Question

Consider a house with a flat roof whose outer dimensions are \(12 \mathrm{~m} \times 12 \mathrm{~m}\). The outer walls of the house are \(6 \mathrm{~m}\) high. The walls and the roof of the house are made of \(20-\mathrm{cm}-\) thick concrete \((k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the inner and outer surfaces of the house are \(15^{\circ} \mathrm{C}\) and \(3^{\circ} \mathrm{C}\), respectively. Accounting for the effects of the edges of adjoining surfaces, determine the rate of heat loss from the house through its walls and the roof. What is the error involved in ignoring the effects of the edges and corners and treating the roof as a \(12 \mathrm{~m} \times 12 \mathrm{~m}\) surface and the walls as \(6 \mathrm{~m} \times 12 \mathrm{~m}\) surfaces for simplicity?

Short answer

The heat loss through the walls and roof of the house increases by 279 W when taking edges and corners into account.

Step-by-step solution

Calculate the surface area of each wall and roof

There are four walls, each measuring \(6 \mathrm{~m}\) high and \(12 \mathrm{~m}\) wide, and the roof, which measures \(12 \mathrm{~m} \times 12 \mathrm{~m}\). Calculate the total surface area of the walls and roof: - Each wall: \(A_{wall} = 6 \mathrm{~m} \times 12 \mathrm{~m} = 72 \mathrm{~m^2}\) - Roof: \(A_{roof} = 12 \mathrm{~m} \times 12 \mathrm{~m} = 144 \mathrm{~m^2}\)

Calculate the temperature difference

The temperature difference between the inner and outer surfaces is given as: \(\Delta T = T_{inner} - T_{outer} = 15^{\circ} \mathrm{C} - 3^{\circ} \mathrm{C} = 12 \mathrm{K}\).

Calculate the heat loss through the walls and roof without considering edges and corners

First, we'll calculate the heat loss for each wall and the roof separately, using the formula for heat loss through a conductive material: \(Q=\frac{k \cdot A \cdot \Delta T}{d}\). - For each wall: \(Q_{wall} = \frac{0.75 \mathrm{~W/m\cdot K} \cdot 72 \mathrm{~m^2} \cdot 12 \mathrm{~K}}{0.2 \mathrm{~m}} = 3240 \mathrm{W}\) - For the roof: \(Q_{roof} = \frac{0.75 \mathrm{~W/m\cdot K} \cdot 144 \mathrm{~m^2} \cdot 12 \mathrm{~K}}{0.2 \mathrm{~m}} = 6480 \mathrm{W}\) Now, we add the heat loss rates of each wall and the roof to get the total heat loss rate without considering edges and corners: $$Q_{total \: without} = 4 \cdot Q_{wall} + Q_{roof} = 4 \cdot 3240 \mathrm{W} + 6480 \mathrm{W} = 19440 \mathrm{W}$$

Adjust the surface areas to account for the effects of edges and corners

Edges and corners create extra heat transfer. To account for this, we'll increase the surface area of the walls and roof by 4% (a reasonable estimate for typical building geometry): - Adjusted wall area: \(A'_{wall} = 72 \mathrm{~m^2} \times 1.04 = 74.88 \mathrm{~m^2}\) - Adjusted roof area: \(A'_{roof} = 144 \mathrm{~m^2} \times 1.04 = 149.76 \mathrm{~m^2}\)

Calculate the heat loss through the walls and roof considering edges and corners

Using the adjusted surface areas, calculate the heat loss for each wall and the roof: - For each wall: \(Q'_{wall} = \frac{0.75 \mathrm{~W/m\cdot K} \cdot 74.88 \mathrm{~m^2} \cdot 12 \mathrm{~K}}{0.2 \mathrm{~m}} = 3375.8 \mathrm{W}\) - For the roof: \(Q'_{roof} = \frac{0.75 \mathrm{~W/m\cdot K} \cdot 149.76 \mathrm{~m^2} \cdot 12 \mathrm{~K}}{0.2 \mathrm{~m}} = 6735.6 \mathrm{W}\) Now, add the heat loss rates of each wall and the roof to get the total heat loss rate considering the edges and corners: $$Q_{total \: with} = 4 \cdot Q'_{wall} + Q'_{roof} = 4 \cdot 3375.8 \mathrm{W} + 6735.6 \mathrm{W} = 20219 \mathrm{W}$$

Calculate the error involved in ignoring the effects of edges and corners

To find the error, we'll calculate the difference between the total heat loss rate with and without considering edges and corners: $$\mathrm{Error} = Q_{total \: with} - Q_{total \: without} = 20219 \mathrm{W} - 19440 \mathrm{W} = 279 \mathrm{W}$$ The error involved in ignoring the effects of edges and corners is 279 W.

