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Chapter 5: Problem 9

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

Short Answer

Expert verified
The time required for the coated sphere temperature to reach \(140^{\circ}\mathrm{C}\) is approximately \(214.41\, \mathrm{s}\).

Step by step solution

01

Identify the properties of the materials

First, we need to gather the properties of AISI 1010 steel and the given dielectric material. The relevant properties are: - Diameter of the steel sphere: \(D = 300 \, \mathrm{mm}\) - Thickness of the dielectric coating: \(\delta = 2\, \mathrm{mm}\) - Thermal conductivity of the dielectric material: \(k = 0.04 \, \mathrm{W/m\cdot K}\) - Bulk oil temperature: \(T_\infty = 100 ^\circ\mathrm{C}\) - Heat transfer coefficient: \(h = 3300 \, \mathrm{W/m^2 \cdot K}\) - Initial temperature of the sphere: \(T_i = 500^\circ\mathrm{C}\) - Desired temperature of the sphere: \(T_f = 140^\circ\mathrm{C}\)
02

Apply heat transfer equations and establish the temperature as a function of time

Next, we apply the lumped capacitance analysis for transient conduction. Since the hint advises us to neglect the effect of energy storage in the dielectric material, we can calculate the temperature as a function of time as: \[ln\left(\frac{T-T_\infty}{T_i - T_\infty}\right) = -\frac{h A t}{\rho c_p V}\] Where \(A\) is the surface area of the coated sphere, \(\rho\) is the density of AISI 1010 steel, \(c_p\) is the specific heat capacity, and \(V\) is the volume of the sphere.
03

Solve for the required time

We want to solve for time (\(t\)) when the temperature (\(T\)) reaches 140掳C. Rearrange the equation for \(t\): \[t = -\frac{ln\left(\frac{T-T_\infty}{T_i - T_\infty}\right)\rho c_p V}{h A}\] Now, plug in the known values and properties of the AISI 1010 steel (\(\rho = 7.85 \times 10^3 \, \mathrm{kg/m^3}\) and \(c_p = 434 \, \mathrm{J/kg\cdot K}\)). Also, calculate the surface area of the coated sphere: \[A = 4\pi\left(\frac{D}{2} + \delta\right)^2 = 4\pi\left(0.151\, \mathrm{m}\right)^2 \approx 0.287 \, \mathrm{m^2}\] And the volume of the steel sphere: \[V = \frac{4}{3}\pi\left(\frac{D}{2}\right)^3 = \frac{4}{3}\pi\left(0.15\, \mathrm{m}\right)^3 \approx 1.41 \times 10^{-2} \, \mathrm{m^3}\] Plug all the values into the equation for \(t\): \[t = -\frac{ln\left(\frac{140-100}{500 - 100}\right)(7.85 \times 10^3 \, \mathrm{kg/m^3})(434 \, \mathrm{J/kg\cdot K})(1.41 \times 10^{-2} \, \mathrm{m^3})}{3300 \, \mathrm{W/m^2 \cdot K}(0.287 \, \mathrm{m^2})}\] Solve for \(t\): \[t \approx 214.41\, \mathrm{s}\] It takes approximately 214.41 seconds for the coated sphere temperature to reach \(140^{\circ}\mathrm{C}\).

