/*! 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

Copper tubing is joined to a solar collector plate of thickness \(t\), and the working fluid maintains the temperature of the plate above the tubes at \(T_{o}\). There is a uniform net radiation heat flux \(q_{\text {rad }}^{\prime \prime}\) to the top surface of the plate, while the bottom surface is well insulated. The top surface is also exposed to a fluid at \(T_{\infty}\) that provides for a uniform convection coefficient \(h\). (a) Derive the differential equation that governs the temperature distribution \(T(x)\) in the plate. (b) Obtain a solution to the differential equation for appropriate boundary conditions.

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catalyst onto a flat surface. The surface and catalyst are heated and simultaneously exposed to a higher- temperature, low-pressure gas that contains a mixture of chemical species from which the nanowire is to be formed. The catalytic liquid slowly absorbs the species from the gas through its top surface and converts these to a solid material that is deposited onto the underlying liquid-solid interface, resulting in construction of the nanowire. The liquid catalyst remains suspended at the tip of the nanowire. Consider the growth of a 15 -nm-diameter silicon carbide nanowire onto a silicon carbide surface. The surface is maintained at a temperature of \(T_{s}=2400 \mathrm{~K}\), and the particular liquid catalyst that is used must be maintained in the range \(2400 \mathrm{~K} \leq T_{c} \leq 3000 \mathrm{~K}\) to perform its function. Determine the maximum length of a nanowire that may be grown for conditions characterized by \(h=10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=8000 \mathrm{~K}\). Assume properties of the nanowire are the same as for bulk silicon carbide.

A composite cylindrical wall is composed of two materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), which are separated by a very thin, electric resistance heater for which interfacial contact resistances are negligible. Liquid pumped through the tube is at a temperature \(T_{\infty, i}\) and provides a convection coefficient \(h_{i}\) at the inner surface of the composite. The outer surface is exposed to ambient air, which is at \(T_{\infty, o}\) and provides a convection coefficient of \(h_{o^{*}}\) Under steady-state conditions, a uniform heat flux of \(q_{h}^{n}\) is dissipated by the heater. (a) Sketch the equivalent thermal circuit of the system and express all resistances in terms of relevant variables. (b) Obtain an expression that may be used to determine the heater temperature, \(T_{h+}\). (c) Obtain an expression for the ratio of heat flows to the outer and inner fluids, \(q_{o}^{\prime} / q_{i}^{\prime}\). How might the variables of the problem be adjusted to minimize this ratio?

Consider the flat plate of Problem \(3.112\), but with the heat sinks at different temperatures, \(T(0)=T_{o}\) and \(T(L)=T_{L}\), and with the bottom surface no longer insulated. Convection heat transfer is now allowed to occur between this surface and a fluid at \(T_{\infty}\), with a convection coefficient \(h\). (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks. (c) For \(q_{o}^{\prime \prime}=20,000 \mathrm{~W} / \mathrm{m}^{2}, T_{o}=100^{\circ} \mathrm{C}, T_{L}=35^{\circ} \mathrm{C}\), \(T_{\infty}=25^{\circ} \mathrm{C}, k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(L=100 \mathrm{~mm}, t=5 \mathrm{~mm}\), and a plate width of \(W=\) \(30 \mathrm{~mm}\), plot the temperature distribution and determine the sink heat rates, \(q_{x}(0)\) and \(q_{x}(L)\). On the same graph, plot three additional temperature distributions corresponding to changes in the following parameters, with the remaining parameters unchanged: (i) \(q_{o}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), (ii) \(h=200\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and (iii) the value of \(q_{o}^{\prime \prime}\) for which \(q_{x}(0)=0\) when \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

In Problem 3.48, the electrical power required to maintain the heater at \(T_{o}=25^{\circ} \mathrm{C}\) depends on the thermal conductivity of the wall material \(k\), the thermal contact resistance \(R_{t, c}^{\prime}\) and the convection coefficient \(h\). Compute and plot the separate effect of changes in \(k\) \((1 \leq k \leq 200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), \quad R_{t, c}^{\prime} \quad\left(0 \leq R_{t, c}^{\prime} \leq 0.1 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\right)\), and \(h\left(10 \leq h \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) on the total heater power requirement, as well as the rate of heat transfer to the inner surface of the tube and to the fluid.
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