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Chapter 3: Problem 31

A commercial grade cubical freezer, \(3 \mathrm{~m}\) on a side, has a composite wall consisting of an exterior sheet of \(6.35-\mathrm{mm}\)-thick plain carbon steel, an intermediate layer of \(100-\mathrm{mm}\)-thick cork insulation, and an inner sheet of \(6.35\)-mm-thick aluminum alloy (2024). Adhesive interfaces between the insulation and the metallic strips are each characterized by a thermal contact resistance of \(R_{t, c}^{\prime \prime}=2.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the steady-state cooling load that must be maintained by the refrigerator under conditions for which the outer and inner surface temperatures are \(22^{\circ} \mathrm{C}\) and \(-6^{\circ} \mathrm{C}\), respectively?

Short Answer

Expert verified
The steady-state cooling load that must be maintained by the refrigerator is approximately \(109.38\mathrm{~W}\).

Step by step solution

01

Determine the surface area of the Layers#

Since the freezer is a cube with sides of length 3 meters, the surface area of each face is: \(A=3\mathrm{~m} \times 3\mathrm{~m}=9\mathrm{~m^2}\)
02

Calculate the thermal resistance of each layer #

We need to determine the thermal resistance of each layer using the formula: \(R_{i} = \frac{L_{i}}{k_i A_i}\) First, we must look up the thermal conductivities for each layer's material: - Plain carbon steel: \(k_{steel} = 54\mathrm{~W / (m \cdot K)}\) - Cork insulation: \(k_{cork} = 0.044\mathrm{~W / (m \cdot K)}\) - Aluminum alloy (2024): \(k_{aluminum} = 186\mathrm{~W / (m \cdot K)}\) Now, we can compute the thermal resistance of each layer: \(R_{steel} = \frac{6.35 \times 10^{-3}\mathrm{~m}}{54\mathrm{~W / (m \cdot K)}\cdot 9\mathrm{~m^2}} = 1.31 \times 10^{-4} \mathrm{~m^2 \cdot K / W}\) \(R_{cork} = \frac{0.1\mathrm{~m}}{0.044\mathrm{~W / (m \cdot K)}\cdot 9\mathrm{~m^2}} = 0.255\mathrm{~m^2 \cdot K / W}\) \(R_{aluminum} = \frac{6.35 \times 10^{-3}\mathrm{~m}}{186\mathrm{~W / (m \cdot K)}\cdot 9\mathrm{~m^2}} = 3.80 \times 10^{-5} \mathrm{~m^2 \cdot K / W}\)
03

Calculate the total thermal resistance#

We will now find the total thermal resistance by adding the resistance of each layer: \(R_{total} = R_{steel} + R_{cork} + R_{aluminum} + 2 \times R_{t, c}^{\prime \prime}\) \(R_{total} = 1.31 \times 10^{-4}\mathrm{~m^2 \cdot K / W} + 0.255\mathrm{~m^2 \cdot K / W} + 3.80 \times 10^{-5} \mathrm{~m^2 \cdot K / W} + 2 \times 2.5 \times 10^{-4} \mathrm{~m^2 \cdot K / W}\) \(R_{total} = 0.256 \mathrm{~m^2 \cdot K / W}\)
04

Calculate the cooling load (heat transfer rate)#

Now, we can find the cooling load (heat transfer rate) using the formula: \(q = \frac{\Delta T}{R_{total}}\) The temperature difference between the inner and outer surfaces is: \(\Delta T = 22 - (-6) = 28\mathrm{~K}\) Now we can calculate the cooling load: \(q = \frac{28\mathrm{~K}}{0.256\mathrm{~m^2 \cdot K / W}} = 109.38\mathrm{~W}\) The steady-state cooling load that must be maintained by the refrigerator is approximately \(109.38\mathrm{~W}\).

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Key Concepts

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

Heat Transfer
Heat transfer is the movement of thermal energy from one object or substance to another. It's an essential concept in understanding how refrigeration systems work. The transfer of heat occurs because of the temperature difference between the interior of the freezer and the warmer external environment.

