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Chapter 5: Problem 8

A solid steel sphere (AISI 1010), \(300 \mathrm{~mm}\) in diameter, is couted with a dielectric material layer of thickness \(2 \mathrm{~mm}\) and thermal conductivity \(0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The couted sphere is initially at a uniform temperabure of \(500^{7} \mathrm{C}\) and is suddenly quenched in a large oil bath for which \(T_{x}=100^{\circ} \mathrm{C}\) and \(h=3300 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\). 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 capocitance \((\rho \in V)\) is small compared to that of the steel sphere.

Short Answer

Expert verified
The sphere will take approximately 2070 seconds to reach 140掳C.

Step by step solution

01

Understand the Given Information

The sphere is made of AISI 1010 steel and has a diameter of 300 mm. It is coated with a dielectric material 2 mm thick, with a thermal conductivity of 0.04 W/m路K. The initial temperature of the sphere is 500掳C and the oil bath temperature is 100掳C. The heat transfer coefficient is h = 3300 W/m虏路K. We are asked to find the time required for the sphere's temperature to reach 140掳C, assuming the thermal capacitance of the dielectric layer is negligible.
02

Calculate Biot Number

First, determine if lumped capacitance can be used by calculating the Biot number using the formula: \[ Bi = \frac{h imes L_c}{k} \] where \(L_c\) is the characteristic length; for a sphere, it is the radius, 150 mm or 0.15 m. The thermal conductivity \(k\) of the sphere is assumed to be high compared to the coating. Thus, \[ Bi = \frac{3300 \times 0.15}{0.04} \approx 12.375 \]. Since Bi > 0.1, using lumped capacitance may not be valid directly for the sphere without simplifying assumptions.
03

Calculate Thermal Resistance of the Coating

Calculate the thermal resistance of the dielectric coating using: \[ R_{th, coating} = \frac{t}{k imes A} \] where \(t = 0.002\, m\) (thickness), \(k = 0.04\, W/m\cdot K\), and the surface area \(A = 4 \pi \times r^2 = 4 \pi \times (0.15)^2\). So, \[ R_{th, coating} = \frac{0.002}{0.04 \times 4 \pi (0.15)^2} \approx 0.883 \times 10^{-3}\, K/W \].
04

Calculate Time using Simplified Heat Transfer Equation

Because the internal dielectric resistance is small compared to the convective resistance, the system can be simplified using a heat transfer equation. Use:\[ \frac{T(t) - T_{\infty}}{T_i - T_{\infty}} = e^{-\frac{hA}{\rho V C_p}t} \] Rearrange for time \(t\):\[ t = \frac{- \rho V C_p}{hA} \ln\left( \frac{T(t) - T_{\infty}}{T_i - T_{\infty}} \right) \]Where: \(\rho\) = 7870 kg/m鲁 (density of steel), \(C_p = 486 J/kg路K\) (specific heat of steel), \(V = \frac{4}{3}\pi r^3\), \(A\) is calculated earlier. Solve for \(t\) with given temperatures.
05

Insert Values and Calculate Time

Insert the known values into the equation:\[ t = \frac{- (7870 \times \frac{4}{3} \pi (0.15)^3 \times 486)}{3300 \times 4 \pi (0.15)^2} \ln\left( \frac{140 - 100}{500 - 100} \right) \]Complete the calculation to find:\[ t \approx 2070\, seconds \]

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Key Concepts

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

Biot Number
The Biot number is a dimensionless parameter that helps determine whether a simplified heat transfer analysis can be used. It's calculated using the formula: \[ Bi = \frac{h \times L_c}{k} \] where \( h \) is the heat transfer coefficient, \( L_c \) is the characteristic length, and \( k \) is the thermal conductivity of the material. A Biot number less than 0.1 suggests that the system can be analyzed using the lumped capacitance approach. In the original exercise, the Biot number was found to be approximately 12.375, indicating that the lumped capacitance method might not directly apply because heat conduction within the solid is not negligible compared to convective heat transfer at the surface. As a result, additional considerations or assumptions, such as focusing only on thermal resistances that dominate, might be necessary for such an analysis.
Thermal Resistance
Thermal resistance quantifies how a material resists heat flow. It is analogous to electrical resistance, but in the context of heat transfer systems. For layers or coatings, it can be calculated using the formula: \[ R_{th} = \frac{t}{k \times A} \] where \( t \) is the thickness, \( k \) is the thermal conductivity, and \( A \) is the surface area. A low thermal resistance means heat can flow easily through the material. In the exercise, the dielectric coating was found to have a relatively low thermal resistance of \( 0.883 \times 10^{-3} \) K/W. This implies that the coating does not add much resistance to heat transfer compared to the convective resistance from the oil bath, validating the assumption made to neglect the thermal capacitance of the dielectric layer.
Lumped Capacitance
The lumped capacitance method is a simplification in thermal analysis used when the Biot number is less than 0.1. It assumes that the temperature within an object is uniform at any given time during heat transfer. This means the thermal resistance within the object is much smaller than the resistance to heat flowing into it. Although in the exercise the Biot number is larger than 0.1, indicating inapplicability of this method directly, it is still interesting to know that using lumped capacitance simplifies calculations significantly. When applicable, this method transforms partial differential equations into ordinary ones, making analysis much more manageable.
Quenching Process
The quenching process is a rapid cooling method used to change the microstructure of a material, often to increase hardness. In the context of heat transfer, it involves submerging an object in a colder fluid, leading to a high heat transfer rate due to the large temperature difference. In the original problem, a steel sphere is quenched in an oil bath at a significantly lower temperature than that of the sphere, facilitating rapid cooling. Understanding the quenching process is essential for efficiently designing industrial processes and machinery that rely on rapid cooling techniques. The process profoundly affects material properties beyond just the thermal analysis.

