/*! 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 A 2-mm-diameter electrical wire ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 54

A 2-mm-diameter electrical wire is insulated by a 2 -mm-thick rubberized sheath \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and the wire/sheath interface is characterized by a thermal contact resistance of \(R_{t, c}^{\prime \prime}=3 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The convection heat transfer coefficient at the outer surface of the sheath is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the temperature of the ambient air is \(20^{\circ} \mathrm{C}\). If the temperature of the insulation may not exceed \(50^{\circ} \mathrm{C}\), what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?

Short Answer

Expert verified
The maximum allowable electrical power that may be dissipated per unit length of the conductor is 15.15 W/m, and the critical radius of insulation is 0.013 m (13 mm).

Step by step solution

01

Determine the parameters for the given system

First, let's list all the known parameters: - Diameter of the electrical wire, \(D_w = 0.002\) m (2 mm) - Thickness of rubberized sheath, \(t = 0.002\) m (2 mm) - Thermal conductivity of the sheath, \(k = 0.13 \ \text{W/mK}\) - Thermal contact resistance at the wire/sheath interface, \(R_{t,c}^{\prime\prime} = 3 \times 10^{-4} \ \text{m}^{2} \ \text{K/W}\) - Convection heat transfer coefficient at the outer surface of the sheath, \(h = 10 \ \text{W/m}^{2} \ \text{K}\) - Ambient air temperature, \(T_\infty = 20^{\circ} \text{C}\) - Maximum allowable temperature of the insulation, \(T_{max} = 50^{\circ} \text{C}\)
02

Calculate the maximum wire temperature

To find the maximum electrical power dissipated per unit length, we must first determine the maximum temperature of the wire, \(T_w\), that maintains the insulation temperature at 50掳C. From the thermal contact resistance, we can calculate the temperature difference between the wire and sheath as: \(\Delta T = R_{t,c}^{\prime\prime} \cdot Q^{\prime}\) where \(Q^{\prime}\) is the electrical power dissipated per unit length (in W/m), and \(\Delta T\) is the temperature difference between the wire and sheath. We know the interface temperature, \(T_s = T_{max} - \Delta T\).
03

Calculate the heat transfer from the wire

The electrical power (heat) dissipated per unit length travels through the sheath by conduction and then through the surrounding air by convection. So, we can calculate the heat transfer through the sheath by conduction: \(Q^{\prime} = 2 \pi k (T_w - T_s)\) where \(k = 0.13\, W/m路K\), and the radius of the wire is \(r_w=0.001\) m and the radius of the sheath is \(r_s = 0.003\) m.
04

Calculate the heat transfer through the surrounding air

Next, we can express the heat transfer through the surrounding air by convection: \(Q^{\prime} = 2 \pi h (T_s - T_\infty)\) where \(h =10\,W/m^2.K\).
05

Calculate the maximum allowable electrical power

Now, we can combine the equations for the heat transfer through conduction and convection and solve for the maximum allowable electrical power: \(2\pi k(T_w - T_s) = 2\pi h(T_s - T_\infty)\) Rearrange the equation and solve for \(Q^{\prime}\): \(Q^{\prime}= 2 \pi k \frac{(T_w - T_\infty)}{(1 + \frac{hk}{hr_w})}\) Substitute the known values: \(Q^{\prime} = 2 \pi (0.13) \frac{(50^{\circ} \text{C} - 20^{\circ} \text{C})}{(1 + \frac{(0.13)(10)}{(0.001)})}\) Calculate \(Q^{\prime}\): \(Q^{\prime} \approx 15.15\ \text{W/m}\) The maximum allowable electrical power that may be dissipated per unit length of the conductor is 15.15 W/m.
06

Calculate the critical radius of insulation

The critical radius of insulation is given by the expression: \(r_{crit} = \frac{k}{h}\) where \(k = 0.13\, W/m.K\) and \(h =10\,W/m^2.K\). Plug in the values to calculate the critical radius: \(r_{crit} = \frac{0.13}{10} = 0.013\ \text{m}\) The critical radius of insulation is 0.013 m (13 mm).

