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Chapter 1: Problem 7

A freezer compartment consists of a cubical cavity that is \(2 \mathrm{~m}\) on a side. Assume the bottom to be perfectly insulated. What is the minimum thickness of styrofoam insulation \((k=0.030 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that muse be applied to the top and side walls to ensure a heat loed of less than \(500 \mathrm{~W}\). when the inner and outer surfaces are \(-10\) and \(35^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The minimum thickness of the styrofoam insulation is 5.4 cm.

Step by step solution

01

Calculate the Temperature Difference

The temperature difference between the inner and outer surfaces of the freezer is given by:\[ \Delta T = T_{\text{outer}} - T_{\text{inner}} = 35 - (-10) = 45^{\circ} \text{C} \]
02

Determine the Surface Area Requiring Insulation

The freezer is a cube with a side length of \(2\) m. The top and side walls need insulation, not the bottom. Calculate the surface area, considering 5 sides:\[ A = 5 \times (2 \times 2) = 20 \text{ m}^2 \]
03

Use Fourier鈥檚 Law for Heat Transfer

Fourier's law for heat conduction is given by:\[ Q = \frac{k \cdot A \cdot \Delta T}{d} \]where \(Q\) is the heat load (\(<500 \text{ W}\)), \(k\) is the thermal conductivity, \(A\) is the area, \(\Delta T\) is the temperature difference, and \(d\) is the insulation thickness. We need to solve for \(d\):\[ 500 = \frac{0.030 \times 20 \times 45}{d} \]
04

Solve for Insulation Thickness

Rearrange the equation from Step 3 to solve for \(d\):\[ d = \frac{0.030 \times 20 \times 45}{500} \]Calculate \(d\):\[ d \approx \frac{27}{500} = 0.054 \text{ m} \]
05

Convert Thickness to Centimeters

Convert the thickness \(d\) from meters to centimeters (since it is often more intuitive):\[ d = 0.054 \times 100 = 5.4 \text{ cm} \]

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Key Concepts

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

Heat Conduction
Heat conduction is a process that explains how heat energy is transferred through a material. The energy moves from a region of higher temperature to a region of lower temperature. Imagine when you place your hand on a cold surface; the heat from your hand transfers into that surface, making it feel cold to the touch. This is a simple example of heat conduction.

In the context of insulation for the freezer, heat conduction is the key factor that determines how much heat enters or leaves the inner compartment. To minimize this unwanted heat transfer, we rely on materials with poor heat conduction properties, such as styrofoam, which is used as an insulating material in our exercise.
  • Heat flows naturally from hot to cold areas.
  • Conduction refers to direct contact heat transfer.
  • Insulating materials slow down the transfer process.
By understanding conduction, we ensure that the insulation efficiently keeps the freezer鈥檚 inner temperature from escaping.
Thermal Conductivity
Thermal conductivity is a property that measures how well a material can conduct heat. It is represented by the symbol \( k \) and is measured in units of watts per meter per kelvin (W/m路K). A low thermal conductivity means that the material is a good insulator, as it does not allow heat to pass through easily.

In the example exercise, the styrofoam's thermal conductivity of \(0.030 \, \text{W/m} \cdot \text{K} \) is quite low, indicating it's suitable for minimizing heat transfer. The formula:
  • \(Q = \frac{k \cdot A \cdot \Delta T}{d}\)
emphasizes the role of thermal conductivity. The formula guides us in calculating the thickness of insulation required by considering the material's ability to conduct heat, the area needing protection, and the temperature difference across the material.
Reducing \( k \) or increasing thickness \( d \) leads to better insulation. In selecting materials for insulation, thermal conductivity is crucial as it determines the effectiveness of the insulating barrier against heat flow.
Temperature Difference
Temperature difference, denoted by \( \Delta T \), is simply the difference in temperature between two points. In heat conduction problems, it is the driving force that determines the rate and direction of heat transfer.
  • \( \Delta T = T_{\text{outer}} - T_{\text{inner}} \)
In our exercise, the temperature difference is \( 45^{\circ} \text{C} \) between the freezer's inside and outside surfaces. This difference motivates the heat to move from the warmer external environment into the cooler inner section, creating a need for insulation.

The larger the temperature difference, the greater the potential for heat transfer. Without proper insulation, this difference would cause the inner temperature to rise, reducing the efficiency of the cooling system. Therefore, understanding and calculating the temperature difference is essential in designing effective thermal protection systems, such as a well-insulated freezer.

