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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

Expert verified
Answer: 40.4 kW

Step by step solution

01

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: \(T_{i}=20^{\circ}\mathrm{C}\)
02

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: \(\Delta T = T_i - T_a\). \(\Delta T = 20 ^{\circ}\mathrm{C} - (-10^{\circ}\mathrm{C}) = 30^{\circ}\mathrm{C}\)
03

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: \(q = \frac{30\mathrm{K}*700\mathrm{~m}^2}{0.52\mathrm{~m}^2\mathrm{K/W}}\)
04

Calculate the final value

Now, calculate the value of the heat transfer rate: \(q = \frac{21000\mathrm{~m}^2\mathrm{K}}{0.52\mathrm{~m}^2\mathrm{K/W}}= 40384.62 \mathrm{W}\)
05

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).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Two 5-cm-diameter, 15-cm-long aluminum bars \((k=\) \(176 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with ground surfaces are pressed against each other with a pressure of \(20 \mathrm{~atm}\). The bars are enclosed in an insulation sleeve and, thus, heat transfer from the lateral surfaces is negligible. If the top and bottom surfaces of the twobar system are maintained at temperatures of \(150^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, determine \((a)\) the rate of heat transfer along the cylinders under steady conditions and (b) the temperature drop at the interface. Answers: (a) \(142.4 \mathrm{~W}\), (b) \(6.4^{\circ} \mathrm{C}\)

In the United States, building insulation is specified by the \(R\)-value (thermal resistance in \(\mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu units). A homeowner decides to save on the cost of heating the home by adding additional insulation in the attic. If the total \(R\)-value is increased from 15 to 25 , the homeowner can expect the heat loss through the ceiling to be reduced by (a) \(25 \%\) (b) \(40 \%\) (c) \(50 \%\) (d) \(60 \%\) (e) \(75 \%\)

A 4-mm-diameter and 10-cm-long aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the percent error in the rate of heat transfer from the fin when the infinitely long fin assumption is used instead of the adiabatic fin tip assumption.

Circular fins of uniform cross section, with diameter of \(10 \mathrm{~mm}\) and length of \(50 \mathrm{~mm}\), are attached to a wall with surface temperature of \(350^{\circ} \mathrm{C}\). The fins are made of material with thermal conductivity of \(240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and they are exposed to an ambient air condition of \(25^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the heat transfer rate and plot the temperature variation of a single fin for the following boundary conditions: (a) Infinitely long fin (b) Adiabatic fin tip (c) Fin with tip temperature of \(250^{\circ} \mathrm{C}\) (d) Convection from the fin tip

A 1-cm-diameter, 30-cm-long fin made of aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface at \(80^{\circ} \mathrm{C}\). The surface is exposed to ambient air at \(22^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the fin can be assumed to be very long, the rate of heat transfer from the fin is (a) \(2.2 \mathrm{~W}\) (b) \(3 \mathrm{~W}\) (c) \(3.7 \mathrm{~W}\) (d) \(4 \mathrm{~W}\) (e) \(4.7 \mathrm{~W}\)

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