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a fundamental property of materials that describes their ability to conduct heat.
It is denoted by the symbol \( k \) and is measured in watts per meter-kelvin (W/m路K).
In practical terms, it tells us how efficiently heat can move through a material, like the rubberized sheath in our exercise. Materials with high thermal conductivity transfer heat quickly and are thus ideal for applications needing efficient heat distribution.
Copper, for example, is commonly used for electrical wires because of its high thermal conductivity.
On the other hand, materials with low thermal conductivity tend to act as insulators, preventing heat flow.
This is why the rubberized sheath, with its low thermal conductivity of \(0.13 \, ext{W/mK}\), is used in the insulation of the electrical wire.
Understanding thermal conductivity is crucial when evaluating heat transfer in materials. It helps in designing systems where heat management is essential. Whether preventing overheating or conserving heat, knowing the thermal conductivity of materials informs these decisions.
Convection Heat Transfer
Convection is a mechanism of heat transfer in fluids and gases, where heat is carried away by the movement of the fluid itself.
It usually occurs when heat moves from a solid surface to a fluid or vice-versa.
In our problem, convection happens at the boundary between the rubberized sheath and the surrounding air.A key parameter in convection is the convection heat transfer coefficient, denoted as \( h \), measured in W/m虏路K.
It quantifies how effectively heat is transferred between a surface and a fluid.
A higher coefficient means more efficient heat transfer.
In this problem, a value of \(10 \, ext{W/m}^2 \, ext{K}\) is given, which signifies moderate heat transfer efficiency.Convection can further be classified into natural and forced convection. Natural convection occurs due to buoyancy forces caused by temperature differences within the fluid.
Forced convection involves external forces, like fans or pumps to enhance the heat transfer process.
Mastering convection heat transfer is important for optimizing thermal systems, as it can affect system performance and energy efficiency.
Thermal Contact Resistance
Thermal contact resistance, denoted as \( R_{t,c}'' \), is a measure of resistance to heat flow across the interface between two materials.
In this exercise, it exists at the boundary between the wire and the rubberized sheath, affecting how heat moves from the wire through the insulation.This resistance results from microscopic surface irregularities that create air gaps at the interface, impeding heat flow.
In this scenario, it's given as \(3 imes 10^{-4} \, ext{m}^2 \, ext{K/W}\).
A higher thermal contact resistance indicates more hindrance to heat transfer at the contact surface.
Knowing this helps determine how much the temperature will drop across the interface.To minimize thermal contact resistance:
  • Use conductive materials with smooth surfaces to ensure good contact.
  • Employ thermal interface materials (TIMs) to fill gaps and improve heat transfer.
Taking into account thermal contact resistance is essential in accurate heat transfer analysis across interfaces.

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Most popular questions from this chapter

A firefighter's protective clothing, referred to as a turnout coat, is typically constructed as an ensemble of three layers separated by air gaps, as shown schematically. The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation exchange through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rad }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right) \approx 4 \sigma T_{\text {avg }}^{3}\), where \(T_{\text {avg }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{\text {rad }}^{\prime \prime}=h_{\text {rad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W}\) ) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\mathrm{avg}}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-ash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the turnout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the turnout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in burn injury?

Turbine blades mounted to a rotating disc in a gas turbine engine are exposed to a gas stream that is at \(T_{\infty}=1200^{\circ} \mathrm{C}\) and maintains a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) over the blade. The blades, which are fabricated from Inconel, \(k \approx 20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), have a length of \(L=50 \mathrm{~mm}\). The blade profile has a uniform cross-sectional area of \(A_{c}=6 \times 10^{-4} \mathrm{~m}^{2}\) and a perimeter of \(P=110 \mathrm{~mm}\). A proposed blade- cooling scheme, which involves routing air through the supporting disc, is able to maintain the base of each blade at a temperature of \(T_{b}=300^{\circ} \mathrm{C}\). (a) If the maximum allowable blade temperature is \(1050^{\circ} \mathrm{C}\) and the blade tip may be assumed to be adiabatic, is the proposed cooling scheme satisfactory? (b) For the proposed cooling scheme, what is the rate at which heat is transferred from each blade to the coolant?

