/*! 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 57 A composite cylindrical wall is ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 57

A composite cylindrical wall is composed of two materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), which are separated by a very thin, electric resistance heater for which interfacial contact resistances are negligible. Liquid pumped through the tube is at a temperature \(T_{\infty, i}\) and provides a convection coefficient \(h_{i}\) at the inner surface of the composite. The outer surface is exposed to ambient air, which is at \(T_{\infty, o}\) and provides a convection coefficient of \(h_{o^{*}}\) Under steady-state conditions, a uniform heat flux of \(q_{h}^{n}\) is dissipated by the heater. (a) Sketch the equivalent thermal circuit of the system and express all resistances in terms of relevant variables. (b) Obtain an expression that may be used to determine the heater temperature, \(T_{h+}\). (c) Obtain an expression for the ratio of heat flows to the outer and inner fluids, \(q_{o}^{\prime} / q_{i}^{\prime}\). How might the variables of the problem be adjusted to minimize this ratio?

Short Answer

Expert verified
The heater temperature expression, given the equivalent thermal circuit, can be found as: \(T_{h+}= \frac{q_{h}^n (R_{k_A}+R_{k_B}) + T_{\infty_o } R_{c_{o^*}}} {R_{c_{o^*}}} \) The ratio of heat flows to the outer and inner fluids is given by: \(\frac{q_{o'}}{q_{i'}} = \frac{(T_{h+}-T_{\infty_o})/R_{c_{o^*}}}{(T_{\infty_i}-T_{h-})/R_{c_i}}\) To minimize this ratio, increase the outer surface convection coefficient \(h_{o^*}\) and decrease the inner surface convection coefficient \(h_i\). Adjusting the thickness and thermal conductivity of the composite materials A and B can also help minimize the ratio.

Step by step solution

01

Understanding the given information

Before diving into the solution, let's briefly understand the given information, which is crucial for the solution. - A composite cylindrical wall is composed of two materials A and B. - Material properties include thermal conductivity (\(k_A\), \(k_B\)) and convection coefficients (\(h_i\), \(h_{o^*}\)). - The thin electric resistance heater is sandwiched between both materials. - The liquid flowing through the tube causes convection on the inner surface, and ambient air causes convection on the outer surface. - All heat losses are uniform and steady-state.
02

Sketch the equivalent thermal circuit

Now, let's create the equivalent thermal circuit for the given information. The thermal circuit represents the temperature and resistance relations for the entire system. The circuit includes the following resistances: 1. Convection resistance (\(R_{c_i}\)) between the inner fluid and material A. 2. Conduction resistance (\(R_{k_A}\)) in material A. 3. Conduction resistance (\(R_{k_B}\)) in material B. 4. Convection resistance (\(R_{c_{o^*}}\)) between material B and the outer fluid. Sketch the above resistances connected in series along with the temperature nodes for \(T_{\infty_i}\), T_h-, T_h+, \(T_{\infty_o}\), and the respective heat fluxes \(q_{i'}\), \(q_{h}^n\), \(q_{o'}\).
03

Express resistances in terms of relevant variables (a)

Use the equations for the convection and conduction resistances in cylindrical coordinates to express each resistance in terms of the given variables. Recall that the convection resistance is given by \(R_{c} = \frac{1}{Ah}\) and the conduction resistance is given by \(R_{k} = \frac{L}{Ak}\). 1. \(R_{c_i} = \frac{1}{2\pi r_i L h_i}\) 2. \(R_{k_A} = \frac{\ln(\frac{r_{A_o}}{r_i})}{2\pi L k_A}\) 3. \(R_{k_B} = \frac{\ln(\frac{r_{o}}{r_{A_o}})}{2\pi L k_B}\) 4. \(R_{c_{o^*}} = \frac{1}{2\pi r_o L h_{o^*}}\)
04

Find the expression for heater temperature T_h+ (b)

Write the heat transfer equations for the entire thermal circuit, starting from the inner fluid to the outer fluid. Assuming energy conservation and that all heat is transferred through the resistance elements, \(q_{i'} = (T_{\infty_i}-T_{h-})/R_{c_i}\) \(q_{h}^n = (T_{h-}-T_{h+})/(R_{k_A}+R_{k_B})\) \(q_{o'} = (T_{h+}-T_{\infty_o})/R_{c_{o^*}}\) Now, solve these equations to get an expression for the heater temperature \(T_{h+}\). \(T_{h+}= \frac{q_{h}^n (R_{k_A}+R_{k_B}) + T_{\infty_o } R_{c_{o^*}}} {R_{c_{o^*}}} \)
05

