/*! 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 29 A long wire of diameter \(D=1 \m... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 29

A long wire of diameter \(D=1 \mathrm{~mm}\) is submerged in an oil bath of temperature \(T_{\infty}=25^{\circ} \mathrm{C}\). The wire has an electrical resistance per unit length of \(R_{c}^{\prime}=0.01 \Omega / \mathrm{m}\). If a current of \(I=100\) A flows through the wire 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 steadystate value? The properties of the wire are \(\rho=\) \(8000 \mathrm{~kg} / \mathrm{m}^{3}, c=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
The steady-state temperature of the wire is \(65.0^{\circ} \mathrm{C}\). It takes approximately \(2.965 \mathrm{~s}\) for the wire to reach a temperature within \(1^{\circ} \mathrm{C}\) of the steady-state value.

Step by step solution

01

1: Calculate Heat Generated and Heat Dissipated

First, we need to find the heat generated by the wire due to the current and the heat dissipated by convection. The heat generated (Q) can be calculated by using Ohm's law: \[Q=I^{2}R\] where \(I\) is the current and \(R\) is the resistance per unit length Since we are given the resistance per unit length, we also have to multiply it by the length. However, since the wire is infinitely long, both heat generated and heat dissipated will also be per unit length. Therefore, we don't need the length in our calculations. Now, calculate the heat generated: \[Q=I^{2}R_c^{'}=100^{2} \times 0.01=100\ \mathrm{W}/\mathrm{m}\] For the heat dissipated, we'll use the convection equation: \[Q=hA(T_{w}-T_{\infty})\] where \(h\) is the convection coefficient, \(A\) is the surface area of the wire per unit length, and \(T_w\) and \(T_{\infty}\) are the wire temperature and oil bath temperature, respectively. Since the wire is infinitely long, the surface area per unit length can be calculated as: \[A = \pi D = \pi \times 0.001 = 0.003141 \mathrm{m}^{2}/\mathrm{m}\]
02

2: Solve for the Steady-State Temperature

Now we need to equate the heat generated with heat dissipated by convection: \[100=500 \times 0.003141 (T_{w}-25)\] Now solve for \(T_w\): \[T_{w} = \frac{100}{500 \times 0.003141}+25 \approx 65.0^{\circ} \mathrm{C}\] So, the steady-state temperature of the wire is 65.0掳C.
03

3: Calculate the Time Constant

Now we need to find the time it takes for the wire to reach a temperature within 1掳C of the steady-state value. First, we need to calculate the time constant, \(\tau\), for the wire. We can find it using the equation: \[\tau = \frac{\rho c r}{3h}\] where \(\rho\) is the density, \(c\) is the specific heat, \(r\) is the radius of the wire, and \(h\) is the convection coefficient. Now, calculate the time constant: \[\tau = \frac{8000 \times 500 \times 0.0005}{3 \times 500} \approx 1.333 \mathrm{~s}\]
04

4: Solve for the Time to reach a Temperature within 1掳C of the Steady-State Value

Now, we can calculate the time it takes for the wire to reach a temperature within 1掳C of the steady-state value: \[t = -\tau \ln \left(1 - \frac{T_w - (T_{\infty} + 1)}{T_w - T_{\infty}}\right)\] Plugging in the values for \(\tau\), \(T_w\), and \(T_{\infty}\): \[t = -1.333 \ln \left(1 - \frac{65.0 - (25 + 1)}{65.0 - 25}\right)\] Now, calculate the time: \[t \approx 2.965 \mathrm{~s}\] Therefore, it takes approximately 2.965 seconds for the wire to reach a temperature within 1掳C of the steady-state value.

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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 a critical phenomenon in understanding how energy moves through different media. It involves the transfer of thermal energy from a hotter region to a cooler one. In the case of the wire submerged in an oil bath, the heat generated within the wire due to the electrical current is transferred to the surrounding oil bath. This process primarily happens through conduction and convection.
Conduction refers to the direct transfer of heat through a material, while convection involves the movement of heat through a fluid, like the oil, due to the fluid's microscopic motions. For the wire scenario, conduction transfers heat along the length of the wire, and convection dissipates it into the oil.
Ohm's Law
Ohm's Law is a fundamental principle in electrical engineering, stating that the current through a conductor between two points is directly proportional to the voltage across the two points. In this context, Ohm's Law is instrumental in calculating the heat generated in the wire:
  • The formula to determine the heat generated is given by: \[Q = I^2 R\] where \(I\) is the current and \(R\) is the resistance.
  • For a wire with an electrical resistance of \(0.01 \Omega/\text{m}\) and a current of \(100 \text{ A}\), the heat generated per unit length is \(100 \text{ W/m}\).
Ohm's Law helps us understand that increasing the current will lead to more heat being generated in the wire, and vice versa.
Convection
Convection is the mode of heat transfer that involves the bulk movement of molecules within fluids such as liquids and gases. It is crucial in designing systems where heat dissipation is needed. In the given exercise, convection plays a crucial role in removing the heat generated in the wire to the surrounding oil bath.
To quantify the heat dissipation by convection, we use the formula:
  • \[Q = h A (T_w - T_\infty)\]
  • Where \(h\) is the convection coefficient, \(A\) is the surface area per unit length of the wire, \(T_w\) is the temperature of the wire, and \(T_\infty\) is the oil bath temperature.
Effective convection ensures that the wire doesn't overheat, maintaining a safe and steady-state temperature.
Electrical Resistance
Electrical resistance is a property of materials that indicates how much they oppose the flow of electric current. It's an essential factor in calculating the heat generated by current flow. For the wire example, the resistance per unit length significantly impacts how much heat is generated:
  • This is expressed in ohms per meter (\(\Omega/\text{m}\)).
  • The resistance value affects the heat generated: \(Q = I^2 R\).
Materials with higher resistance will generate more heat for the same current, while those with lower resistance will generate less. Knowing the resistance is thus crucial for designing circuits and systems to handle electrical current safely.

