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Chapter 1: Problem 81

During its manufacture, plate glass at \(600^{\circ} \mathrm{C}\) is cooled by passing air over its surface such that the convection heat transfer coefficient is \(h=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent cracking, it is known that the temperature gradient must not exceed \(15^{\circ} \mathrm{C} / \mathrm{mm}\) at any point in the glass during the cooling process. If the thermal conductivity of the glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its surface emissivity is \(0.8\), what is the lowest temperature of the air that can initially be used for the cooling? Assume that the temperature of the air equals that of the surroundings.

Short Answer

Expert verified
The lowest air temperature that can initially be used for cooling without causing cracking in the plate glass is approximately \(579^{\circ} \mathrm{C}\).

Step by step solution

01

Find the total heat transfer from the glass

The total heat transfer involves both convection and radiation. The heat transfer due to convection can be calculated using the formula \(q_{conv} = hA(T_{surf} - T_{air})\), where \(q_{conv}\) is the convective heat transfer, h is the heat transfer coefficient, A is the surface area of the plate glass, \(T_{surf}\) is the surface temperature of the glass and \(T_{air}\) is the air temperature. For radiation, we can use the formula \(q_{rad} = \epsilon \sigma A (T_{surf}^4 - T_{surroundings}^4)\), where \(q_{rad}\) is the radiative heat transfer, \(\epsilon\) is the surface emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2 K^4}\)), and \(T_{surroundings}\) is the temperature of the surroundings (which is equal to \(T_{air}\)). The total heat transfer will be the sum of these two, i.e., \(q_{total} = q_{conv} + q_{rad}\).
02

Calculate the temperature gradient

Fourier's law relates the heat transfer in a solid to the temperature gradient: \(q_{total} = -kA\frac{dT}{dx}\), where k is the thermal conductivity of the glass, and \(\frac{dT}{dx}\) is the temperature gradient. Rearranging for the temperature gradient, we have \(\frac{dT}{dx} = -\frac{q_{total}}{kA}\).
03

Ensure the temperature gradient does not exceed the limit

We have to ensure the temperature gradient does not exceed the given limit of \(15^{\circ} \mathrm{C} / \mathrm{mm}\). Set the maximum temperature gradient equal to the limit, and solve for the corresponding air temperature: \(\frac{dT}{dx} = 15^{\circ} \mathrm{C} / \mathrm{mm} \rightarrow T_{air} = T_{surf} - \frac{k_{glass}}{hA}\frac{15^{\circ} \mathrm{C} / \mathrm{mm}}{q_{total}}\)
04

Calculate the lowest air temperature allowed for cooling

Substitute the given values into the equation derived in Step 3 and solve for the air temperature: \(T_{air} = 600^{\circ} \mathrm{C} - \frac{1.4 \mathrm{W/m\cdot K}}{5 \mathrm{W/m^2\cdot K}}\frac{15^{\circ} \mathrm{C} / \mathrm{mm}}{q_{total}}\). Since the glass thickness was not given, we assume A = 1 m虏, that simplifies the expression to \(T_{air} = 600^{\circ} \mathrm{C} - \frac{1.4}{5} \frac{15}{q_{total}}\). Now, the minimum allowed \(T_{air}\) is when \(q_{total}\) is maximum, and since both \(q_{conv}\) and \(q_{rad}\) are positive, then the maximum heat transfer that occurs will be when there are no radiation heat transfer, i.e., \(q_{total} = hA(T_{surf} - T_{air})\). Thus, the expression becomes \(T_{air} = 600^{\circ} \mathrm{C} - \frac{1.4}{5} \frac{15}{h(600^{\circ} \mathrm{C} - T_{air})}\). Solve for \(T_{air}\), you'll get \(T_{air} \approx 579^{\circ} \mathrm{C}\). Therefore, the lowest air temperature that can initially be used for cooling without causing cracking in the plate glass is approximately \(579^{\circ} \mathrm{C}\).

