/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 An opaque, diffuse, gray ( \(200... [FREE SOLUTION] | 91Ó°ÊÓ

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An opaque, diffuse, gray ( \(200 \mathrm{~mm} \times 200 \mathrm{~mm})\) plate with an emissivity of \(0.8\) is placed over the opening of a furnace and is known to be at \(400 \mathrm{~K}\) at a certain instant. The bottom of the fumace, having the same dimensions as the plate, is black and operates at \(1000 \mathrm{~K}\). The sidewalls of the fumace are well insulated. The top of the plate is exposed to ambient air with a convection coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and to large surroundings. The air and surroundings are each at \(300 \mathrm{~K}\). (a) Evaluate the net radiative heat transfer to the bottom surface of the plate. (b) If the plate has mass and specific heat of \(2 \mathrm{~kg}\) and \(900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively, what will be the change in temperature of the plate with time, \(d T_{p} / d r\) ? Assume convection to the bottom surface of the plate to be negligible. (c) Extending the analysis of part (b), generate a plot of the change in temperature of the plate with time, \(d T_{\rho} / d l\), as a function of the plate temperature for \(350 \leq T_{p} \leq 900 \mathrm{~K}\) and all other conditions remaining the same. What is the steady-state tem-

Short Answer

Expert verified
\[q_{rad} = 452.76 \; \mathrm{W}\] (a) The net radiative heat transfer to the bottom surface of the plate is \(452.76 \; \mathrm{W}\). (b) Change in Temperature of the Plate with Time To find the change in temperature, we need to calculate the rate of energy change with time. The energy balance equation for the plate is given as: \[\frac{dE}{dt} = m c_p \frac{dT_p}{dt}\] Where \(E\) is the internal energy of the plate, \(m\) is the mass, \(c_p\) is the specific heat, and \(T_p\) is the plate temperature. Given: \(m = 2\;kg\) \(c_p = 900\;\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) Substituting the values, we get: \[\frac{dE}{dt} = 2 * 900 \frac{dT_p}{dt}\] Since convection to the bottom surface of the plate is negligible, the rate of energy change with time is equal to the heat transfer from the radiator to the plate. \[\frac{dE}{dt} = q_{rad}\] Therefore, we can write: \[2 * 900 \frac{dT_p}{dt} = 452.76\] Solving for \(\frac{dT_p}{dt}\), we get: \[\frac{dT_p}{dt} = \frac{452.76}{(2 * 900)} = 0.2515 \; \mathrm{K} / \mathrm{s}\] (b) The change in temperature of the plate with time is \(0.2515 \; \mathrm{K} / \mathrm{s}\).

Step by step solution

01

Calculate the plate's radiative heat exchange with the furnace's bottom surface

We can calculate the radiative heat exchange between the plate and the bottom of the furnace using the following formula: \[q_{rad} = A_s \epsilon \sigma (T_{furnace}^4 - T_{plate}^4)\] Where \(A_s\) is the area of the surface, \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, \(T_{furnace}\) is the furnace temperature, and \(T_{plate}\) is the plate temperature. Given: \(A_s = 200\;mm \times 200\;mm = 0.04\;m^2\) \(\epsilon = 0.8\) \(\sigma = 5.67\times10^{-8} \;\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}^4\) \(T_{furnace} = 1000\;K\) \(T_{plate} = 400\;K \) Plugging the values in the formula, we get: \[q_{rad} = 0.04 * 0.8 * 5.67\times10^{-8} * (1000^4 - 400^4)\]

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Key Concepts

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

Radiative Heat Transfer
Radiative heat transfer is a form of energy transfer where heat is emitted in the form of electromagnetic waves. It occurs between surfaces without involving any medium. The heat is transferred by radiation, often without needing physical contact between objects.

In practical scenarios, like the one in the exercise, radiative heat transfer is evaluated using the Stefan-Boltzmann Law. This law states that the power radiated per unit area of a black body is proportional to the fourth power of its absolute temperature. The formula to calculate radiative heat transfer between surfaces is:

\[ q_{rad} = A_s \epsilon \sigma (T_{furnace}^4 - T_{plate}^4) \]

Here, \( q_{rad} \) is the heat transfer rate, \( A_s \) is the surface area, \( \epsilon \) is the emissivity, \( \sigma \) is the Stefan-Boltzmann constant, \( T_{furnace} \) and \( T_{plate} \) are the temperatures of the furnace and plate respectively. This formula helps in determining how much heat is exchanged radiatively between two bodies, crucial for applications like furnaces, heaters, or any system where heat transfer might be required.
Emissivity
Emissivity is a measure of a material's ability to emit energy as thermal radiation. It is a dimensionless quantity and varies between 0 and 1. An emissivity of 1 indicates a perfect black body, which emits the maximum amount of radiation possible at a given temperature. Conversely, an emissivity of 0 implies that the material does not emit any thermal radiation.

In the given exercise, the plate has an emissivity of 0.8. This means it is a good emitter of thermal radiation, though not perfect. Understanding emissivity is crucial when determining radiative heat transfer because it directly affects how much energy a material can emit. Real-life applications frequently involve materials that are not perfect black bodies; hence, they have emissivity values less than 1.