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Key Concepts

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

Lumped Capacitance Model
The Lumped Capacitance Model is a simplified approach used in transient heat conduction analysis. This model assumes that the temperature within an object changes uniformly with time. Thus, it allows for the heat conduction problem to be reduced to a function that changes with time only, instead of both time and space. The Lumped Capacitance Model is valid when the Biot number (Bi), defined as the ratio of conductive to convective heat transfer, is much less than 1. The Biot number is given by the formula:\[ \text{Bi} = \frac{hL_c}{k} \] where \(h\) is the heat transfer coefficient, \(L_c\) is the characteristic length of the object, and \(k\) is the thermal conductivity of the material. If the Biot number is small, the object's resistance to heat conduction is negligible compared to its surface resistance to heat transfer. Therefore, the object can be treated as if it has a uniform temperature at any given time. This simplifies the calculation of the time required to cool or heat the object through convection.
Thermal Conductivity
Thermal Conductivity, denoted as \(k\), is a physical property of materials that measures their ability to conduct heat. It is typically expressed in units of \(\mathrm{W/m \, \cdot \, K}\). Materials with high thermal conductivity are good heat conductors and transfer heat quickly, while those with low thermal conductivity are good insulators. In the given exercise, the dielectric material coating has a thermal conductivity of \(0.04 \, \mathrm{W/m \cdot K}\), indicating it is a relatively poor conductor of heat.

This lower thermal conductivity is why energy storage in the dielectric layer can be neglected, as mentioned in the problem's hint. Compared to the high thermal conductivity of carbon steel (AISI 1010), which is around \(51.9 \, \mathrm{W/m \cdot K}\), the dielectric material primarily acts as an insulating barrier. This property impacts the rate at which heat flows through the material and is essential for determining how quickly temperature changes in the system.
  • Understanding thermal conductivity helps in selecting materials for thermal insulation or conduction.
  • Designing systems like heat exchangers requires materials with suitable thermal conductivities to ensure efficient heat transfer.
Heat Transfer Coefficient
The Heat Transfer Coefficient, symbolized as \(h\), is a measure that quantifies the convective heat transfer between a fluid and a solid surface. It reflects how well heat is transferred from the sphere to the surrounding oil bath in the exercise. The heat transfer coefficient depends on factors like fluid properties, flow regime, and surface characteristics.

In the given problem, \(h = 3300 \, \mathrm{W/m^2 \cdot K}\), indicating a high rate of heat transfer due to convection. The convective heat transfer process occurs when the fluid motion increases the contact between the fluid and the surface, enhancing thermal exchange. To apply the lumped capacitance model effectively, a high heat transfer coefficient ensures rapid thermal equilibrium between the solid and the fluid, allowing us to assume a uniform temperature within the sphere.
  • A high heat transfer coefficient is favorable for applications that require fast heating or cooling, such as quenching processes.
  • For comprehensive analysis, factors like the Reynold's number and Nusselt's number, which relate to flow and thermal dynamics, can provide deeper insights.

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Most popular questions from this chapter

Stainless steel (AISI 304) ball bearings, which have uniformly been heated to \(850^{\circ} \mathrm{C}\), are hardened by quenching them in an oil bath that is maintained at \(40^{\circ} \mathrm{C}\). The ball diameter is \(20 \mathrm{~mm}\), and the convection coefficient associated with the oil bath is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If quenching is to occur until the surface temperature of the balls reaches \(100^{\circ} \mathrm{C}\), how long must the balls be kept in the oil? What is the center temperature at the conclusion of the cooling period? (b) If 10,000 balls are to be quenched per hour, what is the rate at which energy must be removed by the oil bath cooling system in order to maintain its temperature at \(40^{\circ} \mathrm{C}\) ?

The operations manager for a metals processing plant anticipates the need to repair a large furnace and has come to you for an estimate of the time required for the furnace interior to cool to a safe working temperature. The furnace is cubical with a \(16-\mathrm{m}\) interior dimension and \(1-\mathrm{m}\) thick walls for which \(\rho=2600 \mathrm{~kg} / \mathrm{m}^{3}, c=960 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=\) \(1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The operating temperature of the furnace is \(900^{\circ} \mathrm{C}\), and the outer surface experiences convection with ambient air at \(25^{\circ} \mathrm{C}\) and a convection coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Use a numerical procedure to estimate the time required for the inner surface of the furnace to cool to a safe working temperature of \(35^{\circ} \mathrm{C}\). Hint: Consider a two-dimensional cross section of the furnace, and perform your analysis on the smallest symmetrical section. (b) Anxious to reduce the furnace downtime, the operations manager also wants to know what effect circulating ambient air through the furnace would have on the cool-down period. Assume equivalent convection conditions for the inner and outer surfaces.

