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Chapter 1: Problem 13

What is the thickness required of a masonry wall having thermal conductivity \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) if the heat rate is to be \(80 \%\) of the heat rate through a composite structural wall having a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m}+\mathrm{K}\) and a thickness of \(100 \mathrm{~mm}\) ? Both walls are subjected to the same surface temperature difference.

Short Answer

Expert verified
The thickness required of the masonry wall is \(375\, \mathrm{mm}\).

Step by step solution

01

Understand the heat conduction equation

We can begin by reviewing the heat conduction equation, which can be expressed as: \(q = k * A * (T_2 - T_1) / d\) where: - \(q\) is the heat rate, measured in watts (W) - \(k\) is the thermal conductivity, measured in \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) - \(A\) is the surface area, measured in \(m^2\) - \(T_2\) and \(T_1\) are the outside and inside temperatures, respectively, measured in Kelvin (K) - \(d\) is the thickness of the wall, measured in meters (m) For this problem, we will be comparing the heat rates of the masonry and composite walls and using their thermal conductivities to determine the required thickness of the masonry wall.
02

Set up the ratio of heat rates

We are given that the heat rate through the masonry wall is supposed to be 80% of the heat rate through the composite wall. We can set up the following equation to represent this fact: \(q_\mathrm{m} = 0.8 * q_\mathrm{c}\) where: - \(q_\mathrm{m}\) is the heat rate through the masonry wall - \(q_\mathrm{c}\) is the heat rate through the composite wall
03

Use the heat conduction equation for both walls

Next, we can express the heat rates for both the masonry and composite walls using the heat conduction equation. Since both walls are subjected to the same surface temperature difference (\(T_2 - T_1\)) and the surface area does not change, we can set up the following equation by dividing the heat conduction equations for both walls: \(\frac{q_\mathrm{m}}{q_\mathrm{c}} = \frac{k_\mathrm{m} (T_2 - T_1) / d_\mathrm{m}}{k_\mathrm{c} (T_2 - T_1) / d_\mathrm{c}}\) where: - \(k_\mathrm{m}\) is the thermal conductivity of the masonry wall, given as 0.75 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) - \(d_\mathrm{m}\) is the thickness of the masonry wall, which we are trying to find - \(k_\mathrm{c}\) is the thermal conductivity of the composite wall, given as 0.25 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) - \(d_\mathrm{c}\) is the thickness of the composite wall, given as 100 mm (0.1 m)
04

Substitute given values and solve for the thickness of masonry wall

We can substitute the given values for thermal conductivities and thickness of the composite wall into the equation from Step 3, along with the heat rate ratio of 0.8 from Step 2: \(\frac{0.8}{1} = \frac{0.75 (T_2 - T_1) / d_\mathrm{m}}{0.25 (T_2 - T_1) / 0.1}\) Since the temperature difference term (\(T_2 - T_1\)) is the same for both walls, we can cancel it out: \(\frac{0.8}{1} = \frac{0.75 / d_\mathrm{m}}{0.25 / 0.1}\) Now, solve for \(d_\mathrm{m}\): \(\frac{0.8}{1} = \frac{0.75 / d_\mathrm{m}}{2.5}\) \(d_\mathrm{m} = 0.75 / (0.8 * 2.5)\) \(d_\mathrm{m} = 0.75 / 2\) \(d_\mathrm{m} = 0.375\, \mathrm{m}\)
05

Convert the result to millimeters

Finally, we can convert the thickness of the masonry wall from meters to millimeters: \(d_\mathrm{m} = 0.375 * 1000\) \(d_\mathrm{m} = 375\, \mathrm{mm}\) The thickness required of the masonry wall is 375 mm.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity, represented by the symbol 'k', is a material property that indicates the ability of a material to conduct heat. In the context of heat transfer through walls, thermal conductivity plays a crucial role. A high thermal conductivity means that the material can transfer heat quickly, while a low thermal conductivity indicates that the material is more of an insulator, transferring heat at a slower rate.

In our exercise, we're provided with the thermal conductivities of two different wall materials. The masonry wall has a thermal conductivity of 0.75 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\), and the composite wall has a thermal conductivity of 0.25 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\). Knowing these values is essential to solving the problem at hand, which involves calculating the necessary thickness to achieve a certain heat transfer rate.
Heat Transfer Rate
The heat transfer rate quantifies the amount of heat passing through a material over a certain period, measured in watts (W). This rate is influenced by the material's thermal conductivity, surface area through which heat is being transferred, the temperature difference across the material, and the thickness of the material.

In our example, the heat transfer rate is a central part of the problem. We are looking to achieve a masonry wall's heat transfer rate that's 80% of the composite wall's rate. By using the formula \(q = k \cdot A \cdot (T_2 - T_1) / d\), where 'q' is the heat rate, we can set up an equation to find the unknown variable, which in this case, is the required thickness of the masonry wall to attain the desired heat transfer rate.
Wall Thickness Calculation
Calculating the thickness of a wall is crucial when designing a structure to meet specific thermal performance criteria. By rearranging the heat conduction equation, we use the known values of thermal conductivity, heat transfer rate, and temperature difference to find the unknown wall thickness.