Obtain the expression for the ratio of heat flows (c)

We need to find the expression for the ratio of heat flows to the outer and inner fluids, defined as \(q_{o'} / q_{i'}\). Using the heat transfer equations obtained in Step 4, we get: \(\frac{q_{o'}}{q_{i'}} = \frac{(T_{h+}-T_{\infty_o})/R_{c_{o^*}}}{(T_{\infty_i}-T_{h-})/R_{c_i}}\) To minimize this ratio, manipulate the variables in the problem so that the resulting ratio is as small as possible. To minimize the ratio, increase the outer surface convection coefficient \(h_{o^*}\) and decrease the inner surface convection coefficient \(h_i\). Adjusting the thickness and thermal conductivity of the composite materials A and B can also help minimize the ratio.

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.

Thermal Conductivity
Thermal conductivity, denoted by the symbol \(k\), is a measure of a material's ability to conduct heat. It quantifies the rate at which heat can pass through a material when there is a temperature difference across that material. Materials with high thermal conductivity, such as metals, can transfer heat quickly, whereas insulating materials, such as wood or foam, have low thermal conductivity and therefore transfer heat more slowly.

To illustrate, if we consider a cylindrical wall made from two different materials, each with their own thermal conductivity values \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), the heat will flow from one side to the other at rates determined by these values. In the exercise, the concept of thermal conductivity is crucial for determining the conduction resistance of each material segment within the thermal circuit, and thus impacts the estimation of how much heat is dissipated by the heater.
When solving for heat transfer problems, it's paramount to accurately assess these conductivities so that the thermal behavior of the system can be properly understood and controlled.
Convection Coefficient
The convection coefficient, symbolized by \(h\), is a metric that describes the effectiveness of convective heat transfer between a surface and a fluid flowing over it. A high convection coefficient means that the fluid is very effective at removing heat from the surface, like the way a fan boosts air movement, thereby increasing heat transfer from your skin. Conversely, a low convection coefficient indicates that the fluid isn't as efficient in transferring heat.

For instance, in the given exercise, the inner surface of the composite cylinder has a convection coefficient \(h_i\), while the outer surface has a different convection coefficient \(h_{o^*}\). These coefficients are central to computing the convection resistance at both the inner and outer surfaces of the cylinder. The greater the convection coefficient, the lower the thermal resistance to heat flow, facilitating more efficient heat transfer from the heater to the respective fluids.
Understanding the convection coefficient is essential for optimizing heat transfer processes, and in practical settings, it can be influenced by the type of fluid, fluid velocity, and surface characteristics.
Thermal Resistance
Thermal resistance is analogous to electrical resistance, but instead of impeding electrical current, it quantifies the resistance to heat flow within a material or between different materials. It is directly related to both the thermal conductivity of the material and the geometry of the thermal path. The higher the thermal resistance, the less heat passes through per unit of time for a given temperature difference.

In the context of the exercise involving a composite cylindrical wall, there are multiple thermal resistances in play: the conduction resistances \(R_{k_A}\) and \(R_{k_B}\) for materials A and B, and the convection resistances \(R_{c_i}\) and \(R_{c_{o^*}}\) at the inner and outer surfaces. Each resistance plays a role in the overall thermal circuit, which models how heat is dissipated from the heater. By breaking down the thermal circuit into individual resistances, one can solve for temperatures and heat fluxes at different points.
  • Conduction Resistance: It hinders the heat flow through a material and depends on the material's thickness, thermal conductivity, and area perpendicular to the heat flow.
  • Convection Resistance: It slows down the heat transfer between a solid surface and the fluid around it and is dependent on the convection coefficient and surface area.
Minimizing thermal resistance is a key goal in the design of thermal systems to ensure efficient heat transfer. This principle applies whether improving the cooling of electronic devices or optimizing insulation in buildings.