There are three main modes of heat transfer:
  • Conduction: This is the most relevant in the context of a freezer's insulation. It involves the transfer of heat through a material without the movement of the material itself. In our problem, heat transfers through the steel, cork, and aluminum layers via conduction.
  • Convection: Involves the transfer of heat by the movement of fluid (liquid or gas). Though convection is not directly highlighted in the given problem, it occurs outside and inside the freezer as air circulates.
  • Radiation: The transfer of heat in the form of electromagnetic waves. In most refrigeration problems, radiation's effect is less significant compared to conduction and convection.
To calculate the heat transfer rate (or cooling load) in the freezer, we subtract the inside temperature from the outside temperature to find the temperature difference. Then, we divide this difference by the total thermal resistance of the freezer's wall.
Thermal Conductivity
Thermal conductivity is a material's ability to conduct heat. It's a crucial property when determining how insulating or conductive a material is in terms of heat flow. Each material in the freezer's construction has a specific thermal conductivity, often denoted by the symbol \( k \).

In our exercise:
  • Plain carbon steel has a thermal conductivity of \( 54 \text{ W/(m路K)} \). It's quite conductive and thus transfers heat readily.
  • Cork insulation exhibits a low thermal conductivity of \( 0.044 \text{ W/(m路K)} \). This makes cork a good insulator, slowing down the heat transfer.
  • Aluminum alloy (2024) has a high thermal conductivity of \( 186 \text{ W/(m路K)} \), indicating it's an excellent conductor of heat.
To compute the thermal resistance of each material layer, we use the formula:\[R_i = \frac{L_i}{k_i A_i}\]where \( R_i \) is the thermal resistance, \( L_i \) is the thickness of the layer, \( k_i \) is the thermal conductivity, and \( A_i \) is the surface area.
Contact Resistance
Contact resistance occurs at the interfaces between two materials. It measures the hindrance to heat flow across these boundary surfaces and is especially important in composite walls like the freezer's barrier.

When different materials meet, like the cork insulation and metal sheets in the freezer, tiny imperfections or air gaps at the interface can reduce the efficiency of heat transfer. This added resistance is known as thermal contact resistance. In our exercise, the contact resistance is given by \( R_{t, c}^{\prime \prime} = 2.5 \times 10^{-4} \text{ m}^2 \cdot \text{K/W} \).

Thermal contact resistance can be reduced by:
  • Using thermal grease or adhesive that fills air gaps at the interface.
  • Applying pressure to press the materials together more effectively.
In our problem, the overall heat transfer analysis includes the sum of the resistances from each material layer plus the contributions from the contact resistances, which are doubled since there are two interfaces (one on each side of the cork insulation).

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Most popular questions from this chapter

When raised to very high temperatures, many conventional liquid fuels dissociate into hydrogen and other components. Thus the advantage of a solid oxide fuel cell is that such a device can internally reform readily available liquid fuels into hydrogen that can then be used to produce electrical power in a manner similar to Example 1.5. Consider a portable solid oxide fuel cell, operating at a temperature of \(T_{\mathrm{fc}}=800^{\circ} \mathrm{C}\). The fuel cell is housed within a cylindrical canister of diameter \(D=\) \(75 \mathrm{~mm}\) and length \(L=120 \mathrm{~mm}\). The outer surface of the canister is insulated with a low-thermal-conductivity material. For a particular application, it is desired that the thermal signature of the canister be small, to avoid its detection by infrared sensors. The degree to which the canister can be detected with an infrared sensor may be estimated by equating the radiation heat flux emitted from the exterior surface of the canister (Equation 1.5; \(E_{s}=\varepsilon_{s} \sigma T_{s}^{4}\) ) to the heat flux emitted from an equivalent black surface, \(\left(E_{b}=\sigma T_{b}^{4}\right)\). If the equivalent black surface temperature \(T_{b}\) is near the surroundings temperature, the thermal signature of the canister is too small to be detected-the canister is indistinguishable from the surroundings. (a) Determine the required thickness of insulation to be applied to the cylindrical wall of the canister to ensure that the canister does not become highly visible to an infrared sensor (i.e., \(T_{b}-T_{\text {sur }}<5 \mathrm{~K}\) ). Consider cases where (i) the outer surface is covered with a very thin layer of \(\operatorname{dirt}\left(\varepsilon_{s}=0.90\right)\) and (ii) the outer surface is comprised of a very thin polished aluminum sheet \(\left(\varepsilon_{s}=0.08\right)\). Calculate the required thicknesses for two types of insulating material, calcium silicate \((k=0.09 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and aerogel \((k=0.006 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the surroundings and the ambient are \(T_{\text {sur }}=300 \mathrm{~K}\) and \(T_{\infty}=298 \mathrm{~K}\), respectively. The outer surface is characterized by a convective heat transfer coefficient of \(h=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) Calculate the outer surface temperature of the canister for the four cases (high and low thermal conductivity; high and low surface emissivity). (c) Calculate the heat loss from the cylindrical walls of the canister for the four cases.

