/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 219 The \(700 \mathrm{~m}^{2}\) ceil... [FREE SOLUTION] | 91Ó°ÊÓ

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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

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

We are interested in steady state heat transfer analysis from a human forearm subjected to certain environmental conditions. For this purpose consider the forearm to be made up of muscle with thickness \(r_{m}\) with a skin/fat layer of thickness \(t_{s f}\) over it, as shown in the Figure P3-138. For simplicity approximate the forearm as a one-dimensional cylinder and ignore the presence of bones. The metabolic heat generation rate \(\left(\dot{e}_{m}\right)\) and perfusion rate \((\dot{p})\) are both constant throughout the muscle. The blood density and specific heat are \(\rho_{b}\) and \(c_{b}\), respectively. The core body temperate \(\left(T_{c}\right)\) and the arterial blood temperature \(\left(T_{a}\right)\) are both assumed to be the same and constant. The muscle and the skin/fat layer thermal conductivities are \(k_{m}\) and \(k_{s f}\), respectively. The skin has an emissivity of \(\varepsilon\) and the forearm is subjected to an air environment with a temperature of \(T_{\infty}\), a convection heat transfer coefficient of \(h_{\text {conv }}\), and a radiation heat transfer coefficient of \(h_{\mathrm{rad}}\). Assuming blood properties and thermal conductivities are all constant, \((a)\) write the bioheat transfer equation in radial coordinates. The boundary conditions for the forearm are specified constant temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and temperature symmetry at the centerline of the forearm. \((b)\) Solve the differential equation and apply the boundary conditions to develop an expression for the temperature distribution in the forearm. (c) Determine the temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and the maximum temperature in the forearm \(\left(T_{\max }\right)\) for the following conditions: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{m}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s} \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\mathrm{rad}}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

A spherical vessel, \(3.0 \mathrm{~m}\) in diameter (and negligible wall thickness), is used for storing a fluid at a temperature of \(0^{\circ} \mathrm{C}\). The vessel is covered with a \(5.0\)-cm-thick layer of an insulation \((k=0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The surrounding air is at \(22^{\circ} \mathrm{C}\). The inside and outside heat transfer coefficients are 40 and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate \((a)\) all thermal resistances, in \(\mathrm{K} / \mathrm{W},(b)\) the steady rate of heat transfer, and \((c)\) the temperature difference across the insulation layer.

Consider two finned surfaces that are identical except that the fins on the first surface are formed by casting or extrusion, whereas they are attached to the second surface afterwards by welding or tight fitting. For which case do you think the fins will provide greater enhancement in heat transfer? Explain.

Consider an insulated pipe exposed to the atmosphere. Will the critical radius of insulation be greater on calm days or on windy days? Why?

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