/*! 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 7 The walls of a refrigerator are ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 7

The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between sheet metal panels. Consider a wall made from fiberglass insulation of thermal conductivity \(k_{i}=0.046 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(L_{i}=50 \mathrm{~mm}\) and steel panels, each of thermal conductivity \(k_{p}=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(L_{p}=3 \mathrm{~mm}\). If the wall separates refrigerated air at \(T_{\infty, i}=4^{\circ} \mathrm{C}\) from ambient air at \(T_{\infty, o}=25^{\circ} \mathrm{C}\), what is the heat gain per unit surface area? Coefficients associated with natural convection at the inner and outer surfaces may be approximated as \(h_{i}=h_{o}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Short Answer

Expert verified
To find the heat gain per unit surface area for the refrigerator wall, follow these steps: 1. Calculate the conduction resistance of the insulation (\(R_{cond, i}\)) and steel panels (\(R_{cond, p}\)) using the formula \(R_{cond} = \frac{L}{kA}\). 2. Calculate the natural convection resistance for the inner (\(R_{conv, i}\)) and outer surfaces (\(R_{conv, o}\)) using the formula \(R_{conv} = \frac{1}{hA}\). 3. Calculate the total thermal resistance of the system (\(R_{total}\)) by summing the conduction resistance of the insulation, the conduction resistance of the steel panels (multiplied by 2 to account for both panels), and the convection resistance of the inner and outer surfaces. 4. Determine the overall heat transfer through the refrigerator wall (\(q\)) using the formula \(q = \frac{\Delta T}{R_{total}}\), where \(\Delta T\) is the difference between the ambient temperature and the refrigerated air temperature. 5. Calculate the heat gain per unit surface area (\(q''\)) by dividing the overall heat transfer by the surface area of the wall using the formula \(q'' = \frac{q}{A} = \frac{\Delta T}{R_{total}}\). Computing these values with the given data will provide the heat gain per unit surface area in W/m虏.

Step by step solution

01

Calculate Conduction Resistance of Insulation and Steel Panels

To calculate the conduction resistance of the insulation and steel panels, we can use the formula: \(R_{cond} = \frac{L}{kA}\), where \(R_{cond}\) is the conduction resistance, \(L\) is the thickness of the material, \(k\) is the thermal conductivity, and \(A\) is the surface area. For insulation: \(R_{cond, i} = \frac{L_{i}}{k_{i}A}\), For steel panels: \(R_{cond, p} = \frac{L_{p}}{k_{p}A}\).
02

Calculate Natural Convection Resistance for Inner and Outer Surfaces

To calculate the natural convection resistance, we use the formula: \(R_{conv} = \frac{1}{hA}\), where \(R_{conv}\) is the convection resistance, and \(h\) is the convection heat transfer coefficient. For inner surface: \(R_{conv, i} = \frac{1}{h_{i}A}\), For outer surface: \(R_{conv, o} = \frac{1}{h_{o}A}\).
03

Calculate Total Resistance

Now, find the total thermal resistance of the system by summing the conduction resistance of the insulation, the conduction resistance of the steel panels, and the convection resistance of the inner and outer surfaces: \(R_{total} = R_{cond, i} + 2\cdot R_{cond, p} + R_{conv, i} + R_{conv, o}\), (The steel panels' resistance has been multiplied by 2 to account for the two panels on either side of the insulation.)
04

Calculate Overall Heat Transfer

With the total resistance calculated, we can now determine the overall heat transfer through the refrigerator wall, using the formula: \(q = \frac{\Delta T}{R_{total}}\), where \(q\) is the heat transfer rate, and \(\Delta T\) is the difference between the ambient temperature and the refrigerated air temperature, \(\Delta T = T_{\infty, o} - T_{\infty, i}\).
05

