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Chapter 9: Problem 46

A 200-mm-square, 10-mm-thick tile has the thermophysical properties of Pyrex \((\varepsilon=0.80)\) and emerges from a curing process at an initial temperature of \(T_{i}=140^{\circ} \mathrm{C}\). The backside of the tile is insulated while the upper surface is exposed to ambient air and surroundings at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The initial heat transfer rate from the Pyrex tile to the ambient air as it emerges from the curing process is approximately 72.50 Watts. This is calculated using the heat transfer through radiation equation, taking into account the emissivity of Pyrex, the initial and ambient temperatures, and the surface area of the exposed tile.

Step by step solution

01

Identify the properties and constants

First, we must identify the properties and constants provided in the problem statement that will be useful in solving the problem. - Tile dimensions: \(200~\text{mm} \times 200~\text{mm} \times 10~\text{mm}\) (note that only the surface area is relevant here, as heat transfer occurs on the exposed surface) - Material: Pyrex, with \(\varepsilon = 0.80\) (emissivity) - Initial temperature: \(T_{i} = 140^{\circ} \mathrm{C}\), which must be converted to Kelvin: \(T_{i, K} = 140 + 273.15 = 413.15~\text{K}\) - Ambient air temperature: \(25^{\circ} C\), which should also be converted to Kelvin: \(T_{\infty, K} = 25 + 273.15 = 298.15~\text{K}\) - Stefan-Boltzmann constant: \(\sigma = 5.670 \times 10^{-8}~\mathrm{W/m^2 \cdot K^4}\)
02

Calculate the heat transfer through radiation

To calculate the heat transfer through radiation, we can use the following equation: \[q = \varepsilon \cdot \sigma \cdot A \cdot (T_{i, K}^4 - T_{\infty, K}^4)\] where: - \(q\) is the heat transfer rate (in Watts) - \(\varepsilon\) is the emissivity of the material - \(\sigma\) is the Stefan-Boltzmann constant - \(A\) is the surface area of the exposed surface (in square meters) - \(T_{i, K}\) is the initial temperature of the tile (in Kelvin) - \(T_{\infty, K}\) is the temperature of the ambient air (in Kelvin) First, we need to calculate the surface area of the exposed surface. Given that the tile is of square shape, with a side length of 200 mm, we can express its surface area as follows: \[A = (0.2~\text{m})^2 = 0.04~\text{m}^2\] Now, we can substitute the values into the equation to find the initial heat transfer rate: \[q = 0.80 \cdot (5.670 \times 10^{-8} \frac{\text{W}}{\text{m}^2 \text{K}^4}) \cdot 0.04~\text{m}^2 \cdot (413.15^4 - 298.15^4) \text{K}^4\]
03

Determine the initial heat transfer rate

Now, we can evaluate the expression to find the initial heat transfer rate: \[q = 0.80 \cdot (5.670 \times 10^{-8} \frac{\text{W}}{\text{m}^2 \text{K}^4}) \cdot 0.04~\text{m}^2 \cdot (2.053 \times 10^8 - 7.875 \times 10^7)~\text{K}^4\] \[q = 0.80 \cdot (5.670 \times 10^{-8} \frac{\text{W}}{\text{m}^2 \text{K}^4}) \cdot 0.04~\text{m}^2 \cdot (1.266 \times 10^8)~\text{K}^4\] \[q \approx 72.50~\text{W}\] The initial heat transfer rate from the Pyrex tile to the ambient air as it emerges from the curing process is approximately 72.50 Watts.

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Key Concepts

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

Radiation Heat Transfer
Radiation heat transfer is one way energy moves from a hotter object to a cooler one without needing a medium in between. Imagine the sun warming your face: that's radiation heat transfer taking place across space!
This type of heat transfer plays a crucial role in many natural and industrial processes.
  • **Direct Transfer:** Unlike conduction and convection, radiation doesn't require molecules to pass on energy. This makes it unique as it can occur through a vacuum.
  • **Electromagnetic Waves:** Energy is emitted in the form of electromagnetic waves, mainly in the infrared spectrum, which is why we feel heat even if there's no physical contact.
In the context of the Pyrex tile, radiation heat transfer occurs as the tile cools down by emitting thermal radiation to its cooler surroundings.
It's an efficient way for heat to dissipate when there鈥檚 a significant temperature difference, like between the tile and the ambient air.
Emissivity
Emissivity is a measure of a material's ability to emit thermal radiation. Think of it as a material's efficiency in cooling itself by radiating energy. In simple terms, it's how well a surface can radiate energy compared to a perfect black body.
  • **Value Range:** Emissivity values range from 0 to 1. A value of 1 means the material is a perfect emitter, while 0 means it emits no radiation.
  • **Material Dependency:** Different materials have different emissivity values; shiny metals might have lower values compared to non-metal surfaces.
For the Pyrex tile in our exercise, the emissivity is 0.80. This value means that Pyrex is quite good at emitting energy, though not as perfectly as a black body. This property is crucial in determining how effectively the tile can lose heat through radiation when it emerges hot from the curing process.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is foundational when calculating radiation heat transfer. It helps us determine how much heat an object can radiate based on its temperature and emissivity.
The law is expressed with the formula: \[ q = \varepsilon \cdot \sigma \cdot A \cdot (T^4 - T_{\infty}^4) \]
  • \( q \) is the heat transfer rate, or how much heat an object emits over time.
  • \( \varepsilon \) represents emissivity, showing a material鈥檚 efficiency in emitting heat.
  • \( \sigma \) is the Stefan-Boltzmann constant, \( 5.670 \times 10^{-8} \mathrm{W/m^2 \cdot K^4} \).
  • \( A \) is the surface area, crucial because larger surfaces emit more heat.
  • The term \( (T^4 - T_{\infty}^4) \) represents the temperature difference's influence, raised to the fourth power.
Understanding this law allows us to calculate the heat transfer for the tile accurately. By applying the given temperatures and taking into account the tile's emissivity, we derived an initial heat transfer rate of approximately 72.50 Watts as the tile cools from curing.

