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Chapter 3: Problem 24

A commercial grade cubical freezer, \(3 \mathrm{~m}\) on a side, has a compesite wall consisting of an exterior sheet of 6.35-mm-thick plain carbon steel, an intermediate layer of 100 -rnm-thick cork insulation, and an inner sheet of 6.35-mm-thick aluminum alloy (2024). Adhesive interfaces berween the insulation and the metallic strips are each characterized by a thermul contact resistance of \(R_{L A}^{\prime \prime}=2.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the steady-state cooling load that must be maintained by the refrigerawor under conditions for which the outer and inner surface temperatures are \(2^{\circ} 2^{\circ} \mathrm{C}\) and \(-6^{\circ} \mathrm{C}\) respectively?

Short Answer

Expert verified
Follow the steps to find the cooling load using thermal resistances and temperature difference.

Step by step solution

01

Understand the Geometry and Material Properties

The freezer is a cube with each side measuring 3 m. It has a composite wall consisting of three layers: an exterior 6.35 mm (0.00635 m) thick plain carbon steel layer, a 100 mm (0.1 m) thick cork insulation layer, and an inner 6.35 mm (0.00635 m) thick aluminum alloy layer.
02

Identify Thermal Resistances

For each layer of material and the interface, calculate the thermal resistance. For a plane wall of thickness L and thermal conductivity k, the resistance is given by \( R = \frac{L}{kA} \) where A is the area. Given the contact thermal resistance \( R_c \), the total thermal resistance for each interface is \( R = R_c/A \). For this problem, you'll want to estimate the thermal resistances for steel, cork, and aluminum and add the contact resistances.
03

Calculate Individual Thermal Resistances

Calculate the thermal resistance for each material using their respective conductivities (let k_steel, k_cork, k_aluminum be the conductivities for steel, cork, and aluminum, respectively). Thus, \( R_{steel} = \frac{0.00635}{k_{steel} imes 3^2} \), \( R_{cork} = \frac{0.1}{k_{cork} imes 3^2} \), and \( R_{aluminum} = \frac{0.00635}{k_{aluminum} imes 3^2} \). Add the contact resistances: \( R_{contact} = \frac{2.5 imes 10^{-4}}{3^2} \) twice, for both interfaces.
04

Sum Up Total Thermal Resistance

Sum all the resistances to get the total resistance \( R_{total} = R_{steel} + R_{cork} + R_{aluminum} + 2 \times R_{contact} \). This gives the composite thermal resistance of the wall system.
05

Calculate the Cooling Load

The cooling load can be calculated using the formula \( Q = \frac{T_{inside} - T_{outside}}{R_{total}} \), where \( T_{inside} = -6^{\circ}C \) and \( T_{outside} = 2^{\circ}C \). Substitute the total resistance value and temperatures to find \( Q \), the steady-state cooling load in watts.

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Key Concepts

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

Thermal Resistance
Thermal resistance is a measure of a material's ability to resist heat flow. It is an important concept when analyzing heat transfer through materials. Thermal resistance is similar to electrical resistance, but instead of resisting electric current, it resists heat flow. When you have materials stacked together, like in a composite wall, each layer provides its own resistance to heat flow.

To calculate thermal resistance, you use the equation:
  • \( R = \frac{L}{kA} \)
Where:
  • \( L \) is the thickness of the material,
  • \( k \) is the thermal conductivity, and
  • \( A \) is the area through which heat is flowing.
The total thermal resistance of a composite wall is the sum of the resistances of all layers, plus any additional contact resistances between layers. This total resistance plays a crucial role in determining how much heat is lost or absorbed, influencing the energy required for heating or cooling.
Composite Wall
A composite wall is a wall made up of multiple layers of different materials. Each material has its unique properties, like thickness and thermal conductivity, contributing to the wall's overall thermal resistance. Composite walls are common in construction, particularly in regions where insulation is crucial.

In our exercise, the composite wall consists of:
  • A 6.35 mm thick plain carbon steel outer layer.
  • A 100 mm thick cork insulation layer.
  • A 6.35 mm thick aluminum alloy inner layer.
  • Adhesive interfaces with additional contact resistance.
The layers are carefully selected to provide a balance between structural integrity and thermal insulation. Each material serves a purpose, such as minimizing heat transfer or providing support. When calculating the cooling load for a system with a composite wall, understanding how these layers interact to resist heat flow is critical.
Cooling Load Calculation
Cooling load calculation is essential for determining how much energy is required to maintain a certain temperature in a space. It involves understanding how much heat the system needs to remove to keep the interior at the desired temperature.

In our exercise, the cooling load is calculated by understanding the temperature difference between the inside and outside of the freezer and the total thermal resistance of the composite wall. The formula used for this calculation is:
  • \( Q = \frac{T_{inside} - T_{outside}}{R_{total}} \)
Where:
  • \( T_{inside} \) is the temperature inside the freezer,
  • \( T_{outside} \) is the temperature outside the freezer, and
  • \( R_{total} \) is the sum of all thermal resistances.
By using this formula, you can determine the rate at which heat must be removed, ensuring the efficient operation of the cooling system. Accurate cooling load calculations help in designing systems that are not only effective but also energy-efficient, reducing operational costs.
Thermal Conductivity
Thermal conductivity is a property that indicates how well a material can conduct heat. It's represented by the symbol \( k \) and is measured in watts per meter-kelvin (W/m路K). Materials with high thermal conductivity transfer heat quickly, making them good conductors, such as metals. Conversely, low thermal conductivity materials, like insulation, are good insulators.