A plane wall of a furnace is fabricated from plain carbon steel \(\left(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7850 \mathrm{~kg} / \mathrm{m}^{3}, c=430 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) and is of thickness \(L=10 \mathrm{~mm}\). To protect it from the corrosive effects of the furnace combustion gases, one surface of the wall is coated with a thin ceramic film that, for a unit surface area, has a thermal resistance of \(R_{t, f}^{\prime \prime}=0.01 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The opposite surface is well insulated from the surroundings. At furnace start-up the wall is at an initial temperature of \(T_{i}=300 \mathrm{~K}\), and combustion gases at \(T_{\infty}=1300 \mathrm{~K}\) enter the furnace, providing a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the ceramic film. Assuming the film to have negligible thermal capacitance, how long will it take for the inner surface of the steel to achieve a temperature of \(T_{s, i}=1200 \mathrm{~K}\) ? What is the temperature \(T_{s, o}\) of the exposed surface of the ceramic film at this time?

Derive the explicit finite-difference equation for an interior node for three- dimensional transient conduction. Also determine the stability criterion. Assume constant properties and equal grid spacing in all three directions.

Common transmission failures result from the glazing of clutch surfaces by deposition of oil oxidation and decomposition products. Both the oxidation and decomposition processes depend on temperature histories of the surfaces. Because it is difficult to measure these surface temperatures during operation, it is useful to develop models to predict clutch-interface thermal behavior. The relative velocity between mating clutch plates, from the initial engagement to the zero-sliding (lock-up) condition, generates heat that is transferred to the plates. The relative velocity decreases at a constant rate during this period, producing a heat flux that is initially very large and decreases linearly with time, until lock-up occurs. Accordingly, \(q_{f}^{\prime \prime}=q_{o}^{\prime \prime}=\left[1-\left(t / t_{\mathrm{lu}}\right)\right]\), where \(q_{o}^{\prime \prime}=1.6 \times 10^{7} \mathrm{~W} / \mathrm{m}^{2}\) and \(t_{1 \mathrm{u}}=100 \mathrm{~ms}\) is the lock-up time. The plates have an initial uniform temperature of \(T_{i}=40^{\circ} \mathrm{C}\), when the prescribed frictional heat flux is suddenly applied to the surfaces. The reaction plate is fabricated from steel, while the composite plate has a thinner steel center section bonded to low- conductivity friction material layers. The thermophysical properties are \(\rho_{s}=\) \(7800 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{s}}=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k_{s}=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the steel and \(\rho_{\mathrm{im}}=1150 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{fm}}=1650 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k_{\mathrm{fm}}=4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the friction material. (a) On \(T-t\) coordinates, sketch the temperature history at the midplane of the reaction plate, at the interface between the clutch pair, and at the midplane of the composite plate. Identify key features. (b) Perform an energy balance on the clutch pair over the time interval \(\Delta t=t_{\mathrm{lu}}\) to determine the steadystate temperature resulting from clutch engagement. Assume negligible heat transfer from the plates to the surroundings. (c) Compute and plot the three temperature histories of interest using the finite-element method of FEHT or the finite-difference method of IHT (with \(\Delta x=0.1 \mathrm{~mm}\) and \(\Delta t=1 \mathrm{~ms}\) ). Calculate and plot the frictional heat fluxes to the reaction and composite plates, \(q_{\mathrm{rp}}^{\prime \prime}\) and \(q_{\mathrm{cp}}^{\prime \prime}\), respectively, as a function of time. Comment on features of the temperature and heat flux histories. Validate your model by comparing predictions with the results from part (b). Note: Use of both \(F E H T\) and \(I H T\) requires creation of a look-up data table for prescribing the heat flux as a function of time.
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