/*! 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 87 A tile-iron consists of a massiv... [FREE SOLUTION] | 91Ó°ÊÓ

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A tile-iron consists of a massive plate maintained at \(150^{\circ} \mathrm{C}\) by an embedded electrical heater. The iron is placed in contact with a tile to soften the adhesive, allowing the tile to be easily lifted from the subflooring. The adhesive will soften sufficiently if heated above \(50^{\circ} \mathrm{C}\) for at least \(2 \mathrm{~min}\), but its temperature should not exceed \(120^{\circ} \mathrm{C}\) to avoid deterioration of the adhesive. Assume the tile and subfloor to have an initial temperature of \(25^{\circ} \mathrm{C}\) and to have equivalent thermophysical properties of \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=1.5 \times 10^{6}\) \(\mathrm{J} / \mathrm{m}^{3} \cdot \mathrm{K}\) Tile, 4-mm thickness Subflooring (a) How long will it take a worker using the tile-iron to lift a tile? Will the adhesive temperature exceed \(120^{\circ} \mathrm{C} ?\) (b) If the tile-iron has a square surface area \(254 \mathrm{~mm}\) to the side, how much energy has been removed from it during the time it has taken to lift the tile?

Short Answer

Expert verified
The worker will require 160 seconds (approximately 2 minutes and 40 seconds) to lift a tile, and the adhesive temperature will not exceed the desired limit of \(120^{\circ} \mathrm{C}\). The energy removed from the tile-iron during the process is approximately 9679.2 J.

Step by step solution

01

Define the given parameters

We are given the following values: Plate temperature: \(150^{\circ} \mathrm{C}\) Desired adhesive temperature range: \(50^{\circ} \mathrm{C} - 120^{\circ} \mathrm{C}\) Initial tile and subfloor temperature: \(25^{\circ} \mathrm{C}\) Thermophysical properties: \(k = 0.15 \ \mathrm{W/m \cdot K}\) and \(\rho c_{p} = 1.5 \times 10^{6} \ \mathrm{J/m^3 \cdot K}\) Tile thickness (L): \(4 \ \mathrm{mm} = 0.004 \ \mathrm{m}\) Tile-iron surface area: \(254 \ \mathrm{mm} \ × 254 \ \mathrm{mm} = 254^2 \times 10^{-6} \ \mathrm{m^2}\)
02

Apply one-dimensional unsteady-state energy balance equation

We will use the one-dimensional unsteady-state energy balance equation to calculate the time it takes for the adhesive to reach the desired temperature range: \[\frac{\partial T}{\partial t} = \frac{k}{\rho c_p} \frac{\partial^2 T}{\partial x^2}\]
03

Find the solution for the equation

To find the time it takes for the adhesive to reach the desired temperature range, we will use the heat transfer equation. Thus, the solution of the problem is given by: \[t = \frac{L^2}{4 \alpha}\] Where \(t\) is the time, \(L\) is the tile thickness, and \(\alpha = \frac{k}{\rho c_p}\) is the thermal diffusivity. Now, we will calculate the thermal diffusivity: \[\alpha = \frac{k}{\rho c_p} = \frac{0.15 \ \mathrm{W/m \cdot K}}{1.5 \times 10^6 \ \mathrm{J/m^3 \cdot K}} = 1 \times 10^{-7} \ \mathrm{m^2/s}\] Now, we will calculate the time: \[t = \frac{L^2}{4 \alpha} = \frac{(0.004 \ \mathrm{m})^2}{4 \times 1 \times 10^{-7} \ \mathrm{m^2/s}} = 160 \ \mathrm{s}\]
04

Determine if the adhesive temperature exceeds the limit

The worker will require 160 seconds (approximately 2 minutes and 40 seconds) to lift a tile. The maximum temperature that the adhesive can reach is \(120^{\circ} \mathrm{C}\), which is below the required limit. Therefore, the adhesive temperature will not exceed the desired limit.
05

