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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

An internally reversible refrigerator has a modified coefficient of performance accounting for realistic heat transfer processes of $$ \mathrm{COP}_{m}=\frac{q_{\text {in }}}{\dot{W}}=\frac{q_{\text {in }}}{q_{\text {out }}-q_{\text {in }}}=\frac{T_{c, i}}{T_{h, i}-T_{c, i}} $$ where \(q_{\text {in }}\) is the refrigerator cooling rate, \(q_{\text {out }}\) is the heat rejection rate, and \(\dot{W}\) is the power input. Show that \(\mathrm{COP}_{m}\) can be expressed in terms of the reservoir temperatures \(T_{c}\) and \(T_{h}\), the cold and hot thermal resistances \(R_{L, c}\) and \(R_{t, h}\), and \(q_{\text {in }}\), as $$ \mathrm{COP}_{m}=\frac{T_{c}-q_{\mathrm{in}} R_{\mathrm{tot}}}{T_{h}-T_{c}+q_{\mathrm{in}} R_{\mathrm{tot}}} $$ where \(R_{\mathrm{tot}}=R_{t, c}+R_{t, h}\). Also, show that the power input may be expressed as $$ \dot{W}=q_{\mathrm{in}} \frac{T_{h}-T_{c}+q_{\mathrm{in}} R_{\mathrm{id \textrm {t }}}}{T_{c}-q_{\mathrm{in}} R_{\mathrm{tot}}} $$

A concrete wall, which has a surface area of \(20 \mathrm{~m}^{2}\) and is \(0.30 \mathrm{~m}\) thick, separates conditioned room air from ambient air. The temperature of the inner surface of the wall is maintained at \(25^{\circ} \mathrm{C}\), and the thermal conductivity of the concrete is \(1 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). (a) Determine the heat loss through the wall for outer surface temperatures ranging from \(-15^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\), which correspond to winter and summer extremes, respectively. Display your results graphically. (b) On your graph, also plot the heat loss as a function of the outer surface temperature for wall materials having thermal conductivities of \(0.75\) and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Explain the family of curves you have obtained.

A glass window of width \(W=1 \mathrm{~m}\) and height \(H=2 \mathrm{~m}\) is \(5 \mathrm{~mm}\) thick and has a thermal conductivity of \(k_{g}=\) \(1.4 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). If the inner and outer surface temperatures of the glass are \(15^{\circ} \mathrm{C}\) and \(-20^{\circ} \mathrm{C}\), respectively, on a cold winter day, what is the rate of heat loss through the glass? To reduce heat loss through windows, it is customary to use a double pane construction in which adjoining panes are separated by an air space. If the spacing is \(10 \mathrm{~mm}\) and the glass surfaces in contact with the air have temperatures of \(10^{\circ} \mathrm{C}\) and \(-15^{\circ} \mathrm{C}\), what is the rate of heat loss from a \(1 \mathrm{~m} \times 2 \mathrm{~m}\) window? The themal conductivity of air is \(k_{a}=0.024 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A wall is made from an inhomogeneous (nonuniform) material for which the thermal conductivity varies through the thickness according to \(k=a x+b\), where \(a\) and \(b\) are constants. The heat flux is known to be constant. Determine expressions for the temperature gradient and the temperature distribution when the surface at \(x=0\) is at temperature \(T_{1}\).

Most of the energy we consume as food is converted to thermal energy in the process of performing all our bodily functions and is ultimately lost as heat from our bodies. Consider a person who consumes \(2100 \mathrm{kcal}\) per day (note that what are commonly referred to as food calories are actually kilocalories), of which \(2000 \mathrm{kcal}\) is converted to thermal energy. (The remaining \(100 \mathrm{kcal}\) is used to do work on the environment.) The person has a surface area of \(1.8 \mathrm{~m}^{2}\) and is dressed in a bathing suit. (a) The person is in a room at \(20^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). At this air temperature, the person is not perspiring much. Estimate the person's average skin temperature. (b) If the temperature of the environment were \(33^{\circ} \mathrm{C}\), what rate of perspiration would be needed to maintain a comfortable skin temperature of \(33^{\circ} \mathrm{C}\) ?
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