In the given exercise:
  • Carbon steel and aluminum alloy typically have higher thermal conductivities, so they transfer heat more efficiently.
  • Cork, used for insulation, has a lower thermal conductivity, making it effective at minimizing heat transfer.
Knowing the thermal conductivity of each layer in a composite wall is essential for calculating the thermal resistance and ultimately determining the cooling load. This property allows engineers to tailor materials to specific needs, whether you're aiming to conserve energy or quickly dissipate heat in applications.

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Most popular questions from this chapter

One modality for destroying malignant tissue involves imbedding a small spherical heat source of radius \(r_{e}\) within the tissue and maintaining local temperatures above a critical value \(T_{z}\) for an extended period. Tissue that is well removed from the source may be assumed to remain at nomal 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 \(r_{e}=0.5 \mathrm{~mm}\), what heat rate must be supplied to maintain a tissue temperature of \(T \geq T_{e}=42^{\circ} \mathrm{C}\) in the domain \(0.5 \leq r \leq\) \(5 \mathrm{~mm}\) ? The tissue thermal cceductivity is approximately \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assume negligible perfusion.

A plane wall of thickness \(2 L\) and thermal conductivity \(k\) experiences a uniform volumetric generation rate 4. As shown in the sketch for Case I. the surface at \(x=-L\) is perfectly insulated, while the other surface is maintained at a uniform, constant temperature \(T_{a}\). For Case 2, a very thin dielectric strip is inserted at the midpoint of the wall \((x=0)\) in order to electrically isolate the two sections, \(A\) and B. The thermal resistance of the strip is \(R_{7}^{*}=0.0005 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The parameters associated with the wall are \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=20 \mathrm{~mm}\), \(q=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\), and \(T_{e}=50^{\circ} \mathrm{C}\). (a) Sketch the temperature distribution for case I on \(T-x\) coordinates. Deseribe the key features of this distribution. Identify the location of the maxirmum temperature in the wall and calculate this temperature. (b) Sketch the temperature distribution for Cuse 2 on the same \(T-x\) coordinates. Describe the key features of this distribution. (c) What is the temperature difference between the two walls at \(x=0\) for Case 2? (d) What is the location of the maximum temperature in the composite wall of Case 2? Calculate this temperature.

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worm. Treat the eye as a splerical system and assume the system to be at steady state. The convection coefficient \(h_{e}\) 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: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{-a}=21^{\circ} \mathrm{C} \\ T_{s 1}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{n}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{4}=12 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K} & \end{array}\) (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.

As a means of enhancing heat transfer from highperformance logic chips, it is common to attach a heot sink to the chip surface in order to increase the surface area available for convection heat transfer. Because of the ease with which it may be manufactured (by taking orthogonal sawcuts in a block of material), an attmctive option is to use a heat sink consisting of an atmy of square fins of width w on s side. The spacing between adjoining fins would be determined by the width of the sawblade, with the sum of this spacing and the fan width designated as the fin pitch \(S\). The method by which the hent sink is joined to the chip would detesmine the interfacial contact resistance, \(R_{\mathcal{U}^{*}}^{*}\) Consider a square chip of width \(W_{c}=16 \mathrm{~mm}\) and conditions for which cooling is provided by a diclectric liquid with \(T_{w}=25^{\circ} \mathrm{C}\) and \(h=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat sink is fahricated from copper \((k=400\) Whe \(\mathrm{K})\), and its characteristic dimensions are \(w=0.25\) \(\operatorname{mmn}_{,} S=0.50 \mathrm{~mm}, L_{y}=6 \mathrm{~mm}\), and \(L_{1}=3 \mathrm{~mm}\). The preccribed values of \(w\) and \(S\) represent minima imposed by manufacturing constraints and the need to maintain adequate flow in the passages between fins. (a) If a metallurgical joint provides a contact resistance of \(R_{\mathrm{W}}^{*}=5 \times 10^{-6} \mathrm{~m}^{2}+\mathrm{K} / \mathrm{W}\) and the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\), what is the maximum allowable chip power dissipation \(q, ?\) Assume all of the heat to be transferred through the heat sink. (b) It may be possible to increase the heat dissipation by increasing \(w\), subject to the constraint that \((S-w) \geq 0.25 \mathrm{~mm}\), and/or increasing \(L_{f}\) (suhject to manufacturing constraints that \(L_{y} \leq 10 \mathrm{~mm}\). Assess the effect of such changes.

Consider a plane composite wall that is composed of two naterials of thermal conductivities \(k_{A}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{B}=0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thicknesses \(L_{A}=10 \mathrm{~mm}\) and \(L_{\mathrm{B}}=20 \mathrm{~mm}\). The contact resistance at the interface between the two materials is known to be \(0.30 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Material A adjoins a fluid at \(200^{\circ} \mathrm{C}\) for which \(h=10\) W/m虏 \(\cdot \mathrm{K}\), and material B adjoins a fluid at \(40^{\circ} \mathrm{C}\) for which \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the rate of heat transfer through a wall that is \(2 \mathrm{~m}\) high by \(2.5 \mathrm{~m}\) wide? (b) Sketch the temperature distribution.
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