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In a manufacturing process, long rods of different diameters are at a uniform temperature of \(400^{\circ} \mathrm{C}\) in a curing oven, from which they are removed and cooled by forced convection in air at \(25^{\circ} \mathrm{C}\). One of the line operators has observed that it takes \(280 \mathrm{~s}\) for a \(40-\mathrm{mm}\) diameter rod to cool to a safe-to-handle temperature of \(60^{\circ} \mathrm{C}\). For an equivalent convection coefficient, how long will it take for an 80 -mm-diameter rod to cool to the same temperature? The thermophysical properties of the rod are \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3}, c=900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Comment on your result. Did you anticipate this outcome?

Short Answer

Expert verified
It will take approximately 1120 seconds for an 80-mm-diameter rod to cool down to 60°C, which is longer than the time it takes for the 40-mm-diameter rod (280 seconds). This can be anticipated because the 80-mm-diameter rod has a greater volume and mass, requiring more time for the heat to dissipate effectively.

Step by step solution

01

Calculate Biot number for 40-mm rod

To find the Biot number, we will use the formula: \( Bi = \frac{hD}{k} \), where \(h\) is the convection coefficient, \(D\) is the diameter of the rod, and \(k\) is the thermal conductivity. We are given the diameter (\(D = 0.04 \mathrm{m}\)), the thermal conductivity (\(k = 15 \mathrm{~W}/\mathrm{m} \cdot \mathrm{K}\)), and the initial time (\(t = 280 \mathrm{~s}\)) for the 40-mm rod to cool down from 400°C to 60°C. First, we need to calculate the time constant (\(\tau\)) using the formula: \( \tau = \frac{\rho cD^2}{12k} \). Substituting the given values, we get: \(\tau = \frac{2500 \times 900 \times (0.04)^2}{12 \times 15} = 20 \mathrm{~s}\) Now, we can find the Biot number by rearranging the formula: \( Bi = \frac{t}{\tau} \) \( Bi = \frac{280}{20} = 14 \)
02

Calculate Fourier number for 80-mm rod

To find the Fourier number (\(Fo\)) for the 80-mm rod, we have to calculate the time constant (\(\tau_{80}\)) for the 80-mm diameter rod using the formula \( \tau_{80} = \frac{\rho c(80 \times 10^{-3})^2}{12k} \). \(\tau_{80} = \frac{2500 \times 900 \times (0.08)^2}{12 \times 15} = 80 \mathrm{~s}\) Next, we will find the Fourier number using the formula: \( Fo = \frac{t}{\tau_{80}} \). We already know that \(Bi = 14\), so we have \( Fo = \frac{t_{80}}{80} = 14 \), where \(t_{80}\) is the time taken for the 80-mm rod to cool down to 60°C.
03

Calculate time for 80-mm rod to cool down to 60°C

Now, we can calculate the time taken for the 80-mm rod to cool down to 60°C using the Fourier number: \(t_{80} = Fo \times \tau_{80}\) \(t_{80} = 14 \times 80 = 1120 \mathrm{~s}\) So, it will take approximately 1120 seconds for an 80-mm-diameter rod to cool down to 60°C. The 80-mm-diameter rod takes more time to cool down compared to the 40-mm-diameter rod (1120 seconds vs. 280 seconds). This is because the 80-mm-diameter rod has a greater volume and mass, which means it takes longer for the heat to dissipate from the rod effectively. The result may be anticipated, as increasing the diameter of the rod requires more time to cool it down due to the larger surface area and increased mass.

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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 the phenomenon of thermal energy moving from a region of higher temperature to one of lower temperature. It occurs via three main mechanisms: conduction, convection, and radiation. In forced convection cooling, like in our manufacturing process example, an external factor, such as a fan or pump, is used to enhance the cooling effect by increasing the fluid flow over the surface of the heated object, in this case, the rods. The rate of heat transfer by convection is proportional to the temperature difference between the rod's surface and the air, as well as the convection coefficient, which is a measure of how effectively the cooling medium removes heat from the surface.

In practical terms for the exercise, as the air is moved over the surface of the hot rods, it carries away heat, reducing the rod's temperature from a uniform high temperature to a level that is considered safe-to-handle. The forced convection accelerates this process compared to natural convection, which only relies on the buoyancy-induced movement caused by changes in density with temperature.
Biot Number
The Biot number (Bi) is a dimensionless quantity used in heat transfer calculations. It compares the rate of heat conduction within an object to the rate of heat convection from its surface to the surrounding fluid. It is calculated with the formula:
\( Bi = \frac{hD}{k} \)

Where \( h \) is the convection coefficient, \( D \) is the characteristic length (such as the diameter of the rod), and \( k \) is the thermal conductivity of the material. In the given exercise, a lower Biot number would imply that conduction within the rod is much faster than convection from the rod's surface to the air, which may suggest that the surface temperature is representative of the rod's overall temperature. For a higher Biot number, the opposite is true, and there might exist a larger temperature gradient within the object.

