Question

The \(700 \mathrm{~m}^{2}\) ceiling of a building has a thermal resistance of \(0.52 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The rate at which heat is lost through this ceiling on a cold winter day when the ambient temperature is \(-10^{\circ} \mathrm{C}\) and the interior is at \(20^{\circ} \mathrm{C}\) is (a) \(23.1 \mathrm{~kW} \quad\) (b) \(40.4 \mathrm{~kW}\) (c) \(55.6 \mathrm{~kW}\) (d) \(68.1 \mathrm{~kW}\) (e) \(88.6 \mathrm{~kW}\)

Short answer

Answer: 40.4 kW

Step-by-step solution

Identify the necessary information

We need the following information from the problem: 1. Ceiling area: \(A=700\mathrm{~m}^2\) 2. Thermal resistance: \(R=0.52\mathrm{~m^2}\cdot\mathrm{K/W}\) 3. Ambient temperature: \(T_{a}=-10^{\circ}\mathrm{C}\) 4. Interior temperature: \(T_{i}=20^{\circ}\mathrm{C}\)

Compute the temperature difference

To find the difference in temperature between the interior and ambient temperatures, we subtract the ambient temperature from the interior temperature: \(\Delta T = T_i - T_a\). \(\Delta T = 20 ^{\circ}\mathrm{C} - (-10^{\circ}\mathrm{C}) = 30^{\circ}\mathrm{C}\)

Compute the heat transfer rate

Using the formula for heat transfer rate, \(q = \frac{\Delta T * A}{R}\), we can compute the heat transfer rate q: \(q = \frac{30\mathrm{K}*700\mathrm{~m}^2}{0.52\mathrm{~m}^2\mathrm{K/W}}\)

Calculate the final value

Now, calculate the value of the heat transfer rate: \(q = \frac{21000\mathrm{~m}^2\mathrm{K}}{0.52\mathrm{~m}^2\mathrm{K/W}}= 40384.62 \mathrm{W}\)

Convert the heat transfer rate to kW

Finally, convert the heat transfer rate from Watts to kilowatts and round the result: \(q = 40.38462 \mathrm{kW} \approx 40.4 \mathrm{kW}\) The rate at which heat is lost through the ceiling is approximately \(40.4\mathrm{kW}\), which corresponds to option (b).

Key concepts

Thermal Resistance

In the realm of building physics, thermal resistance is a critical concept that governs how effectively heat is retained or released in structures. It is, in essence, a measure of a material's ability to resist the flow of heat. Higher thermal resistance means better insulation, as the material opposes the movement of heat more significantly. In our exercise, the ceiling's thermal resistance was given as 0.52 . This number signifies the ceiling's ability to insulate the interior from external temperatures. When calculating heat transfer, a high thermal resistance indicates less heat will be lost through the surface, making the building more energy-efficient. Thermal resistance is key when selecting materials for building insulation, as the goal is often to minimize energy costs while maintaining comfortable indoor temperatures.

Temperature Difference

The driving force behind heat transfer in buildings is the temperature difference between the inside and the outside. This temperature difference, noted as , is what motivates heat to flow from a region of higher temperature to a region of lower temperature. In our problem, the inside of the building was a cozy , while the cold outside air was a chilly . The resulting temperature difference of is what causes the heat to flow outwards. Heat transfer rate calculations require this differential as a fundamental input, determining how quickly heat will pass through a building envelope like the ceiling in question. This concept is essential in energy conservation practices as it helps assess the energy needed to maintain thermal comfort within a space.

Heat Loss

Heat loss is an inescapable phenomenon in any structure, where heat escapes from warmer to cooler areas. In the context of our exercise, heat loss through the building's ceiling was calculated to understand how much energy is being expended to maintain indoor temperatures. The rate of heat loss is influenced by various factors, including the materials' thermal resistance, the area of the surface, and the temperature difference across it. Our calculation shows the importance of these factors, as they enable us to quantify the energy loss in terms of kilowatts. With the calculated heat loss rate of approximately , we get a clearer picture of the building's energy efficiency, helping us make informed decisions about heating requirements and potential improvements to insulation.

Building Thermal Insulation

Building thermal insulation is the application of materials designed to significantly slow down the transfer of heat between the interior and exterior of a building. Good insulation can make a building more comfortable and drastically reduce energy consumption by maintaining a stable indoor temperature regardless of external fluctuations. In our ceiling example, the thermal resistance provided by the insulation materials directly impacts the calculated heat loss rate. The goal when selecting building thermal insulation is to achieve a balance between adequate thermal resistance and cost-effectiveness. Insulation is often the first detail examined when aiming to increase energy efficiency, as it plays a pivotal role in minimizing unnecessary heating or cooling expenses. Understanding how insulation works in tandem with factors like temperature difference and surface area is fundamental for both designing new buildings and upgrading existing structures.