/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 An inexpensive food and beverage... [FREE SOLUTION] | 91Ó°ÊÓ

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An inexpensive food and beverage container is fabricated from 25 -mm-thick polystyrene \((k=0.023 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) 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 \(2^{\circ} \mathrm{C}\) 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 heat flux through the container wall is \(16.56\ \mathrm{W/m^2}\), and the total heat load, excluding the base, is approximately \(43.72\ \mathrm{W}\).

Step by step solution

01

Identify the variables and constants

- \(k = 0.023 \mathrm{~W/m \cdot K}\) (thermal conductivity) - \(t = 0.025 \mathrm{~m}\) (thickness) - Inner surface temperature, \(T_{in} = 2^{\circ} \mathrm{C}\) - Outer surface temperature, \(T_{out} = 20^{\circ} \mathrm{C}\) - Interior dimensions: \(0.8 \mathrm{~m} \times 0.6 \mathrm{~m} \times 0.6 \mathrm{~m}\)
02

Calculate the temperature difference

To find the temperature difference across the wall, we will subtract the inner surface temperature from the outer surface temperature: \(ΔT = T_{out} - T_{in} = 20 - 2 = 18^{\circ} \mathrm{C}\)
03

Apply Fourier's law of heat conduction

Fourier's law states that the heat flux (q) through a material is proportional to the temperature gradient and the material's thermal conductivity: \(q = -k \cdot \frac{ΔT}{t}\) Now, plugging in the values provided: \(q = -0.023 \frac{18}{0.025}\) \(q = -16.56\ \mathrm{W/m^2}\) We have a negative result because the heat flux is directed from high temperature to low temperature. In this case, the heat is flowing from the outer surface of the container to the inner surface. However, for calculation purposes, we can consider the magnitude of the heat flux, which is \(16.56\ \mathrm{W/m^2}\).
04

Calculate the surface area excluding the base

To find the area through which the heat flows, we first need to calculate the surface area excluding the base. The container has five faces: two measuring \(0.8\ \mathrm{m} \times 0.6\ \mathrm{m}\), and three measuring \(0.6\ \mathrm{m} \times 0.6\ \mathrm{m}\). So, \(A = 2(0.8 \times 0.6) + 3(0.6 \times 0.6)\) \(A = 2.64\ \mathrm{m^2}\)
05

Calculate the total heat load

Now we can find the total heat load (Q) by multiplying the heat flux by the surface area: \(Q = q \cdot A\) \(Q = 16.56 \cdot 2.64\) \(Q = 43.72\ \mathrm{W}\) The total heat load through the container wall, excluding the base, is approximately \(43.72\ \mathrm{W}\).

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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 crucial concept in understanding how heat transfers through materials. It is defined as the ability of a material to conduct heat. Each material has its own value of thermal conductivity, denoted by the symbol, \( k \). This is measured in watts per meter Kelvin (\( \text{W/m} \cdot \text{K} \)), and highlights how well the material can carry thermal energy.
In the context of our problem, polystyrene has a thermal conductivity (\( k = 0.023 \, ext{W/m} \cdot \text{K}\)), which is quite low. This means that polystyrene is a poor conductor of heat, making it an ideal candidate for insulating purposes.
When heat conduction occurs, it is driven by the temperature difference across the material. In simpler terms, the greater the difference in temperature from one side of the material to the other, the more heat will flow. Here, a higher conductivity would mean more heat carries through the polystyrene, leading to potential loss of insulation capabilities.
Heat Flux
Heat flux is the rate at which heat energy passes through a surface. Think of it as the 'speed' of heat transfer through a certain area. It is expressed in watts per square meter (\( \text{W/m}^2 \)). To calculate the heat flux for our container, we used Fourier’s Law of Heat Conduction. The equation \( q = -k \cdot \frac{\Delta T}{t} \) helps in determining how much heat moves across the container wall.
Let's break down this formula:
  • \( k \) represents the material's thermal conductivity.
  • \( \Delta T \) is the temperature difference between the inner and outer surfaces, given as \( 18^{\circ} \, \text{C} \).
  • \( t \) is the thickness of the container wall, which is \( 0.025 \, \text{m} \).
The minus sign in the equation is indicative of the heat naturally moving from the hotter side to the cooler side, which aligns with heat's inherent tendency to flow towards lower energy levels. The calculated heat flux here is \( 16.56 \, \text{W/m}^2 \), indicating how fast heat is transferred from the container's exterior to the cooler interior.
Surface Area Calculation
Calculating the surface area of an object is essential to understanding how much of it can exchange heat with its surroundings. Here, we're looking at a container without accounting for its base, which is assumed to have negligible heat transfer.
The container in question has five faces; two with dimensions of \( 0.8 \, \text{m} \times 0.6 \, \text{m} \) and three with dimensions of \( 0.6 \, \text{m} \times 0.6 \, \text{m}\). By calculating and summing these individual surface areas, we derived the total surface area excluding the base, \( A = 2(0.8 \times 0.6) + 3(0.6 \times 0.6) = 2.64 \, \text{m}^2 \).
This calculation tells us the total area that is actively contributing to heat loss. It's an important aspect to ensure that energy efficiency is maintained, especially for keeping food and beverages insulated from external temperature conditions. By understanding the area involved in heat transfer, engineers can design better insulating methods and materials.

