/*! 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 49 Liquid oxygen, which has a boili... [FREE SOLUTION] | 91Ó°ÊÓ

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Liquid oxygen, which has a boiling point of \(90 \mathrm{~K}\) and a latent heat of vaporization of \(214 \mathrm{~kJ} / \mathrm{kg}\), is stored in a spherical container whose outer surface is of \(500-\mathrm{mm}\) diameter and at a temperature of \(-10^{\circ} \mathrm{C}\). The container is housed in a laboratory whose air and walls are at \(25^{\circ} \mathrm{C}\). (a) If the surface emissivity is \(0.20\) and the heat transfer coefficient associated with free convection at the outer surface of the container is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the rate, in \(\mathrm{kg} / \mathrm{s}\), at which oxygen vapor must be vented from the system? (b) Moisture in the ambient air will result in frost formation on the container, causing the surface emissivity to increase. Assuming the surface temperature and convection coefficient to remain at \(-10^{\circ} \mathrm{C}\) and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, compute the oxygen evaporation rate \((\mathrm{kg} / \mathrm{s})\) as a function of surface emissivity over the range \(0.2 \leq \varepsilon \leq 0.94\).

Short Answer

Expert verified
For part (a), the rate at which oxygen vapor must be vented from the system is \(m_{rate} = -1.39\times10^{-3}\,kg/s\). For part (b), the oxygen evaporation rate as a function of surface emissivity can be calculated using the equation \(m_{rate}(\varepsilon) = \frac{\varepsilon \sigma A (T_s^4 - T_a^4) + hA(T_s - T_a)}{L_v}\), where \(\varepsilon\) ranges from 0.2 to 0.94.

Step by step solution

01

Constants and Conversions

First, let's note down the given information and convert it to SI units if necessary. Boiling point of liquid oxygen: \(T_b = 90K\) Latent heat of vaporization: \(L_v = 214\times10^3\,J/kg\) Diameter of the container: \(D = 0.5\,m\) Outer surface temperature: \(T_s = -10^\circ C = 263.15K\) Ambient Temperature: \(T_a= 25^\circ C= 298.15K\) Emissivity: \(\varepsilon = 0.20\) Heat transfer coefficient: \(h = 10\,W/ m^2\cdot K\)
02

Calculate Heat Transfer due to Radiation

We will use the Stefan-Boltzmann law to calculate the heat transfer due to radiation: \(q_r = \varepsilon \sigma A (T_s^4 - T_a^4)\) Here, \(q_r\) is the heat transfer due to radiation, \(\sigma\) is the Stefan-Boltzmann constant \((5.67\times10^{-8} W/m^2\cdot K^4)\) and \(A\) is the outer surface area of the container. We can calculate the outer surface area as: \(A = 4\pi (\frac{D}{2})^2 = 4\pi (0.25)^2 = 0.7854\,m^2\) Now, calculating the heat transfer due to radiation: \(q_r = 0.20 \times 5.67\times10^{-8} \times 0.7854 (263.15^4 - 298.15^4) = -22.41 \,W\)
03

Calculate Heat Transfer due to Convection

We can calculate the heat transfer due to convection using Newton's law: \(q_c = hA(T_s - T_a)\) Now calculating the heat transfer due to convection: \(q_c = 10\times 0.7854 (263.15 - 298.15) = -274.5 \,W\)
04

Calculate the Total Heat Transfer

The total heat transfer to the container is the sum of the heat transfers due to radiation and convection: \(q_t = q_r + q_c = -22.41 + (-274.5) = -296.91\,W\)
05

Calculate Oxygen Evaporation Rate (Part a)

We can calculate the rate at which oxygen evaporates using the total heat transfer and the latent heat of vaporization: \(m_{rate} = \frac{q_t}{L_v}\) Now calculating the oxygen evaporation rate: \(m_{rate} = \frac{-296.91}{214\times10^3} = -1.39\times10^{-3}\,kg/s\)
06

Calculate Oxygen Evaporation Rate as a function of Emissivity (Part b)

For this step, we need to calculate the oxygen evaporation rate considering the different values of surface emissivities in the range (0.2 to 0.94). To do that, we will use the following equation derived from combining the expressions for \(q_r\), \(q_c\), and \(q_t\), and the relationship between \(m_{rate}\) and \(q_t\): \(m_{rate}(\varepsilon) = \frac{\varepsilon \sigma A (T_s^4 - T_a^4) + hA(T_s - T_a)}{L_v}\) Using this equation, we can find the oxygen evaporation rate for any value of emissivity \(\varepsilon\) between 0.2 and 0.94. For completing this exercise, we have successfully broken down the problem into a step-by-step solution with each step clearly explaining the concept and calculation.

