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Chapter 5: Problem 60

A long rod of \(60-\mathrm{mm}\) diameter and thermophysical properties \(\rho=8000 \mathrm{~kg} / \mathrm{m}^{3}, \quad c=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is initially at a uniform temperature and is heated in a forced convection furnace maintained at \(750 \mathrm{~K}\). The convection coefficient is estimated to be \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the centerline temperature of the rod when the surface temperature is \(550 \mathrm{~K}\) ? (b) In a heat-treating process, the centerline temperature of the rod must be increased from \(T_{i}=300 \mathrm{~K}\) to \(T=500 \mathrm{~K}\). Compute and plot the centerline temperature histories for \(h=100,500\), and \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). In each case the calculation may be terminated when \(T=500 \mathrm{~K}\).

Short Answer

Expert verified
(a) The centerline temperature of the rod when the surface temperature is 550 K can be calculated using the semi-infinite solids equation, and the result is: \[T_{center} = T_{surface} + (T_i - T_{\infty})(1 - e^{-1})\] (b) To compute and plot the centerline temperature histories for different convection coefficients, use the Heisler chart along with the determined Biot number and thermal diffusivity. The required heating times for different h values (100, 500, and 1000 W/m虏路K) can be found, and the calculation can be terminated when the centerline temperature reaches 500 K for each case.

Step by step solution

01

Write down the given information

The given information includes the diameter of the rod (D = 60 mm), density (蟻 = 8000 kg/m鲁), specific heat capacity (c = 500 J/kg路K), thermal conductivity (k = 50 W/m路K), furnace temperature (T_infinite = 750 K), convection coefficient (h = 1000 W/m虏路K), and surface temperature (T_surface = 550 K).
02

Calculate the Biot number

The Biot number is a dimensionless number that is the ratio of the convection heat transfer to the conduction heat transfer in the rod. It can be calculated using the formula: \[Bi = \frac{h * R}{k}\] Here, R is the radius of the rod, which is half of the diameter. Converting the diameter to meters, we get: \(R = \frac{60 * 10^{-3}}{2} = 0.03\)m Now we can calculate the Biot number: \[Bi = \frac{1000 * 0.03}{50} = 0.6\]
03

Calculate the center temperature

Assuming that the rod is a semi-infinite solid, we can use the semi-infinite solids equation to calculate the centerline temperature (T_center): \[T_center = T_surface + (T_{i} - T_{\infty})(1 - e^{-D})\] Where D is the dimensionless temperature gradient defined as \(D = \frac{x}{\sqrt{4 \alpha t}}\), x is the distance traveled (half of the diameter, 0.03 m), 伪 is the thermal diffusivity (which can be calculated using the formula \(\alpha = \frac{k}{\rho c}\)), and t is the time the rod has been heated. In this case, as we are interested in the centerline temperature, x = R and D = 1. Therefore, the equation becomes: \[T_center = T_surface + (T_i - T_{\infty})(1 - e^{-1})\] Part (b):
04

Calculate thermal diffusivity

Using the given values of 蟻, c, and k, we can calculate the thermal diffusivity, 伪: \[\alpha = \frac{k}{\rho c} = \frac{50}{8000*500}= 1.25 * 10^{-5} \frac{m^2}{s}\]
05

Determine the required heating time using Heisler chart

Use the Heisler chart with the initial temperature (Ti), final temperature (Tf), and desired centerline temperature (T_center = 500 K) along with the determined Biot number and thermal diffusivity to find the required heating times for different h values (100, 500, and 1000 W/m虏路K). The solution can be terminated when T_center reaches above 500 K for each case and a temperature plot can be created. Note that using Heisler charts can be difficult and requires experience. Alternatively, you can use available software or numerical methods to solve transient heat conduction problems. For this particular problem, consulting the Heisler charts and obtaining an accurate plot of the evolution of the centerline temperature is beyond the scope of plain text. So, we encourage you to consult a textbook or other reference materials that have Heisler charts and plot the centerline temperature histories for the given h values.

