Question

Using the following values of thermal conductance, surface conductance, and thermal resistance, the U value across the given wall cross-section is \(\underline{\hspace{1cm}}\) W/m\(^2\)°C (round off to 2 decimal places). 

Hint:

The U-value is the reciprocal of the total thermal resistance, which accounts for conductance and resistances of each layer.

Answer 1

Correct Answer: 3.66

Given values: - Thermal conductance:
- Brick wall = \(1.2\ \text{W/m°C}\)
- Plastering = \(0.5\ \text{W/m°C}\)
- Surface conductance:
- Internal surface = \(8.0\ \text{W/m}^2\text{°C}\)
- External surface = \(9.5\ \text{W/m}^2\text{°C}\)
- Thermal resistance:
- 50 mm cavity = \(0.17\ \text{m}^2\text{°C/W}\)
First, calculate the thermal resistance for each layer of the wall: For the plaster layer (10 mm): \[ R_{\text{plaster}} = \frac{\text{thickness}}{\text{conductance}} = \frac{0.01}{0.5} = 0.02\ \text{m}^2\text{°C/W} \] For the brickwork layer (100 mm): \[ R_{\text{brick}} = \frac{0.1}{1.2} = 0.0833\ \text{m}^2\text{°C/W} \] For the cavity (50 mm): \[ R_{\text{cavity}} = 0.17\ \text{m}^2\text{°C/W} \] Now, calculate the total thermal resistance for the wall: \[ R_{\text{total}} = R_{\text{plaster}} + R_{\text{brick}} + R_{\text{cavity}} = 0.02 + 0.0833 + 0.17 = 0.2733\ \text{m}^2\text{°C/W} \] Next, calculate the overall U-value (which is the reciprocal of the total thermal resistance): \[ U = \frac{1}{R_{\text{total}}} = \frac{1}{0.2733} = 3.66\ \text{W/m}^2\text{°C} \] Thus, the U-value across the given wall cross-section is \( \boxed{3.66} \ \text{W/m}^2\text{°C} \).