/*! 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

When raised to very high temperatures, many conventional liquid fuels dissociate into hydrogen and other components. Thus the advantage of a solid oxide fuel cell is that such a device can internally reform readily available liquid fuels into hydrogen that can then be used to produce electrical power in a manner similar to Example 1.5. Consider a portable solid oxide fuel cell, operating at a temperature of \(T_{\mathrm{fc}}=800^{\circ} \mathrm{C}\). The fuel cell is housed within a cylindrical canister of diameter \(D=\) \(75 \mathrm{~mm}\) and length \(L=120 \mathrm{~mm}\). The outer surface of the canister is insulated with a low-thermal-conductivity material. For a particular application, it is desired that the thermal signature of the canister be small, to avoid its detection by infrared sensors. The degree to which the canister can be detected with an infrared sensor may be estimated by equating the radiation heat flux emitted from the exterior surface of the canister (Equation 1.5; \(E_{s}=\varepsilon_{s} \sigma T_{s}^{4}\) ) to the heat flux emitted from an equivalent black surface, \(\left(E_{b}=\sigma T_{b}^{4}\right)\). If the equivalent black surface temperature \(T_{b}\) is near the surroundings temperature, the thermal signature of the canister is too small to be detected-the canister is indistinguishable from the surroundings. (a) Determine the required thickness of insulation to be applied to the cylindrical wall of the canister to ensure that the canister does not become highly visible to an infrared sensor (i.e., \(T_{b}-T_{\text {sur }}<5 \mathrm{~K}\) ). Consider cases where (i) the outer surface is covered with a very thin layer of \(\operatorname{dirt}\left(\varepsilon_{s}=0.90\right)\) and (ii) the outer surface is comprised of a very thin polished aluminum sheet \(\left(\varepsilon_{s}=0.08\right)\). Calculate the required thicknesses for two types of insulating material, calcium silicate \((k=0.09 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and aerogel \((k=0.006 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the surroundings and the ambient are \(T_{\text {sur }}=300 \mathrm{~K}\) and \(T_{\infty}=298 \mathrm{~K}\), respectively. The outer surface is characterized by a convective heat transfer coefficient of \(h=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) Calculate the outer surface temperature of the canister for the four cases (high and low thermal conductivity; high and low surface emissivity). (c) Calculate the heat loss from the cylindrical walls of the canister for the four cases.

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?

A thin flat plate of length \(L\), thickness \(t\), and width \(W \geqslant L\) is thermally joined to two large heat sinks that are maintained at a temperature \(T_{o}\). The bottom of the plate is well insulated, while the net heat flux to the top surface of the plate is known to have a uniform value of \(q_{o}^{\prime \prime}\) (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks.

The outer surface of a hollow sphere of radius \(r_{2}\) is subjected to a uniform heat flux \(q_{2}^{\prime \prime}\). The inner surface at \(r_{1}\) is held at a constant temperature \(T_{s, 1}\). (a) Develop an expression for the temperature distribution \(T(r)\) in the sphere wall in terms of \(q_{2}^{\prime \prime}, T_{s, 1}, r_{1}, r_{2}\), and the thermal conductivity of the wall material \(k\). (b) If the inner and outer tube radii are \(r_{1}=50 \mathrm{~mm}\) and \(r_{2}=100 \mathrm{~mm}\), what heat flux \(q_{2}^{\prime \prime}\) is required to maintain the outer surface at \(T_{s, 2}=50^{\circ} \mathrm{C}\), while the inner surface is at \(T_{s, 1}=20^{\circ} \mathrm{C}\) ? The thermal conductivity of the wall material is \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Rows of the thermoelectric modules of Example \(3.13\) are attached to the flat absorber plate of Problem 3.108. The rows of modules are separated by \(L_{\text {sep }}=0.5 \mathrm{~m}\) and the backs of the modules are cooled by water at a temperature of \(T_{w}=40^{\circ} \mathrm{C}\), with \(h=45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the electric power produced by one row of thermoelectric modules connected in series electrically with a load resistance of \(60 \Omega\). Calculate the heat transfer rate to the flowing water. Assume rows of 20 immediately adjacent modules, with the lengths of both the module rows and water tubing to be \(L_{\text {row }}=20 W\) where \(W=54 \mathrm{~mm}\) is the module dimension taken from Example 3.13. Neglect thermal contact resistances and the temperature drop across the tube wall, and assume that the high thermal conductivity tube wall creates a uniform temperature around the tube perimeter. Because of the thermal resistance provided by the thermoelectric modules, it is no longer appropriate to assume that the temperature of the absorber plate directly above a tube is equal to that of the water.
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