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Most popular questions from this chapter

A spherical interplanetary probe of \(0.5\) - m diameter contains electronics that dissipate \(150 \mathrm{~W}\), If the probe surface has an emissivity of \(0.8\) and the probe does not receive radiation from other surfaces, as, for example, from the sun, what is its surface temperature?

In considcring the following jroblems involving heat transfer in the natural environment (outdoors), recog. nize that solar radiation is comprised of long and shon wavelength components. If this radiation is incident on a semitrarsporent medium, such as water or glass, two things will happen to the noneflecied porticn of the radiation. The long wavelength component will be aboubed at the surface of the medium, whereas the short wavelength component will be transmitted by the surface. (a) The number of panes in a window can strongly inftuence the heat loss from a heated room to the outside ambient air, Compare the single- and doublepaned units shown by identifying relevant healt transfer processes for each case. (b) In a typical fat-plate solar collector, energy is collected by a working fluid thut is circulated through fubes that are in good contact with the hack face of an absorber plate. The back. face is insulated frum the surroundings, and the absuber plate receives solar radiation on its front face, which is typically covered by one or more transparent plates. Idensify the relevant heat transfer processes, first for the absorber plate with no cover plate and then for the absorber plate with a single cover plate. (c) The solar energy collector design stoun below has been used for agricultural applications. Air is blown through a long duct whone cross section is in the form of an equilateral triangle. One side of the triangle is comprised of a double-paned, semitransparent cover, while the ceher two sides are constructed from aluminum sheets painted flat black on the inside and covered on the outside with a Layer of styroform insulation. During sunny periods, air entering the system is heated for delivery to cither a greenhouse, grain drying unit, or a storage system. Identify all heat transfer processes associated with the cover plates, the absorter plate(s), and the air. (d) Evacuated-tube solar collectors are capable of improved performance relative to flat-plate collectors. The design consists of an inner tube enclosed in an outer tube that is transparent to solar radiation. The annular space between the tubes is evacuated. The ouler, opaque surface of the inner tube absorbs solar raciation, and a working fluid is passed through the tube to collect the solar energy. The collector design generally consists of a row of such tubes arranged in frunt of a reflecting panel. Identify all heat transfer processes relevant to the performance of this device.

A hair dryer may be idealized as a circular doct through which a mmall fan draws ambient air and withie which the air is heated as it flows over a coiled electric resistance wire. (a) If a dryer is designed to operate with an electric power consumption of \(P_{\text {tesc }}=500 \mathrm{~W}\) and to heat air from an ambient temperature of \(T_{i}=20^{\circ} \mathrm{C}\) to a discharge temperature of \(T_{n}=45^{\circ} \mathrm{C}\), at what volumetric flow rate \(\forall\) should the fan operate? Heat loss from the casing to the ambient air and the surroundings nuy be neglected. If the duct has a diameler of \(D=70 \mathrm{~mm}\), what is the discharge velocity \(V_{0}\) of the uir? The density and specific heat of the air maly be approximated as \(\rho=1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=1007\) \(\mathrm{J} / \mathrm{Kg}+\mathrm{K}\), respectively. (b) Consider a dryer duct length of \(L=150 \mathrm{~mm}\) and a surface cmisuivity of \(e=0.8\). If the coeflicient associaled with heat transfer by natural convection from the casing to the ambient air is \(h=4 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\) and the temperature of the air and the surroundings is \(T_{\mathrm{w}}=\) \(T_{w e}=20^{\circ} \mathrm{C}\), confirm that the heat loss from the casing is, in fact, negligitle. The casing may be assumed to have an average surface temperature of \(T_{3}=40^{\circ} \mathrm{C}\).

A thin electrical heating element provides a uniform air flows. The duct wall has a thickness of \(10 \mathrm{~mm}\) and a thermal conductivity of \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) At a particular location, the air temperature is \(30^{-} \mathrm{C}\) and the convection hest transfer coefficient between the air and inner surface of the duct is \(100 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\). What heat flux \(q_{0}^{\prime \prime}\) is required to maintain the inner surface of the duct at \(T_{i}=85^{\circ} \mathrm{C}\) ? (b) For the conditions of purt (a), what is the temperature \(\left(T_{a}\right)\) of the duct surface next to the heater? (c) With \(T_{1}=85^{\circ} \mathrm{C}\), compute and plot \(q_{n}^{\prime \prime}\) and \(T_{e}\) as a function of the air-side convection coefficient \(h\) for the range \(10 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Briefly discuss your results.

The free convection heat transfer cocfficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cuoks. Ascuming the plate is iscthermal and radiation exchange with its surroundings is negligible, evaluate the convection cocfficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\prime} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).
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