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Most popular questions from this chapter

The maximum surface temperature of the \(20-\mathrm{mm}-\) diameter shaft of a motor operating in ambient air at \(27^{\circ} \mathrm{C}\) should not exceed \(87^{\circ} \mathrm{C}\). Because of power dissipation within the motor housing, it is desirable to reject as much heat as possible through the shaft to the ambient air. In this problem, we will investigate several methods for heat removal. (a) For rotating cylinders, a suitable correlation for estimating the convection coefficient is of the form $$ \begin{gathered} \overline{N u}_{D}=0.133 \operatorname{Re}_{D}^{2 / 3} \operatorname{Pr}^{1 / 3} \\ \left(\operatorname{Re}_{D}<4.3 \times 10^{5}, \quad 0.7<\operatorname{Pr}<670\right) \end{gathered} $$ where \(R e_{D} \equiv \Omega D^{2} / \nu\) and \(\Omega\) is the rotational velocity (rad/s). Determine the convection coefficient and the maximum heat rate per unit length as a function of rotational speed in the range from 5000 to \(15,000 \mathrm{rpm}\). (b) Estimate the free convection coefficient and the maximum heat rate per unit length for the stationary shaft. Mixed free and forced convection effects may become significant for \(R e_{D}<4.7\left(G r_{D}^{3} / P r\right)^{0.137}\). Are free convection effects important for the range of rotational speeds designated in part (a)? (c) Assuming the emissivity of the shaft is \(0.8\) and the surroundings are at the ambient air temperature, is radiation exchange important? (d) If ambient air is in cross flow over the shaft, what air velocities are required to remove the heat rates determined in part (a)?

In the analytical treatment of the fin with uniform cross-sectional area, it was assumed that the convection heat transfer coefficient is constant along the length of the fin. Consider an AISI 316 steel fin of 6-mm diameter and 50-mm length (with insulated tip) operating under conditions for which \(T_{b}=125^{\circ} \mathrm{C}, T_{\infty}=27^{\circ} \mathrm{C}\), \(T_{\text {sur }}=27^{\circ} \mathrm{C}\), and \(\varepsilon=0.6\). (a) Estimate average values of the fin heat transfer coefficients for free convection \(\left(h_{c}\right)\) and radiation exchange \(\left(h_{r}\right)\). Use these values to predict the tip temperature and fin effectiveness. (b) Use a numerical method of solution to estimate the foregoing parameters when the convection and radiation coefficients are based on local, rather than average, values for the fin.

Square panels \((250 \mathrm{~mm} \times 250 \mathrm{~mm}\) ) with a decorative, highly reflective plastic finish are cured in an oven at \(125^{\circ} \mathrm{C}\) and cooled in quiescent air at \(29^{\circ} \mathrm{C}\). Quality considerations dictate that the panels remain horizontal and that the cooling rate be controlled. To increase productivity in the plant, it is proposed to replace the batch cooling method with a conveyor system having a velocity of \(0.5 \mathrm{~m} / \mathrm{s}\).

Liquid nitrogen is stored in a thin-walled spherical vessel of diameter \(D_{i}=1 \mathrm{~m}\). The vessel is positioned concentrically within a larger, thin-walled spherical container of diameter \(D_{o}=1.10 \mathrm{~m}\), and the intervening cavity is filled with atmospheric helium. Under normal operating conditions, the inner and outer surface temperatures are \(T_{i}=77 \mathrm{~K}\) and \(T_{o}=283 \mathrm{~K}\). If the latent heat of vaporization of nitrogen is \(2 \times 10^{5} \mathrm{~J} / \mathrm{kg}\), what is the mass rate \(m(\mathrm{~kg} / \mathrm{s})\) at which gaseous nitrogen is vented from the system?

Beginning with the free convection correlation of the form given by Equation 9.24, show that for air at atmospheric pressure and a film temperature of \(400 \mathrm{~K}\), the average heat transfer coefficient for a vertical plate can be expressed as $$ \begin{array}{ll} \bar{h}_{L}=1.40\left(\frac{\Delta T}{L}\right)^{1 / 4} & 10^{4}
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