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?

Step by step solution

01

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: \(A_{roof} = 12 \mathrm{~m} \times 12 \mathrm{~m} = 144 \mathrm{~m^2}\)

02

Calculate the temperature difference

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

03

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: $$Q_{total \: without} = 4 \cdot Q_{wall} + Q_{roof} = 4 \cdot 3240 \mathrm{W} + 6480 \mathrm{W} = 19440 \mathrm{W}$$

04

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: \(A'_{roof} = 144 \mathrm{~m^2} \times 1.04 = 149.76 \mathrm{~m^2}\)

05

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: $$Q_{total \: with} = 4 \cdot Q'_{wall} + Q'_{roof} = 4 \cdot 3375.8 \mathrm{W} + 6735.6 \mathrm{W} = 20219 \mathrm{W}$$

06

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.

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