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A 30 -cm-long cylindrical aluminum block $\left(\rho=2702 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.896 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=236 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\(, and \)\left.\alpha=9.75 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right), 15 \mathrm{~cm}$ in diameter, is initially at a uniform temperature of \(20^{\circ} \mathrm{C}\). The block is to be heated in a furnace at \(1200^{\circ} \mathrm{C}\) until its center temperature rises to \(300^{\circ} \mathrm{C}\). If the heat transfer coefficient on all surfaces of the block is $80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine how long the block should be kept in the furnace. Also, determine the amount of heat transfer from the aluminum block if it is allowed to cool in the room until its temperature drops to $20^{\circ} \mathrm{C}$ throughout. Solve this problem using the analytical one-term approximation method.

Short Answer

Expert verified
Question: Calculate the time required for the center temperature of an aluminum block to rise to 300°C when it is heated in a furnace at 1200°C. Also, determine the amount of heat transfer when the block is allowed to cool down again. The properties of the aluminum block and its surroundings are given. Use the analytical one-term approximation method. Answer: The time required for the center temperature of the aluminum block to rise to 300°C is _____ seconds, and the amount of heat transferred during the cooling process is _____ Joules.

Step by step solution

01

List the given information

We have the following given information about the block and its surroundings: - Aluminum block properties: - Length (L) = 30 cm - Diameter (D) = 15 cm - Density (ρ) = 2702 kg/m³ - Specific heat (c_p) = 0.896 kJ/kg·K - Thermal conductivity (k) = 236 W/m·K - Alpha (α) = 9.75 x 10⁻⁵ m²/s - Initial temperature (T_initial) = 20°C - Furnace temperature (T_furnace) = 1200°C - Required center temperature (T_center) = 300°C - Heat transfer coefficient (h) = 80 W/m²·K Since our answers need to be in seconds and Joules, we'll also convert some given units. c_p = 896 J/kg·K
02

Calculate the Bi number and find the approximation for the given scenario

Using the analytical one-term approximation method, calculate the Biot number to decide the suitable estimation. Bi = (h * L) / k Now, if Bi < 0.1, we can use the one-term approximation.
03

Calculate the time required for the center temperature to rise to 300°C

Using analytical one-term approximation, the time required for the center of the block to reach the target temperature can be calculated using the following equation: t = (L² * θ) / (π² * α) where θ = [(T_center - T_initial) / (T_furnace - T_initial)]
04

Calculate the heat transferred during the cooling process

When the aluminum block is cooled to the room temperature, we need to find the amount of heat transferred. We'll use the following equation: Q = m * c_p * (T_center - T_initial) First, find the mass of the block using its volume and density: m = ρ * V Then, calculate the heat transferred (Q) using the mass and the difference in temperature.
05

Summarize the results

Summarize the found time required for the center of the block to reach the target temperature, and the amount of heat transferred during the cooling process.

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Most popular questions from this chapter

A long 18-cm-diameter bar made of hardwood \((k=\) $\left.0.159 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=1.75 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\( is exposed to air at \)30^{\circ} \mathrm{C}$ with a heat transfer coefficient of \(8.83 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the center temperature of the bar is measured to be \(15^{\circ} \mathrm{C}\) after \(3 \mathrm{~h}\), the initial temperature of the bar is (a) \(11.9^{\circ} \mathrm{C}\) (b) \(4.9^{\circ} \mathrm{C}\) (c) \(1.7^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-9.2^{\circ} \mathrm{C}\)

A 6-cm-diameter, 13-cm-high canned drink \((\rho=977\) $\left.\mathrm{kg} / \mathrm{m}^{3}, k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( initially at \)25^{\circ} \mathrm{C}\( is to be cooled to \)5^{\circ} \mathrm{C}$ by dropping it into iced water at \(0^{\circ} \mathrm{C}\). Total surface area and volume of the drink are \(A_{s}=301.6 \mathrm{~cm}^{2}\) and \(V=367.6 \mathrm{~cm}^{3}\). If the heat transfer coefficient is \(120 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\), determine how long it will take for the drink to cool to $5^{\circ} \mathrm{C}$. Assume the can is agitated in water, and thus the temperature of the drink changes uniformly with time. (a) \(1.5 \mathrm{~min}\) (b) \(8.7 \mathrm{~min}\) (c) \(11.1 \mathrm{~min}\) (d) \(26.6 \mathrm{~min}\) (e) \(6.7 \mathrm{~min}\)

Can the one-term approximate solutions for a plane wall exposed to convection on both sides be used for a plane wall with one side exposed to convection while the other side is insulated? Explain.

Citrus trees are very susceptible to cold weather, and extended exposure to subfreezing temperatures can destroy the crop. In order to protect the trees from occasional cold fronts with subfreezing temperatures, tree growers in Florida usually install water sprinklers on the trees. When the temperature drops below a certain level, the sprinklers spray water on the trees and their fruits to protect them against the damage the subfreezing temperatures can cause. Explain the basic mechanism behind this protection measure, and write an essay on how the system works in practice.

Copper balls $\left(\rho=8933 \mathrm{~kg} / \mathrm{m}^{3}, \quad k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\(, \)c_{p}=385 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, \alpha=1.166 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\( ) initially at \)180^{\circ} \mathrm{C}$ are allowed to cool in air at \(30^{\circ} \mathrm{C}\) for a period of $2 \mathrm{~min}\(. If the balls have a diameter of \)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) \(78^{\circ} \mathrm{C}\) (b) \(95^{\circ} \mathrm{C}\) (c) \(118^{\circ} \mathrm{C}\) (d) \(134^{\circ} \mathrm{C}\) (e) \(151^{\circ} \mathrm{C}\)

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