/*! 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 18 A long wire of diameter \(D=1 \m... [FREE SOLUTION] | 91Ó°ÊÓ

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A long wire of diameter \(D=1 \mathrm{~mm}\) is submerged it an oil bath of temperature \(T_{s}=25^{\circ} \mathrm{C}\). The wire has at electrical resistance per unit length of \(R_{e}^{2}=0.01 \mathrm{f} \mathrm{Nm}\) If a current of \(I=100 \mathrm{~A}\) flows throegh the wirt and the convection coefficient is \(h=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). what is the steady-state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature that is within \(1^{\circ} \mathrm{C}\) of the steady-state value? The properties of the wire are \(\rho\) a \(8000 \mathrm{~kg} / \mathrm{m}^{3}, c=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(\mathrm{k}=20 \mathrm{~W} / \mathrm{m}-\mathrm{K}\).

Short Answer

Expert verified
The steady-state temperature is approximately 88.94 °C, and it takes about 1.2 seconds to reach within 1°C of this temperature.

Step by step solution

01

Calculate Power Generated

Use the formula for electrical power as \( P = I^2 R_e L \) to find the power generated per unit length of the wire. Given \( I = 100 \) A and \( R_e = 0.01 \) \( \ P = (100)^2 \times 0.01 = 100 \) W/m.
02

Calculate Steady-State Temperature

In steady state, the power loss due to convection equals the power generated. Use the formula \( q = hA(T_w - T_s) \), where \( q = 100 \) W/m, \( h = 500 \) W/m²·K, and \( A = \pi D \). Rewrite as \( 100 = 500 \times \pi \times 0.001 \times (T_w - 25) \), and solve for \( T_w \). \( T_w = \frac{100}{500 \times \pi \times 0.001} + 25 \approx 88.94 \) °C.
03

Calculate Time to Reach Within 1°C of Steady-State

Use the lumped capacitance method: \( \frac{{dT}}{{dt}} = \frac{{hA}}{{\rho c V}} (T_w - T_s) \). Substitute \( T_w \) with \( 88.94 \), \( T_s \) with 25, \( \Delta T \approx 1 \) makes \( T(t) \approx 87.94 \). Calculate \( V = A \times L = \frac{\pi D^2}{4} \times L \), then find \( t \). Calculate \( \tau = \frac{{\rho c V}}{{h A}} = \frac{{8000 \times 500 \times \frac{{\pi \times (0.001)^2}}{{4}}}}{{500 \times \pi \times 0.001}} \approx 0.4 \) s.The time \( t \) to reach temperature within 1°C of steady-state can be approximated using \( t \approx 3 \tau \approx 1.2 \) s.

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Key Concepts

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

Steady-State Temperature
The concept of steady-state temperature is crucial for understanding how heat transfers within systems that achieve equilibrium over time, like our wire in this example. When we talk about steady-state, we mean that the temperature remains constant over time because the heat entering the system is balanced by the heat leaving it. This condition implies that all parts of the wire have uniform temperature and there's no net heat accumulation. In our particular scenario, the energy supplied to the wire via electrical current equals the energy lost to the surrounding oil bath, maintaining the wire's temperature at constant level. This is achieved when the heat generated by electrical resistance is entirely dissipated through convection into the bath at a specific temperature, calculated to be approximately 88.94 °C in this case. Understanding this state helps in designing systems that operate safely and efficiently.
Electrical Resistance Heating
Electrical resistance heating is a key principle that explains how electrical energy is converted into heat. In our example, when electricity flows through the wire, the resistance it encounters transforms some of this electrical energy into thermal energy. This transformation happens because the moving electrons collide with atoms within the wire, causing the atoms to vibrate and transfer heat. The resistance value ( 0.01 Ω/m in our case) and the amount of current ( 100 A) are directly related to how much heat is generated, quantified by the power formula, P = I^2 R . The higher the resistance and current, the more heat is produced. This principle is the foundation for many heating applications, from household heaters to industrial processes.
Convection Heat Transfer
Convection heat transfer is the process by which heat is transferred from a solid surface to a fluid (or vice versa) through the motion of fluid particles. It is one of the primary ways by which thermal energy is redistributed. In our problem, convection happens between the heated wire surface and the surrounding oil bath. The key variables affecting convection include the convection heat transfer coefficient ( 500 W/m²K in our example), the surface area through which heat is transferred, and the temperature difference between the wire and the oil. The role of convection is to dissipate the heat generated by the wire into the oil to maintain steady-state conditions. The efficiency of this process dictates how quickly and effectively heat can be removed from the wire, thus preventing overheating and determining operational stability.
Lumped Capacitance Method
The lumped capacitance method is a simplified approach to evaluate transient heat conduction problems, where temperature changes with time. It assumes that the temperature of the object is uniform throughout its volume at any given time. This assumption holds when the heat transfers quickly relative to the object’s size, leading to uniform temperature profiles across it. In our scenario, this method helps us determine the time it takes for the wire to reach near steady-state temperature once the current is applied. By approximating the response time as a multiple of the time constant, \( \tau \), we can predict that it will take about 1.2 seconds for the wire to get within 1 °C of the steady-state temperature. This method is particularly useful in engineering applications due to its simplicity and sufficiency in many practical situations.

