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Chapter 1: Problem 10

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 must be applied to the top and side walls to ensure a heat load 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 styrofoam insulation required for the top and side walls of the freezer compartment to ensure a heat load of less than 500W is \(5.4 cm\).

Step by step solution

01

List Given Information

Cubical Cavity Dimensions: \(2m \times 2m \times 2m\) Thermal conductivity (k) of Styrofoam: \(0.030 W/m \cdot K\) Temperature difference: \((35 - (-10)) ^{\circ}C = 45^{\circ}C\) Heat Load: Less than 500W
02

Determine Area of Passage for Heat Transfer

In this case, the heat transfer will occur through the side walls and the top of the freezer, which together make a total of 5 walls that are \(2m \times 2m\) each. So, calculate the total area: Area = 5 walls 脳 \(area_{wall}\) Area = 5 脳 \((2m)(2m)\) Area = \(20m^2\)
03

Calculate Minimum Thickness of Insulation Using Heat Conduction Formula

The heat conduction formula for plane walls is: \(Q = k \cdot \frac{A \cdot \Delta T}{d}\) Where Q is the heat transfer rate (less than 500W), A is the area through which heat is being transferred, 螖T is the temperature difference, and d is the thickness of insulation. For our given values, we must find the minimum d, thickness of insulation, to ensure the heat transfer rate is less than 500W. \(500 = 0.030 \cdot \frac{20 \times 45}{d}\)
04

Solve for Insulation Thickness (d)

Rearrange the equation and solve for d: \(d = \frac{0.030 \cdot (20 \times 45)}{500}\) Calculate the value: \(d = \frac{0.030 \cdot 900}{500}\) \(d = \frac{27}{500}\)
05

Convert Result to Appropriate Unit

The value obtained is in meters. Converting it to centimeters will provide a more practical unit. \(d = \frac{27}{500} m = \frac{27}{5} cm = 5.4 cm\)
06

State the Result

The minimum thickness of styrofoam insulation required for the top and side walls of the freezer compartment to ensure a heat load of less than 500W is \(5.4 cm\).

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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 key concept when discussing heat transfer. It defines how well a material can conduct heat. A higher thermal conductivity value means that the material is better at conducting heat. In the exercise given, the styrofoam insulation has a thermal conductivity value of 0.030 W/m路K. This is relatively low, signifying that styrofoam is a poor conductor of heat and thus a good insulator.
Materials used for insulation ideally should have low thermal conductivity. This property is crucial when designing systems needing efficient thermal regulation like freezers.
  • It ensures minimal heat flow, keeping unwanted heat out of the freezer.
  • Calculating the correct thickness of the insulating material ensures the system operates effectively and efficiently.
Styrofoam Insulation
Styrofoam, also known as expanded polystyrene, is widely utilized for insulation due to its excellent insulating properties. It consists of small plastic foam beads, providing it with low thermal conductivity, as discussed earlier.
To determine the thickness needed for effective insulation, formulas such as the heat conduction formula are essential.
This material is often preferred in scenarios where keeping heat transfer to a minimum is vital because:
  • It is lightweight, which means it does not add much weight to structures.
  • It is a cost-effective solution for thermal insulation.
In the given problem, styrofoam's role is to maintain the freezer鈥檚 low temperature by limiting heat flow from the outside environment. Finding the right thickness minimizes heat load effectively, which is crucial for maintaining energy efficiency.
Temperature Difference
Temperature difference, often denoted as 螖T in formulas, is a driving factor for heat transfer. It is calculated by subtracting the cooler temperature from the warmer one. In simple terms, the greater the temperature difference, the greater the potential for heat transfer.
  • The exercise gives us a temperature difference of 45掳C, which is significant.
  • Larger differences result in more heat flowing unless counteracted by adequate insulation.
Understanding this factor is central when assessing the necessary insulation thickness. Greater temperature differences mean more insulation may be required to maintain desired conditions, in this case, to keep the heat load below 500 W.
Using tools like the heat conduction formula allows accurate planning and material usage, optimizing the freezer's performance.

