/*! 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 129 A fumace having a spherical cavi... [FREE SOLUTION] | 91Ó°ÊÓ

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A fumace having a spherical cavity of \(0.5-\mathrm{m}\) diameter contains a gas mixture at 1 atm and \(1400 \mathrm{~K}\). The mixture consists of \(\mathrm{CO}_{2}\), with a partial pressure of \(0.25 \mathrm{~atm}\) and nitrogen with a partial pressure of \(0.75 \mathrm{arm}\). If the cavity wall is black, what is the cooling rate needed to maintain its temperature at \(500 \mathrm{~K}\) ?

Short Answer

Expert verified
The required cooling rate to maintain the temperature at 500 K is approximately 149444 W (watts).

Step by step solution

01

Calculate the surface area of the cavity

To find the cooling rate, we first need to know the surface area of the cavity. Since it's a sphere, we can use the formula for the surface area of a sphere, which is \(A = 4 * \pi * r^2\), where \(r\) is the radius of the sphere. We are given the diameter (0.5 m), so the radius will be half of that. Radius: \(r = \frac{0.5}{2} = 0.25 \mathrm{m}\) Now, calculate the surface area: \(A = 4 * \pi * (0.25)^2\) \(A = 4 * \pi * 0.0625 \approx 0.785 \mathrm{m}^2\)
02

Calculate the initial and final emission rate

We can now calculate the difference in emitted power per unit area in the cavity before and after cooling. We use the Stefan-Boltzmann law, which says that the power radiated per unit area is proportional to the fourth power of the temperature: \(E = \sigma * T^4\) Here, E is the emission rate, \(\sigma\) is the Stefan-Boltzmann constant (approximately \(5.67 * 10^{-8} W / (m^2 K^4)\)), and T is the temperature in Kelvin. Initial temperature: \(T_{initial} = 1400 \mathrm{K}\) Final temperature: \(T_{final} = 500 \mathrm{K}\) Calculate the initial and final emission rates: \(E_{initial} = \sigma * T_{initial}^4\) \(E_{final} = \sigma * T_{final}^4\)
03

Calculate the cooling rate

To maintain the temperature at 500 K, we need to remove the difference in emission rates from the cavity. The cooling rate will be the difference in emitted power per unit area multiplied by the surface area of the cavity. Cooling rate: \((E_{initial} - E_{final}) * A\) Plug in the values we calculated before: Cooling rate: \((\sigma * T_{initial}^4 - \sigma * T_{final}^4) * A\) \(= \sigma * (T_{initial}^4 - T_{final}^4) * A\) \(= (5.67 * 10^{-8}) * (1400^4 - 500^4) * 0.785\) \(= (5.67 * 10^{-8}) * (34300000000 - 625000000) * 0.785\) \(= (5.67 * 10^{-8}) * 33675000000 * 0.785\) \(= 149444 \mathrm{W}\) The required cooling rate to maintain the temperature at 500 K is approximately 149444 W (watts).

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Key Concepts

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

Emission Rate
The emission rate is a vital concept in understanding the energy emitted by objects due to their temperature. It is directly related to the Stefan-Boltzmann Law, which states that the power radiated per unit area of a surface is proportional to the fourth power of its absolute temperature. The formula used for calculating the emission rate is:

\[ E = \sigma \times T^4 \]

Where:
  • \( E \) is the emission rate (W/m²)
  • \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \text{ W/m}^2 \text{K}^4 \)
  • \( T \) is the temperature in Kelvin (K)
This tells us that as temperature increases, the emission rate grows at a rapid, exponential pace. Objects at higher temperatures emit significantly more radiation. Understanding this relationship helps in calculating how much energy a body emits when it cools or heats, which in turn guides the analysis of heat exchange processes.
Spherical Cavity
A spherical cavity, like the one described in the exercise, refers to a hollow, ball-shaped space. It can be part of different engineering systems. In this context, the spherical cavity contains a gas mixture.

When dealing with spheres, the surface area plays a crucial role in thermal calculations. To find the surface area of a sphere, the formula is:
  • \[ A = 4\pi r^2 \]
where \( r \) is the radius of the sphere.

Knowing the surface area is essential for calculating the amount of radiative heat transfer between the cavity and its surroundings. In this exercise, the surface area helps determine how much energy needs to be removed to cool the cavity consistently at 500 K. By reducing the temperature of the gas mixture in the cavity, we have a clear application of using geometry and physics to solve practical heat transfer problems.
Cooling Rate
The cooling rate is about how quickly an object or system loses heat. Understanding this helps in designing systems that maintain specific temperatures efficiently. In our exercise scenario, the cooling rate refers to the amount of energy that must be extracted from the cavity to sustain a lowered temperature of 500 K despite its initial high temperature of 1400 K.