Calculate Heat Gain per Unit Surface Area

Finally, find the heat gain per unit surface area by dividing the heat transfer rate calculated in step 4 by the surface area of the wall. Since the area cancels out when calculating the total resistance, the formula simplifies to: \(q'' = \frac{q}{A} = \frac{\Delta T}{R_{total}}\), where \(q''\) denotes the heat gain per unit surface area. Computing these values with the given data will provide the heat gain per unit surface area in W/m虏.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a fundamental property of materials that indicates how well they conduct heat. It is denoted by the symbol \( k \) and is measured in watts per meter per degree Kelvin (W/m路K). This property is crucial when analyzing heat transfer through materials like those in refrigerator walls.
Materials with high thermal conductivity, such as steel (\( k_p = 60 \text{ W/m} \cdot \text{K} \)), are efficient at conducting heat and are typically used in applications where heat distribution is desired. Conversely, materials with low thermal conductivity, like fiberglass insulation (\( k_i = 0.046 \text{ W/m} \cdot \text{K} \)), are preferred for insulation purposes to minimize heat transfer.
In the case of the refrigerator wall, understanding thermal conductivity helps in determining how much heat is transferred through different layers. This affects the appliance's energy efficiency and operational costs. The key takeaway is that thermal conductivity helps us analyze heat flow direction, speed, and overall effectiveness of different materials.
Conduction Resistance
Conduction resistance quantifies a material's ability to resist heat flow. It complements thermal conductivity by providing a straightforward measure of how challenging it is for heat to pass through a given material thickness. The formula for conduction resistance \( R_{cond} \) is given by:
\[ R_{cond} = \frac{L}{kA} \]
where \( L \) is the thickness of the material, \( k \) is the thermal conductivity, and \( A \) is the surface area.
For the refrigerator wall, conduction resistance helps assess how effectively the insulation (fiberglass) and steel panels block heat. A higher conduction resistance means that less heat is transferred through the material, making it an effective barrier.
In practical scenarios, materials with desired conduction resistance values are chosen based on their performance in the overall heat resistance pathway within a system. For example, the insulation must have a high conduction resistance to minimize heat transfer from the outside into the refrigerated space.
Convection Resistance
Convection resistance deals with the opposition to heat flow at the boundary layer where a solid surface meets a fluid, such as air. It is particularly important in natural convection applications, where buoyancy-driven flow moves fluid past a surface without any forced mechanisms like fans or pumps.
The formula for calculating convection resistance \( R_{conv} \) is:
\[ R_{conv} = \frac{1}{hA} \]
where \( h \) is the convective heat transfer coefficient, and \( A \) is the area of the surface.
For the refrigerator wall, the convection resistances of both the inner and outer surfaces are taken into account. The convection coefficients given (\( h_i = h_o = 5 \text{ W/m}^2 \cdot \text{K} \)) specify how efficiently heat is transferred from the solid walls to the air surrounding them.
This component of thermal resistance is vital because it affects how quickly temperature changes occur at the surfaces in contact with air, ultimately influencing the overall heat transfer rate across the wall.
Natural Convection
Natural convection is the process by which heat is transferred through a fluid (liquid or gas) without any mechanical aids, relying solely on the fluid's natural motion driven by temperature differences. In the context of the refrigerator wall, natural convection occurs at the interfaces where the air surrounding the wall works as the fluid.
This mode of heat transfer is less efficient compared to forced convection but is crucial in scenarios where mechanical systems (like fans) cannot be used or are not energy-efficient.
The convective heat transfer coefficient \( h \) is key to understanding natural convection. It reflects how well the surface exchanges thermal energy with the neighboring fluid. For natural convection, this value tends to be lower compared to forced convection since the fluid velocity is determined by buoyancy effects alone.
Understanding this concept is vital for designing and optimizing systems involving heat transfer through natural convection, as it can greatly influence the thermal comfort and efficiency in applications such as refrigeration.

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Most popular questions from this chapter

Finned passages are frequently formed between parallel plates to enhance convection heat transfer in compact heat exchanger cores. An important application is in electronic equipment cooling, where one or more air-cooled stacks are placed between heat-dissipating electrical components. Consider a single stack of rectangular fins of length \(L\) and thickness \(t\), with convection conditions corresponding to \(h\) and \(T_{\infty}\). (a) Obtain expressions for the fin heat transfer rates, \(q_{f, o}\) and \(q_{f, L}\), in terms of the base temperatures, \(T_{o}\) and \(T_{L}\). (b) In a specific application, a stack that is \(200 \mathrm{~mm}\) wide and \(100 \mathrm{~mm}\) deep contains 50 fins, each of length \(L=12 \mathrm{~mm}\). The entire stack is made from aluminum, which is everywhere \(1.0 \mathrm{~mm}\) thick. If temperature limitations associated with electrical components joined to opposite plates dictate maximum allowable plate temperatures of \(T_{o}=400 \mathrm{~K}\) and \(T_{L}=350 \mathrm{~K}\), what are the corresponding maximum power dissipations if \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=300 \mathrm{~K} ?\)

Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of \(\dot{q}=\) \(2 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The rod is concentric with a hollow ceramic cylinder, creating an enclosure that is filled with air. The thermal resistance per unit length due to radiation between the enclosure surfaces is \(R_{\mathrm{rad}}^{\prime}=0.30 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\), and the coefficient associated with free convection in the enclosure is \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Construct a thermal circuit that can be used to calculate the surface temperature of the rod, \(T_{r}\). Label all temperatures, heat rates, and thermal resistances, and evaluate each thermal resistance. (b) Calculate the surface temperature of the rod for the prescribed conditions.

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.

A nuclear fuel element of thickness \(2 L\) is covered with a steel cladding of thickness \(b\). Heat generated within the nuclear fuel at a rate \(\dot{q}\) is removed by a fluid at \(T_{\infty}\), which adjoins one surface and is characterized by a convection coefficient \(h\). The other surface is well insulated, and the fuel and steel have thermal conductivities of \(k_{f}\) and \(k_{s}\), respectively. (a) Obtain an equation for the temperature distribution \(T(x)\) in the nuclear fuel. Express your results in terms of \(\dot{q}, k_{f}, L, b, k_{s}, h\), and \(T_{\infty}\). (b) Sketch the temperature distribution \(T(x)\) for the entire system.

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.
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