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Key Concepts

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

Latent Heat of Vaporization
Latent heat of vaporization is a fundamental concept in thermodynamics, particularly when we're dealing with phase changes between liquid and gaseous states. It represents the amount of energy required to convert one kilogram of a substance from a liquid to a gas without changing its temperature.

In the given exercise, liquid oxygen has a latent heat of vaporization of \(214 \times 10^3 \, J/kg\). This high energy requirement is due to the strong molecular forces that must be overcome to transition from the densely packed liquid state to the more dispersed gaseous state. To calculate the rate of oxygen evaporation, one must consider the total heat being transferred from the surroundings to the oxygen within the container. This heat transfer is directly related to the latent heat as it provides the necessary energy for phase change.

Diving into the practical implications, our day-to-day interactions with evaporative processes, like boiling water for tea, involve latent heat. Recognizing the significant energy involved in such transitions can lead to a better understanding of energy efficiency and conservation in both industrial and environmental processes.
Stefan-Boltzmann Law
The Stefan-Boltzmann law is a cornerstone in the field of thermodynamics, particularly in the analysis of thermal radiation. It states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black-body radiant exitance or emissive power) is directly proportional to the fourth power of the black body's thermodynamic temperature. Mathematically, it's expressed as \( q_r = \varepsilon \sigma A (T_s^4 - T_a^4) \), where:\
    \
  • \( q_r \) is the radiative heat transfer\
  • \( A \) is the surface area\
  • \( T_s \) is the surface temperature\
  • \( T_a \) is the ambient temperature\
  • \( \varepsilon \) is the emissivity of the surface\
  • \( \sigma \) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} W/m^2 \cdot K^4)\)\
\
In our exercise, the Stefan-Boltzmann law was used to determine the heat loss due to radiation from the surface of the spherical container housing the liquid oxygen. It's important to note that this law is only strictly accurate for ideal black bodies, but it can be applied to real-world objects by incorporating the emissivity factor, which accounts for how closely a surface's radiative properties approximate those of a black body. In practical applications, understanding radiative heat transfer and the law's dependence on temperature differences is essential for engineers and scientists designing systems like satellites, furnaces, and even climate models.
Free Convection Heat Transfer
Free convection is a type of heat transfer that occurs in fluids (liquids and gases) without any external force, driven instead by buoyancy forces that result from density variations due to variations in temperature within the fluid. When a surface is heated, it transfers heat to the adjacent fluid layers. This causes the fluid to expand, decrease in density, and rise, being replaced by cooler fluid that is then heated, creating a convection current.

Newton's law of cooling is keenly applied here, represented by the equation \( q_c = hA(T_s - T_a) \). The heat transfer coefficient \( h \) is an empirical value that encompasses the properties and conditions of the surface and the fluid and the nature of the convective flow. In the scenario outlined in the textbook problem, the heat transfer coefficient associated with free convection at the outer surface of the oxygen container is \(10 \, W/m^2 \cdot K\).

Being aware of how free convection works is valuable in various practical applications, such as designing cooling systems for electronic devices, optimizing heating and ventilation in buildings, and even in meteorological phenomena like the formation of wind. Knowledge of free convection can help anticipate and control the heat transfer to maintain desired temperatures in different systems.

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Most popular questions from this chapter