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Most popular questions from this chapter

You've experienced convection cooling if you've ever extended your hand out the window of a moving vehicle or into a flowing water stream. With the surface of your hand at a temperature of \(30^{\circ} \mathrm{C}\), determine the convection heat flux for (a) a vehicle speed of \(35 \mathrm{~km} / \mathrm{h}\) in air at \(-5^{\circ} \mathrm{C}\) with a convection coefficient of 40 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and (b) a velocity of \(0.2 \mathrm{~m} / \mathrm{s}\) in a water stream at \(10^{\circ} \mathrm{C}\) with a convection coefficient of \(900 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Which condition would feel colder? Contrast these results with a heat loss of approximately \(30 \mathrm{~W} / \mathrm{m}^{2}\) under normal room conditions.

The inner and outer surface temperatures of a glass window \(5 \mathrm{~mm}\) thick are 15 and \(5^{\circ} \mathrm{C}\). What is the heat loss through a \(1 \mathrm{~m} \times 3 \mathrm{~m}\) window? The thermal conductivity of glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

1.42 One method for growing thin silicon sheets for photovoltaic solar panels is to pass two thin strings of high melting temperature material upward through a bath of molten silicon. The silicon solidifies on the strings near the surface of the molten pool, and the solid silicon sheet is pulled slowly upward out of the pool. The silicon is replenished by supplying the molten pool with solid silicon powder. Consider a silicon sheet that is \(W_{\mathrm{si}}=85 \mathrm{~mm}\) wide and \(t_{\mathrm{si}}=150 \mu \mathrm{m}\) thick that is pulled at a velocity of \(V_{\mathrm{si}}=20 \mathrm{~mm} / \mathrm{min}\). The silicon is melted by supplying electric power to the cylindrical growth chamber of height \(H=350 \mathrm{~mm}\) and diameter \(D=300 \mathrm{~mm}\). The exposed surfaces of the growth chamber are at \(T_{s}=\) \(320 \mathrm{~K}\), the corresponding convection coefficient at the exposed surface is \(h=8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the surface is characterized by an emissivity of \(\varepsilon_{s}=0.9\). The solid silicon powder is at \(T_{\mathrm{s}, i}=298 \mathrm{~K}\), and the solid silicon sheet exits the chamber at \(T_{\text {si, } o}=420 \mathrm{~K}\). Both the surroundings and ambient temperatures are \(T_{\infty}=T_{\text {sur }}=298 \mathrm{~K}\). (a) Determine the electric power, \(P_{\text {elec }}\), needed to operate the system at steady state. (b) If the photovoltaic panel absorbs a time-averaged solar flux of \(q_{\text {sol }}^{\prime \prime}=180 \mathrm{~W} / \mathrm{m}^{2}\) and the panel has a conversion efficiency (the ratio of solar power absorbed to electric power produced) of \(\eta=0.20\), how long must the solar panel be operated to produce enough electric energy to offset the electric energy that was consumed in its manufacture?

A vertical slab of Wood's metal is joined to a substrate on one surface and is melted as it is uniformly irradiated by a laser source on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\), and the melt runs off by gravity as soon as it is formed. The absorptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{s f}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantaneous rate of melting in \(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m}^{2}\) if the laser irradiation is \(5 \mathrm{~kW} / \mathrm{m}^{2}\). How much material is removed if irradiation is maintained for a period of \(2 \mathrm{~s}\) ? (b) Allowing for convection to ambient air, with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \((\varepsilon=0.4\), \(T_{\text {sur }}=20^{\circ} \mathrm{C}\) ), determine the instantaneous rate of melting during irradiation.

A common procedure for measuring the velocity of an airstream involves the insertion of an electrically heated wire (called a hot-wire anemometer) into the airflow, with the axis of the wire oriented perpendicular to the flow direction. The electrical energy dissipated in the wire is assumed to be transferred to the air by forced convection. Hence, for a prescribed electrical power, the temperature of the wire depends on the convection coefficient, which, in turn, depends on the velocity of the air. Consider a wire of length \(L=20 \mathrm{~mm}\) and diameter \(D=0.5 \mathrm{~mm}\), for which a calibration of the form \(V=6.25 \times 10^{-5} h^{2}\) has been determined. The velocity \(V\) and the convection coefficient \(h\) have units of \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{W} / \mathrm{m}^{2}+\mathrm{K}\), respectively. In an application involving air at a temperature of \(T_{\infty}=25^{\circ} \mathrm{C}\), the surface temperature of the anemometer is maintained at \(T_{s}=75^{\circ} \mathrm{C}\) with a voltage drop of \(5 \mathrm{~V}\) and an electric current of \(0.1 \mathrm{~A}\). What is the velocity of the air?
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