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In a production facility, 3-cm-thick large brass plates \(\left(k=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), and \(\left.\alpha=33.9 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) that are initially at a uniform temperature of \(25^{\circ} \mathrm{C}\) are heated by passing them through oven maintained at \(700^{\circ} \mathrm{C}\). The plates remain in the oven for a period of \(10 \mathrm{~min}\). Taking the convection heat transfer coefficient to be \(h=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the surface temperature of the plates when they come out of the oven. Solve this problem using analytical one-term approximation method (not the Heisler charts). Can this problem be solved using lumped system analysis? Justify your answer.

Short Answer

Expert verified
Answer: No, this problem cannot be solved using lumped system analysis, as the Biot number is greater than 0.1. The surface temperature of the brass plate after being heated in the oven for 10 minutes is approximately 507.56 °C.

Step by step solution

01

Fourier number and Biot number

We need to first calculate the Fourier number (Fo) and the Biot number (Bi) using the given properties. The Fourier number (Fo) is calculated as: $$ \text{Fo} = \frac{\alpha t}{L^2} $$ Here, α is the thermal diffusivity, t is the heating time, and L is the thickness of the plate. The Biot number (Bi) is calculated as: $$ \text{Bi} = \frac{hL}{k} $$ Here, h is the convection heat transfer coefficient and k is the thermal conductivity.
02

Calculate Fourier number and Biot number

Now, we will calculate the values of Fourier number and Biot number using the given values. For the Fourier number: $$ \text{Fo} = \frac{33.9 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s} \cdot 600 \mathrm{~s}}{(0.03 \mathrm{~m})^2} = 2.274 $$ For the Biot number: $$ \text{Bi} = \frac{80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot 0.03 \mathrm{~m}}{110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} = 0.2182 $$
03

One-term approximation method

In the one-term approximation method, the expression for the temperature ratio (θ) for a situation such as this is given as: $$ \theta = 1 - C_1 \exp{(-z_1^2 Bi \cdot Fo)} $$ Here, C1 and z1 are the first coefficients for a one-term approximation which can be determined from tables. For a semi-infinite solid with convective boundary, C1 is 1 and z1 is 0.4604.
04

Calculate temperature ratio

Now, we will find the temperature ratio (θ) using the calculated values of Bi, Fo, and the coefficients C1 and z1. $$ \theta = 1 - 1 \exp{(-0.4604^2 \cdot 0.2182 \cdot 2.274)} = 0.7098 $$
05

Calculate the surface temperature

With the calculated temperature ratio (θ), we can find the surface temperature (Ts) of the plate as it exits the oven using the equation: $$ \text{T}_\text{s} = \text{T}_\text{i} + \theta(\text{T}_\text{o} - \text{T}_\text{i}) $$ Where Ti is the initial temperature and To is the oven temperature. $$ \text{T}_\text{s} = 25 + 0.7098(700 - 25) = 507.56^{\circ} \mathrm{C} $$ The surface temperature of the plates when they come out of the oven is approximately 507.56 °C.
06

Lumped system analysis applicability

To determine if this problem can be solved using lumped system analysis, we need to check if Bi < 0.1. In this case, the Biot number is 0.2182, which is greater than 0.1. Therefore, this problem cannot be solved using lumped system analysis.

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Key Concepts

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

Fourier Number
The Fourier number (Fo) is a dimensionless number that is crucial for analyzing heat conduction. It is defined as the ratio of diffusive/conductive heat transfer rate to the heat storage rate. In the context of the problem, we calculated the Fourier number using the formula \[\text{Fo} = \frac{\alpha t}{L^2}\] where \(\alpha\) represents the thermal diffusivity of the material, \(t\) is the time during which heat transfer takes place, and \(L\) is the characteristic length, which usually is the thickness of the object in question.

In practical terms, the Fourier number provides an estimate of how the temperature distribution within an object changes with time. A high Fourier number suggests that the heat transfer through the material is quick relative to the rate at which the material stores heat. In the given exercise, with a calculated Fourier number of 2.274, we observe a situation where considerable temperature change within the brass plate has occurred over time due to the prolonged exposure to the high oven temperature.
Biot Number
The Biot number (Bi) is another essential dimensionless number in the heat transfer analysis, especially when assessing the temperature gradients within a solid object. We calculate it using the formula \[\text{Bi} = \frac{hL}{k}\] where \(h\) is the convection heat transfer coefficient, \(L\) is the characteristic length as defined earlier, and \(k\) is the thermal conductivity of the material.