To compute the cooling rate, it's essential to find the difference between the emission rates at the initial and final temperatures and then multiply it by the surface area:
  • Cooling rate: \( (E_{initial} - E_{final}) \times A \)
  • Substituting values, it becomes: \( \sigma \times (T_{initial}^4 - T_{final}^4) \times A \)
This calculation reveals how much power needs to be removed in watts (W) to maintain the cavity at its desired lower temperature. By mastering this concept, students can better understand the dynamics of energy exchange in thermally regulated environments.
Partial Pressure
Partial pressure describes the pressure a single gas in a mixture contributes to the total pressure. In a mixed atmosphere, like our spherical cavity containing \( \text{CO}_2 \) and nitrogen, each gas has its own partial pressure, adding up to the total pressure inside the system.

The total pressure is the sum of all individual partial pressures according to Dalton's Law. Given that the total pressure inside the cavity is 1 atm, with \( \text{CO}_2 \) at 0.25 atm and nitrogen at 0.75 atm, this law is a crucial piece of understanding how gases behave under shared container conditions.
  • It explains thermodynamic behavior in terms of individual gases despite being part of a mixture.
  • Helps calculate properties like density and temperature-specific energy exchange.
By grasping how partial pressures impact temperature and energy exchanges, one can better model and predict the behavior of complex gas mixtures in thermal systems.

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Most popular questions from this chapter

The arrangement shown is to be used to calibrate a heat flux gage. The gage has a black surface that is \(10 \mathrm{~mm}\) in diameter and is maintained at \(17^{\circ} \mathrm{C}\) by means of a water-cooled backing plate. The heater, \(200 \mathrm{~mm}\) in diameter, has a black surface that is maintained at \(800 \mathrm{~K}\) and is located \(0.5 \mathrm{~m}\) from the gage. The surroundings and the air are at \(27^{\circ} \mathrm{C}\) and the convection heat transfer coefficient between the gage and the air is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the net radiation exchange between the heater and the gage. (b) Determine the net transfer of radiation to the gage per unit area of the gage. (c) What is the net heat transfer rate to the gage per unit area of the gage? (d) If the gage is constructed according to the description of Problem 3.107, what heat flux will it indicate?

A very long electrical conductor \(10 \mathrm{~mm}\) in diameter is concentric with a cooled cylindrical tube \(50 \mathrm{~mm}\) in diameter whose surface is diffuse and gray with an emissivity of \(0.9\) and temperature of \(27^{\circ} \mathrm{C}\). The electrical conductor has a diffuse, gray surface with an emissivity of \(0.6\) and is dissipating \(6.0 \mathrm{~W}\) per meter of length. Assuming that the space between the two surfaces is evacuated, calculate the surface temperature of the conductor.

A row of regularly spaced, cylindrical heating elements is used to maintain an insulated furnace wall at \(500 \mathrm{~K}\). The opposite wall is at a uniform temperature of \(300 \mathrm{~K}\). The insulated wall experiences convection with air at \(450 \mathrm{~K}\) and a convection coefficient of \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the walls and elements are black, estimate the required operating temperature for the elements.

Consider two large, diffuse, gray, parallel surfaces separated by a small distance. If the surface emissivities are \(0.8\), what emissivity should a thin radiation shield have to reduce the radiation heat transfer rate between the two surfaces by a factor of 10 ?

Long, cylindrical bars are heat-treated in an infrared oven. The bars, of diameter \(D=50 \mathrm{~mm}\), are placed on an insulated tray and are heated with an overhead infrared panel maintained at temperature \(T_{p}=800 \mathrm{~K}\) with \(\varepsilon_{p}=0.85\). The bars are at \(T_{b}=300 \mathrm{~K}\) and have an emissivity of \(\varepsilon_{b}=0.92\). (a) For a product spacing of \(s=100 \mathrm{~mm}\) and a product length of \(L=1 \mathrm{~m}\), determine the radiation heat flux delivered to the product. Determine the heat flux at the surface of the panel heater. (b) Plot the radiation heat flux experienced by the product and the panel heater radiation heat flux over the range \(50 \mathrm{~mm} \leq s \leq 250 \mathrm{~mm}\).

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