For the 40-mm rod, as calculated in the solution steps, the Biot number was found to be 14, indicating that convection is significant in this scenario due to a sizable temperature difference across the rod from the center to the surface.
Fourier Number
The Fourier number (Fo) is another dimensionless quantity, which is used to describe heat conduction. It relates the rate of heat conduction to the energy stored, providing an indication of how temperature changes with time within an object. The Fourier number is defined as:\( Fo = \frac{\alpha t}{L^2} \),where \( \alpha \) is the thermal diffusivity which equals \( \frac{k}{\rho c} \), \( t \) is the time, and \( L \) is the characteristic length. In the example given, it is used to scale time for the cooling process of different diameter rods. A higher Fourier number indicates a longer cool-down process, which is expected when dealing with larger diameters since these have a higher volume-to-surface-area ratio, meaning there's more material to cool in relation to the area that is losing heat. Thus, the 80-mm rod displays a higher Fourier number in respect to the cool-down time compared to the 40-mm rod.
Thermal Conductivity
Thermal conductivity (\( k \)) is a property of a material that indicates its ability to conduct heat. Materials with high thermal conductivity, such as metals, can quickly transfer heat from one part to another, whereas insulating materials with low thermal conductivity do not conduct heat well. The value of \( k \) is crucial in evaluating how efficiently an object can equilibrate its temperature with its surroundings.In our example, the rod's thermal conductivity affects both the Biot and Fourier numbers. As an intrinsic property of the rod's material, it plays a central role in determining how quickly the rod can reach the safe-to-handle temperature. The given value for thermal conductivity in the exercise is \( k = 15 \) W/mK, which indicates the rod material conducts heat fairly well, thus influencing the time required for the rods to cool down under forced convection.
Convection Coefficient
The convection coefficient (\( h \)) quantifies the convective heat transfer between a solid surface and a fluid flowing over it. It depends on various factors including the fluid's properties, flow velocity, and the nature of the surface. A higher convection coefficient suggests a more efficient heat transfer, which often is the aim in cooling processes like the one described in our exercise.

In the context of the exercise, the convection coefficient is a given constant for both rods — however, in practical applications, it varies with changes in flow conditions and the physical setup. Estimating the convection coefficient correctly is critical for accurate predictions of cooling times. As it relates directly to the Biot number calculation, it provides insight into the efficiency of the forced convection cooling process employed in the manufacturing scenario.

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Most popular questions from this chapter

Circuit boards are treated by heating a stack of them under high pressure, as illustrated in Problem 5.45. The platens at the top and bottom of the stack are maintained at a uniform temperature by a circulating fluid. The purpose of the pressing-heating operation is to cure the epoxy, which bonds the fiberglass sheets, and impart stiffness to the boards. The cure condition is achieved when the epoxy has been maintained at or above \(170^{\circ} \mathrm{C}\) for at least \(5 \mathrm{~min}\). The effective thermophysical properties of the stack or book (boards and metal pressing plates) are \(k=0.613 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=2.73 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3} \cdot \mathrm{K}\) (a) If the book is initially at \(15^{\circ} \mathrm{C}\) and, following application of pressure, the platens are suddenly brought to a uniform temperature of \(190^{\circ} \mathrm{C}\), calculate the elapsed time \(t_{e}\) required for the midplane of the book to reach the cure temperature of \(170^{\circ} \mathrm{C}\). (b) If, at this instant of time, \(t=t_{e}\), the platen temperature were reduced suddenly to \(15^{\circ} \mathrm{C}\), how much energy would have to be removed from the book by the coolant circulating in the platen, in order to return the stack to its initial uniform temperature?

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.

A soda lime glass sphere of diameter \(D_{1}=25 \mathrm{~mm}\) is encased in a bakelite spherical shell of thickness \(L=\) \(10 \mathrm{~mm}\). The composite sphere is initially at a uniform temperature, \(T_{i}=40^{\circ} \mathrm{C}\), and is exposed to a fluid at \(T_{\infty}=10^{\circ} \mathrm{C}\) with \(h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the center temperature of the glass at \(t=200 \mathrm{~s}\). Neglect the thermal contact resistance at the interface between the two materials.

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 \(40^{\circ} \mathrm{C}\) ?

A microwave oven operates on the principle that application of a high- frequency field causes electrically polarized molecules in food to oscillate. The net effect is a nearly uniform generation of thermal energy within the food. Consider the process of cooking a slab of beef of thickness \(2 L\) in a microwave oven and compare it with cooking in a conventional oven, where each side of the slab is heated by radiation. In each case the meat is to be heated from \(0^{\circ} \mathrm{C}\) to a minimum temperature of \(90^{\circ} \mathrm{C}\). Base your comparison on a sketch of the temperature distribution at selected times for each of the cooking processes. In particular, consider the time \(t_{0}\) at which heating is initiated, a time \(t_{1}\) during the heating process, the time \(t_{2}\) corresponding to the conclusion of heating, and a time \(t_{3}\) well into the subsequent cooling process.

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