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Chapter 13: Problem 54

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 \(2.13 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).)

Short Answer

Expert verified
The mass rate of evaporation (\(m\)) is found by calculating the radiative heat transfer (\(Q\)) between the surfaces using the given emissivities, radii, and temperatures, and then dividing \(Q\) by the latent heat of vaporization of oxygen (\(L\)). The final formula is: \[m = \frac{4 \pi (0.4)(0.6) (0.05 \times 5.67 \times 10^{-8} (95^4 - 280^4))}{(0.6 - 0.4)(2.13 \times 10^5)}\]

Step by step solution

01

Find the radiative heat transfer between the surfaces

Use the following formula for radiation heat transfer between two concentric spheres: \[Q = \dfrac{4 \pi r_1 r_2 (\epsilon_1 \sigma T_1^4 - \epsilon_2 \sigma T_2^4)}{r_2 - r_1}\] Where: - \(Q\) is the radiative heat transfer - \(r_1\) is the radius of the inner sphere - \(r_2\) is the radius of the outer sphere - \(\epsilon_1\) is the emissivity of the inner surface - \(\epsilon_2\) is the emissivity of the outer surface - \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} W/(m^2 K^4)\)) - \(T_1\) is the temperature of the inner surface - \(T_2\) is the temperature of the outer surface Since both surfaces have the same emissivity, we have: \[\epsilon_1 = \epsilon_2 = 0.05\] Given, diameter of inner sphere \(d_1\) = 0.8m, diameter of outer sphere \(d_2\) = 1.2m, so, \[r_1 = \frac{d_1}{2} = 0.4~m\] \[r_2 = \frac{d_2}{2} = 0.6~m\] Now, plug in the given values and solve for the radiative heat transfer: \(Q = \dfrac{4 \pi (0.4)(0.6) (0.05 \times 5.67 \times 10^{-8} (95^4 - 280^4))}{0.6 - 0.4}\)
02

Calculate the mass rate of evaporation

Now that we have the radiation heat transfer between the surfaces, we can find the mass rate of evaporation by dividing it by the latent heat of vaporization of oxygen. The formula for the mass rate of evaporation (\(m\)) is: \[m = \frac{Q}{L}\] Where: - \(m\) is the mass rate of evaporation (kg/s) - \(Q\) is the radiative heat transfer (W) - \(L\) is the latent heat of vaporization of oxygen (\(2.13 \times 10^5 J/kg\)) Plug in the values and solve for the mass rate of evaporation: \(m = \frac{Q}{2.13 \times 10^5}\)

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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 process where heat is transferred through electromagnetic waves. In a vacuum or any transparent medium (like the evacuated space between the two spherical containers in our exercise), this is the primary method of heat transport, as there are no particles to conduct heat.

Every object with a temperature above absolute zero emits thermal radiation. The quantity of heat transferred per unit time can be calculated using the Stefan-Boltzmann law for black bodies, which states that the energy radiated per unit area is proportional to the fourth power of the temperature. However, real-world materials are not perfect black bodies, so their emission is described by their emissivity, a factor that quantifies how effectively they emit thermal radiation.

In the context of our exercise, the rate of radiative heat transfer between the inner and outer surfaces of the spheres is found by using the modified Stefan-Boltzmann law, which accounts for the geometry of the system and the emissivity of the surfaces involved.
Mass Rate of Evaporation
The mass rate of evaporation refers to the amount of mass lost by a liquid as it transforms from the liquid to the gaseous phase over a given amount of time. This process is driven by the latent heat of vaporization, the energy required to change a unit mass of a substance from liquid to gas at a constant temperature.

To quantify this in practical terms, if you have a system where energy is transferred to a liquid, such as the oxygen in our exercise, that energy can cause the liquid to evaporate. The mass rate of evaporation can be determined by the amount of energy (heat transfer) delivered to the liquid divided by the latent heat of vaporization of that liquid, as per the formula provided in the step-by-step solution.

In applications like cryogenic storage, proper insulation and minimal heat transfer are crucial to reduce the mass rate of evaporation and thus, conserve valuable resources like liquid oxygen.
Stefan-Boltzmann Constant
The Stefan-Boltzmann constant, represented by the symbol \(\sigma\), is a physical constant that plays a pivotal role in the realm of thermodynamics and specifically in the context of radiative heat transfer. It originates from the Stefan-Boltzmann law, which describes the power radiated from a black body in terms of its temperature.

Specifically, the constant has a value of \(5.67 \times 10^{-8} W/(m^2 K^4)\), and it allows us to calculate the total energy radiated per unit surface area of a black body across all wavelengths per unit time. In the exercise provided, the Stefan-Boltzmann constant is used to calculate the heat transfer between the two spherical containers, illustrating how fundamental this constant is in predicting the behavior of systems involving thermal radiation.

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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?

Two plane coaxial disks are separated by a distance \(L=0.20 \mathrm{~m}\). The lower disk \(\left(A_{1}\right)\) is solid with a diameter \(D_{\mathrm{o}}=0.80 \mathrm{~m}\) and a temperature \(T_{1}=300 \mathrm{~K}\). The upper disk \(\left(A_{2}\right)\), at temperature \(T_{2}=1000 \mathrm{~K}\), has the same outer diameter but is ring-shaped with an inner diameter \(D_{i}=0.40 \mathrm{~m}\). Assuming the disks to be blackbodies, calculate the net radiative heat exchange between them.

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.

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 \(50 \mathrm{~mm} \leq s \leq 250 \mathrm{~mm}\).

A flue gas at 1 -atm total pressure and a temperature of \(1400 \mathrm{~K}\) contains \(\mathrm{CO}_{2}\) and water vapor at partial pressures of \(0.05\) and \(0.10 \mathrm{~atm}\), respectively. If the gas flows through a long flue of \(1-\mathrm{m}\) diameter and \(400 \mathrm{~K}\) surface temperature, determine the net radiative heat flux from the gas to the surface. Blackbody behavior may be assumed for the surface.
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