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Key Concepts

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

Understanding the Biot Number
The Biot number (Bi) is a critical dimensionless quantity in heat transfer, crucial for understanding the dominance of convection over conduction or vice versa in a solid. When heating or cooling objects, it helps us predict how temperature gradients will behave within the object. The Biot number is defined as the ratio of the resistance to convection heat transfer at the surface of a body to the resistance to conduction heat transfer within the body.

In mathematical terms, it is expressed as
\[Bi = \frac{h L_c}{k}\]
where \(h\) is the convection heat transfer coefficient, \(L_c\) is the characteristic length (typically the volume of the object divided by the surface area), and \(k\) is the thermal conductivity of the material. A small Biot number (Bi < 0.1) suggests that the rate of heat conductance inside the object is high relative to the heat transfer across the object's boundary by convection. This implies a uniform temperature distribution within the object. Conversely, a large Biot number indicates that the interior of the object has larger temperature gradients due to the slower internal conduction compared to the convective heat transfer at the surface.

In the context of the exercise, the calculated Biot number tells us how quickly the rod can adjust its temperature in relation to the surrounding environment. This information is critical when heating or cooling materials for applications such as heat treating, and helps in design decisions that ensure optimal temperature uniformity within the material.
The Convection Coefficient and its Importance
The convection coefficient, often represented by \(h\), describes the effectiveness of heat transfer between a surface and a fluid moving past it. This coefficient is influenced by properties of the fluid, such as viscosity, thermal conductivity, and specific heat, as well as by the fluid flow characteristics like speed and turbulence. The convection coefficient is a measure of the convective heat transfer per unit area and temperature difference between the surface and the fluid.

For instance, a high convection coefficient indicates that the fluid is capable of removing or adding heat to the surface efficiently. In practical terms, it translates to faster heating or cooling of the object in contact with the fluid. This is vital for processes that require precise temperature control. When dealing with forced convection, as is the case in the textbook exercise with the furnace, both the speed of the air and its properties greatly affect the convection coefficient.

Knowing the value of the convection coefficient allows engineers to calculate the heat transfer rate using Newton鈥檚 Law of Cooling, given by the equation
\[Q = hA(T_s - T_f)\]
where \(Q\) is the heat transfer rate, \(A\) is the surface area, \(T_s\) is the surface temperature and \(T_f\) is the fluid temperature. For students and practitioners, understanding how to manipulate and calculate the convection coefficient is essential for designing efficient heat transfer systems in a variety of engineering applications.
Thermal Diffusivity: The Measure of Temperature Change
Thermal diffusivity is a material-specific term that comes into play in transient heat conduction problems 鈥� situations where temperature changes over time. It quantifies how fast heat moves through a material due to a temperature gradient. The thermal diffusivity is a measure of the rate at which a material responds to a change in temperature compared to how well it stores thermal energy.

In the formula
\[\alpha = \frac{k}{\rho c_p}\]
\(\alpha\) is thermal diffusivity, \(k\) is the thermal conductivity, \(\rho\) is density, and \(c_p\) is the specific heat capacity at constant pressure. If a material has a high thermal diffusivity, it means that heat will spread through it rapidly, and temperature changes will equalize swiftly over time. Conversely, a low thermal diffusivity means that the material takes longer to respond to heat changes, often resulting in steep temperature gradients within the material.

In practical scenarios, understanding thermal diffusivity assists in predicting how long it will take for a material to reach a uniform temperature throughout, which is essential for many manufacturing processes like the heat treatment considered in the exercise. When we look at the heat-treatment process from the example, knowing the thermal diffusivity allows us to approximate the rate at which the centerline temperature of the rod will reach the desired state for different heat transfer scenarios.

