/*! 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 63 A cryogenic fluid flows through ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 63

A cryogenic fluid flows through a tube \(20 \mathrm{~mm}\) in diameter, the outer surface of which is diffuse and gray with an emissivity of \(0.02\) and temperature of \(77 \mathrm{~K}\). This tube is concentric with a larger tube of 50 \(\mathrm{mm}\) diameter, the inner surface of which is diffuse and gray with an emissivity of \(0.05\) and temperature of \(300 \mathrm{~K}\). The space between the surfaces is evacuated. Determine the heat gain by the cryogenic fluid per unit length of the inner tube. If a thin-walled radiation shield that is diffuse and gray with an emissivity of \(0.02\) (both sides) is inserted midway between the inner and outer surfaces, calculate the change (percentage) in heat gain per unit length of the inner tube.

Short Answer

Expert verified
The heat gain by the cryogenic fluid per unit length of the inner tube without the radiation shield, \(Q\), and with the radiation shield, \(Q_{shield}\), can be calculated using the radiative heat transfer formula. After computing these values, the percentage change in heat gain can be found using the formula: \[% change = \frac{Q_{shield} - Q}{Q} \times 100\%\] By solving for these values, we can find the change in heat gain per unit length of the inner tube when the radiation shield is inserted.

Step by step solution

01

Determining surface areas

To calculate the surface areas of the tubes, we can use the formula for the area of a cylinder without the top and bottom parts: \[A = 2\pi r L\] Where: - r is the radius - L is the length of the tube (we can consider it as a unit length) For the inner tube: \[A_1 = 2\pi \frac{0.01}{1}\] For the outer tube: \[A_2 = 2\pi \frac{0.025}{1}\]
02

Determining the view factor

Since the space between the inner and outer surfaces is evacuated, we can assume that the view factor \(F_{1 \to 2}\) is equal to 1.
03

Calculate the heat transfer without the radiation shield

Now we have all the parameters needed to find the heat transfer without the radiation shield. We can use the given formula and plug in the values: \[Q = \frac{2\pi \frac{0.01}{1}(1) \sigma (77^4 - 300^4)}{(((1-0.02)/0.02)2\pi \frac{0.01}{1}) + (1 - 1) + (((1-0.05)/0.05)2\pi \frac{0.025}{1})}\] Solve this equation to obtain the heat transfer.
04

Calculate the heat transfer with the radiation shield

To account for the radiation shield, we need to modify the radiative heat transfer formula to include the shield. We have to divide the space between the inner and outer surfaces into two sections. The heat transfer equation now becomes: \[Q_{shield} = A_1 F_{1 \to s} \sigma \frac{T_1^4 - T_s^4}{((1-e_1)/e_1A_1) + (\frac{1}{F_{1 \to s}} - 1) + ((1-e_3)/e_3 A_s)} + A_s F_{s \to 2} \sigma \frac{T_s^4 - T_2^4}{((1-e_3)/e_3A_s) + (\frac{1}{F_{s \to 2}} - 1) + ((1-e_2)/e_2 A_2)}\] Where: - \(T_s\) is the temperature of the shield - \(e_3\) is the emissivity of the shield - \(A_s\) is the surface area of the shield Since the shield is thin-walled and located midway between the inner and outer surfaces, we can assume that its surface area is the average of the surface areas of the inner and outer surfaces. Also, the shield's emissivity is given as 0.02 on both sides. Calculate the heat transfer with the shield by solving the modified equation using the assumption and given parameters.
05

Calculate the percentage change in heat gain

To calculate the percentage change in heat gain per unit length of the inner tube, use the following formula: \[% change = \frac{Q_{shield} - Q}{Q} \times 100\%\] Use the values of heat transfer with and without the radiation shield to calculate the percentage change.

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.

Radiation Shield Efficiency
In the context of thermal engineering, when dealing with sensitive substances like cryogenic fluids, maintaining low temperatures is essential. One effective method used in these scenarios is the introduction of a radiation shield. A radiation shield is a barrier that reflects or absorbs radiant energy, thereby reducing the heat exchange between objects.
A radiation shield's efficiency is touted based on its emissivity, which is a measure of how much radiation a surface emits compared to a perfect black body at the same temperature. To improve radiation shield efficiency, materials with low emissivity are chosen. The shield acts by reflecting radiant energy or by absorbing and re-emitting it at a lower intensity.
In our exercise, the shield, with an emissivity of 0.02, is inserted between two concentric tubes. The efficiency of this addition is quantified by calculating the percentage change in heat gain by the cryogenic fluid. The more significant the reduction, the higher the shield's efficiency. It's found by comparing heat transfer rates before and after inserting the shield and expressed as a percentage.
Cryogenic Fluid Heat Gain
Cryogenic fluid heat gain is a critical parameter in the design and operation of systems involving very low temperature substances, such as liquid nitrogen or helium. These fluids are susceptible to external heat, which can cause them to rapidly gain energy and increase in temperature.
In our example, the inner tube carries a cryogenic fluid. It's crucial to understand and minimize the heat gain to prevent the warming of the fluid. Various environmental factors can contribute to heat gain, but given the vacuum between the inner and outer tube, radiative heat transfer is the dominant mode.