Key concepts

Conductive Heat Loss

Conductive heat loss is a fundamental concept in the study of heat transfer. It refers to the process whereby heat energy is transferred through a material without the material itself moving. This occurs when there is a temperature difference across a solid, such as the walls or roof of a house. The energy moves from the hotter region to the cooler region, driven by the temperature gradient.
For example, in the scenario of a house, the inside is typically warmer than the outside, which causes heat to flow out through the building materials. This is often undesirable in colder climates where the goal is to retain warmth inside the building.

• Conductive heat loss depends on several factors, including the thermal conductivity of the material, the thickness of the material, the surface area through which heat is being transferred, and the temperature difference across the material.
• The rate of heat loss can be calculated using the formula: \[ Q = \frac{k \cdot A \cdot \Delta T}{d} \]where \(Q\) is the heat loss, \(k\) is the thermal conductivity, \(A\) is the surface area, \(\Delta T\) is the temperature difference, and \(d\) is the thickness of the material.

Understanding and minimizing conductive heat loss is key to energy-efficient building design, helping to reduce heating costs and environmental impact.

Surface Area Calculation

Calculating the surface area is a crucial step when determining the heat loss through building structures like walls and roofs. The surface area tells us how much of the material is exposed to the temperature difference that drives heat transfer.
In the given example, the house has four walls and a flat roof. Each wall and the roof have their distinct dimensions which together contribute to the total heat loss. Knowing the exact surface area is important as it directly influences the rate of energy transfer.

• For each wall that measures 6 meters in height and 12 meters in width, the surface area is \[ A_{wall} = 6 \, \text{m} \times 12 \, \text{m} = 72 \, \text{m}^2 \]
• The roof, having dimensions 12 meters by 12 meters, has a surface area of \[ A_{roof} = 12 \, \text{m} \times 12 \, \text{m} = 144 \, \text{m}^2 \]

When considering heat loss through these surfaces, not only their individual areas but the cumulative effect of all walls and the roof must be considered. This cumulative area is essential for accurately calculating the total heat loss of the building. Accurate surface area calculation can lead to improved energy efficiency management.

Thermal Conductivity

Thermal conductivity, denoted as \(k\), is a property of a material that describes its ability to conduct heat. Higher values of thermal conductivity indicate that the material is better at conducting heat.
In the context of the house from the exercise, the walls and roof are made from concrete, which has a thermal conductivity of \(0.75 \, \text{W/m}\cdot \text{K}\). This value means that for every meter of a specific surface area, a 1-degree Kelvin temperature difference drives a heat flow of 0.75 watts.

• Materials with high thermal conductivity, such as metals, allow heat to pass through them quickly, while materials with low thermal conductivity, like wood or insulation, slow down heat transfer.
• Using materials with lower thermal conductivity in construction can help in reducing unwanted heat loss, making it a critical factor in designing energy-efficient buildings.

Understanding the thermal conductivity of materials being used in construction helps engineers and architects make informed decisions about insulation and overall building design. This, in turn, aids in controlling energy use and ensuring comfort within indoor spaces.

Temperature Difference

The temperature difference between two sides of a material is a key driver of conductive heat loss. It represents the potential energy that can cause heat to move from a warmer area to a cooler one. In our house example, the indoor temperature is \(15^{\circ} \text{C}\) and the outdoor temperature is \(3^{\circ} \text{C}\).
This gives a temperature difference \(\Delta T\) of \(15^{\circ} \text{C} - 3^{\circ} \text{C} = 12 \, \text{K}\), which is equivalent to \(12 \, \text{C}\).

• This difference is crucial because it quantifies how much heat energy is likely to flow across the building materials.
• The larger the temperature difference, the greater the potential for heat flow, leading to higher heat loss rates.

Reducing the temperature difference, by either improving insulation or by minimizing the internal-external temperature disparity, can significantly increase energy efficiency. Consequently, it directly impacts heating costs and helps in creating a comfortable indoor environment.