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Chapter 5: Problem 57

In a tempering process, glass plate, which is initially at a uniform temperature \(T_{i}\), is cooled by suddenly reducing the temperature of both surfaces to \(T_{s}\). The plate is \(20 \mathrm{~mm}\) thick, and the glass has a thermal diffusivity of \(6 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). (a) How long will it take for the midplane temperature to achieve \(50 \%\) of its maximum possible temperature reduction? (b) If \(\left(T_{i}-T_{s}\right)=300^{\circ} \mathrm{C}\), what is the maximum temperature gradient in the glass at the time calculated in part (a)?

Short Answer

Expert verified
The time it takes for the midplane temperature to achieve 50% of its maximum possible temperature reduction is approximately \(119.5 \,s\). The maximum temperature gradient in the glass at that time is approximately \(-1336.4 \, ^\circ\text{C/m}\).

Step by step solution

01

(Step 1) Understand the temperature distribution equation.

Our primary tool in this problem is the temperature distribution equation for one-dimensional, unsteady-state heat conduction. For a plate of infinite width, like in this problem, the equation can be written as: \[ T(x,t) = T_s + (T_i - T_s) erf \left( \frac{x}{2 \sqrt{\alpha t}} \right)\] where \(t\) is the time, \(x\) is the position from the surface, \(\alpha\) is the thermal diffusivity, and erf is the error function defined as: \[erf(z) = \frac{2}{\sqrt{\pi}} \int_{0}^{z} e^{-u^2} du \]
02

(Step 2) Substitute variables and find midplane temperature.

To find the time it takes for the midplane temperature to achieve 50% of its maximum possible temperature reduction, we need to find the temperature at the midpoint of the plate \((x = w/2)\), where \(w = 20 \times 10^{-3} \text{m}\): \[T\left(\frac{w}{2}, t\right) = T_s + (T_i - T_s) erf \left( \frac{w}{4\sqrt{\alpha t}} \right)\] Because we're looking for the time it takes for the temperature to decrease by 50%, we can set the temperature in this equation as: \[T\left(\frac{w}{2}, t\right) = T_s + \frac{T_i - T_s}{2}\] Now let's find the time \(t\) for this condition to be satisfied.
03

(Step 3) Solve for the time.

We can see that: \[\frac{T_i - T_s}{2} = (T_i - T_s) erf \left( \frac{w}{4\sqrt{\alpha t}} \right)\] Divide both sides by \((T_i - T_s)\): \[0.5 = erf \left( \frac{w}{4\sqrt{\alpha t}} \right)\] Now, finding the inverse error function, we get: \[erf^{-1}(0.5) = \frac{w}{4\sqrt{\alpha t}}\] And solving for the time, \(t\): \[t = \frac{w^2}{16\alpha erf^{-1}(0.5)^2}\] Plugging in the given values, \(w = 20 \times 10^{-3} \text{m}\) and \(\alpha = 6 \times 10^{-7} \text{m}^2/\text{s}\). Also, erf^{-1}(0.5) 鈮� 0.4769. We can now calculate the time: \[t \approx \frac{(20 \times 10^{-3})^2}{16(6 \times 10^{-7})(0.4769)^2} \approx 119.5 \text{s}\] The required time is approximately 119.5 seconds.
04

(Step 4) Find the maximum temperature gradient.

In part (b), we need to find the maximum temperature gradient in the glass at the time calculated in part (a). First, find the equation for the temperature derivative with respect to x: \[\frac{dT(x,t)}{dx} = -\frac{(T_i - T_s)}{\sqrt{\pi \alpha t}}e^{-x^2/4\alpha t}\] Because the temperature gradient is highest at the surface (\(x = 0\)), we can calculate the maximum temperature gradient as: \[\frac{dT(0,t)}{dx} = -\frac{(T_i - T_s)}{\sqrt{\pi \alpha t}}\] Now plug in the values from the given information and from part (a): \[\frac{dT}{dx} = -\frac{300}{\sqrt{\pi (6 \times 10^{-7})(119.5)}} \approx -1336.4 \frac{^{\circ}\text{C}}{\text{m}}\] The maximum temperature gradient in the glass at the time calculated in part (a) is approximately -1336.4 掳C/m

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Key Concepts

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

Temperature Distribution
In heat conduction problems like the tempering process, understanding how temperature changes over time and space is crucial. Temperature distribution refers to how temperature values vary across a medium, such as the glass plate mentioned in the exercise.

For heat conduction in an infinitely wide plate, we use the unsteady-state heat conduction equation. This is given by:\[T(x,t) = T_s + (T_i - T_s) \cdot \text{erf} \left( \frac{x}{2 \sqrt{\alpha t}} \right)\]

Here,
  • \(T(x,t)\) is the temperature at a position \(x\) and time \(t\).
  • \(T_s\) is the surface temperature.
  • \(T_i\) is the initial temperature of the plate.
  • \(\alpha\) represents thermal diffusivity.
  • The term \(\frac{x}{2 \sqrt{\alpha t}}\) represents the normalized position and time variable.
Calculating the exact temperature at any given time or position allows us to predict how the material will behave under thermal stresses, crucial for designing processes like glass tempering.
Thermal Diffusivity
Thermal diffusivity reflects how quickly heat can spread throughout a material. It's a property of the material and plays a critical role in heat conduction problems.

