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Chapter 5: Problem 87

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: \(254 \ \mathrm{mm} \ 脳 254 \ \mathrm{mm} = 254^2 \times 10^{-6} \ \mathrm{m^2}\)
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 Section \(5.2\) we noted that the value of the Biot number significantly influences the nature of the temperature distribution in a solid during a transient conduction process. Reinforce your understanding of this important concept by using the IHT model for one-dimensional transient conduction to determine radial temperature distributions in a 30 -mm-diameter, stainless steel rod \(\left(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8000 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=475 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\), as it is cooled from an initial uniform temperature of \(325^{\circ} \mathrm{C}\) by a fluid at \(25^{\circ} \mathrm{C}\). For the following values of the convection coefficient and the designated times, determine the radial temperature distribution: \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) \((t=0,100,500 \mathrm{~s}) ; h=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}(t=0,10,50 \mathrm{~s})\); \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}(t=0,1,5,25 \mathrm{~s})\). Prepare a separate graph for each convection coefficient, on which temperature is plotted as a function of dimensionless radius at the designated times.

A chip that is of length \(L=5 \mathrm{~mm}\) on a side and thickness \(t=1 \mathrm{~mm}\) is encased in a ceramic substrate, and its exposed surface is convectively cooled by a dielectric liquid for which \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=20^{\circ} \mathrm{C}\). In the off-mode the chip is in thermal equilibrium with the coolant \(\left(T_{i}=T_{\infty}\right)\). When the chip is energized, however, its temperature increases until a new steady state is established. For purposes of analysis, the energized chip is characterized by uniform volumetric heating with \(\dot{q}=9 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Assuming an infinite contact resistance between the chip and substrate and negligible conduction resistance within the chip, determine the steady-state chip temperature \(T_{f}\). Following activation of the chip, how long does it take to come within \(1^{\circ} \mathrm{C}\) of this temperature? The chip density and specific heat are \(\rho=2000 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=700 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

Consider the fuel element of Example 5.11. Initially, the element is at a uniform temperature of \(250^{\circ} \mathrm{C}\) with no heat generation. Suddenly, the element is inserted into the reactor core, causing a uniform volumetric heat generation rate of \(\dot{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\). The surfaces are convectively cooled with \(T_{\infty}=250^{\circ} \mathrm{C}\) and \(h=1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the explicit method with a space increment of \(2 \mathrm{~mm}\), determine the temperature distribution \(1.5 \mathrm{~s}\) after the element is inserted into the core.

Steel balls \(12 \mathrm{~mm}\) in diameter are annealed by heating to \(1150 \mathrm{~K}\) and then slowly cooling to \(400 \mathrm{~K}\) in an air environment for which \(T_{\infty}=325 \mathrm{~K}\) and \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the properties of the steel to be \(k=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c=600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), estimate the time required for the cooling process. .

For each of the following cases, determine an appropriate characteristic length \(L_{c}\) and the corresponding Biot number \(B i\) that is associated with the transient thermal response of the solid object. State whether the lumped capacitance approximation is valid. If temperature information is not provided, evaluate properties at \(T=300 \mathrm{~K}\). (a) A toroidal shape of diameter \(D=50 \mathrm{~mm}\) and cross-sectional area \(A_{c}=5 \mathrm{~mm}^{2}\) is of thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The surface of the torus is exposed to a coolant corresponding to a convection coefficient of \(h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) A long, hot AISI 304 stainless steel bar of rectangular cross section has dimensions \(w=3 \mathrm{~mm}\), \(W=5 \mathrm{~mm}\), and \(L=100 \mathrm{~mm}\). The bar is subjected to a coolant that provides a heat transfer coefficient of \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at all exposed surfaces. (c) A long extruded aluminum (Alloy 2024) tube of inner and outer dimensions \(w=20 \mathrm{~mm}\) and \(W=24 \mathrm{~mm}\), respectively, is suddenly submerged in water, resulting in a convection coefficient of \(h=37 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the four exterior tube surfaces. The tube is plugged at both ends, trapping stagnant air inside the tube. (d) An \(L=300-m m\)-long solid stainless steel rod of diameter \(D=13 \mathrm{~mm}\) and mass \(M=0.328 \mathrm{~kg}\) is exposed to a convection coefficient of \(h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (e) A solid sphere of diameter \(D=12 \mathrm{~mm}\) and thermal conductivity \(k=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is suspended in a large vacuum oven with internal wall temperatures of \(T_{\text {sur }}=20^{\circ} \mathrm{C}\). The initial sphere temperature is \(T_{i}=100^{\circ} \mathrm{C}\), and its emissivity is \(\varepsilon=0.73\). (f) A long cylindrical rod of diameter \(D=20 \mathrm{~mm}\), density \(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\), specific heat \(c_{p}=1750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and thermal conductivity \(k=16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is suddenly exposed to convective conditions with \(T_{\infty}=20^{\circ} \mathrm{C}\). The rod is initially at a uniform temperature of \(T_{i}=200^{\circ} \mathrm{C}\) and reaches a spatially averaged temperature of \(T=100^{\circ} \mathrm{C}\) at \(t=225 \mathrm{~s}\). (g) Repeat part (f) but now consider a rod diameter of \(D=200 \mathrm{~mm}\).
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