/*! 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 12 A thermopane window consists of ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 12

A thermopane window consists of two pieces of glass \(7 \mathrm{~mm}\) thick that enclose an air space \(7 \mathrm{~mm}\) thick. The window separates room air at \(20^{\circ} \mathrm{C}\) from outside ambient air at \(-10^{\circ} \mathrm{C}\). The convection coefficient associated with the inner (room-side) surface is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the convection coefficient associated with the outer (ambient) air is \(h_{o}=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the heat loss through a window that is \(0.8 \mathrm{~m}\) long by \(0.5 \mathrm{~m}\) wide? Neglect radiation, and assume the air enclosed between the panes to be stagnant. (b) Compute and plot the effect of \(h_{o}\) on the heat loss for \(10 \leq h_{o} \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Repeat this calculation for a triple-pane construction in which a third pane and a second air space of equivalent thickness are added.

Short Answer

Expert verified
The heat loss through the double-pane window with given dimensions and properties can be calculated using the total thermal resistance, considering convection at the inner and outer surfaces and conduction through the glass panes and air space. The total thermal resistance is found to be \(R_{T} = R_{conv,i} + R_{cond,1} + R_{cond,air} + R_{cond,2} + R_{conv,o}\). Using the formula \(Q = \frac{\Delta T}{R_{T}}\), we can calculate the heat loss for different values of the outer convection coefficient, \(h_o\). By plotting the heat loss for double-pane and triple-pane windows as a function of \(h_o\), we can analyze the effect of different convection coefficients on heat loss.

Step by step solution

01

Find the thermal resistance of the window components

For the window, we have to determine the thermal resistance of each component. The components we need to consider are: 1. Convection resistance on the inside surface, 2. Conduction resistance of the two glass panes, 3. Radiation resistance at the inner surface of the outer pane, and 4. Convection resistance on the outside surface. Thermal resistance for convection is given by \(R_{conv} = \frac{1}{hA}\), where h is the convection coefficient, and A is the surface area. Thermal resistance for conduction is given by \(R_{cond} = \frac{l}{kA}\), where l is the thickness, k is the thermal conductivity, and A is the surface area. For simplicity, we assume the thermal conductivity of the glass to be \(k_{glass} = 0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and neglect radiation in this problem. 1) Convection resistance at the inner surface, \(R_{conv,i} = \frac{1}{h_{i}A}\), 2) Conduction resistance of the first glass pane, \(R_{cond,1} = \frac{l_{glass}}{k_{glass}A}\), 3) Conduction resistance of the air space, \(R_{cond,air} = \frac{l_{air}}{k_{air}A}\), (Neglecting convection within the air gap which is assumed stagnant), 4) Conduction resistance of the second glass pane, \(R_{cond,2} = \frac{l_{glass}}{k_{glass}A}\), 5) Convection resistance at the outer surface, \(R_{conv,o} = \frac{1}{h_{o}A}\).
02

Calculate the total thermal resistance

The total thermal resistance, \(R_{T}\), is obtained by summing up the individual resistances as: \(R_{T} = R_{conv,i} + R_{cond,1} + R_{cond,air} + R_{cond,2} + R_{conv,o}\).
03

Calculate the total heat loss

Calculate the heat loss through the window using the formula \(Q = \frac{\Delta T}{R_{T}}\), where \(\Delta T\) is the temperature difference between the inside and outside air. #b) Plotting the effect of different outer convection coefficients on heat loss#
04

Set up a range of convection coefficients

Set up a range of convection coefficients for the outer ambient air, \(h_o\), where \(10 \leq h_o \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).
05

Calculate heat loss for each convection coefficient

For each value of \(h_{o}\), repeat steps 1, 2, and 3 to obtain the corresponding heat loss.
06

Plot the result

Plot the heat loss as a function of \(h_{o}\).
07

Perform the same calculations for a triple-pane window

Repeat steps 1 through 6 for a triple-pane window with an additional pane and air space. Plot the results on the same graph.

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Key Concepts

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

Thermal Resistance
Understanding thermal resistance is key to evaluating how well a material insulates against heat flow. It's a measure of a material's ability to resist the transfer of heat.

In the case of a window, thermal resistance comes into play in multiple ways. For the inner surface of the window, the resistance is related to convection, which is heat transfer due to fluid motion, in this instance, the indoor air moving over the surface. The equation for convection resistance is \( R_{conv} = \frac{1}{hA} \) where \( h \) is the convection coefficient and \( A \) is the area through which heat is being transferred.

A similar concept applies to conduction, which is how heat moves through solid materials, like the window glass and the trapped air between panes. Conduction resistance is calculated using \( R_{cond} = \frac{l}{kA} \) where \( l \) is the thickness of the material, \( k \) is the thermal conductivity, and, again, \( A \) is the area. Higher thermal resistance implies better insulation, leading to lower heat loss through the material.
Conduction in Building Materials
Heat transfer through conduction in building materials like glass is a key aspect of thermal management in structures. Conduction is the process by which heat energy is transmitted through collisions between neighboring atoms or molecules in a material.

The thermal conductivity, \( k \) of a material, is a measure of how easily heat can pass through it. In the case of the thermopane window from the exercise, we have two panes of glass with a known thermal conductivity. The resistance to heat flow through the glass is governed by the thickness of the panes and their intrinsic thermal conductivity.