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Most popular questions from this chapter

The free convection heat transfer coefficient 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 cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d t)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\circ} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

In analyzing the performance of a thermal system, the engineer must be able to identify the relevant heat transfer processes. Only then can the system behavior be properly quantified. For the following systems, identify the pertinent processes, designating them by appropriately labeled arrows on a sketch of the system. Answer additional questions that appear in the problem statement. (a) Identify the heat transfer processes that determine the temperature of an asphalt pavement on a summer day. Write an energy balance for the surface of the pavement. (b) Microwave radiation is known to be transmitted by plastics, glass, and ceramics but to be absorbed by materials having polar molecules such as water. Water molecules exposed to microwave radiation align and reverse alignment with the microwave radiation at frequencies up to \(10^{9} \mathrm{~s}^{-1}\), causing heat to be generated. Contrast cooking in a microwave oven with cooking in a conventional radiant or convection oven. In each case, what is the physical mechanism responsible for heating the food? Which oven has the greater energy utilization efficiency? Why? Microwave heating is being considered for drying clothes. How would the operation of a microwave clothes dryer differ from a conventional dryer? Which is likely to have the greater energy utilization efficiency? Why? (c) To prevent freezing of the liquid water inside the fuel cell of an automobile, the water is drained to an onboard storage tank when the automobile is not in use. (The water is transferred from the tank back to the fuel cell when the automobile is turned on.) Consider a fuel cell-powered automobile that is parked outside on a very cold evening with \(T_{\infty}=-20^{\circ} \mathrm{C}\). The storage tank is initially empty at \(T_{i, t}=-20^{\circ} \mathrm{C}\), when liquid water, at atmospheric pressure and temperature \(T_{i, w}=50^{\circ} \mathrm{C}\), is introduced into the tank. The tank has a wall thickness \(t_{t}\) and is blanketed with insulation of thickness \(t_{\text {ins }}\). Identify the heat transfer processes that will promote freezing of the water. Will the likelihood of freezing change as the insulation thickness is modified? Will the likelihood of freezing depend on the tank wall's thickness and material? Would freezing of the water be more likely if plastic (low thermal conductivity) or stainless steel (moderate thermal conductivity) tubing is used to transfer the water to and from the tank? Is there an optimal tank shape that would minimize the probability of the water freezing? Would freezing be more likely or less likely to occur if a thin sheet of aluminum foil (high thermal conductivity, low emissivity) is applied to the outside of the insulation? (d) Your grandmother is concerned about reducing her winter heating bills. Her strategy is to loosely fit rigid polystyrene sheets of insulation over her double-pane windows right after the first freezing weather arrives in the autumn. Identify the relevant heat transfer processes on a cold winter night when the foamed insulation sheet is placed (i) on the inner surface and (ii) on the outer surface of her window. To avoid condensation damage, which configuration is preferred? Condensation on the window pane does not occur when the foamed insulation is not in place. (e) There is considerable interest in developing building materials with improved insulating qualities. The development of such materials would do much to enhance energy conservation by reducing space heating requirements. It has been suggested that superior structural and insulating qualities could be obtained by using the composite shown. The material consists of a honeycomb, with cells of square cross section, sandwiched between solid slabs. The cells are filled with air, and the slabs, as well as the honeycomb matrix, are fabricated from plastics of low thermal conductivity. For heat transfer normal to the slabs, identify all heat transfer processes pertinent to the performance of the composite. Suggest ways in which this performance could be enhanced. (f) A thermocouple junction (bead) is used to measure the temperature of a hot gas stream flowing through a channel by inserting the junction into the mainstream of the gas. The surface of the channel is cooled such that its temperature is well below that of the gas. Identify the heat transfer processes associated with the junction surface. Will the junction sense a temperature that is less than, equal to, or greater than the gas temperature? A radiation shield is a small, openended tube that encloses the thermocouple junction, yet allows for passage of the gas through the tube. How does use of such a shield improve the accuracy of the temperature measurement? (g) A double-glazed, glass fire screen is inserted between a wood-burning fireplace and the interior of a room. The screen consists of two vertical glass plates that are separated by a space through which room air may flow (the space is open at the top and bottom). Identify the heat transfer processes associated with the fire screen. (h) A thermocouple junction is used to measure the temperature of a solid material. The junction is inserted into a small circular hole and is held in place by epoxy. Identify the heat transfer processes associated with the junction. Will the junction sense a temperature less than, equal to, or greater than the solid temperature? How will the thermal conductivity of the epoxy affect the junction temperature?

