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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

Stainless steel (AISI 304) ball bearings, which have uniformly been heated to \(850^{\circ} \mathrm{C}\), are hardened by quenching them in an oil bath that is maintained at \(40^{\circ} \mathrm{C}\). The ball diameter is \(20 \mathrm{~mm}\), and the convection coefficient associated with the oil bath is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If quenching is to occur until the surface temperature of the balls reaches \(100^{\circ} \mathrm{C}\), how long must the balls be kept in the oil? What is the center temperature at the conclusion of the cooling period? (b) If 10,000 balls are to be quenched per hour, what is the rate at which energy must be removed by the oil bath cooling system in order to maintain its temperature at \(40^{\circ} \mathrm{C}\) ?

A one-dimensional slab of thickness \(2 L\) is initially at a uniform temperature \(T_{i}\). Suddenly, electric current is passed through the slab causing uniform volumetric heating \(\dot{q}\left(\mathrm{~W} / \mathrm{m}^{3}\right)\). At the same time, both outer surfaces \((x=\pm L)\) are subjected to a convection process at \(T_{\infty}\) with a heat transfer coefficient \(h\). Write the finite-difference equation expressing conservation of energy for node 0 located on the outer surface at \(x=-L\). Rearrange your equation and identify any important dimensionless coefficients.

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 \(37^{\circ} \mathrm{C}\).

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 \(T_{i}=40^{\circ} \mathrm{C}\).

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.
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