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

A microwave oven operates on the principle that application of a high- frequency field causes electrically polarized molecules in food to oscillate. The net effect is a nearly uniform generation of thermal energy within the food. Consider the process of cooking a slab of beef of thickness \(2 L\) in a microwave oven and compare it with cooking in a conventional oven, where each side of the slab is heated by radiation. In each case the meat is to be heated from \(0^{\circ} \mathrm{C}\) to a minimum temperature of \(90^{\circ} \mathrm{C}\). Base your comparison on a sketch of the temperature distribution at selected times for each of the cooking processes. In particular, consider the time \(t_{0}\) at which heating is initiated, a time \(t_{1}\) during the heating process, the time \(t_{2}\) corresponding to the conclusion of heating, and a time \(t_{3}\) well into the subsequent cooling process.

An electronic device, such as a power transistor mounted on a finned heat sink, can be modeled as a spatially isothermal object with internal heat generation and an external convection resistance. (a) Consider such a system of mass \(M\), specific heat \(c\), and surface area \(A_{s}\), which is initially in equilibrium with the environment at \(T_{\infty}\). Suddenly, the electronic device is energized such that a constant heat generation \(\dot{E}_{g}(\mathrm{~W})\) occurs. Show that the temperature response of the device is $$ \frac{\theta}{\theta_{i}}=\exp \left(-\frac{t}{R C}\right) $$ where \(\theta \equiv T-T(\infty)\) and \(T(\infty)\) is the steady-state temperature corresponding to \(t \rightarrow \infty ; \theta_{i}=T_{i}-T(\infty)\); \(T_{i}=\) initial temperature of device; \(R=\) thermal resistance \(1 / \bar{h} A_{s} ;\) and \(C=\) thermal capacitance \(M c\). (b) An electronic device, which generates \(60 \mathrm{~W}\) of heat, is mounted on an aluminum heat sink weighing \(0.31 \mathrm{~kg}\) and reaches a temperature of \(100^{\circ} \mathrm{C}\) in ambient air at \(20^{\circ} \mathrm{C}\) under steady-state conditions. If the device is initially at \(20^{\circ} \mathrm{C}\), what temperature will it reach \(5 \mathrm{~min}\) after the power is switched on?

Common transmission failures result from the glazing of clutch surfaces by deposition of oil oxidation and decomposition products. Both the oxidation and decomposition processes depend on temperature histories of the surfaces. Because it is difficult to measure these surface temperatures during operation, it is useful to develop models to predict clutch-interface thermal behavior. The relative velocity between mating clutch plates, from the initial engagement to the zero-sliding (lock-up) condition, generates heat that is transferred to the plates. The relative velocity decreases at a constant rate during this period, producing a heat flux that is initially very large and decreases linearly with time, until lock-up occurs. Accordingly, \(q_{f}^{\prime \prime}=q_{o}^{\prime \prime}=\left[1-\left(t / t_{\mathrm{lu}}\right)\right]\), where \(q_{o}^{\prime \prime}=1.6 \times 10^{7} \mathrm{~W} / \mathrm{m}^{2}\) and \(t_{1 \mathrm{u}}=100 \mathrm{~ms}\) is the lock-up time. The plates have an initial uniform temperature of \(T_{i}=40^{\circ} \mathrm{C}\), when the prescribed frictional heat flux is suddenly applied to the surfaces. The reaction plate is fabricated from steel, while the composite plate has a thinner steel center section bonded to low- conductivity friction material layers. The thermophysical properties are \(\rho_{s}=\) \(7800 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{s}}=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k_{s}=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the steel and \(\rho_{\mathrm{im}}=1150 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{fm}}=1650 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k_{\mathrm{fm}}=4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the friction material. (a) On \(T-t\) coordinates, sketch the temperature history at the midplane of the reaction plate, at the interface between the clutch pair, and at the midplane of the composite plate. Identify key features. (b) Perform an energy balance on the clutch pair over the time interval \(\Delta t=t_{\mathrm{lu}}\) to determine the steadystate temperature resulting from clutch engagement. Assume negligible heat transfer from the plates to the surroundings. (c) Compute and plot the three temperature histories of interest using the finite-element method of FEHT or the finite-difference method of IHT (with \(\Delta x=0.1 \mathrm{~mm}\) and \(\Delta t=1 \mathrm{~ms}\) ). Calculate and plot the frictional heat fluxes to the reaction and composite plates, \(q_{\mathrm{rp}}^{\prime \prime}\) and \(q_{\mathrm{cp}}^{\prime \prime}\), respectively, as a function of time. Comment on features of the temperature and heat flux histories. Validate your model by comparing predictions with the results from part (b). Note: Use of both \(F E H T\) and \(I H T\) requires creation of a look-up data table for prescribing the heat flux as a function of time.

A very thick slab with thermal diffusivity \(5.6 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) and thermal conductivity \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is initially at a uniform temperature of \(325^{\circ} \mathrm{C}\). Suddenly, the surface is exposed to a coolant at \(15^{\circ} \mathrm{C}\) for which the convection heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine temperatures at the surface and at a depth of \(45 \mathrm{~mm}\) after \(3 \mathrm{~min}\) have elapsed. (b) Compute and plot temperature histories \((0 \leq t \leq\) \(300 \mathrm{~s}\) ) at \(x=0\) and \(x=45 \mathrm{~mm}\) for the following parametric variations: (i) \(\alpha=5.6 \times 10^{-7}, 5.6 \times\) \(10^{-6}\), and \(5.6 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\); and (ii) \(k=2,20\), and \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).
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