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Key Concepts

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

Temperature Gradient
A temperature gradient represents the rate of temperature change along a specific direction in a material. When you have a high temperature on one side of a glass pane and a lower temperature on the other, the gradient is created as the heat travels through the glass. In the context of this exercise, the temperature gradient is crucial because exceeding a certain value may cause thermal stress, leading to cracking. For the plate glass in question, the gradient must remain under \( 15^{\circ} \text{C/mm} \) during cooling. The gradient is essentially a measure of how quickly temperature decreases per millimeter of glass thickness, ensuring safe cooling processes.
Thermal Conductivity
Thermal conductivity is a material's ability to conduct heat. It is a fundamental parameter when analyzing heat transfer through solids. For the glass plate, thermal conductivity is given as \( 1.4 \, \text{W/m} \cdot \text{K} \). This value indicates how effectively heat can pass through the glass. A higher thermal conductivity means heat flows more easily, while a lower value would resist heat flow, maintaining a greater temperature difference over the same distance. Understanding thermal conductivity helps determine how quickly or slowly the glass will respond to temperature changes, which is key to managing the temperature gradient during cooling.
Radiative Heat Transfer
Radiative heat transfer is the process by which energy is transferred in the form of electromagnetic radiation. In simple terms, anything that is warm emits heat energy as radiation. The exercise mentions the surface emissivity of the glass, which is \(0.8\). Emissivity is a measure of how effectively a surface emits thermal radiation compared to a perfect emitter, also known as a blackbody. The higher the emissivity, the more heat energy it radiates away. This concept comes into play when calculating the total heat transfer from the glass鈥攃ombining the effects of both convection and radiation ensures accurate assessment of the cooling process.
Fourier's Law
Fourier's Law of Heat Conduction is crucial for understanding how heat energy moves through materials due to temperature differences. The law is based on the premise that heat flows from regions of higher temperature to regions of lower temperature. Mathematically, it is expressed as \( q = -kA \frac{dT}{dx} \). This formula states that the heat transfer rate \( q \) is proportional to the negative of the temperature gradient \( \frac{dT}{dx} \), multiplied by the thermal conductivity \( k \) and the area \( A \). It provides a clear relationship between material properties, the actual rate of heat flow, and differences in temperature, offering a powerful tool for analyzing heat transfer scenarios like the one in the plate glass exercise.

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Most popular questions from this chapter

In the thermal processing of semiconductor materials, annealing is accomplished by heating a silicon wafer according to a temperature-time recipe and then maintaining a fixed elevated temperature for a prescribed period of time. For the process tool arrangement shown as follows, the wafer is in an evacuated chamber whose walls are maintained at \(27^{\circ} \mathrm{C}\) and within which heating lamps maintain a radiant flux \(q_{s}^{\prime \prime}\) at its upper surface. The wafer is \(0.78 \mathrm{~mm}\) thick, has a thermal conductivity of \(30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and an emissivity that equals its absorptivity to the radiant flux \(\left(\varepsilon=\alpha_{l}=0.65\right.\) ). For \(q_{s}^{\prime \prime}=3.0 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2}\), the temperature on its lower surface is measured by a radiation thermometer and found to have a value of \(T_{w, l}=997^{\circ} \mathrm{C}\). To avoid warping the wafer and inducing slip planes in the crystal structure, the temperature difference across the thickness of the wafer must be less than \(2^{\circ} \mathrm{C}\). Is this condition being met?

A freezer compartment consists of a cubical cavity that is \(2 \mathrm{~m}\) on a side. Assume the bottom to be perfectly insulated. What is the minimum thickness of styrofoam insulation \((k=0.030 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that must be applied to the top and side walls to ensure a heat load of less than \(500 \mathrm{~W}\), when the inner and outer surfaces are \(-10\) and \(35^{\circ} \mathrm{C}\) ?

The heat flux through a wood slab \(50 \mathrm{~mm}\) thick, whose inner and outer surface temperatures are 40 and \(20^{\circ} \mathrm{C}\), respectively, has been determined to be \(40 \mathrm{~W} / \mathrm{m}^{2}\). What is the thermal conductivity of the wood?

The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d t)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\circ} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

An inexpensive food and beverage container is fabricated from 25 -mm-thick polystyrene \((k=0.023 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and has interior dimensions of \(0.8 \mathrm{~m} \times 0.6 \mathrm{~m} \times 0.6 \mathrm{~m}\). Under conditions for which an inner surface temperature of approximately \(2^{\circ} \mathrm{C}\) is maintained by an ice-water mixture and an outer surface temperature of \(20^{\circ} \mathrm{C}\) is maintained by the ambient, what is the heat flux through the container wall? Assuming negligible heat gain through the \(0.8 \mathrm{~m} \times\) \(0.6 \mathrm{~m}\) base of the cooler, what is the total heat load for the prescribed conditions?
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