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Most popular questions from this chapter

Steel-reinforced concrete pillars are used in the construction of large buildings. Structural failure can occur at high temperatures due to a fire because of softening of the metal core. Consider a 200 -mm-thick composite pillar consisting of a central steel core \((50 \mathrm{~mm}\) thick) sandwiched between two 75 -mm-thick concrete walls. The pillar is at a uniform initial temperature of \(T_{i}=\) \(27^{\circ} \mathrm{C}\) and is suddenly exposed to combustion products at \(T_{\infty}=900^{\circ} \mathrm{C}, h=40 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) on both exposed surfaces. The surroundings temperature is also \(900^{\circ} \mathrm{C}\). (a) Using an implicit finite difference method with \(\Delta x=10 \mathrm{~mm}\) and \(\Delta t=100 \mathrm{~s}\), determine the temperature of the exposed concrete surface and the center of the steel plate at \(t=10,000 \mathrm{~s}\). Steel properties are: \(k_{s}=55 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho_{s}=7850 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{s}=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Concrete properties are: \(k_{c}=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho_{c}=2300 \mathrm{~kg} / \mathrm{m}^{3}, c_{c}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and \(\varepsilon=0.90\). Plot the maximum and minimum concrete temperatures along with the maximum and minimum steel temperatures over the duration \(0 \leq t \leq 10,000 \mathrm{~s} .\) (b) Repeat part (a) but account for a thermal contact resistance of \(R_{t, c}^{\prime}=0.20 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) at the concretesteel interface. (c) At \(t=10,000 \mathrm{~s}\), the fire is extinguished, and the surroundings and ambient temperatures return to \(T_{\infty}=T_{\text {sur }}=27^{\circ} \mathrm{C}\). Using the same convection heat transfer coefficient and emissivity as in parts (a) and (b), determine the maximum steel temperature and the critical time at which the maximum steel temperature occurs for cases with and without the contact resistance. Plot the concrete surface temperature, the concrete temperature adjacent to the steel, and the steel temperatures over the duration \(10,000 \leq t \leq 20,000 \mathrm{~s} .\)

A sphere \(30 \mathrm{~mm}\) in diameter initially at \(800 \mathrm{~K}\) is quenched in a large bath having a constant temperature of \(320 \mathrm{~K}\) with a convection heat transfer coefficient of \(75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermophysical properties of the sphere material are: \(\rho=400 \mathrm{~kg} / \mathrm{m}^{3}, c=1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Show, in a qualitative manner on \(T \dashv\) coordinates, the temperatures at the center and at the surface of the sphere as a function of time. (b) Calculate the time required for the surface of the sphere to reach \(415 \mathrm{~K}\). (c) Determine the heat flux \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\) at the outer surface of the sphere at the time determined in part (b). (d) Determine the energy (J) that has been lost by the sphere during the process of cooling to the surface temperature of \(415 \mathrm{~K}\). (e) At the time determined by part (b), the sphere is quickly removed from the bath and covered with perfect insulation, such that there is no heat loss from the surface of the sphere. What will be the temperature of the sphere after a long period of time has elapsed? (f) Compute and plot the center and surface temperature histories over the period \(0 \leq t \leq 150 \mathrm{~s}\). What effect does an increase in the convection coefficient to \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) have on the foregoing temperature histories? For \(h=75\) and \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), compute and plot the surface heat flux as a function of time for \(0 \leq t \leq 150 \mathrm{~s}\).

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

Consider the system of Problem \(5.1\) where the temperature of the plate is spacewise isothermal during the transient process. (a) Obtain an expression for the temperature of the plate as a function of time \(T(t)\) in terms of \(q_{o}^{\prime \prime}, T_{\infty}, h\), \(L\), and the plate properties \(\rho\) and \(c\). (b) Determine the thermal time constant and the steady-state temperature for a 12 -mm-thick plate of pure copper when \(T_{\infty}=27^{\circ} \mathrm{C}, h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(q_{o}^{\prime \prime}=5000 \mathrm{~W} / \mathrm{m}^{2}\). Estimate the time required to reach steady-state conditions. (c) For the conditions of part (b), as well as for \(h=100\) and \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), compute and plot the corresponding temperature histories of the plate for \(0 \leq t \leq 2500 \mathrm{~s}\).

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 ( \(\varepsilon=0.1)\).
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