The wind chill, which is experienced on a cold, windy day, is related to increased heat transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty tissue that is \(3 \mathrm{~mm}\) thick and whose interior surface is maintained at a temperature of \(36^{\circ} \mathrm{C}\). On a calm day the convection heat transfer coefficient at the outer surface is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), but with \(30 \mathrm{~km} / \mathrm{h}\) winds it reaches \(65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). In both cases the ambient air temperature is \(-15^{\circ} \mathrm{C}\). (a) What is the ratio of the heat loss per unit area from the skin for the calm day to that for the windy day? (b) What will be the skin outer surface temperature for the calm day? For the windy day? (c) What temperature would the air have to assume on the calm day to produce the same heat loss occurring with the air temperature at \(-15^{\circ} \mathrm{C}\) on the windy day?

An uninsulated, thin-walled pipe of \(100-\mathrm{mm}\) diameter is used to transport water to equipment that operates outdoors and uses the water as a coolant. During particularly harsh winter conditions, the pipe wall achieves a temperature of \(-15^{\circ} \mathrm{C}\) and a cylindrical layer of ice forms on the inner surface of the wall. If the mean water temperature is \(3^{\circ} \mathrm{C}\) and a convection coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained at the inner surface of the ice, which is at \(0^{\circ} \mathrm{C}\), what is the thickness of the ice layer?

A technique for measuring convection heat transfer coefficients involves bonding one surface of a thin metallic foil to an insulating material and exposing the other surface to the fluid flow conditions of interest. By passing an electric current through the foil, heat is dissipated uniformly within the foil and the corresponding flux, \(P_{\text {elec }}^{\prime \prime}\), may be inferred from related voltage and current measurements. If the insulation thickness \(L\) and thermal conductivity \(k\) are known and the fluid, foil, and insulation temperatures \(\left(T_{\infty}, T_{s}, T_{b}\right)\) are measured, the convection coefficient may be determined. Consider conditions for which \(T_{\infty}=T_{b}=25^{\circ} \mathrm{C}, P_{\text {elec }}^{\prime \prime}=2000\) \(\mathrm{W} / \mathrm{m}^{2}, L=10 \mathrm{~mm}\), and \(k=0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) With water flow over the surface, the foil temperature measurement yields \(T_{s}=27^{\circ} \mathrm{C}\). Determine the convection coefficient. What error would be incurred by assuming all of the dissipated power to be transferred to the water by convection? (b) If, instead, air flows over the surface and the temperature measurement yields \(T_{s}=125^{\circ} \mathrm{C}\), what is the convection coefficient? The foil has an emissivity of \(0.15\) and is exposed to large surroundings at \(25^{\circ} \mathrm{C}\). What error would be incurred by assuming all of the dissipated power to be transferred to the air by convection? (c) Typically, heat flux gages are operated at a fixed temperature \(\left(T_{s}\right)\), in which case the power dissipation provides a direct measure of the convection coefficient. For \(T_{s}=27^{\circ} \mathrm{C}\), plot \(P_{\text {elec }}^{\prime \prime}\) as a function of \(h_{o}\) for \(10 \leq h_{o} \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What effect does \(h_{o}\) have on the error associated with neglecting conduction through the insulation?

As a means of enhancing heat transfer from highperformance logic chips, it is common to attach a heat \(\sin k\) 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 attractive option is to use a heat sink consisting of an array of square fins of width \(w\) on a side. The spacing between adjoining fins would be determined by the width of the sawblade, with the sum of this spacing and the fin width designated as the fin pitch \(S\). The method by which the heat sink is joined to the chip would determine the interfacial contact resistance, \(R_{t, c^{*}}^{n}\) Consider a square chip of width \(W_{c}=16 \mathrm{~mm}\) and conditions for which cooling is provided by a dielectric liquid with \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(h=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat \(\operatorname{sink}\) is fabricated from copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and its characteristic dimensions are \(w=0.25 \mathrm{~mm}\), \(S=0.50 \mathrm{~mm}, L_{f}=6 \mathrm{~mm}\), and \(L_{b}=3 \mathrm{~mm}\). The prescribed 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 \(R_{t, c}^{\prime \prime}=5 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) and the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\), what is the maximum allowable chip power dissipation \(q_{c} ?\) 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 \(L_{f}\) (subject to manufacturing constraints that \(L_{f} \leq 10 \mathrm{~mm}\) ). Assess the effect of such changes.
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