/*! 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 54 Liquid oxygen is stored in a thi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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: \(m = \frac{Q}{2.13 \times 10^5}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At the bottom of a very large vacuum chamber whose walls are at \(300 \mathrm{~K}\), a black panel \(0.1 \mathrm{~m}\) in diameter is maintained at \(77 \mathrm{~K}\). To reduce the heat gain to this panel, a radiation shield of the same diameter \(D\) and an emissivity of \(0.05\) is placed very close to the panel. Calculate the net heat gain to the panel.

Consider two very large parallel plates. The botton plate is warmer than the top plate, which is held at a constant temperature of \(T_{\mathrm{1}}=330 \mathrm{~K}\). The plates are separated by \(L=0.1 \mathrm{~m}\), and the gap between the two surfaces is filled wath air at atmospheric pressure. The heat flux from the bottom plate is \(q^{*}=250 \mathrm{~W} \mathrm{~m}^{2}\). (a) Determine the temperature of the bottom plate and the ratio of the convective to radiative heat fluxes for \(\varepsilon_{1}=\varepsilon_{2}=0.5\). Evaluate air properties at \(T=350 \mathrm{~K}\). (b) Repeat part (a) for \(\varepsilon_{1}=\varepsilon_{2}=0.25\) and \(0.75\).

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.

A flat-bottomed hole \(6 \mathrm{~mm}\) in diameter is bored to a depth of \(24 \mathrm{~mm}\) in a diffuse, gray material having an emissivity of \(0.8\) and a uniform temperature of \(1000 \mathrm{~K} .\) (a) Determine the radiant power leaving the opening of the cavity. (b) The effective emissivity \(\varepsilon_{e}\) of a cavity is defined as the ratio of the radiant power leaving the cavity to that from a blackbody having the area of the cavity opening and a temperature of the inner surfaces of the cavity. Calculate the effective emissivity of the cavity described above. (c) If the depth of the hole were increased, would \(\varepsilon_{e}\) increase or decrease? What is the limit of \(s_{\epsilon}\) as the depth increases?

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.
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Electricity

Read Explanation

Electricity and Magnetism

Read Explanation

Circular Motion and Gravitation

Read Explanation

Geometrical and Physical Optics

Read Explanation

Particle Model of Matter

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.