/*! 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 122 A rod of diameter \(D=25 \mathrm... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 122

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 \(T_{o}\).

Short Answer

Expert verified
In this problem, we derived an expression for the exposed surface temperature, \(T_o\), for a rod protruding from a furnace wall as a function of given thermal and geometrical parameters. The expression was found to be \(T_o = T_\infty + Q \cdot R_{conv}\), where \(Q\) is the heat transfer rate, and \(R_{conv}\) is the resistance due to convection. By substituting the given values, we can evaluate if the rod with an exposed length of \(L_o = 200mm\) meets the specified operating limit of \(T_{max} = 100^{\circ}C\). If it does not, alternative design parameters or changes to the current design need to be considered, such as using a different material, increasing insulation thickness, increasing rod length, or attaching the base of the rod differently to the furnace wall.

Step by step solution

01

Calculate the Thermal Resistances

First, we need to determine the resistance due to conduction through the insulation and convection from the exposed surface. Resistance due to conduction, \(R_{cond} = \frac{L_{ins}}{k \cdot A}\), where \(A\) is the cross-sectional area of the rod, which can be found by \(A = \pi \frac{D^2}{4}\). Resistance due to convection, \(R_{conv} = \frac{1}{h \cdot A}\).
02

Calculate the Heat Transfer Rate

We'll use the concept of thermal resistance to calculate the heat transfer rate through the system. The total resistance of the system, \(R_{total} = R_{cond} + R_{conv}\). The heat transfer rate, \(Q = \frac{T_w - T_\infty}{R_{total}}\).
03

Derive Expression for Exposed Surface Temperature

Now, we'll derive an expression for the exposed surface temperature, \(T_o\). Using the heat transfer rate and resistance due to convection, \(Q = \frac{T_o - T_\infty}{R_{conv}}\). Rearranging the equation to make \(T_o\) the subject gives us: \(T_o = T_\infty + Q \cdot R_{conv}\).
04

Evaluate the Design Parameters

We'll substitute the given values of \(D\), \(k\), \(T_w\), \(T_\infty\), \(L_{ins}\), \(h\), and \(L_o = 200mm\) into the expression for \(T_o\) to see if it meets the specified operating limit. If the calculated \(T_o > T_{max}\), then we'll consider alternative design parameters or changes to the current design, such as using a different material, increasing the insulation thickness, increasing the rod length, or attaching the base of the rod differently to the furnace wall. #Phase 1: Conclusion# In this exercise, we derived an expression for the exposed surface temperature of a rod as a function of the given thermal and geometrical parameters. We also evaluated if the given rod with a specific exposed length fulfills the boundary conditions regarding the maximum allowable temperature (\(T_{max}\)). If the given design does not meet the requirements, alternative design parameters should be considered.

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Key Concepts

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

Heat Transfer
Heat transfer is the process by which heat energy is exchanged between materials or systems. In this example, the rod connected to the furnace wall transfers heat from the furnace to the surrounding air.
The process can occur through conduction, convection, and radiation. Here, we'll focus on conduction and convection, which are the main forms of heat transfer in the rod scenario.
  • Conduction: This happens when heat moves through a solid, such as the rod itself and the insulation. The heat flows from areas of higher temperature (the furnace side) to areas of lower temperature (the exposed surface).
  • Convection: It comes into play when heat is transferred from the rod's surface to the air. The surrounding air absorbs heat and carries it away from the surface of the rod.
Understanding these two forms of heat transfer helps calculate the temperature at the rod's exposed surface and ensures the temperature is kept below a safe limit to prevent cable damage.
Thermal Resistance
Thermal resistance is a measure of how well a material resists the flow of heat. It acts much like electrical resistance, helping us understand how heat flows through different layers in the system.
  • Conduction Resistance: This is due to the insulation around the rod. The thermal resistance due to conduction, given by \( R_{cond} = \frac{L_{ins}}{k \cdot A} \), shows how much the material opposes the heat flow due to its thermal conductivity \( k \), insulation thickness \( L_{ins} \), and area \( A \).
  • Convection Resistance: Refers to the resistance on the surface of the rod, expressed as \( R_{conv} = \frac{1}{h \cdot A} \). It depends on the convection coefficient \( h \) and the surface area \( A \).
By summing these resistances, we get the total thermal resistance. It helps predict the heat transfer rate, which is crucial for finding the exposed surface temperature \( T_o \). Proper management of thermal resistance ensures the temperature limits are not exceeded.
Convection Coefficient
The convection coefficient \( h \) is a key factor in the rate of convective heat transfer, which describes how efficiently heat is transferred from a surface to a fluid. A higher \( h \) implies better heat transfer as the air absorbs the heat more effectively.
For a rod exposed to air, which in this case refers to the ambient air surrounding the rod, the value \( h = 15 \text{ W/m}^2 \cdot \text{K} \) was given, indicating how heat flows to the air from the rod.

This coefficient is essential in tailoring the design to control temperature effectively. If the calculated exposed surface temperature \( T_o \) exceeds \( T_{max} \), altering \( h \) by changing environmental conditions or surface treatments can often help achieve desired heat transfer.
Thermal Insulation
Thermal insulation plays a crucial role in reducing heat flow from the furnace wall to the environment. By adding an insulating layer, you lower the rate at which heat is conducted along the rod, keeping the exposed surface temperature in check.
Insulation is quantified by its thickness \( L_{ins} \) and material thermal conductivity \( k \). In this exercise, a thickness of 200 mm is specified. A thicker or more conductive insulating material can help keep the temperature lower.
  • Using thicker insulation can improve thermal resistance, thereby lowering the heat transfer rate and protecting against excess heating.
  • Choosing a material with low thermal conductivity enhances insulation capabilities, effectively acting as a barrier to heat flow.
By optimizing insulation properties, you can maintain safe temperature conditions for the rod, ensuring it doesn't exceed operating limits.

