/*! 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 34 A metal sphere of diameter \(D\)... [FREE SOLUTION] | 91Ó°ÊÓ

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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 ( \(\varepsilon=0.1)\).

Short Answer

Expert verified
In summary, we derived expressions for cooling time due to convection and radiation separately: For convection, we found \(T = T_\infty + (T_i-T_\infty)e^{-\frac{hA}{mc}t}\). For radiation, we determined that the general solution would be given as an implicit formula due to the nonlinearity of the problem. We also mentioned that to determine the cooling time if both convection and radiation have the same magnitude, we need to solve the combined equation numerically using methods like Euler's method, Runge-Kutta method, or software like MATLAB or Python. Finally, we discussed solving for the specific conditions given for an anodized aluminum sphere and a polished aluminum sphere. This involves using the derived expressions and numerical approaches to find the cooling times and temperature profiles in each case.

Step by step solution

01

Part (a): Time for cooling due to convection only

To derive the expression for cooling due to convection, we need Newton's law of cooling: \(q = hA(T-T_\infty)\). Here, q is the rate of heat transfer, h is the convection coefficient, A is the surface area, and T is the sphere's temperature. Now, the rate of heat transfer due to convection is related to the change in the sphere's internal energy and can be written as \(q = mc\frac{dT}{dt}\), where m is the mass of the sphere, c is the specific heat capacity, and dt is the infinitesimal time change. Combining these two equations, we have \(mc\frac{dT}{dt} = hA(T - T_\infty)\) Now, separate the variables: \(\frac{dT}{T - T_\infty} = \frac{hA}{mc} dt\) Integrating both sides with limits from Ti to T (for temperature) and 0 to t (for time), we have: \(\int_{T_i}^T \frac{dT}{T - T_\infty} = \int_0^t \frac{hA}{mc} dt\) Now, solving the integrals we get: \(-\ln{(T-T_\infty)} |_{T_i}^T = \frac{hA}{mc}t |^t_0\) Now, \(-\ln{\frac{T-T_\infty}{T_i-T_\infty}} = \frac{hA}{mc}t\) Taking the natural exponential of both sides, we get: \(T = T_\infty + (T_i-T_\infty)e^{-\frac{hA}{mc}t}\)
02

Part (b): Time for cooling due to radiation only

For radiation, we can use the Stefan-Boltzmann Law: \(q = \varepsilon \sigma A(T^4 - T_{sur}^4)\), where q is the rate of heat transfer, A is the surface area, \(\sigma\) is the Stefan-Boltzmann constant, \(T\) is the sphere's temperature, and \(T_{sur}\) is the temperature of surroundings. As before, we can relate this to the rate of change in the internal energy, and write: \(mc\frac{dT}{dt} = \varepsilon \sigma A (T^4 - T_{sur}^4)\) Now, separate the variables: \(\frac{dT}{T^4 - T_{sur}^4} = \frac{\varepsilon\sigma A}{mc} dt\) To solve this, we can integrate both sides with limits from Ti to T (for temperature) and 0 to t (for time), but the integration becomes a bit more complicated because of the nonlinearity of the problem. The general solution would be given as an implicit formula.
03

Part (c): Cooling time if both convection and radiation have the same magnitude

To solve this part, we need to solve the combined equation of both convection and radiation. \(mc\frac{dT}{dt} = hA(T - T_\infty) + \varepsilon \sigma A (T^4 - T_{sur}^4) \) Now, we would solve this numerically to obtain the cooling time and corresponding temperature profiles. We can employ numerical methods like Euler's method, Runge-Kutta method, or even use software like MATLAB or Python for solving it.
04

Part (d): Finding the cooling times and temperature histories for the given conditions

We have the initial values and required quantities for an anodized aluminum sphere and the corresponding values for a polished aluminum sphere. The required quantities are the time to reach 400K and the temperature profiles during the cooling process. To find the cooling times and temperature profiles for both cases, we need to solve the separate situations: cooling due to convection, cooling due to radiation, and combined in parts (a), (b), and (c). We can use the expressions derived in part (a) and (b), and the technique outlined in part (c). Solving these numerically would give us the cooling times and temperature profiles for both situations. Since this involves solving equations like those in part (a), especially part (b), and the numerical approach in part (c), we would need a computer to analyze the results over all stages, and temperature histories can be plotted on a graph where time is on the X-axis and sphere temperature is on the Y-axis. The same exercise can be repeated for a polished aluminum sphere with \(\varepsilon = 0.1\).

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Key Concepts

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

Understanding Newton's Law of Cooling
Newton's law of cooling is fundamental when exploring how objects cool down in a given environment. It describes the rate at which an object exchanges heat with its surroundings. The law asserts that the rate of heat loss of a body is directly proportional to the difference in temperatures between the body and its environment, and is expressed mathematically as

\( q = hA(T - T_{\text{env}}) \).