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

A rod of diameter \(D=25 \mathrm{~mm}\) and thermal conductivity \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) protrudes normally from a furnace wall that is at \(T_{w}=200^{\circ} \mathrm{C}\) and is covered by insulation of thickness \(L_{\text {ins }}=200 \mathrm{~mm}\). The rod is welded to the furnace wall and is used as a hanger for supporting instrumentation cables. To avoid damaging the cables, the temperature of the rod at its exposed surface, \(T_{o}\), must be maintained below a specified operating limit of \(T_{\max }=100^{\circ} \mathrm{C}\). The ambient air temperature is \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the convection coefficient is \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Derive an expression for the exposed surface temperature \(T_{o}\) as a function of the prescribed thermal and geometrical parameters. The rod has an exposed length \(L_{o}\), and its tip is well insulated. (b) Will a rod with \(L_{o}=200 \mathrm{~mm}\) meet the specified operating limit? If not, what design parameters would you change? Consider another material, increasing the thickness of the insulation, and increasing the rod length. Also, consider how you might attach the base of the rod to the furnace wall as a means to reduce \(T_{o}\).

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.

Circular copper rods of diameter \(D=1 \mathrm{~mm}\) and length \(L=25 \mathrm{~mm}\) are used to enhance heat transfer from a surface that is maintained at \(T_{s, 1}=100^{\circ} \mathrm{C}\). One end of the rod is attached to this surface (at \(x=0\) ), while the other end \((x=25 \mathrm{~mm})\) is joined to a second surface, which is maintained at \(T_{s, 2}=0^{\circ} \mathrm{C}\). Air flowing between the surfaces (and over the rods) is also at a temperature of \(T_{\infty}=0^{\circ} \mathrm{C}\), and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained. (a) What is the rate of heat transfer by convection from a single copper rod to the air? (b) What is the total rate of heat transfer from a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) section of the surface at \(100^{\circ} \mathrm{C}\), if a bundle of the rods is installed on 4 -mm centers?

Radioactive wastes \(\left(k_{\mathrm{rw}}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) are stored in a spherical, stainless steel \(\left(k_{\mathrm{ss}}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) container of inner and outer radii equal to \(r_{i}=0.5 \mathrm{~m}\) and \(r_{o}=0.6 \mathrm{~m}\). Heat is generated volumetrically within the wastes at a uniform rate of \(\dot{q}=10^{5} \mathrm{~W} / \mathrm{m}^{3}\), and the outer surface of the container is exposed to a water flow for which \(h=\) \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\). (a) Evaluate the steady-state outer surface temperature, \(T_{s, o}\) (b) Evaluate the steady-state inner surface temperature, \(T_{s, i^{*}}\) (c) Obtain an expression for the temperature distribution, \(T(r)\), in the radioactive wastes. Express your result in terms of \(r_{i}, T_{s, i}, k_{\mathrm{rw}}\), and \(\dot{q}\). Evaluate the temperature at \(r=0\). (d) A proposed extension of the foregoing design involves storing waste materials having the same thermal conductivity but twice the heat generation \(\left(\dot{q}=2 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\right)\) in a stainless steel container of equivalent inner radius \(\left(r_{i}=0.5 \mathrm{~m}\right)\). Safety considerations dictate that the maximum system temperature not exceed \(475^{\circ} \mathrm{C}\) and that the container wall thickness be no less than \(t=0.04 \mathrm{~m}\) and preferably at or close to the original design \((t=0.1 \mathrm{~m})\). Assess the effect of varying the outside convection coefficient to a maximum achievable value of \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (by increasing the water velocity) and the container wall thickness. Is the proposed extension feasible? If so, recommend suitable operating and design conditions for \(h\) and \(t\), respectively.

Consider the oven of Problem 1.54. The walls of the oven consist of \(L=30\)-mm- thick layers of insulation characterized by \(k_{\text {ins }}=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) that are sandwiched between two thin layers of sheet metal. The exterior surface of the oven is exposed to air at \(23^{\circ} \mathrm{C}\) with \(h_{\text {ext }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The interior oven air temperature is \(180^{\circ} \mathrm{C}\). Neglecting radiation heat transfer, determine the steady-state heat flux through the oven walls when the convection mode is disabled and the free convection coefficient at the inner oven surface is \(h_{\mathrm{fr}}=3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the heat flux through the oven walls when the convection mode is activated, in which case the forced convection coefficient at the inner oven surface is \(h_{\mathrm{fo}}=27 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Does operation of the oven in its convection mode result in significantly increased heat losses from the oven to the kitchen? Would your conclusion change if radiation were included in your analysis?
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Magnetism

Read Explanation

Fluids

Read Explanation

Radiation

Read Explanation

Particle Model of Matter

Read Explanation

Kinematics 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.