In considering the following problems involving heat transfer in the natural environment (outdoors), recognize that solar radiation is comprised of long and short wavelength components. If this radiation is incident on a semitransparent medium, such as water or glass, two things will happen to the nonreflected portion of the radiation. The long wavelength component will be absorbed at the surface of the medium, whereas the short wavelength component will be transmitted by the surface. (a) The number of panes in a window can strongly influence the heat loss from a heated room to the outside ambient air. Compare the single- and double-paned units shown by identifying relevant heat transfer processes for each case. (b) In a typical flat-plate solar collector, energy is collected by a working fluid that is circulated through tubes that are in good contact with the back face of an absorber plate. The back face is insulated from the surroundings, and the absorber plate receives solar radiation on its front face, which is typically covered by one or more transparent plates. Identify the relevant heat transfer processes, first for the absorber plate with no cover plate and then for the absorber plate with a single cover plate. (c) The solar energy collector design shown in the schematic has been used for agricultural applications. Air is blown through a long duct whose cross section is in the form of an equilateral triangle. One side of the triangle is comprised of a double-paned, semitransparent cover; the other two sides are constructed from aluminum sheets painted flat black on the inside and covered on the outside with a layer of styrofoam insulation. During sunny periods, air entering the system is heated for delivery to either a greenhouse, grain drying unit, or storage system. Identify all heat transfer processes associated with the cover plates, the absorber plate(s), and the air. (d) Evacuated-tube solar collectors are capable of improved performance relative to flat-plate collectors. The design consists of an inner tube enclosed in an outer tube that is transparent to solar radiation. The annular space between the tubes is evacuated. The outer, opaque surface of the inner tube absorbs solar radiation, and a working fluid is passed through the tube to collect the solar energy. The collector design generally consists of a row of such tubes arranged in front of a reflecting panel. Identify all heat transfer processes relevant to the performance of this device.

Bus bars proposed for use in a power transmission station have a rectangular cross section of height \(H=600 \mathrm{~mm}\) and width \(W=200 \mathrm{~mm}\). The electrical resistivity, \(\rho_{e}(\mu \Omega \cdot \mathrm{m})\), of the bar material is a function of temperature, \(\rho_{e}=\rho_{e, o}\left[1+\alpha\left(T-T_{o}\right)\right]\), where \(\rho_{e, a}=\) \(0.0828 \mu \Omega \cdot \mathrm{m}, T_{o}=25^{\circ} \mathrm{C}\), and \(\alpha=0.0040 \mathrm{~K}^{-1}\). The emissivity of the bar's painted surface is \(0.8\), and the temperature of the surroundings is \(30^{\circ} \mathrm{C}\). The convection coefficient between the bar and the ambient air at \(30^{\circ} \mathrm{C}\) is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Assuming the bar has a uniform temperature \(T\), calculate the steady-state temperature when a current of \(60,000 \mathrm{~A}\) passes through the bar. (b) Compute and plot the steady-state temperature of the bar as a function of the convection coefficient for \(10 \leq h \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What minimum convection coefficient is required to maintain a safe-operating temperature below \(120^{\circ} \mathrm{C}\) ? Will increasing the emissivity significantly affect this result?

A computer consists of an array of five printed circuit boards (PCBs), each dissipating \(P_{b}=20 \mathrm{~W}\) of power. Cooling of the electronic components on a board is provided by the forced flow of air, equally distributed in passages formed by adjoining boards, and the convection coefficient associated with heat transfer from the components to the air is approximately \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Air enters the computer console at a temperature of \(T_{i}=20^{\circ} \mathrm{C}\), and flow is driven by a fan whose power consumption is \(P_{f}=25 \mathrm{~W}\). (a) If the temperature rise of the airflow, \(\left(T_{o}-T_{i}\right)\), is not to exceed \(15^{\circ} \mathrm{C}\), what is the minimum allowable volumetric flow rate \(\dot{\forall}\) of the air? The density and specific heat of the air may be approximated as \(\rho=1.161\) \(\mathrm{kg} / \mathrm{m}^{3}\) and \(c_{p}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (b) The component that is most susceptible to thermal failure dissipates \(1 \mathrm{~W} / \mathrm{cm}^{2}\) of surface area. To minimize the potential for thermal failure, where should the component be installed on a PCB? What is its surface temperature at this location?

A \(50 \mathrm{~mm} \times 45 \mathrm{~mm} \times 20 \mathrm{~mm}\) cell phone charger has a surface temperature of \(T_{s}=33^{\circ} \mathrm{C}\) when plugged into an electrical wall outlet but not in use. The surface of the charger is of emissivity \(\varepsilon=0.92\) and is subject to a free convection heat transfer coefficient of \(h=4.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The room air and wall temperatures are \(T_{\infty}=22^{\circ} \mathrm{C}\) and \(T_{\text {sur }}=20^{\circ} \mathrm{C}\), respectively. If electricity costs \(C=\$ 0.18 / \mathrm{kW} \cdot \mathrm{h}\), determine the daily cost of leaving the charger plugged in when not in use.

A spherical interplanetary probe of \(0.5-\mathrm{m}\) diameter contains electronics that dissipate \(150 \mathrm{~W}\). If the probe surface has an emissivity of \(0.8\) and the probe does not receive radiation from other surfaces, as, for example, from the sun, what is its surface temperature?

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