A one-dimensional plane wall of thickness \(L\) is constructed of a solid material with a linear, nonuniform porosity distribution described by \(\varepsilon(x)=\varepsilon_{\max }(x / L)\). Plot the steady-state temperature distribution, \(T(x)\), for \(k_{s}=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad k_{f}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=1 \mathrm{~m}, \quad \varepsilon_{\max }=\) \(0.25, T(x=0)=30^{\circ} \mathrm{C}\) and \(q_{x}^{\prime \prime}=100 \mathrm{~W} / \mathrm{m}^{2}\) using the expression for the minimum effective thermal conductivity of a porous medium, the expression for the maximum effective thermal conductivity of a porous medium, Maxwell's expression, and for the case where \(k_{\mathrm{efl}}(x)=k_{s}\).

A scheme for concurrently heating separate water and air streams involves passing them through and over an array of tubes, respectively, while the tube wall is heated electrically. To enhance gas-side heat transfer, annular fins of rectangular profile are attached to the outer tube surface. Attachment is facilitated with a dielectric adhesive that electrically isolates the fins from the current-carrying tube wall. (a) Assuming uniform volumetric heat generation within the tube wall, obtain expressions for the heat rate per unit tube length \((\mathrm{W} / \mathrm{m})\) at the inner \(\left(r_{i}\right)\) and outer \(\left(r_{o}\right)\) surfaces of the wall. Express your results in terms of the tube inner and outer surface temperatures, \(T_{s, i}\) and \(T_{s, e}\), and other pertinent parameters. (b) Obtain expressions that could be used to determine \(T_{s, i}\) and \(T_{s, o}\) in terms of parameters associated with the water- and air-side conditions. (c) Consider conditions for which the water and air are at \(T_{\infty, i}=T_{\infty, o}=300 \mathrm{~K}\), with corresponding convection coefficients of \(h_{i}=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Heat is uniformly dissipated in a stainless steel tube \(\left(k_{w}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), having inner and outer radii of \(r_{i}=25 \mathrm{~mm}\) and \(r_{o}=30\) \(\mathrm{mm}\), and aluminum fins \(\left(t=\delta=2 \mathrm{~mm}, r_{t}=55\right.\) \(\mathrm{mm}\) ) are attached to the outer surface, with \(R_{t, c}^{\prime \prime}=\) \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Determine the heat rates and temperatures at the inner and outer surfaces as a function of the rate of volumetric heating \(\dot{q}\). The upper limit to \(\dot{q}\) will be determined by the constraints that \(T_{s, i}\) not exceed the boiling point of water \(\left(100^{\circ} \mathrm{C}\right)\) and \(T_{s, o}\) not exceed the decomposition temperature of the adhesive \(\left(250^{\circ} \mathrm{C}\right)\).

The cross section of a long cylindrical fuel element in a nuclear reactor is shown. Energy generation occurs uniformly in the thorium fuel rod, which is of diameter \(D=25 \mathrm{~mm}\) and is wrapped in a thin aluminum cladding. (a) It is proposed that, under steady-state conditions, the system operates with a generation rate of \(\dot{q}=\) \(7 \times 10^{8} \mathrm{~W} / \mathrm{m}^{3}\) and cooling system characteristics of \(T_{\infty}=95^{\circ} \mathrm{C}\) and \(h=7000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Is this proposal satisfactory? (b) Explore the effect of variations in \(\dot{q}\) and \(h\) by plotting temperature distributions \(T(r)\) for a range of parameter values. Suggest an envelope of acceptable operating conditions.
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