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Most popular questions from this chapter

Consider the thick slab of copper in Example 5.12, which is initially at a uniform temperature of \(20^{\circ} \mathrm{C}\) and is suddenly exposed to a net radiant flux of \(3 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2}\). Use the Finite-Difference Equations/ One-Dimensional/Transient conduction model builder of \(I H T\) to obtain the implicit form of the finite-difference equations for the interior nodes. In your analysis, use a space increment of \(\Delta x=37.5 \mathrm{~mm}\) with a total of 17 nodes (00-16), and a time increment of \(\Delta t=1.2 \mathrm{~s}\). For the surface node 00 , use the finite-difference equation derived in Section 2 of the Example. (a) Calculate the 00 and 04 nodal temperatures at \(t=120 \mathrm{~s}\), that is, \(T(0,120 \mathrm{~s})\) and \(T(0.15 \mathrm{~m}, 120 \mathrm{~s})\), and compare the results with those given in Comment 1 for the exact solution. Will a time increment of \(0.12 \mathrm{~s}\) provide more accurate results? (b) Plot temperature histories for \(x=0,150\), and \(600 \mathrm{~mm}\), and explain key features of your results.

A metal sphere of diameter \(D\), which is at a uniform temperature \(T_{i}\), is suddenly removed from a furnace and suspended from a fine wire in a large room with air at a uniform temperature \(T_{\infty}\) and the surrounding walls at a temperature \(T_{\text {sur }}\) (a) Neglecting heat transfer by radiation, obtain an expression for the time required to cool the sphere to some temperature \(T\). (b) Neglecting heat transfer by convection, obtain an expression for the time required to cool the sphere to the temperature \(T\). (c) How would you go about determining the time required for the sphere to cool to the temperature \(T\) if both convection and radiation are of the same order of magnitude? (d) Consider an anodized aluminum sphere ( \(\varepsilon=0.75\) ) \(50 \mathrm{~mm}\) in diameter, which is at an initial temperature of \(T_{i}=800 \mathrm{~K}\). Both the air and surroundings are at \(300 \mathrm{~K}\), and the convection coefficient is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For the conditions of parts (a), (b), and (c), determine the time required for the sphere to cool to \(400 \mathrm{~K}\). Plot the corresponding temperature histories. Repeat the calculations for a polished aluminum sphere ( \(\varepsilon=0.1)\).

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a \(100-\mathrm{mm}\)-thick steel plate \(\left(\rho=7830 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c=550 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is initially at a uniform temperature of \(T_{i}=200^{\circ} \mathrm{C}\) and is to be heated to a minimum temperature of \(550^{\circ} \mathrm{C}\). Heating is effected in a gas-fired furnace, where products of combustion at \(T_{\infty}=800^{\circ} \mathrm{C}\) maintain a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on both surfaces of the plate. How long should the plate be left in the furnace?

Steel is sequentially heated and cooled (annealed) to relieve stresses and to make it less brittle. Consider a 100 -mm-thick plate \(\left(k=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7800 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c_{p}=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) that is initially at a uniform temperature of \(300^{\circ} \mathrm{C}\) and is heated (on both sides) in a gas-fired furnace for which \(T_{\infty}=700^{\circ} \mathrm{C}\) and \(h=\) \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). How long will it take for a minimum temperature of \(550^{\circ} \mathrm{C}\) to be reached in the plate?

5.53 Stone mix concrete slabs are used to absorb thermal energy from flowing air that is carried from a large concentrating solar collector. The slabs are heated during the day and release their heat to cooler air at night. If the daytime airflow is characterized by a temperature and convection heat transfer coefficient of \(T_{\infty}=200^{\circ} \mathrm{C}\) and \(h=35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine the slab thickness \(2 L\) required to transfer a total amount of energy such that \(Q / Q_{o}=0.90\) over a \(t=8\)-h period. The initial concrete temperature is \(T_{i}=40^{\circ} \mathrm{C}\).
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