Here, \(q\) is the rate of heat loss, \(h\) is the heat transfer coefficient, \(A\) is the surface area of the object, \(T\) is the object's temperature, and \(T_{\text{env}}\) is the environmental temperature. For a student attempting to calculate the cooling time of an object, the challenge is to integrate this equation over the range of object temperatures from its initial state to the final temperature. In the exercise, integrating this equation with respect to time yields a logarithmic function that can be solved for the time \(t\), showing how temperature changes over time. Understanding Newton's law of cooling provides a powerful tool for predicting how long it will take for an object to reach a certain temperature when cooling by convection.
Delving into the Stefan-Boltzmann Law
The Stefan-Boltzmann law plays a critical role in thermal physics, particularly when dealing with heat transfer by radiation. This law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time \( (q) \) is directly proportional to the fourth power of the black body's absolute temperature \( (T^4) \). Mathematically, it's described as

\( q = \varepsilon \sigma A(T^4 - T_{\text{sur}}^4) \),

where \( \varepsilon \) is the emissivity of the object's surface, \( \sigma \) is the Stefan-Boltzmann constant, \(A\) is the surface area, \(T\) is the absolute temperature of the object, and \(T_{\text{sur}}\) is the temperature of the surroundings. In practical terms, this law is used to calculate the cooling time when thermal radiation is the dominant heat transfer mode, such as in the second part of the given exercise. The complexity arises due to the non-linearity of temperature with respect to time, which typically leads to numerical methods being employed for solutions.
Exploring Numerical Integration Methods
Numerical integration methods are indispensable tools for solving equations that cannot be solved analytically. These methods approximate the value of an integral, which is particularly useful when dealing with complex functions, non-linear relationships, or when the equation involves variables that are difficult to integrate directly. Common methods include the Trapezoidal Rule, Simpson's Rule, Euler's method, and the Runge-Kutta methods. The choice of method depends on factors such as the desired accuracy and the complexity of the problem.
In the context of heat transfer analysis, as in the exercise, numerical integration comes into play notably in part (c), where both convection and radiation contribute to cooling, leading to a set of differential equations that are often solved numerically. Tools like MATLAB or Python can be used to implement these numerical methods and provide practical solutions for time-dependent temperature profiles.
Interpreting Temperature Profiles
Temperature profiles provide a graphical representation of how the temperature of an object changes with time. They are an essential aspect of heat transfer studies because they give a visual understanding of the cooling or heating process over time. Interpreting these profiles helps in understanding how rapidly an object reaches thermal equilibrium with its environment.

In the exercise, the profile would show the exponential decrease of temperature of the sphere as it cools down due to convection and radiation. By plotting temperature on the Y-axis against time on the X-axis, one can visually analyze the rate of cooling. Different materials with distinct properties, such as emissivities in anodized and polished aluminum spheres, exhibit unique curves on this graph, offering insights into their thermal behaviors. For students dealing with heat transfer problems, constructing and understanding these temperature profiles is crucial to comprehend the dynamics of cooling.

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Most popular questions from this chapter

In a tempering process, glass plate, which is initially at a uniform temperature \(T_{i}\), is cooled by suddenly reducing the temperature of both surfaces to \(T_{s}\). The plate is \(20 \mathrm{~mm}\) thick, and the glass has a thermal diffusivity of \(6 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). (a) How long will it take for the midplane temperature to achieve \(50 \%\) of its maximum possible temperature reduction? (b) If \(\left(T_{i}-T_{s}\right)=300^{\circ} \mathrm{C}\), what is the maximum temperature gradient in the glass at the time calculated in part (a)?

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 \(T_{i}=80^{\circ} \mathrm{C}\). The air jets are at a temperature of \(T_{\infty}=20^{\circ} \mathrm{C}\) and provide a uniform convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) 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 \(1 \mathrm{~h}\) of exposure to the gas jets.

A horizontal structure consists of an \(L_{\mathrm{A}}=10\)-mm-thick layer of copper and an \(L_{\mathrm{B}}=10\)-mm-thick layer of aluminum. The bottom surface of the composite structure receives a heat flux of \(q^{\prime \prime}=100 \mathrm{~kW} / \mathrm{m}^{2}\), while the top surface is exposed to convective conditions characterized by \(h=40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, T_{\infty}=25^{\circ} \mathrm{C}\). The initial temperature of both materials is \(T_{i, \mathrm{~A}}=T_{i, \mathrm{~B}}=25^{\circ} \mathrm{C}\), and a contact resistance of \(R_{t, c}^{\prime \prime}=400 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) exists at the interface between the two materials. (a) Determine the times at which the copper and aluminum each reach a temperature of \(T_{f}=90^{\circ} \mathrm{C}\). The copper layer is on the bottom. (b) Repeat part (a) with the copper layer on the top. Hint: Modify Equation \(5.15\) to include a term associated with heat transfer across the contact resistance. Apply the modified form of Equation \(5.15\) to each of the two slabs. See Comment 3 of Example 5.2.

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 \(k=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Joints of high quality can be formed by friction welding. Consider the friction welding of two 40 -mm-diameter Inconel rods. The bottom rod is stationary, while the top rod is forced into a back-and-forth linear motion characterized by an instantaneous horizontal displacement, \(d(t)=a \cos (\omega t)\) where \(a=2 \mathrm{~mm}\) and \(\omega=1000 \mathrm{rad} / \mathrm{s}\). The coefficient of sliding friction between the two pieces is \(\mu=0.3\). Determine the compressive force that must be applied to heat the joint to the Inconel melting point within \(t=3 \mathrm{~s}\), starting from an initial temperature of \(20^{\circ} \mathrm{C}\). Hint: The frequency of the motion and resulting heat rate are very high. The temperature response can be approximated as if the heating rate were constant in time, equal to its average value.

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