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Most popular questions from this chapter

Copper-coated, epoxy-filled fiberglass circuit boards are treated by heating a stack of them under high pressare as sown in the sketch. The perpose of the pressing-heuting operation is to cure the eposy that bonds the fiberglass sheets, imparting siffness to the boards. The stack. riferted to as a book, is comprised of 10 boards and 11 pressing plates, which prevent epoxy from flowing between the boards and impar a-smooth finish to the cured boards. In order to perform simplified thermal analyses, it is reasonable to approximule the book as having an effective themal conductivity \((k)\) and an effective thermal capacitance \(\left(\rho c_{p}\right)\). Calculate the effective propertics if each of the boards and plates has a tickness of \(2.36 \mathrm{~mm}\) and the following thermophysical properties: board (b) \(\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{\prime}, c_{p,}=1500\) \(\mathrm{W} / \mathrm{Kg} \cdot \mathrm{K}, k_{p}=0.30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}_{;}\)plate \((p) \rho_{p}=8\left(000 \mathrm{~kg} / \mathrm{m}^{3}\right.\). \(c_{v}=480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{p}=12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Consider a thin electrical heater artached to a plate and hacked by insulation. Initially, the heater and plate are at the temperafure of the ambient air, \(T\) - Suddenly, the power to the heater is activated, yielding a constant heat fux \(q_{e}^{*}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) at the inner surface of the plate. (a) Sketch and label, on \(T-x\) cocrdinates, the temperature distributions: initial, steady-state, and at two intermediate times. (b) Sketch the heat flux at the outer surface \(q_{z}^{\prime \prime}(L, t)\) as a function of time.

A molded plastic product \(\left(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3} . c=\right.\) \(1500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=0.30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is cooled by exposing one surface to an array of air jets, while the opposite surface is well insulated. The product may be approximated as a slab of thickness \(L=60 \mathrm{~mm}\). which is initially at a uniform temperature of \(T_{i}=\) \(80 \% \mathrm{C}\). The air jets are at a temperature of \(T_{2}=20^{\circ} \mathrm{C}\) and provide a uniform convection coefficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the cooled surface. Using a finite-difference solution with a space increment of \(\Delta x=6 \mathrm{~mm}\), determine temperatures at the cooled and insulated surfaces after 1 hour of exposure to the gas jets.

A cold air chamber is proposed for quenching steel ball bearings of dinmeter \(D=0.2 \mathrm{~m}\) and initial temperature \(T_{i}=400^{\circ} \mathrm{C}\). Air in the chamber is maintained at \(-15^{\circ} \mathrm{C}\) by a refrigeration system, and the steel balls pass through the chamber on a conveyor belt. Optimum bearing production requires that \(70 \%\) of the initial thermal energy content of the ball above \(-15^{\circ} \mathrm{C}\) be removed. Radiation effects muy be neglected, and the convection heut transfer coefficient within the chamber is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimute the residence time of the balls within the chamber, und recommend a drive velocity of the conveyor. The following properties may be used for the steel: \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=2 \times 10^{-3} \mathrm{~m}^{2} / \mathrm{s}\), and \(c=450 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\).

Batch processes are often used in chemical and pharmaceutical operations to achicve a desired chemical composition for the final product and typically involve a transient heating operation to take the product from room temperature to the desired process temperature. Consider a situation for which a chemical of density \(\rho=\) \(1200 \mathrm{~kg} / \mathrm{m}^{3}\) and specific heat \(c=2200 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\) occupies a volume of \(V=2.25 \mathrm{~m}^{3}\) in an insulated vessel. The chemical is to be heated from room temperature, \(T_{f}=\) \(300 \mathrm{~K}\), to a process temperature of \(T=450 \mathrm{~K}\) by passing saturated steam at \(T_{h}=500 \mathrm{~K}\) through a coiled, thinwalled, 20 -mm-diameter tube in the vessel. Steam condensation within the tube maintains an interior convection cocfficient of \(h_{y}=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the highly agitated liquid in the stirned vessel maintains an outside convection coefficient of \(h_{e}=2000 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\). If the chemical is to be heated from 300 to \(450 \mathrm{~K}\) in 60 minutes, what is the required length \(L\) of the submerged tubing?

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