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Most popular questions from this chapter

The inner surface of a plane wall is insulated while the cuter surface is exposed to an airstream at \(T_{\text {a }}\). The wall is af a uniform temperature corresponding to that of the ainstream, Sudilenly, a radiation heat source is switched cn applying a uniform fiux \(q_{e}^{*}\) to the outer surface. (a) Shetch and label, on \(T-x\) coordinates, the temperature distributions: initial, steady-state, and at two intermediate times. (b) Sketch the heat flux at the outer surface \(q_{,}^{\prime \prime}\left(L_{n} t\right)\) as a function of time.

A cold air chamber is proposed for quenching steel ball bearings of dinmeter \(D=0.2 \mathrm{~m}\) and initial temperature \(T_{i}=400^{\circ} \mathrm{C}\). Air in the chamber is maintained at \(-15^{\circ} \mathrm{C}\) by a refrigeration system, and the steel balls pass through the chamber on a conveyor belt. Optimum bearing production requires that \(70 \%\) of the initial thermal energy content of the ball above \(-15^{\circ} \mathrm{C}\) be removed. Radiation effects muy be neglected, and the convection heut transfer coefficient within the chamber is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimute the residence time of the balls within the chamber, und recommend a drive velocity of the conveyor. The following properties may be used for the steel: \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=2 \times 10^{-3} \mathrm{~m}^{2} / \mathrm{s}\), and \(c=450 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\).

A thick steel slab \(\left(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, c=480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is initially at \(300^{\circ} \mathrm{C}\) and is cooled by water jets impinging on one of its surfaces. The temperature of the water is \(25^{\circ} \mathrm{C}\), and the jets maintain an extremely large, approximately uniform convection coefficient at the surface. Assuming that the surface is maintained at the temperature of the water throughout the cooling. how long will it take for the temperature to reich \(50^{\circ} \mathrm{C}\) at a distance of \(25 \mathrm{~mm}\) from the surface?

The convection coefficient for flow over a solid sphere may be determined by submerging the sphere, which is initially at \(25^{\circ} \mathrm{C}\), into the flow, which is at \(75^{\circ} \mathrm{C}\), and measuring its surface temperature at some time during the transient heating process. (a) If the sphere has a diameter of \(0.1 \mathrm{~m}\), a thermal conductivity of \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and a thermul diffusivity of \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), at what time will a surface tempertiture of \(60^{\circ} \mathrm{C}\) be recorded if the convection coefficient is \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? (b) Assess the effect of thermal diffusivity on the thermal response of the material by computing center and surface temperature histories for \(\alpha=10^{-6}\). \(10^{-5}\), and \(10^{-4} \mathrm{~m}^{2} / \mathrm{s}\). Plot your results for the perid \(0 \leq t \leq 300 \mathrm{~s}\). In a similar manner, assess tte effect of themal conductivity by considering vak ues of \(k=1.5,15\), and \(150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Consider a thin electrical heater artached to a plate and hacked by insulation. Initially, the heater and plate are at the temperafure of the ambient air, \(T\) - Suddenly, the power to the heater is activated, yielding a constant heat fux \(q_{e}^{*}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) at the inner surface of the plate. (a) Sketch and label, on \(T-x\) cocrdinates, the temperature distributions: initial, steady-state, and at two intermediate times. (b) Sketch the heat flux at the outer surface \(q_{z}^{\prime \prime}(L, t)\) as a function of time.
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