Calculate the energy removed from the tile-iron during the process

We can determine the energy removed from the tile-iron using the following formula: \[Q = \rho c_p \Delta T \times V\] Where \(Q\) is the energy removed, \(\Delta T\) is the temperature difference, and \(V\) is the volume of the tile. The temperature difference, \(\Delta T\), is the initial temperature of the tile (\(25^{\circ} \mathrm{C}\)) and the temperature achieved by the adhesive (\(50^{\circ} \mathrm{C}\)). Thus, \(\Delta T = 25 \ \mathrm{K}\). Now, we will find the volume of the tile: \[V = A \times L\] \[V = (254^2 \times 10^{-6} \ \mathrm{m^2}) \times 0.004 \ \mathrm{m} = 2.58112 \times 10^{-4} \ \mathrm{m^3}\] Now, we will calculate the energy removed: \[Q = \rho c_p \Delta T \times V = (1.5 \times 10^6 \ \mathrm{J/m^3 \cdot K}) \times 25 \ \mathrm{K} \times 2.58112 \times 10^{-4} \ \mathrm{m^3} = 9679.2 \ \mathrm{J}\] So, the energy removed from the tile-iron during the process is approximately 9679.2 J.

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Key Concepts

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

Unsteady-State Energy Balance
Understanding the unsteady-state energy balance is crucial when it comes to problems involving heat transfer over time. This concept deals with scenarios where the temperature within a system changes as a function of time. Unlike steady-state conditions, where temperatures remain constant over time, unsteady-state or transient heat transfer describes how temperatures evolve.

For the case of the tile-iron exercise, we use a simplified model: a one-dimensional heat transfer. This is a reasonable assumption given the flatness and uniformity of the tile and the iron. The equation that characterizes this scenario is the heat equation, which in its partial differential form reads as \[\frac{\partial T}{\partial t} = \frac{k}{\rho c_p} \frac{\partial^2 T}{\partial x^2}\] where \(T\) is temperature, \(t\) is time, and \(x\) is the position along the thickness of the material. This equation is a statement of energy conservation, asserting that any change of heat in a region is due to heat entering or leaving the region via conduction.

When solving the tile-iron problem, we find that it takes approximately 160 seconds for the adhesive to reach the desired temperature, ensuring it is soft enough to lift the tile without causing deterioration.
Thermal Diffusivity
Thermal diffusivity is a measure of how quickly heat diffuses through a material. It plays a significant role in unsteady-state heat transfer as it influences the rate at which temperature changes within a material. The thermal diffusivity, \(\alpha\), is defined as the ratio of thermal conductivity, \(k\), to the product of density, \(\rho\), and specific heat capacity, \(c_{p}\): \[\alpha = \frac{k}{\rho c_p}\] Higher values of thermal diffusivity indicate that the material can conduct heat more rapidly compared to its capacity to store energy.

In the exercise, we calculated the thermal diffusivity of the tile and subfloor, which was found to be \(1 \times 10^{-7} \mathrm{m^2/s}\). This value was then used to determine the necessary time for the tile to reach a temperature that would soften the adhesive. Understanding thermal diffusivity is crucial for accurately predicting the heating or cooling behavior of materials in real-world applications.
Energy Removal Calculation
Calculating the amount of energy removed from a system is essential in energy management and efficiency studies. In the context of the tile-iron problem, this calculation lets us understand how much energy the tile-iron loses while it raises the temperature of the tile to soften the adhesive. The energy removed, \(Q\), can be calculated using the formula: \[Q = \rho c_p \Delta T \times V\] where \(\Delta T\) is the change in temperature and \(V\) is the volume of the material that the heat is being transferred into.

The purposeful calculation of energy removal can influence decisions on power consumption, which is especially pertinent to industrial processes or everyday tools such as a tile-iron. For our example, the energy removed while the tile reached the required temperature to soften the adhesive was 9679.2 Joules, indicating the amount of energy the electrical heater had to supply for the operation to be successful.

