/*! 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 128 Consider an extended surface of ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider an extended surface of rectangular cross section with heat flow in the longitudinal direction. In this problem we seek to determine conditions for which the transverse ( \(y\)-direction) temperature difference within the extended surface is negligible compared to the temperature difference between the surface and the environment, such that the one-dimensional analysis of Section 3.6.1 is valid. (a) Assume that the transverse temperature distribution is parabolic and of the form $$ \frac{T(y)-T_{o}(x)}{T_{s}(x)-T_{o}(x)}=\left(\frac{y}{t}\right)^{2} $$ where \(T_{s}(x)\) is the surface temperature and \(T_{o}(x)\) is the centerline temperature at any \(x\)-location. Using Fourier's law, write an expression for the conduction heat flux at the surface, \(q_{y}^{\prime \prime}(t)\), in terms of \(T_{s}\) and \(T_{a^{+}}\) (b) Write an expression for the convection heat flux at the surface for the \(x\)-location. Equating the two expressions for the heat flux by conduction and convection, identify the parameter that determines the ratio \(\left(T_{o}-T_{s}\right) /\left(T_{s}-T_{\infty}\right)\). (c) From the foregoing analysis, develop a criterion for establishing the validity of the onedimensional assumption used to model an extended surface.

Short Answer

Expert verified
In summary, the validity of the one-dimensional analysis for an extended surface can be determined by analyzing the Biot number (Bi). If \(Bi \ll 1\), the temperature difference in the transverse direction is negligible, and the one-dimensional analysis is valid. However, if \(Bi \gtrsim 1\), the one-dimensional analysis is less accurate, as heat transfer in the extended surface is influenced by both longitudinal heat conduction and transverse heat transfer due to convection.

Step by step solution

01

a) Find the conduction heat flux at the surface using Fourier's law

First, recall Fourier's law of heat conduction in the y-direction: \(q_y^{\prime \prime} = -k\frac{dT(y)}{dy}\) We are given the following equation for the transverse temperature distribution: \[\frac{T(y)-T_{o}(x)}{T_{s}(x)-T_{o}(x)}=\left(\frac{y}{t}\right)^{2}\] First, we need to find the derivative of the temperature function with respect to y. To do this, we will isolate T(y) in the above equation and take its derivative: \( (T(y) - T_o(x)) = \left(\frac{y}{t}\right)^2 (T_s(x) - T_o(x)) \) \( T(y) = T_o(x) + \left(\frac{y}{t}\right)^2 (T_s(x) - T_o(x)) \) Now, we can take the derivative of T(y) with respect to y: \(\frac{dT(y)}{dy} = \frac{2y}{t^2} (T_s(x) - T_o(x)) \) We can now substitute this derivative into Fourier's law to find the conduction heat flux at the surface: \(q_y^{\prime\prime}(t) = -k \frac{dT(y)}{dy} \) at \(y = t\) \(q_y^{\prime\prime}(t) = -k \frac{2t}{t^2} (T_s(x) - T_o(x)) \) Finally, we have an expression for the conduction heat flux at the surface: \(q_y^{\prime\prime}(t) = -2k \frac{(T_s(x) - T_o(x))}{t} \)
02

b) Find the convection heat flux at the surface and equate it with conduction heat flux

We are given Newton's law of cooling for convection heat transfer: \(q_x^{\prime\prime}(x) = h(T_s(x) - T_\infty)\) Now, we need to equate the conduction heat flux and convection heat flux: \( -2k \frac{(T_s(x) - T_o(x))}{t} = h(T_s(x) - T_\infty) \) We are asked to identify the parameter that determines the ratio: \(\frac{(T_o(x) - T_s(x))}{(T_s(x) - T_\infty)}\) This parameter can be identified from the equation we just obtained after equating the heat fluxes. Rearranging the terms, we obtain: \(\frac{T_s(x) - T_o(x)}{t} = \frac{h}{2k}(T_s(x) - T_\infty)\) Dividing both sides by \(T_s(x) - T_\infty\), we get: \(\frac{T_o(x) - T_s(x)}{T_s(x) - T_\infty} = \frac{t}{2} \frac{h}{k}\) Hence, the parameter that determines the given ratio is: \(Bi = \frac{t}{2}\frac{h}{k}\), which is known as the Biot number
03

c) Establish the validity of one-dimensional analysis using the Biot number

The Biot number characterizes the conditions under which the one-dimensional analysis is valid. When the Biot number is small, the temperature difference across the surface is negligible compared to the difference between the surface and the environment. In such cases, the one-dimensional assumption is valid. Analyzing the Biot number, we can define a criterion for the validity of the one-dimensional analysis of an extended surface as follows: - If \(Bi \ll 1\), then the temperature difference in the y-direction (transverse direction) can be considered negligible. This means that the one-dimensional analysis is valid since the heat transfer within the extended surface is dominated by the longitudinal (x-direction) heat conduction. - Conversely, if \(Bi \gtrsim 1\), the one-dimensional analysis is less accurate, as the heat transfer in the extended surface is a combination of both longitudinal heat conduction and transverse heat transfer due to convection.

