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Chapter 1: Problem 53

The inner and outer surfaces of a 4-m \(\times 7-\mathrm{m}\) brick wall of thickness \(30 \mathrm{~cm}\) and thermal conductivity \(0.69 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) are maintained at temperatures of \(26^{\circ} \mathrm{C}\) and \(8^{\circ} \mathrm{C}\), respectively. Determine the rate of heat transfer through the wall, in W.

Short Answer

Expert verified
Answer: The rate of heat transfer through the wall is approximately 9476 W.

Step by step solution

01

Identify given values and constants

Given values are: - Wall dimensions: 4 m x 7 m - Wall thickness: 30 cm - Thermal conductivity: \(0.69 \frac{\text{W}}{\text{m} \cdot \text{K}}\) - Inner surface temperature \(T_1\): \(26^{\circ}\text{C}\) - Outer surface temperature \(T_2\): \(8^{\circ}\text{C}\)
02

Use Fourier's law of heat conduction

The rate of heat transfer through a plane wall can be determined through Fourier's law of heat conduction: $$ Q = kA\frac{T_1-T_2}{L} $$ where \(Q\) is the rate of heat transfer, \(k\) is the thermal conductivity, \(A\) is the area of the wall, \(T_1\) and \(T_2\) are the temperatures of the inner and outer surfaces of the wall, and \(L\) is the thickness of the wall.
03

Calculate wall area

The area of the wall is the product of its height and length: $$ A = 4\;\text{m} \cdot 7\;\text{m} = 28\;\text{m}^2 $$
04

Convert wall thickness into meters

The wall thickness is given in centimeters, which we need to convert into meters for consistency in our calculations: $$ L = 30\;\text{cm} \cdot \frac{1\;\text{m}}{100\;\text{cm}} = 0.3\;\text{m} $$
05

Calculate the rate of heat transfer

Using the values from the previous steps, substitute them into the Fourier's law equation: $$ Q = 0.69 \frac{\text{W}}{\text{m} \cdot \text{K}} \cdot 28\;\text{m}^2 \cdot \frac{26^{\circ}\text{C} - 8^{\circ}\text{C}}{0.3\;\text{m}} $$ Calculate the result: $$ Q = 0.69 \frac{\text{W}}{\text{m} \cdot \text{K}} \cdot 28\;\text{m}^2 \cdot \frac{18 \text{K}}{0.3\;\text{m}} = 9476\;\text{W} $$ So, the rate of heat transfer through the wall is approximately \(9476\;\text{W}\).

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Most popular questions from this chapter

An electronic package in the shape of a sphere with an outer diameter of \(100 \mathrm{~mm}\) is placed in a large laboratory room. The surface emissivity of the package can assume three different values \((0.2,0.25\), and \(0.3)\). The walls of the room are maintained at a constant temperature of \(77 \mathrm{~K}\). The electronics in this package can only operate in the surface temperature range of \(40^{\circ} \mathrm{C} \leq T_{s} \leq 85^{\circ} \mathrm{C}\). Determine the range of power dissipation \((\dot{W})\) for the electronic package over this temperature range for the three surface emissivity values \((\varepsilon)\). Plot the results in terms of \(\dot{W}(\mathrm{~W})\) vs. \(T_{s}\left({ }^{\circ} \mathrm{C}\right)\) for the three different values of emissivity over a surface temperature range of 40 to \(85^{\circ} \mathrm{C}\) with temperature increments of \(5^{\circ} \mathrm{C}\) (total of 10 data points for each \(\varepsilon\) value). Provide a computer generated graph for the display of your results and tabulate the data used for the graph. Comment on the results obtained.

A hollow spherical iron container with outer diameter \(20 \mathrm{~cm}\) and thickness \(0.2 \mathrm{~cm}\) is filled with iced water at \(0^{\circ} \mathrm{C}\). If the outer surface temperature is \(5^{\circ} \mathrm{C}\), determine the approximate rate of heat loss from the sphere, in \(\mathrm{kW}\), and the rate at which ice melts in the container. The heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\).

Using the conversion factors between W and Btu/h, m and \(\mathrm{ft}\), and \(\mathrm{K}\) and \(\mathrm{R}\), express the Stefan-Boltzmann constant \(\sigma=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\) in the English unit \(\mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot \mathrm{R}^{4}\)

The rate of heat loss through a unit surface area of a window per unit temperature difference between the indoors and the outdoors is called the \(U\)-factor. The value of the \(U\)-factor ranges from about \(1.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or \(0.22 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) ) for low-e coated, argon-filled, quadruple-pane windows to \(6.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or \(1.1 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) ) for a single-pane window with aluminum frames. Determine the range for the rate of heat loss through a \(1.2-\mathrm{m} \times 1.8-\mathrm{m}\) window of a house that is maintained at \(20^{\circ} \mathrm{C}\) when the outdoor air temperature is \(-8^{\circ} \mathrm{C}\).

Consider a person standing in a room at \(18^{\circ} \mathrm{C}\). Determine the total rate of heat transfer from this person if the exposed surface area and the skin temperature of the person are \(1.7 \mathrm{~m}^{2}\) and \(32^{\circ} \mathrm{C}\), respectively, and the convection heat transfer coefficient is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Take the emissivity of the skin and the clothes to be \(0.9\), and assume the temperature of the inner surfaces of the room to be the same as the air temperature.
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