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Chapter 13: Problem 94

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

Consider two very large metal parallel plates. The top plate is at a temperature \(T_{t}=400 \mathrm{~K}\) while the bottom plate is at \(T_{b}=300 \mathrm{~K}\). The desired net radiation hear flux between the two plates is \(q^{\prime \prime}=330 \mathrm{~W} / \mathrm{m}^{2}\). (a) If the two surfaces have the same radiative properties, show that the required surface emissavity is \(\varepsilon=0.5\). (b) Metal surfaces at relatively low temperatures tend to have emissivities much less than \(0.5\) (see Table A.11). An engineer proposes to apply a checker pattern, similar to that of Problem \(12.132\), onto each of the metal surfaces so that half of each surface is characterized by the low emissivity of the bare metal and the other half is covered with the high-emissivity paint. If the average of the high and los emissavities is 0.5, will the net radiative heat flux between the surfaces be the desired value?

Consider the very long, inclined black surfaces \(\left(A_{1}, A_{2}\right)\) maintained at uniform temperatures of \(T_{1}=1000 \mathrm{~K}\) and \(T_{2}=800 \mathrm{~K}\).

Waste heat recovery from the exhaust (flue) gas of a melting furnace is accomplished by passing the gas through a vertical metallic tube and introducing saturated water (liquid) at the bottom of an annular region around the tube. The tube length and inside diameter are 7 and \(1 \mathrm{~m}\), respectively, and the tube inner surface is black. The gas in the tube is at atmospheric pressure, with \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{v})\) partial pressures of \(0.1\) and \(0.2 \mathrm{~atm}\), respectively, and its mean temperature may be approximated as \(T_{g}=1400 \mathrm{~K}\). The gas flow rate is \(\dot{m}=2 \mathrm{~kg} / \mathrm{s}\). If saturated water is introduced at a pressure of \(2.455 \mathrm{~ b a r s , ~ e s t i m a t e ~ t h e ~ w a t e r ~ f l o w ~ r a t e ~ s i t}\) which there is complete conversion from saturated liquid at the inlet to saturated vapor at the outlet. Thermophysical properties of the gas may be approximated as \(\mu=530 \times 10^{-7} \mathrm{~kg} / \mathrm{s}-\mathrm{m}, \mathrm{k}=0.091 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). and \(P_{r}=0.70\).

An electrically heated sample is maintained at a surface temperature of \(T_{s}=500 \mathrm{~K}\). The sample coating is diffuse but spectrally selective, with the spectral emissivity distribution shown schematically. The sample is irradiated by a furnace located coaxially at a distance of \(L_{\mathrm{s}}=750 \mathrm{~mm}\). The furnace has isothermal walls with an emissivity of \(s_{f}=0.7\) and a uniform temperature of \(T_{f}-3000 \mathrm{~K}\). A radiation detector of area \(A_{d}=8 \times 10^{-5} \mathrm{~m}^{2}\) is positioned at a distance of \(L_{\mathrm{dd}}=1.0 \mathrm{~m}\) from the sample along a direction that is \(45^{\circ}\) from the sample normal. The detector is sensitive to spectral radiant power only in the spectral region from 3 to \(5 \mu \mathrm{m}\). The sample surface experiences convection with a gas for which \(T_{\infty}=300 \mathrm{~K}\) and \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surroundings of the sample mount are large and at a uniform temperature of \(T_{\text {sur }}=300 \mathrm{~K}\). (a) Determine the electrical power, \(P_{c}\), required to maintain the sample at \(T_{s}=500 \mathrm{~K}\). (b) Considering both emission and reflected irradiation from the sample, determine the radiant power that is incident on the detector within the spectral region from 3 to \(5 \mu \mathrm{m}\).

2 A composite wall is comprised of two large plates separated by sheets of refractory insulation, as shown in the schematic. In the installation process, the sheets of thickness \(L=50 \mathrm{~mm}\) and thermal conductivity \(k=0.05 \mathrm{~W} / \mathrm{m}+\mathrm{K}\) are separated at 1 -m intervals by gaps of width \(w=10 \mathrm{~mm}\). The hot and cold plates have temperatures and emissivities of \(T_{1}=400^{\circ} \mathrm{C}\). \(\varepsilon_{1}=0.85\) and \(T_{2}=35^{\circ} \mathrm{C}, \varepsilon_{2}=0.5\), respectively. Assume that the plates and insulation are diffuse-gray surfaces. (a) Determine the heat loss by radiation through the gap per unit length of the composite wall (normal to the page). (b) Recognizing that the gaps are located on a 1-m spacing, determine what fraction of the total heat loss through the composite wall is due to transfer by radiation through the insulation gap.
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