A vacuum system, as used in sputtering electrically conducting thin films on microcircuits, is comprised of a baseplate maintained by an electrical heater at \(300 \mathrm{~K}\) and a shroud within the enclosure maintained at \(77 \mathrm{~K}\) by a liquid-nitrogen coolant loop. The circular baseplate, insulated on the lower side, is \(0.3 \mathrm{~m}\) in diameter and has an emissivity of \(0.25\). (a) How much electrical power must be provided to the baseplate heater? (b) At what rate must liquid nitrogen be supplied to the shroud if its heat of vaporization is \(125 \mathrm{~kJ} / \mathrm{kg}\) ? (c) To reduce the liquid nitrogen consumption, it is proposed to bond a thin sheet of aluminum foil \((\varepsilon=0.09)\) to the baseplate. Will this have the desired effect?

The heat flux through a wood slab \(50 \mathrm{~mm}\) thick, whose inner and outer surface temperatures are 40 and \(20^{\circ} \mathrm{C}\), respectively, has been determined to be \(40 \mathrm{~W} / \mathrm{m}^{2}\). What is the thermal conductivity of the wood?

Annealing, an important step in semiconductor materials processing, can be accomplished by rapidly heating the silicon wafer to a high temperature for a short period of time. The schematic shows a method involving the use of a hot plate operating at an elevated temperature \(T_{h}\). The wafer, initially at a temperature of \(T_{w, i}\), is suddenly positioned at a gap separation distance \(L\) from the hot plate. The purpose of the analysis is to compare the heat fluxes by conduction through the gas within the gap and by radiation exchange between the hot plate and the cool wafer. The initial time rate of change in the temperature of the wafer, \(\left(d T_{w} / d t\right)_{i}\), is also of interest. Approximating the surfaces of the hot plate and the wafer as blackbodies and assuming their diameter \(D\) to be much larger than the spacing \(L\), the radiative heat flux may be expressed as \(q_{\text {rad }}^{\prime \prime}=\sigma\left(T_{h}^{4}-T_{w}^{4}\right)\). The silicon wafer has a thickness of \(d=0.78 \mathrm{~mm}\), a density of \(2700 \mathrm{~kg} / \mathrm{m}^{3}\), and a specific heat of 875 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\). The thermal conductivity of the gas in the gap is \(0.0436 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) For \(T_{h}=600^{\circ} \mathrm{C}\) and \(T_{w, i}=20^{\circ} \mathrm{C}\), calculate the radiative heat flux and the heat flux by conduction across a gap distance of \(L=0.2 \mathrm{~mm}\). Also determine the value of \(\left(d T_{\mathrm{w}} / d t\right)_{i}\), resulting from each of the heating modes. (b) For gap distances of \(0.2,0.5\), and \(1.0 \mathrm{~mm}\), determine the heat fluxes and temperature-time change as a function of the hot plate temperature for \(300 \leq\) \(T_{h} \leq 1300^{\circ} \mathrm{C}\). Display your results graphically. Comment on the relative importance of the two heat

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