Understanding Radiative Heat Transfer

The rate of heat exchange due to radiation is proportional to the temperature difference to the fourth power, as described by the Stefan-Boltzmann law, and the emissivity of the materials involved. By calculating the heat gain per unit length of the inner tube, engineers can design effective insulation or choose the appropriate materials to reduce heat input and maintain the cryogenic state of the fluid.
Concentric Tube Radiative Exchange
The concentric tube radiative exchange comes into play when two cylindrical surfaces are aligned coaxially with a vacuum or a negligible medium in between. In heat transfer analysis, this configuration is often used because of its practical applications in many engineering fields, particularly in cases where minimizing heat exchange is vital.
In the absence of conduction and convection due to the vacuum, the radiative heat transfer equation simplifies the calculations. We consider factors such as the surface areas, temperatures, and emissivities of the tubes and the shield. The view factor, which in our evacuated setup is assumed to be unity, indicating that the entirety of the radiative energy leaving the inner surface reaches the outer surface, influences the rate of heat exchange in this configuration.

Impact of Shield

The introduction of a radiation shield changes the dynamics of this exchange. It adds resistance to the radiative heat flow and typically results in a significant reduction of heat transfer. This protective measure is often crucial in designing systems meant to transport or store cryogenic fluids, contributing to the effectiveness and safety of the operation.

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

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 \(0.75\).

Options for thermally shielding the top ceiling of a large furnace include the use of an insulating material of thickness \(L\) and thermal conductivity \(k\), case (a), or an air space of equ盲valent thickness formed by installing a steel sheet above the ceiling, case \((b)\). (a) Develop mathematical models that could be used to assess which of the two approaches is better. In both cases the interior surface is maintained at the same temperature, \(T_{s,}\) and the ambient air and surroundings are at equivalent temperatures \(\left(T_{w}=T_{\text {ue }}\right)\). (b) If \(k=0.090 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=25 \mathrm{~mm}, h_{a}=25\) W/ \(\mathrm{m}^{2} \cdot \mathrm{K}\), the surfaces are diffuse and gray with \(\varepsilon_{i}=\varepsilon_{o}=0.50, T_{s u}=900 \mathrm{~K}\), and \(T_{w}=T_{\mathrm{ar}}=\) \(300 \mathrm{~K}\), what is the outer surface temperature \(T_{\text {a }}\) and the heat loss per unit surface area associated with each option? (c) For each case, assess the effect of surface radiative properties on the outer surface temperature and the heat loss per unit area for values of

Consider the perpendicular rectangles shown schematically. (a) Determine the shape factor \(F_{12}\). (b) For rectangle widths of \(X=0.5,1.5\), and \(5 \mathrm{~m}\), plot \(F_{12}\) as a function of \(Z_{b}\) for \(0.05 \leq Z_{b} \leq 0.4 \mathrm{~m}\).] Compare your results with the view factor obtained from the two-dimensional relation for perpendicular plates with a common edge (Table 13.1).

Heat transfer by radiation occurs between two large parallel plates, which are maintained at temperatures \(T_{1}\) and \(T_{2}\), with \(T_{1}>T_{2}\). To reduce the rate of heat transfer between the plates, it is proposed that they be separated by a thin shield that has different emissivities on opposite surfaces. In particular, one surface has the emissivity \(\varepsilon_{s}<0.5\), while the opposite surface has an emissivity of \(2 \varepsilon_{s}\). (a) How should the shield be oriented to provide the larger reduction in heat transfer between the plates? That is, should the surface of emissivity \(\varepsilon_{s}\) or that of emissivity \(2 s_{s}\) be oriented toward the plate at \(T_{1}\) ? (b) What orientation will result in the larger value of the shield temperature \(T_{s}\) ?

A grain dryer consists of a long semicircular duct of radius \(R=1 \mathrm{~m}\). One-half of the base surface consists of an electrically heated plate of emissivity \(z_{p}=0.8\), while the other half supports the grain to be dried, which has an emissivity of \(\varepsilon_{g}=0.9\). In a batch drying process for which the temperature of the grain is \(T_{s}=330 \mathrm{~K}, 2.50 \mathrm{~kg}\) of water are to be removed per meter of duct length over a 1-h period. (a) Neglecting convection heat transfer, determine the required temperature \(T_{p}\) of the heater plate. (b) If the water vapor is swept from the duct by the flow of dry air, what convection mass transfer coefficient \(h_{w}\) must be maintained by the flow? (c) If the air is at \(300 \mathrm{~K}\), is the assumption of negligible convection justified?
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Quantum Physics

Read Explanation

Particle Model of Matter

Read Explanation

Measurements

Read Explanation

Fields in Physics

Read Explanation

Space Physics

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.