculate the cost of the heat lost annually through this wall, using the following data: Area of external wall = 150 m² Average internal temperature = 18 °C Average external temperature = 6 °C Heating period = 60 days each year, every day for 38 weeks each annum Calorific value of oil = 36,000 kJ per litre Cost of oil = 1.10 € per litre (c) It is proposed to upgrade the thermal properties of the above wall to meet the Passive House standard by fixing expanded polystyrene to the external surface. Given the thermal conductivity (k) of expanded polystyrene is 0.037 W/m°C, calculate the thickness of expanded polystyrene required to achieve a U-value of 0.15 W/m²°C.

Leaving Cert Construction Studies Heat & U-Value Calculations: The external wall of a house of

The external wall of a house of timber frame construction has the following specification: External layer Thickness 120 mm Concrete block outer leaf Thickness 100 mm Timber Strand Board (OSB) sheeting Thickness 18 mm Mineral wool insulation between studs Thickness 120 mm Plasterboard Thickness 12.5 mm Thermal data of outer leaf and cavity: Resistance of the external surface (R) = 0.0483 m²°C/W Resistivity of the concrete block Thickness (1): Conductivity of OSB sheeting (k) = 0.130 W/m°C Conductivity of mineral wool (k) = 0.040 W/m°C Conductivity of plasterboard (k) = 0.160 W/m°C Resistance of inner surface (R) = 0.104 m²°C/W Thermal data of inner leaf: Conductivity of OSB sheeting Conductivity of mineral wool Conductivity of plasterboard Resistance of the interior surface - Leaving Cert Construction Studies - Question 5 - 2016

Question 5

The external wall of a house of timber frame construction has the following specification: External layer Thickness 120 mm Concrete block outer leaf Thickness 100 m... show full transcript

Worked Solution & Example Answer:The external wall of a house of timber frame construction has the following specification: External layer Thickness 120 mm Concrete block outer leaf Thickness 100 mm Timber Strand Board (OSB) sheeting Thickness 18 mm Mineral wool insulation between studs Thickness 120 mm Plasterboard Thickness 12.5 mm Thermal data of outer leaf and cavity: Resistance of the external surface (R) = 0.0483 m²°C/W Resistivity of the concrete block Thickness (1): Conductivity of OSB sheeting (k) = 0.130 W/m°C Conductivity of mineral wool (k) = 0.040 W/m°C Conductivity of plasterboard (k) = 0.160 W/m°C Resistance of inner surface (R) = 0.104 m²°C/W Thermal data of inner leaf: Conductivity of OSB sheeting Conductivity of mineral wool Conductivity of plasterboard Resistance of the interior surface - Leaving Cert Construction Studies - Question 5 - 2016

Step 1

Calculate the U-value of the above external wall.

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Answer

To calculate the U-value, we first need to find the total resistance of the wall. The U-value is then calculated as the reciprocal of the total resistance.

Calculate the resistance of each layer:
- External layer: $R_{ext} = \frac{t}{k} = \frac{0.120 \text{ m}}{2.170 \text{ W/m°C}} = 0.0554 \text{ m²°C/W}$
- Concrete block outer leaf: $R_{block} = \frac{0.100 \text{ m}}{1.320 \text{ W/m°C}} = 0.0758 \text{ m²°C/W}$
- OSB sheeting: $R_{osb} = \frac{0.018 \text{ m}}{0.130 \text{ W/m°C}} = 0.1385 \text{ m²°C/W}$
- Mineral wool insulation: $R_{wool} = \frac{0.120 \text{ m}}{0.040 \text{ W/m°C}} = 3.0000 \text{ m²°C/W}$
- Plasterboard: $R_{plaster} = \frac{0.0125 \text{ m}}{0.160 \text{ W/m°C}} = 0.0781 \text{ m²°C/W}$
- Internal surface: $R_{internal} = 0.1040 ext{ m²°C/W}$
Sum the resistances: $R_{total} = R_{ext} + R_{block} + R_{osb} + R_{wool} + R_{plaster} + R_{internal}$ $R_{total} = 0.0554 + 0.0758 + 0.1385 + 3.0000 + 0.0781 + 0.1040 = 3.4530 ext{ m²°C/W}$
Calculate the U-value: $U = \frac{1}{R_{total}} = \frac{1}{3.4530} \approx 0.289 ext{ W/m²°C}$

Step 2

Calculate the cost of the heat lost annually through the wall.

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Answer

Formula to calculate the heat loss: $Q = U \cdot A \cdot \Delta T \cdot t$ Where:
- $U$ = U-value
- $A$ = Area
- $"){\Delta T}$ = Temperature difference
- $t$ = time (in hours)
Given:
- $A = 150 \text{ m²}$
- Internal temperature = 18 °C
- External temperature = 6 °C
- Heating period = 60 days = 1440 hours
- $\Delta T = 18 - 6 = 12\text{ °C}$
Substitute values into the formula: $Q = 0.289 \cdot 150 \cdot 12 \cdot 1440$ Calculate to find $Q$ : $Q \approx 6163680 \text{ W}$
To convert Watts to kilojoules:
- 1 W = 1 J/s = 3600 J/hr
- $Q_{kJ} = \frac{6163680}{1000} \cdot 3600 \approx 22209000 \text{ kJ}$
Cost calculation:
- Calorific value of oil = 36,000 kJ per litre
- $\text{Ch} = \frac{22209000}{36000} \approx 616.08 \text{ litres}$
- Cost = $1.10 \cdot 616.08 \approx €678.69$

Step 3

Find thickness of expanded polystyrene required to give U-value of 0.15 W/m²°C.

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Answer

Calculate required resistance for a U-value of 0.15: $R_{required} = \frac{1}{0.15} = 6.6667\text{ m²°C/W}$
Current total resistance = 3.4530 m²°C/W, calculate additional resistance needed: $R_{additional} = R_{required} - R_{total} = 6.6667 - 3.4530 = 3.2137 \text{ m²°C/W}$
Calculate thickness of expanded polystyrene, given conducting value k = 0.037: $R_{eps} = \frac{t_{eps}}{k} \implies t_{eps} = R_{eps} \cdot k = 3.2137 \cdot 0.037$ Rearranging gives: $t_{eps} = 0.1198 \text{ m} = 119.8 \text{ mm}$

Calculate the U-value of the above external wall.

Calculate the cost of the heat lost annually through the wall.

Find thickness of expanded polystyrene required to give U-value of 0.15 W/m²°C.

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