As an educational tip, using simpler and more visual explanations can enhance understanding. For instance, comparing glass to a sponge can help; where a sponge easily soaks up water, glass 'soaks' up heat at a much slower rate due to higher conduction resistance, thus serving as a better insulator.
Convection Heat Transfer
The convection heat transfer is pertinent to the conversation about windows and heat loss. Convection is the movement of heat through a fluid, which can be a liquid or a gas, driven by the motion of the fluid itself.

In the context of a window, the indoor air at a higher temperature tends to rise and come in contact with the cooler surface of the window. Here, we assess the convection heat transfer using a convection coefficient, denoted by \( h \). This coefficient encapsulates the nature of the air movement and just how well the air can transport heat away from the window surface.

To apply this to real-life examples and improve retention, consider how a fan improves the cooling effect in a room. It doesn't lower the temperature but rather increases convection by moving air around, leading to a faster heat transfer away from your skin, thereby cooling you more efficiently. Similarly, the convection around a window influences the overall heat loss, adjusting the room's temperature.

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Most popular questions from this chapter

A spherical vessel used as a reactor for producing pharmaceuticals has a 10 -mm-thick stainless steel wall \((k=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(1 \mathrm{~m}\). The exterior surface of the vessel is exposed to ambient air \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) for which a convection coefficient of \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) may be assumed. (a) During steady-state operation, an inner surface temperature of \(50^{\circ} \mathrm{C}\) is maintained by energy generated within the reactor. What is the heat loss from the vessel? (b) If a 20 -mm-thick layer of fiberglass insulation \((k=0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the exterior of the vessel and the rate of thermal energy generation is unchanged, what is the inner surface temperature of the vessel?

A truncated solid cone is of circular cross section, and its diameter is related to the axial coordinate by an expression of the form \(D=a x^{3 / 2}\), where \(a=1.0 \mathrm{~m}^{-1 / 2}\). The sides are well insulated, while the top surface of the cone at \(x_{1}\) is maintained at \(T_{1}\) and the bottom surface at \(x_{2}\) is maintained at \(T_{2}\). (a) Obtain an expression for the temperature distribution \(T(x)\). (b) What is the rate of heat transfer across the cone if it is constructed of pure aluminum with \(x_{1}=0.075 \mathrm{~m}\), \(T_{1}=100^{\circ} \mathrm{C}, x_{2}=0.225 \mathrm{~m}\), and \(T_{2}=20^{\circ} \mathrm{C}\) ?

In a test to determine the friction coefficient \(\mu\) associated with a disk brake, one disk and its shaft are rotated at a constant angular velocity \(\omega\), while an equivalent disk/shaft assembly is stationary. Each disk has an outer radius of \(r_{2}=180 \mathrm{~mm}\), a shaft radius of \(r_{1}=20 \mathrm{~mm}\), a thickness of \(t=12 \mathrm{~mm}\), and a thermal conductivity of \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). A known force \(F\) is applied to the system, and the corresponding torque \(\tau\) required to maintain rotation is measured. The disk contact pressure may be assumed to be uniform (i.e., independent of location on the interface), and the disks may be assumed to be well insulated from the surroundings. (a) Obtain an expression that may be used to evaluate \(\mu\) from known quantities. (b) For the region \(r_{1} \leq r \leq r_{2}\), determine the radial temperature distribution \(T(r)\) in the disk, where \(T\left(r_{1}\right)=T_{1}\) is presumed to be known. (c) Consider test conditions for which \(F=200 \mathrm{~N}\), \(\omega=40 \mathrm{rad} / \mathrm{s}, \tau=8 \mathrm{~N} \cdot \mathrm{m}\), and \(T_{1}=80^{\circ} \mathrm{C}\). Evaluate the friction coefficient and the maximum disk temperature.

A rod of diameter \(D=25 \mathrm{~mm}\) and thermal conductivity \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) protrudes normally from a furnace wall that is at \(T_{w}=200^{\circ} \mathrm{C}\) and is covered by insulation of thickness \(L_{\text {ins }}=200 \mathrm{~mm}\). The rod is welded to the furnace wall and is used as a hanger for supporting instrumentation cables. To avoid damaging the cables, the temperature of the rod at its exposed surface, \(T_{o}\), must be maintained below a specified operating limit of \(T_{\max }=100^{\circ} \mathrm{C}\). The ambient air temperature is \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the convection coefficient is \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Derive an expression for the exposed surface temperature \(T_{o}\) as a function of the prescribed thermal and geometrical parameters. The rod has an exposed length \(L_{o}\), and its tip is well insulated. (b) Will a rod with \(L_{o}=200 \mathrm{~mm}\) meet the specified operating limit? If not, what design parameters would you change? Consider another material, increasing the thickness of the insulation, and increasing the rod length. Also, consider how you might attach the base of the rod to the furnace wall as a means to reduce \(T_{o}\).

One modality for destroying malignant tissue involves imbedding a small spherical heat source of radius \(r_{o}\) within the tissue and maintaining local temperatures above a critical value \(T_{c}\) for an extended period. Tissue that is well removed from the source may be assumed to remain at normal body temperature \(\left(T_{b}=37^{\circ} \mathrm{C}\right)\). Obtain a general expression for the radial temperature distribution in the tissue under steady- state conditions for which heat is dissipated at a rate \(q\). If \(r_{o}=0.5 \mathrm{~mm}\), what heat rate must be supplied to maintain a tissue temperature of \(T \geq T_{c}=42^{\circ} \mathrm{C}\) in the domain \(0.5 \leq r \leq\) \(5 \mathrm{~mm}\) ? The tissue thermal conductivity is approximately \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assume negligible perfusion.
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