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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

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 \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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?

A plane wall of thickness \(0.1 \mathrm{~m}\) and thermal conductivity \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) having uniform volumetric heat generation of \(0.3 \mathrm{MW} / \mathrm{m}^{3}\) is insulated on one side, while the other side is exposed to a fluid at \(92^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the wall and the fluid is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the maximum temperature in the wall.

An electrical current of 700 A flows through a stainless steel cable having a diameter of \(5 \mathrm{~mm}\) and an electrical resistance of \(6 \times 10^{-4} \mathrm{\Omega} / \mathrm{m}\) (i.e., per meter of cable length). The cable is in an environment having a temperature of \(30^{\circ} \mathrm{C}\), and the total coefficient associated with convection and radiation between the cable and the environment is approximately \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the cable is bare, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of \(0.02 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), what are the insulation and cable surface temperatures? (c) There is some concern about the ability of the insulation to withstand elevated temperatures. What thickness of this insulation \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) will yield the lowest value of the maximum insulation temperature? What is the value of the maximum temperature when this thickness is used?

Consider the oven of Problem 1.54. The walls of the oven consist of \(L=30\)-mm- thick layers of insulation characterized by \(k_{\text {ins }}=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) that are sandwiched between two thin layers of sheet metal. The exterior surface of the oven is exposed to air at \(23^{\circ} \mathrm{C}\) with \(h_{\text {ext }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The interior oven air temperature is \(180^{\circ} \mathrm{C}\). Neglecting radiation heat transfer, determine the steady-state heat flux through the oven walls when the convection mode is disabled and the free convection coefficient at the inner oven surface is \(h_{\mathrm{fr}}=3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the heat flux through the oven walls when the convection mode is activated, in which case the forced convection coefficient at the inner oven surface is \(h_{\mathrm{fo}}=27 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Does operation of the oven in its convection mode result in significantly increased heat losses from the oven to the kitchen? Would your conclusion change if radiation were included in your analysis?
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