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Most popular questions from this chapter

In a manufacturing process, long rods of different diameters are at a uniform temperature of \(400^{\circ} \mathrm{C}\) in a curing oven, from which they are removed and cooled by forced convection in air at \(25^{\circ} \mathrm{C}\). One of the line operators has observed that it takes \(280 \mathrm{~s}\) for a \(40-\mathrm{mm}\) diameter rod to cool to a safe-to-handle temperature of \(60^{\circ} \mathrm{C}\). For an equivalent convection coefficient, how long will it take for an 80 -mm-diameter rod to cool to the same temperature? The thermophysical properties of the rod are \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3}, c=900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Comment on your result. Did you anticipate this outcome?

The convection coefficient for flow over a solid sphere may be determined by submerging the sphere, which is initially at \(25^{\circ} \mathrm{C}\), into the flow, which is at \(75^{\circ} \mathrm{C}\), and measuring its surface temperature at some time during the transient heating process. (a) If the sphere has a diameter of \(0.1 \mathrm{~m}\), a thermal conductivity of \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and a thermal diffusivity of \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), at what time will a surface temperature of \(60^{\circ} \mathrm{C}\) be recorded if the convection coefficient is \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? (b) Assess the effect of thermal diffusivity on the thermal response of the material by computing center and surface temperature histories for \(\alpha=10^{-6}\), \(10^{-5}\), and \(10^{-4} \mathrm{~m}^{2} / \mathrm{s}\). Plot your results for the period \(0 \leq t \leq 300 \mathrm{~s}\). In a similar manner, assess the effect of thermal conductivity by considering values of \(k=1.5,15\), and \(150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Circuit boards are treated by heating a stack of them under high pressure, as illustrated in Problem 5.45. The platens at the top and bottom of the stack are maintained at a uniform temperature by a circulating fluid. The purpose of the pressing-heating operation is to cure the epoxy, which bonds the fiberglass sheets, and impart stiffness to the boards. The cure condition is achieved when the epoxy has been maintained at or above \(170^{\circ} \mathrm{C}\) for at least \(5 \mathrm{~min}\). The effective thermophysical properties of the stack or book (boards and metal pressing plates) are \(k=0.613 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=2.73 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3} \cdot \mathrm{K}\) (a) If the book is initially at \(15^{\circ} \mathrm{C}\) and, following application of pressure, the platens are suddenly brought to a uniform temperature of \(190^{\circ} \mathrm{C}\), calculate the elapsed time \(t_{e}\) required for the midplane of the book to reach the cure temperature of \(170^{\circ} \mathrm{C}\). (b) If, at this instant of time, \(t=t_{e}\), the platen temperature were reduced suddenly to \(15^{\circ} \mathrm{C}\), how much energy would have to be removed from the book by the coolant circulating in the platen, in order to return the stack to its initial uniform temperature?

Two large blocks of different materials, such as copper and concrete, have been sitting in a room \(\left(23^{\circ} \mathrm{C}\right)\) for a very long time. Which of the two blocks, if either, will feel colder to the touch? Assume the blocks to be semi-infinite solids and your hand to be at a temperature of \(37^{\circ} \mathrm{C}\).

A plane wall of a furnace is fabricated from plain carbon steel \(\left(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7850 \mathrm{~kg} / \mathrm{m}^{3}, c=430 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) and is of thickness \(L=10 \mathrm{~mm}\). To protect it from the corrosive effects of the furnace combustion gases, one surface of the wall is coated with a thin ceramic film that, for a unit surface area, has a thermal resistance of \(R_{t, f}^{\prime \prime}=0.01 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The opposite surface is well insulated from the surroundings. At furnace start-up the wall is at an initial temperature of \(T_{i}=300 \mathrm{~K}\), and combustion gases at \(T_{\infty}=1300 \mathrm{~K}\) enter the furnace, providing a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the ceramic film. Assuming the film to have negligible thermal capacitance, how long will it take for the inner surface of the steel to achieve a temperature of \(T_{s, i}=1200 \mathrm{~K}\) ? What is the temperature \(T_{s, o}\) of the exposed surface of the ceramic film at this time?

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