Denoted by \(\alpha\), thermal diffusivity is defined as:\[\alpha = \frac{k}{\rho c_p}\]
  • \(k\) is the thermal conductivity of the material.
  • \(\rho\) is the density.
  • \(c_p\) represents the specific heat capacity.
It tells us how easily a material can change its temperature, given a thermal input. In practical terms, a higher diffusivity means the material heats up and cools down faster since heat moves through it more quickly.

In the exercise, the glass has a thermal diffusivity of \(6 \times 10^{-7} \: \text{m}^2/\text{s}\). This helps in computing how long it takes for temperature changes, such as reaching a certain percentage of the maximum temperature reduction at the midplane.
Error Function
The error function, often abbreviated as erf, is a mathematical function used in heat conduction and diffusion problems. It helps us understand how a temperature or concentration profile changes over space and time.

The error function is defined as:\[\text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_{0}^{z} e^{-u^2} du\]
  • It arises naturally in problems involving heat conduction, like in the exercise.
  • The error function provides a solution to differential equations describing the diffusion process.
  • In the unsteady-state heat conduction equation, the error function provides a way to account for the transient nature of heat flow within the medium.
For practical applications, we often use tables or computational tools to find specific values of \(\text{erf}(z)\), rather than calculate the integral directly. In our problem, solving for the time required for a midplane temperature change involves finding specific values of the error function and its inverse, aiding in determining how the temperature evolves with time.

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Most popular questions from this chapter

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 \(D=200 \mathrm{~mm}\).

When a molten metal is cast in a mold that is a poor conductor, the dominant resistance to heat flow is within the mold wall. Consider conditions for which a liquid metal is solidifying in a thick-walled mold of thermal conductivity \(k_{v}\) and thermal diffusivity \(\alpha_{w}\). The density and latent heat of fusion of the metal are designated as \(\rho\) and \(h_{s f}\), respectively, and in both its molten and solid states, the thermal conductivity of the metal is very much larger than that of the mold. Just before the start of solidification \((S=0)\), the mold wall is everywhere at an initial uniform temperature \(T_{i}\) and the molten metal is everywhere at its fusion (melting point) temperature of \(T_{f}\). Following the start of solidification, there is conduction heat transfer into the mold wall and the thickness of the solidified metal \(S\) increases with time \(t\). (a) Sketch the one-dimensional temperature distribution, \(T(x)\), in the mold wall and the metal at \(t=0\) and at two subsequent times during the solidification. Clearly indicate any underlying assumptions. (b) Obtain a relation for the variation of the solid layer thickness \(S\) with time \(t\), expressing your result in terms of appropriate parameters of the system.

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a \(100-\mathrm{mm}\)-thick steel plate \(\left(\rho=7830 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c=550 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is initially at a uniform temperature of \(T_{i}=200^{\circ} \mathrm{C}\) and is to be heated to a minimum temperature of \(550^{\circ} \mathrm{C}\). Heating is effected in a gas-fired furnace, where products of combustion at \(T_{\infty}=800^{\circ} \mathrm{C}\) maintain a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on both surfaces of the plate. How long should the plate be left in the furnace?

5.53 Stone mix concrete slabs are used to absorb thermal energy from flowing air that is carried from a large concentrating solar collector. The slabs are heated during the day and release their heat to cooler air at night. If the daytime airflow is characterized by a temperature and convection heat transfer coefficient of \(T_{\infty}=200^{\circ} \mathrm{C}\) and \(h=35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine the slab thickness \(2 L\) required to transfer a total amount of energy such that \(Q / Q_{o}=0.90\) over a \(t=8\)-h period. The initial concrete temperature is \(T_{i}=40^{\circ} \mathrm{C}\).

A steel strip of thickness \(\delta=12 \mathrm{~mm}\) is annealed by passing it through a large furnace whose walls are maintained at a temperature \(T_{w}\) corresponding to that of combustion gases flowing through the furnace \(\left(T_{w}=T_{\infty}\right)\). The strip, whose density, specific heat, thermal conductivity, and emissivity are \(\rho=7900 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=640 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\varepsilon=0.7\), respectively, is to be heated from \(300^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\). (a) For a uniform convection coefficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{w}=T_{\infty}=700^{\circ} \mathrm{C}\), determine the time required to heat the strip. If the strip is moving at \(0.5 \mathrm{~m} / \mathrm{s}\), how long must the furnace be? (b) The annealing process may be accelerated (the strip speed increased) by increasing the environmental temperatures. For the furnace length obtained in part (a), determine the strip speed for \(T_{w}=T_{\infty}=\) \(850^{\circ} \mathrm{C}\) and \(T_{w}=T_{\infty}=1000^{\circ} \mathrm{C}\). For each set of environmental temperatures \(\left(700,850\right.\), and \(\left.1000^{\circ} \mathrm{C}\right)\), plot the strip temperature as a function of time over the range \(25^{\circ} \mathrm{C} \leq T \leq 600^{\circ} \mathrm{C}\). Over this range, also plot the radiation heat transfer coefficient, \(h_{r}\), as a function of time.
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