In the textbook solution, we see how this calculation unfolds by setting the desired heat transfer rate and equating it with the heat conduction formula, subsequently isolating the variable representing the masonry wall's thickness. As shown in the calculation, the thickness required for the masonry wall to have an 80% heat transfer rate of the composite wall is 375 mm. Such precise calculations ensure buildings are designed with walls that sufficiently regulate heat flow, contributing to energy efficiency and comfort.

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Most popular questions from this chapter

You've experienced convection cooling if you've ever extended your hand out the window of a moving vehicle or into a flowing water stream. With the surface of your hand at a temperature of \(30^{\circ} \mathrm{C}\), determine the convection heat flux for (a) a vehicle speed of \(35 \mathrm{~km} / \mathrm{h}\) in air at \(-5^{\circ} \mathrm{C}\) with a convection coefficient of 40 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and (b) a velocity of \(0.2 \mathrm{~m} / \mathrm{s}\) in a water stream at \(10^{\circ} \mathrm{C}\) with a convection coefficient of \(900 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Which condition would feel colder? Contrast these results with a heat loss of approximately \(30 \mathrm{~W} / \mathrm{m}^{2}\) under normal room conditions.

A rectangular forced air heating duct is suspended from the ceiling of a basement whose air and walls are at a temperature of \(T_{\infty}=T_{\text {sur }}=5^{\circ} \mathrm{C}\). The duct is \(15 \mathrm{~m}\) long, and its cross section is \(350 \mathrm{~mm} \times 200 \mathrm{~mm}\). (a) For an uninsulated duct whose average surface temperature is \(50^{\circ} \mathrm{C}\), estimate the rate of heat loss from the duct. The surface emissivity and convection coefficient are approximately \(0.5\) and \(4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (b) If heated air enters the duct at \(58^{\circ} \mathrm{C}\) and a velocity of \(4 \mathrm{~m} / \mathrm{s}\) and the heat loss corresponds to the result of part (a), what is the outlet temperature? The density and specific heat of the air may be assumed to be \(\rho=1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=1008 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

A cartridge electrical heater is shaped as a cylinder of length \(L=200 \mathrm{~mm}\) and outer diameter \(D=20 \mathrm{~mm}\). Under normal operating conditions, the heater dissipates \(2 \mathrm{~kW}\) while submerged in a water flow that is at \(20^{\circ} \mathrm{C}\) and provides a convection heat transfer coefficient of \(h=5000 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). Neglecting heat transfer from the ends of the heater, determine its surface temperature \(T_{s}\). If the water flow is inadvertently terminated while the heater continues to operate, the heater surface is exposed to air that is also at \(20^{\circ} \mathrm{C}\) but for which \(h=50\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the corresponding surface temperature? What are the consequences of such an event?

A solar flux of \(700 \mathrm{~W} / \mathrm{m}^{2}\) is incident on a flat-plate solar collector used to heat water. The area of the collector is \(3 \mathrm{~m}^{2}\), and \(90 \%\) of the solar radiation passes through the cover glass and is absorbed by the absorber plate. The remaining \(10 \%\) is reflected away from the collector. Water flows through the tube passages on the back side of the absorber plate and is heated from an inlet temperature \(T_{i}\) to an outlet temperature \(T_{o}\). The cover glass, operating at a temperature of \(30^{\circ} \mathrm{C}\), has an emissivity of \(0.94\) and experiences radiation exchange with the sky at \(-10^{\circ} \mathrm{C}\). The convection coefficient between the cover glass and the ambient air at \(25^{\circ} \mathrm{C}\) is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Perform an overall energy balance on the collector to obtain an expression for the rate at which useful heat is collected per unit area of the collector, \(q_{11}^{\prime \prime}\). Determine the value of \(q_{u r^{\prime \prime}}\). (b) Calculate the temperature rise of the water, \(T_{o}-T_{i}\), if the flow rate is \(0.01 \mathrm{~kg} / \mathrm{s}\). Assume the specific heat of the water to be \(4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (c) The collector efficiency \(\eta\) is defined as the ratio of the useful heat collected to the rate at which solar energy is incident on the collector. What is the value of \(\eta\) ?

A \(50 \mathrm{~mm} \times 45 \mathrm{~mm} \times 20 \mathrm{~mm}\) cell phone charger has a surface temperature of \(T_{s}=33^{\circ} \mathrm{C}\) when plugged into an electrical wall outlet but not in use. The surface of the charger is of emissivity \(\varepsilon=0.92\) and is subject to a free convection heat transfer coefficient of \(h=4.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The room air and wall temperatures are \(T_{\infty}=22^{\circ} \mathrm{C}\) and \(T_{\text {sur }}=20^{\circ} \mathrm{C}\), respectively. If electricity costs \(C=\$ 0.18 / \mathrm{kW} \cdot \mathrm{h}\), determine the daily cost of leaving the charger plugged in when not in use.
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