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Chapter 1: Problem 8

An inexpensive food and beverage container is fabricuted from 25-mm-thick polystyrene \((k=0.023 \mathrm{~W} / \mathrm{m}-\mathrm{K})\) and has interior dimensions of \(0.8 \mathrm{~m} \times 0.6 \mathrm{~m} \times 0.6 \mathrm{~m}\). Under conditions for which an inner surface temperature of approximately \(2^{*} \mathrm{C}\) is maintained by an ice-water mixture and an outer surface temperature of \(20^{\circ} \mathrm{C}\) is maintained by the ambient, what is the heat flux through the container wall? Assuming negligible heat gain through the \(0.8 \mathrm{~m} \times\) \(0.6 \mathrm{~m}\) base of the cooler, what is the total heat load for the prescribed conditions?

Short Answer

Expert verified
The total heat load is approximately 31.79 W.

Step by step solution

01

Calculate Surface Area

The cooler's heat flux will occur through the top and side walls, ignoring the base. The dimensions for the internal walls are given as 0.8m, 0.6m, and 0.6m. Calculate the area for each surface:- Two sides with dimensions 0.8m x 0.6m: \( A_{1} = 2 \times 0.8 \times 0.6 = 0.96 \text{ m}^2 \)- The top with dimensions 0.8m x 0.6m: \( A_{2} = 0.8 \times 0.6 = 0.48 \text{ m}^2 \)Total Surface Area, \( A = 0.96 + 0.48 + 0.48 = 1.92 \text{ m}^2 \).
02

Calculate Heat Flux Through the Container Wall

The heat flux, \( q \), through the wall is given by Fourier's law:\[ q = \frac{k \cdot \Delta T}{d} \]where:- \( k = 0.023 \text{ W/m-K} \) is the thermal conductivity of the polystyrene material,- \( \Delta T = 20 - 2 = 18 \text{ K} \) is the temperature difference between the inner and outer walls,- \( d = 0.025 \text{ m} \) is the thickness of the wall.Substituting values, we get:\[ q = \frac{0.023 \cdot 18}{0.025} = 16.56 \text{ W/m}^2 \].
03

Calculate Total Heat Load

The total heat load, \( Q \), is given by the product of heat flux and total surface area:\[ Q = q \times A \]Substituting the previously found values:\[ Q = 16.56 \times 1.92 = 31.79 \text{ W} \].

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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 property that indicates how well a material can conduct heat. Think of it as a pathway for thermal energy to move through a material. The higher the thermal conductivity, the better the material is at transferring heat. Polystyrene, a type of insulating material, has a thermal conductivity of \( 0.023 \ \text{W/m-K} \). This low value suggests it鈥檚 a good insulator, reducing the rate of heat transfer.
Metals generally have high thermal conductivity, meaning they are efficient heat conductors, whereas materials like wood or polystyrene have much lower values.
  • Thermal conductivity (\(k\)) is measured in watts per meter per Kelvin (\(\text{W/m-K}\)).
  • Materials with low thermal conductivity are ideal for insulation purposes.
In our exercise, the polystyrene container is effective at maintaining the temperature inside, given that the low \(k\) value means slower heat loss from its contents to the outside.
Fourier's Law
Fourier's Law provides a way to calculate the heat transfer through a material. It states that the rate at which heat is transferred through a material is proportional to the negative gradient of temperature and the area through which it is being transferred.
Here's the formula used: \[ q = \frac{k \cdot \Delta T}{d} \] Where \(q\) is the heat flux (amount of heat transferred per unit area), \(k\) is the thermal conductivity, \( \Delta T \) is the temperature difference, and \(d\) is the material thickness.
  • Fourier's Law highlights that with greater temperature differences between surfaces, the greater the heat transfer.
  • The formula indicates that thinner materials (decreasing \(d\)), increase the rate of heat transfer, and vice versa.
For the example in the question, using Fourier鈥檚 law allows us to calculate the heat flux through the polystyrene walls of the container, which is consistent with real-world applications where controlling heat transfer is essential.
Heat Flux
Heat flux is a term used to describe the rate of heat transfer per unit area. It's an essential aspect of understanding how heat moves through different materials. Measured in watts per square meter (\( \text{W/m}^2 \)), it tells us how effectively heat is being transferred through a surface.
In this exercise, the heat flux was calculated using Fourier's Law, revealing a value of \( 16.56 \ \text{W/m}^2 \) for the polystyrene cooler.
  • Heat flux helps decide how well a container or material can keep temperature consistent inside it.
  • It determines how much thermal energy is entering or leaving a system, like our cooler.
Understanding heat flux is vital in applications like refrigeration, building insulation, and even clothing materials where maintaining temperatures is crucial. In this scenario, layering more material or using materials with lower thermal conductivity could decrease the heat flux, thus improving the cooler's insulation properties.

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Most popular questions from this chapter

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 \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

The concrete slab of a basement is \(11 \mathrm{~m}\) long. \(8 \mathrm{~m}\) wide, and \(0.20 \mathrm{~m}\) thick. During the winter, temperatures are nominally \(17^{\circ} \mathrm{C}\) and \(10^{\circ} \mathrm{C}\) at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what is the rate of heat loss through the slab? If the basemient is heated by a gas furnace operating at an cfficiency of \(\eta_{r}=0.90\) and natural gas is priced at \(C_{r}=50.01 \mathrm{MJ}\), what is the daily cost of the heat loss?

A furnace for processing semiconductor materials is formed by a silicon carbide chamber that is zone heated on the top section and cooled on the lower section. With the elevator in the lowest position, a robot arm inserts the silicon wafer on the mounting pins. In a production operation, the wafer is rapidly moved toward the hot zone to achieve the temperature-time history required for the process recipe. In this position the top and boltom surfaces of the wafer exchange radiation with the hot and cool rones, respectively, of the chamber. The zone temperatures are \(T_{\mathrm{a}}=1500 \mathrm{~K}\) and \(T_{c}=\) \(330 \mathrm{~K}\), and the emissivity and thickness of the wafer are \(e=0.65\) and \(d=0.78 \mathrm{~mm}\), respectively. With the ambient gas at \(T_{w}=700 \mathrm{~K}\), convection coefficients at the upper and lower surfaces of the wafer are 8 and \(4 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\). respectively. The silicon wafer has a den. sity of \(2700 \mathrm{~kg} / \mathrm{m}^{3}\) and a specific heat of \(875 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (a) For an initial condition corresponding fo a wafer tempersture of \(T_{w i}=300 \mathrm{~K}\) and the position of the wafer shawn schematically. determine the corresponding time rate of change of the wafer tempereture, \(\left(d T_{w} / d r\right)_{0}\). (b) Determine the steady state temperature reached by the wafer if it remains in this position. How significant is convection heat transfer for this situation? Sketch how you would expect the wafer temperature to vary as a function of vertical distance.

An overhead 25-m-long, uninsulated industrial steam pipe of \(100 \mathrm{~mm}\) diameter is routed through a building whose walls and air are at \(25^{\circ} \mathrm{C}\). Pressurized steam maintains a pipe surface temperature of \(150^{7} \mathrm{C}\), and the coeffcient associated with natural convection is \(h=10\) \(W / m^{2} \cdot K\). The surface emissivity is \(\varepsilon=0.8\). (a) What is the rate of heat loss from the steam line? (b) If the steatm is generated in a gas-fired boiler operating at an efficiency of \(\eta f=0.90\) and natural gas is priced at \(C_{\varepsilon}=50.01\) per MJ, what is the annual cost of heat loss from the line?

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.
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