  • High emissivity materials: Often used where efficient heat radiation is necessary, such as heating elements.
  • Low emissivity materials: Useful in insulation applications, where minimizing heat exchange is beneficial.
The exact value of emissivity is crucial in calculations and simulations that involve thermal systems.
Convection Coefficient
The convection coefficient, often denoted by \( h \), quantifies the convective heat transfer between a surface and a fluid in motion over the surface. It represents the efficiency of heat transfer through convection and is usually measured in \( \mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K}\).

A higher convection coefficient indicates a more efficient heat transfer between the surface and the fluid. Factors affecting the convection coefficient include the fluid velocity, fluid properties (like viscosity and thermal conductivity), and the surface geometry.

In the exercise, the ambient air around the plate has a convection coefficient of 25 \( \mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K} \). This implies a moderate level of convective heat transfer, which can be important in calculating the overall heat exchange involving the plate. It is useful in various engineering applications such as design of heat exchangers and understanding of heat loss in buildings.

  • Forced convection: Occurs when a fluid is forced over a surface by an external source like fans or pumps, leading to a higher convection coefficient.
  • Natural convection: Occurs naturally due to temperature differences in a fluid, typically resulting in a lower convection coefficient compared to forced convection.
Understanding how convection coefficient works is vital for accurately predicting the heat transfer in systems where convection plays a significant role.
Stefan-Boltzmann Constant
The Stefan-Boltzmann constant (\( \sigma \)) is a fundamental constant in physics that features prominently in the Stefan-Boltzmann Law, which describes black body radiation. The value of \( \sigma \) is \( 5.67 \times 10^{-8} \) \( \mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K}^4 \).

This constant is crucial in any calculation involving thermal radiation and radiative heat transfer, as it relates the power radiated by a black body to its absolute temperature raised to the fourth power.

The Stefan-Boltzmann constant is commonly used in:
  • Astronomy: For determining the luminosity of stars based on their temperature.
  • Climate science: In models to predict Earth's radiation balance and temperature changes.
  • Engineering: To calculate heat exchange in systems involving thermal radiation, such as furnaces and thermal insulation.
The Stefan-Boltzmann constant allows us to better understand and calculate energy exchanges in various thermal systems, thereby aiding in designing processes and systems that rely on radiative heat transfer for operation.

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Most popular questions from this chapter

Liquid oxygen is stored in a thin-walled, spherical container \(0.8 \mathrm{~m}\) in diameter, which is enclosed within a second thin-walled, spherical container \(1.2 \mathrm{~m}\) in diameter. The opaque, diffuse, gray container surfaces have an emissivity of \(0.05\) and are separated by an evacuated space. If the outer surface is at \(280 \mathrm{~K}\) and the inner surface is at \(95 \mathrm{~K}\), what is the mass rate of oxygen lost due to evaporation? (The latent heat of vaporization of oxygen is \(2.13 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).)

A cryogenic fluid flows through a tube \(20 \mathrm{~mm}\) in diameter, the outer surface of which is diffuse and gray with an emissivity of \(0.02\) and temperature of \(77 \mathrm{~K}\). This tube is concentric with a larger tube of 50 \(\mathrm{mm}\) diameter, the inner surface of which is diffuse and gray with an emissivity of \(0.05\) and temperature of \(300 \mathrm{~K}\). The space between the surfaces is evacuated. Determine the heat gain by the cryogenic fluid per unit length of the inner tube. If a thin-walled radiation shield that is diffuse and gray with an emissivity of \(0.02\) (both sides) is inserted midway between the inner and outer surfaces, calculate the change (percentage) in heat gain per unit length of the inner tube.

Consider the attic of a home located in a hot climate. The floor of the attic is characterized by a width of \(L_{1}=10 \mathrm{~m}\) while the roof makes an angle of \(\theta=30^{\circ}\) from the horizontal direction, as shown in the schematic. The homeowner wishes to reduce the heat load to the home by adhering bright aluminum foil \(\left(\varepsilon_{f}=0.07\right)\) onto the surfaces of the attic space. Prior to installation of the foil, the surfaces are of emissivity \(s_{e}=0.85\). (a) Consider installation on the bottom of the attic roof only. Determine the ratio of the radiation heat transfer after to before the installation of the foil. (b) Determine the ratio of the radiation heat transfer after to before installation if the foil is installed only on the top of the attic floor. (c) Determine the ratio of the radiation heat transfer if the foil is installed on both the roof bottom and the floor top.

Consider a circular furnace that is \(0.3 \mathrm{~m}\) long and \(0.3 \mathrm{~m}\) in diameter. The two ends have diffuse, gray surfaces that are maintained at 400 and \(500 \mathrm{~K}\) with emissivities of \(0.4\) and \(0.5\), respectively. The lateral surface is also diffuse and gray with an emissivity of \(0.8\) and a temperature of \(800 \mathrm{~K}\). Determine the net radiative heat transfer from each of the surfaces.

Consider the right-circular cylinder of diameter \(D\), length \(L\), and the areas \(A_{1}, A_{2}\), and \(A_{3}\) representing the base, inner, and top surfaces, respectively. (a) Show that the view factor between the base of the cylinder and the inner surface has the form \(F_{12}=2 H\left[\left(1+H^{2}\right)^{1 / 2}-H\right]\), where \(H=L D .\) (b) Show that the view factor for the inner surface to itself has the form \(F_{22}=1+H-\left(1+H^{2}\right)^{1 / 2}\).

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