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Key Concepts

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

Fourier's Law
Fourier's law is a fundamental principle that explains how heat is conducted through materials. It states that the rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area through which the heat flows. Mathematically, it is expressed as
\( q = -kA\frac{dT}{dx} \),
where \( q \) is the heat transfer rate, \( k \) is the thermal conductivity of the material, \( A \) is the cross-sectional area, and \( \frac{dT}{dx} \) is the temperature gradient along the direction of heat flow.
In the extended surface heat transfer problem, we apply Fourier's law to determine the conduction heat flux at the surface. This law is crucial in one-dimensional heat transfer analysis, where heat flow in the transverse direction (y-direction) is considered negligible.
Conduction Heat Flux
Conduction heat flux refers to the rate at which heat energy is transferred through a material due to a temperature gradient. In the context of the exercise,
\( q_y^{'{'}}{'{'}'} = -k\frac{dT(y)}{dy} \)
is the formula used to calculate the conduction heat flux at a given point on the surface of the extended material. In layman's terms, the heat flux is essentially the amount of heat passing through a unit area of the material per unit time.
For the parabolic temperature distribution given in the problem, calculation involves finding the temperature gradient at the boundary layer (
y = t
) and applying Fourier's law. Understanding how this heat flux is calculated is crucial for assessing the efficiency of heat transfer in materials and for validating the one-dimensional analysis assumption in heat transfer problems.
Convection Heat Flux
Convection heat flux is the heat transferred per unit area due to the movement of fluid (liquid or gas) over a surface. In the exercise, we consider Newton's law of cooling which stipulates that
\( q_x^{'{'}}{'{'}'} = h(T_s(x) - T_{\infty}) \),
where \( h \) is the convective heat transfer coefficient, \( T_s(x) \) is the surface temperature, and \( T_{\infty} \) represents the temperature of the fluid far from the surface. Equating the conduction and convection heat fluxes allows the identification of dimensionless parameters that govern heat transfer phenomena, such as the Biot number. Convective heat transfer plays a pivotal role in various engineering applications, including HVAC systems, cooling of electronic equipment, and the design of heat exchangers.
Biot Number
The Biot number (Bi) is a dimensionless number that compares the internal resistance to heat conduction within a body to the external resistance to heat convection from the surface of the body to the surrounding fluid.
\( Bi = \frac{t}{2}\frac{h}{k} \)
represents this parameter in the context of extended surface heat transfer. The Biot number helps to assess the validity of one-dimensional heat transfer analysis. If the Biot number is much less than one (
Bi \ll 1
), then the temperature gradient within the material is small compared to the temperature difference between the surface and the ambient environment, indicating that one-dimensional analysis is appropriate. This concept is central to simplifying complex heat transfer problems and is widely used in the design of fins and heat exchangers.
One-dimensional Heat Transfer Analysis
One-dimensional heat transfer analysis assumes that temperature variation occurs in only one dimension, simplifying the complex, multi-dimensional reality of heat transfer into a manageable calculation.
In the textbook exercise, ensuring that the surface temperature variation in the direction perpendicular to heat flow is negligible makes a one-dimensional approach valid.
Criteria such as the Biot number are used to determine the appropriateness of this assumption. If the criteria are met, the analysis simplifies significantly, providing a more straightforward approach to understanding and calculating heat transfer in long, thin objects like fins. Hence, in engineering, selecting the right circumstances under which to apply one-dimensional heat transfer analysis is vital for accurate and efficient thermal designs.

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Most popular questions from this chapter

Two stainless steel plates \(10 \mathrm{~mm}\) thick are subjected to a contact pressure of 1 bar under vacuum conditions for which there is an overall temperature drop of \(100^{\circ} \mathrm{C}\) across the plates. What is the heat flux through the plates? What is the temperature drop across the contact plane?

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.

A spherical tank for storing liquid oxygen on the space shuttle is to be made from stainless steel of \(0.80-\mathrm{m}\) outer diameter and 5 -mm wall thickness. The boiling point and latent heat of vaporization of liquid oxygen are \(90 \mathrm{~K}\) and \(213 \mathrm{~kJ} / \mathrm{kg}\), respectively. The tank is to be installed in a large compartment whose temperature is to be maintained at \(240 \mathrm{~K}\). Design a thermal insulation system that will maintain oxygen losses due to boiling below \(1 \mathrm{~kg} /\) day.

A brass rod \(100 \mathrm{~mm}\) long and \(5 \mathrm{~mm}\) in diameter extends horizontally from a casting at \(200^{\circ} \mathrm{C}\). The rod is in an air environment with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=30\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the temperature of the rod 25,50 , and \(100 \mathrm{~mm}\) from the casting?

To maximize production and minimize pumping costs, crude oil is heated to reduce its viscosity during transportation from a production field. (a) Consider a pipe-in-pipe configuration consisting of concentric steel tubes with an intervening insulating material. The inner tube is used to transport warm crude oil through cold ocean water. The inner steel pipe \(\left(k_{s}=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) has an inside diameter of \(D_{i, 1}=150 \mathrm{~mm}\) and wall thickness \(t_{i}=10 \mathrm{~mm}\) while the outer steel pipe has an inside diameter of \(D_{i, 2}=250 \mathrm{~mm}\) and wall thickness \(t_{o}=t_{i}\). Determine the maximum allowable crude oil temperature to ensure the polyurethane foam insulation \(\left(k_{p}=\right.\) \(0.075 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) between the two pipes does not exceed its maximum service temperature of \(T_{p, \max }=\) \(70^{\circ} \mathrm{C}\). The ocean water is at \(T_{\infty, o}=-5^{\circ} \mathrm{C}\) and provides an external convection heat transfer coefficient of \(h_{o}=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection coefficient associated with the flowing crude oil is \(h_{i}=450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) It is proposed to enhance the performance of the pipe-in-pipe device by replacing a thin \(\left(t_{a}=5 \mathrm{~mm}\right)\) section of polyurethane located at the outside of the inner pipe with an aerogel insulation material \(\left(k_{a}=0.012 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). Determine the maximum allowable crude oil temperature to ensure maximum polyurethane temperatures are below \(T_{p, \max }=70^{\circ} \mathrm{C}\).

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