The Biot number helps to establish whether an object has uniform temperature within (lumped system analysis applies) or significant temperature variation across its volume. A small Biot number (Bi < 0.1) implies a uniform temperature distribution, suitable for lumped system analysis. Conversely, a larger Biot number indicates that temperature gradients within the object must be accounted for. In our example, the Biot number is calculated as 0.2182, which confirms the presence of temperature gradients within the brass plate, thereby invalidating the use of lumped system analysis for this problem.
Convection Heat Transfer Coefficient
The convection heat transfer coefficient \(h\) is significant in determining the rate at which heat is transferred by convection between a solid surface and a fluid in contact with it. It is a measure of the convection heat transfer ability of the fluid and surface in question, with a higher \(h\) indicating more efficient heat transfer.

In the context of our problem, the convection heat transfer coefficient affects the calculation of the Biot number and subsequently, the determination of whether lumped system analysis is applicable. The given value of \(h = 80 \mathrm{~W} / \mathrm{m}^2 \. \. \mathrm{K}\) significantly influences the temperature profile of the brass plates as they pass through the oven. Since convection is the mode of heat transfer to the plates, this coefficient plays a central role in the one-term approximation method used to approximate the surface temperature of the plates upon exiting the oven.

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Most popular questions from this chapter

A long cylindrical wood \(\log (k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=1.28 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) is \(10 \mathrm{~cm}\) in diameter and is initially at a uniform temperature of \(15^{\circ} \mathrm{C}\). It is exposed to hot gases at \(550^{\circ} \mathrm{C}\) in a fireplace with a heat transfer coefficient of \(13.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the surface. If the ignition temperature of the wood is \(420^{\circ} \mathrm{C}\), determine how long it will be before the log ignites. Solve this problem using analytical one-term approximation method (not the Heisler charts).

A steel casting cools to 90 percent of the original temperature difference in \(30 \mathrm{~min}\) in still air. The time it takes to cool this same casting to 90 percent of the original temperature difference in a moving air stream whose convective heat transfer coefficient is 5 times that of still air is (a) \(3 \mathrm{~min}\) (b) \(6 \mathrm{~min}\) (c) \(9 \mathrm{~min}\) (d) \(12 \mathrm{~min}\) (e) \(15 \mathrm{~min}\)

When water, as in a pond or lake, is heated by warm air above it, it remains stable, does not move, and forms a warm layer of water on top of a cold layer. Consider a deep lake \(\left(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that is initially at a uniform temperature of \(2^{\circ} \mathrm{C}\) and has its surface temperature suddenly increased to \(20^{\circ} \mathrm{C}\) by a spring weather front. The temperature of the water \(1 \mathrm{~m}\) below the surface 400 hours after this change is (a) \(2.1^{\circ} \mathrm{C}\) (b) \(4.2^{\circ} \mathrm{C}\) (c) \(6.3^{\circ} \mathrm{C}\) (d) \(8.4^{\circ} \mathrm{C}\) (e) \(10.2^{\circ} \mathrm{C}\)

Copper balls \(\left(\rho=8933 \mathrm{~kg} / \mathrm{m}^{3}, k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=\right.\) \(\left.385 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, \alpha=1.166 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\right)\) initially at \(200^{\circ} \mathrm{C}\) are allowed to cool in air at \(30^{\circ} \mathrm{C}\) for a period of 2 minutes. If the balls have a diameter of \(2 \mathrm{~cm}\) and the heat transfer coefficient is \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the center temperature of the balls at the end of cooling is (a) \(104^{\circ} \mathrm{C}\) (b) \(87^{\circ} \mathrm{C}\) (c) \(198^{\circ} \mathrm{C}\) (d) \(126^{\circ} \mathrm{C}\) (e) \(152^{\circ} \mathrm{C}\)

A 10-cm-inner diameter, 30-cm-long can filled with water initially at \(25^{\circ} \mathrm{C}\) is put into a household refrigerator at \(3^{\circ} \mathrm{C}\). The heat transfer coefficient on the surface of the can is \(14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming that the temperature of the water remains uniform during the cooling process, the time it takes for the water temperature to drop to \(5^{\circ} \mathrm{C}\) is (a) \(0.55 \mathrm{~h}\) (b) \(1.17 \mathrm{~h}\) (c) \(2.09 \mathrm{~h}\) (d) \(3.60 \mathrm{~h}\) (e) \(4.97 \mathrm{~h}\)

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