/*! 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

Special coatings are often formed by depositing thin layers of a molten material on a solid substrate. Solidification begins at the substrate surface and proceeds until the thickness \(S\) of the solid layer becomes equal to the thickness \(\delta\) of the deposit. (a) Consider conditions for which molten material at its fusion temperature \(T_{f}\) is deposited on a large substrate that is at an initial uniform temperature \(T_{i}\). With \(S=0\) at \(t=0\), develop an expression for estimating the time \(t_{d}\) required to completely solidify the deposit if it remains at \(T_{f}\) throughout the solidification process. Express your result in terms of the substrate thermal conductivity and thermal diffusivity \(\left(k_{s}, \alpha_{s}\right)\), the density and latent heat of fusion of the deposit \(\left(\rho, h_{s f}\right)\), the deposit thickness \(\delta\), and the relevant temperatures \(\left(T_{f}, T_{i}\right)\). (b) The plasma spray deposition process of Problem \(5.25\) is used to apply a thin \((\delta=2 \mathrm{~mm})\) alumina coating on a thick tungsten substrate. The substrate has a uniform initial temperature of \(T_{i}=300 \mathrm{~K}\), and its thermal conductivity and thermal diffusivity may be approximated as \(k_{s}=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha_{s}=4.0 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. The density and latent heat of fusion of the alumina are \(\rho=3970 \mathrm{~kg} / \mathrm{m}^{3}\) and \(h_{s f}=3577 \mathrm{~kJ} / \mathrm{kg}\), respectively, and the alumina solidifies at its fusion temperature \(\left(T_{f}=2318 \mathrm{~K}\right)\). Assuming that the molten layer is instantaneously deposited on the substrate, estimate the time required for the deposit to solidify.

As permanent space stations increase in size, there is an attendant increase in the amount of electrical power they dissipate. To keep station compartment temperatures from exceeding prescribed limits, it is necessary to transfer the dissipated heat to space. A novel heat rejection scheme that has been proposed for this purpose is termed a Liquid Droplet Radiator (LDR). The heat is first transferred to a high vacuum oil, which is then injected into outer space as a stream of small droplets. The stream is allowed to traverse a distance \(L\), over which it cools by radiating energy to outer space at absolute zero temperature. The droplets are then collected and routed back to the space station. Consider conditions for which droplets of emissivity \(\varepsilon=0.95\) and diameter \(D=0.5 \mathrm{~mm}\) are injected at a temperature of \(T_{i}=500 \mathrm{~K}\) and a velocity of \(V=0.1 \mathrm{~m} / \mathrm{s}\). Properties of the oil are \(\rho=885 \mathrm{~kg} / \mathrm{m}^{3}, c=1900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=0.145 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming each drop to radiate to deep space at \(T_{\text {sur }}=0 \mathrm{~K}\), determine the distance \(L\) required for the droplets to impact the collector at a final temperature of \(T_{f}=300 \mathrm{~K}\). What is the amount of thermal energy rejected by each droplet?

Plasma spray-coating processes are often used to provide surface protection for materials exposed to hostile environments, which induce degradation through factors such as wear, corrosion, or outright thermal failure. Ceramic coatings are commonly used for this purpose. By injecting ceramic powder through the nozzle (anode) of a plasma torch, the particles are entrained by the plasma jet, within which they are then accelerated and heated. During their time-in-fbht, the ceramic particles must be heated to their melting point and experience complete conversion to the liquid state. The coating is formed as the molten droplets impinge (splat) on the substrate material and experience rapid solidification. Consider conditions for which spherical alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right.\) ) particles of diameter \(D_{p}=50 \mu \mathrm{m}\), density \(\rho_{p}=\) \(3970 \mathrm{~kg} / \mathrm{m}^{3}\), thermal conductivity \(k_{p}=10.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and specific heat \(c_{p}=1560 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) are injected into an arc plasma, which is at \(T_{\infty}=10,000 \mathrm{~K}\) and provides a coefficient of \(h=30,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for convective heating of the particles. The melting point and latent heat of fusion of alumina are \(T_{\text {mp }}=2318 \mathrm{~K}\) and \(h_{s f}=3577 \mathrm{~kJ} / \mathrm{kg}\), respectively. (a) Neglecting radiation, obtain an expression for the time-in-flight, \(t_{i-f}\), required to heat a particle from its initial temperature \(T_{i}\) to its melting point \(T_{\text {mp }}\), and, once at the melting point, for the particle to experience complete melting. Evaluate \(t_{i-f}\) for \(T_{i}=300 \mathrm{~K}\) and the prescribed heating conditions. (b) Assuming alumina to have an emissivity of \(\varepsilon_{p}=0.4\) and the particles to exchange radiation with large surroundings at \(T_{\text {sur }}=300 \mathrm{~K}\), assess the validity of neglecting radiation.

In a material processing experiment conducted aboard the space shuttle, a coated niobium sphere of \(10-\mathrm{mm}\) diameter is removed from a furnace at \(900^{\circ} \mathrm{C}\) and cooled to a temperature of \(300^{\circ} \mathrm{C}\). Although properties of the niobium vary over this temperature range, constant values may be assumed to a reasonable approximation, with \(\rho=8600 \mathrm{~kg} / \mathrm{m}^{3}, c=290 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=\) \(63 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) If cooling is implemented in a large evacuated chamber whose walls are at \(25^{\circ} \mathrm{C}\), determine the time required to reach the final temperature if the coating is polished and has an emissivity of \(\varepsilon=0.1\). How long would it take if the coating is oxidized and \(\varepsilon=0.6\) ? (b) To reduce the time required for cooling, consideration is given to immersion of the sphere in an inert gas stream for which \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(h=\) \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Neglecting radiation, what is the time required for cooling? (c) Considering the effect of both radiation and convection, what is the time required for cooling if \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\varepsilon=0.6\) ? Explore the effect on the cooling time of independently varying \(h\) and \(\varepsilon\).

A solid steel sphere (AISI 1010 ), \(300 \mathrm{~mm}\) in diameter, is coated with a dielectric material layer of thickness \(2 \mathrm{~mm}\) and thermal conductivity \(0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The coated sphere is initially at a uniform temperature of \(500^{\circ} \mathrm{C}\) and is suddenly quenched in a large oil bath for which \(T_{\infty}=100^{\circ} \mathrm{C}\) and \(h=3300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimate the time required for the coated sphere temperature to reach \(140^{\circ} \mathrm{C}\). Hint: Neglect the effect of energy storage in the dielectric material, since its thermal capacitance \((\rho c V)\) is small compared to that of the steel sphere
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