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Most popular questions from this chapter

Superheated steam at \(575^{\circ} \mathrm{C}\) is routed from a boiler to the turbine of an electric power plant through steel tubes \((k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of \(300-\mathrm{mm}\) inner diameter and \(30-\mathrm{mm}\) wall thickness. To reduce heat loss to the surroundings and to maintain a safe-to-touch outer surface temperature, a layer of calcium silicate insulation \((k=0.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the tubes, while degradation of the insulation is reduced by wrapping it in a thin sheet of aluminum having an emissivity of \(\varepsilon=0.20\). The air and wall temperatures of the power plant are \(27^{\circ} \mathrm{C}\). (a) Assuming that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient outside the aluminum sheet is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the minimum insulation thickness needed to ensure that the temperature of the aluminum does not exceed \(50^{\circ} \mathrm{C}\) ? What is the corresponding heat loss(b) Explore the effect of the insulation thickness on the temperature of the aluminum and the heat loss per unit tube length. per meter of tube length?

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catalyst onto a flat surface. The surface and catalyst are heated and simultaneously exposed to a higher- temperature, low-pressure gas that contains a mixture of chemical species from which the nanowire is to be formed. The catalytic liquid slowly absorbs the species from the gas through its top surface and converts these to a solid material that is deposited onto the underlying liquid-solid interface, resulting in construction of the nanowire. The liquid catalyst remains suspended at the tip of the nanowire. Consider the growth of a 15 -nm-diameter silicon carbide nanowire onto a silicon carbide surface. The surface is maintained at a temperature of \(T_{s}=2400 \mathrm{~K}\), and the particular liquid catalyst that is used must be maintained in the range \(2400 \mathrm{~K} \leq T_{c} \leq 3000 \mathrm{~K}\) to perform its function. Determine the maximum length of a nanowire that may be grown for conditions characterized by \(h=10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=8000 \mathrm{~K}\). Assume properties of the nanowire are the same as for bulk silicon carbide.

The air inside a chamber at \(T_{\infty, i}=50^{\circ} \mathrm{C}\) is heated convectively with \(h_{i}=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) by a 200 -mm-thick wall having a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and a uniform heat generation of \(1000 \mathrm{~W} / \mathrm{m}^{3}\). To prevent any heat generated within the wall from being lost to the outside of the chamber at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) with \(h_{o}=5\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), a very thin electrical strip heater is placed on the outer wall to provide a uniform heat flux, \(q_{\sigma^{\prime}}\) (a) Sketch the temperature distribution in the wall on \(T-x\) coordinates for the condition where no heat generated within the wall is lost to the outside of the chamber. (b) What are the temperatures at the wall boundaries, \(T(0)\) and \(T(L)\), for the conditions of part (a)? (c) Determine the value of \(q_{o}^{\prime \prime}\) that must be supplied by the strip heater so that all heat generated within the wall is transferred to the inside of the chamber. (d) If the heat generation in the wall were switched off while the heat flux to the strip heater remained constant, what would be the steady-state temperature, \(T(0)\), of the outer wall surface?

A thermopane window consists of two pieces of glass \(7 \mathrm{~mm}\) thick that enclose an air space \(7 \mathrm{~mm}\) thick. The window separates room air at \(20^{\circ} \mathrm{C}\) from outside ambient air at \(-10^{\circ} \mathrm{C}\). The convection coefficient associated with the inner (room-side) surface is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the convection coefficient associated with the outer (ambient) air is \(h_{o}=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the heat loss through a window that is \(0.8 \mathrm{~m}\) long by \(0.5 \mathrm{~m}\) wide? Neglect radiation, and assume the air enclosed between the panes to be stagnant. (b) Compute and plot the effect of \(h_{o}\) on the heat loss for \(10 \leq h_{o} \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Repeat this calculation for a triple-pane construction in which a third pane and a second air space of equivalent thickness are added.

As a means of enhancing heat transfer from highperformance logic chips, it is common to attach a heat \(\sin k\) to the chip surface in order to increase the surface area available for convection heat transfer. Because of the ease with which it may be manufactured (by taking orthogonal sawcuts in a block of material), an attractive option is to use a heat sink consisting of an array of square fins of width \(w\) on a side. The spacing between adjoining fins would be determined by the width of the sawblade, with the sum of this spacing and the fin width designated as the fin pitch \(S\). The method by which the heat sink is joined to the chip would determine the interfacial contact resistance, \(R_{t, c^{*}}^{n}\) Consider a square chip of width \(W_{c}=16 \mathrm{~mm}\) and conditions for which cooling is provided by a dielectric liquid with \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(h=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat \(\operatorname{sink}\) is fabricated from copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and its characteristic dimensions are \(w=0.25 \mathrm{~mm}\), \(S=0.50 \mathrm{~mm}, L_{f}=6 \mathrm{~mm}\), and \(L_{b}=3 \mathrm{~mm}\). The prescribed values of \(w\) and \(S\) represent minima imposed by manufacturing constraints and the need to maintain adequate flow in the passages between fins. (a) If a metallurgical joint provides a contact resistance of \(R_{t, c}^{\prime \prime}=5 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) and the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\), what is the maximum allowable chip power dissipation \(q_{c} ?\) Assume all of the heat to be transferred through the heat sink. (b) It may be possible to increase the heat dissipation by increasing \(w\), subject to the constraint that \((S-w) \geq 0.25 \mathrm{~mm}\), and/or increasing \(L_{f}\) (subject to manufacturing constraints that \(L_{f} \leq 10 \mathrm{~mm}\) ). Assess the effect of such changes.
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