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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

Consider a carton of milk that is refrigerated at a temperature of \(T_{m \mathrm{r}}=5^{\circ} \mathrm{C}\). The kitchen temperature on a hot summer day is \(T_{\infty}=30^{\circ} \mathrm{C}\). If the four sides of the carton are of height and width \(L=200 \mathrm{~mm}\) and \(w=100 \mathrm{~mm}\), respectively, determine the heat transferred to the milk carton as it sits on the kitchen counter for durations of \(t=10 \mathrm{~s}, 60 \mathrm{~s}\), and \(300 \mathrm{~s}\) before it is returned to the refrigerator. The convection coefficient associated with natural convection on the sides of the carton is \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface emissivity is \(0.90\). Assume the milk carton temperature remains at \(5^{\circ} \mathrm{C}\) during the process. Your parents have taught you the importance of refrigerating certain foods from the food safety perspective. Comment on the importance of quickly returning the milk carton to the refrigerator from an energy conservation point of view.

The temperature controller for a clothes dryer consists of a bimetallic switch mounted on an electrical heater attached to a wall-mounted insulation pad. The switch is set to open at \(70^{\circ} \mathrm{C}\), the maximum dryer air temperature. To operate the dryer at a lower air temperature, sufficient power is supplied to the heater such that the switch reaches \(70^{\circ} \mathrm{C}\left(T_{\text {set }}\right)\) when the air temperature \(T\) is less than \(T_{\text {set. }}\). If the convection heat transfer coefficient between the air and the exposed switch surface of \(30 \mathrm{~mm}^{2}\) is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how much heater power \(P_{e}\) is required when the desired dryer air temperature is \(T_{\infty}=50^{\circ} \mathrm{C}\) ?

Under conditions for which the same room temperature is maintained by a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) by considering a room whose air temperature is maintained at \(20^{\circ} \mathrm{C}\) throughout the year, while the walls of the room are nominally at \(27^{\circ} \mathrm{C}\) and \(14^{\circ} \mathrm{C}\) in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of \(32^{\circ} \mathrm{C}\) throughout the year and to have an emissivity of \(0.90\). The coefficient associated with heat transfer by natural convection between the person and the room air is approximately \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Consider a surface-mount type transistor on a circuit board whose temperature is maintained at \(35^{\circ} \mathrm{C}\). Air at \(20^{\circ} \mathrm{C}\) flows over the upper surface of dimensions \(4 \mathrm{~mm} \times\) \(8 \mathrm{~mm}\) with a convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Three wire leads, each of cross section \(1 \mathrm{~mm} \times 0.25 \mathrm{~mm}\) and length \(4 \mathrm{~mm}\), conduct heat from the case to the circuit board. The gap between the case and the board is \(0.2 \mathrm{~mm}\). (a) Assuming the case is isothermal and neglecting radiation, estimate the case temperature when \(150 \mathrm{~mW}\) is dissipated by the transistor and (i) stagnant air or (ii) a conductive paste fills the gap. The thermal conductivities of the wire leads, air, and conductive paste are \(25,0.0263\), and \(0.12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. (b) Using the conductive paste to fill the gap, we wish to determine the extent to which increased heat dissipation may be accommodated, subject to the constraint that the case temperature not exceed \(40^{\circ} \mathrm{C}\). Options include increasing the air speed to achieve a larger convection coefficient \(h\) and/or changing the lead wire material to one of larger thermal conductivity. Independently considering leads fabricated from materials with thermal conductivities of 200 and \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), compute and plot the maximum allowable heat dissipation for variations in \(h\) over the range \(50 \leq h \leq 250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A thermodynamic analysis of a proposed Brayton cycle gas turbine yields \(P=5 \mathrm{MW}\) of net power production. The compressor, at an average temperature of \(T_{c}=400^{\circ} \mathrm{C}\), is driven by the turbine at an average temperature of \(T_{h}=1000^{\circ} \mathrm{C}\) by way of an \(L=1\)-m-long, \(d=70-\mathrm{mm}-\) diameter shaft of thermal conductivity \(k=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Compare the steady-state conduction rate through the shaft connecting the hot turbine to the warm compressor to the net power predicted by the thermodynamics-based analysis. (b) A research team proposes to scale down the gas turbine of part (a), keeping all dimensions in the same proportions. The team assumes that the same hot and cold temperatures exist as in part (a) and that the net power output of the gas turbine is proportional to the overall volume of the device. Plot the ratio of the conduction through the shaft to the net power output of the turbine over the range \(0.005 \mathrm{~m} \leq L \leq 1 \mathrm{~m}\). Is a scaled-down device with \(L=0.005 \mathrm{~m}\) feasible?

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