/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 An alloy cylinder \(3 \mathrm{~c... [FREE SOLUTION] | 91Ó°ÊÓ

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An alloy cylinder \(3 \mathrm{~cm}\) in diameter and \(2 \mathrm{~m}\) high is removed from an oven at \(200^{\circ} \mathrm{C}\) and stood on its end to cool in air at \(20^{\circ} \mathrm{C}\). Give an estimate of the time for the cylinder to cool to \(100^{\circ} \mathrm{C}\) if the convective heat transfer coefficient is \(80 \mathrm{~W} / \mathrm{m}^{2}\) \(\mathrm{K}\). For the alloy, take \(\rho=8600 \mathrm{~kg} / \mathrm{m}^{3}, c=340 \mathrm{~J} / \mathrm{kg} \mathrm{K}\), and \(k=110 \mathrm{~W} / \mathrm{m} \mathrm{K}\).

Short Answer

Expert verified
The cylinder will take approximately 5.29 minutes to cool to 100°C.

Step by step solution

01

Determine Surface Area and Volume

First, calculate the surface area and volume of the cylinder. The diameter is given as \(3 \text{ cm} = 0.03 \text{ m}\), so the radius \(r = 0.015 \text{ m}\).* **Surface Area:** The cylinder has a lateral surface area and two circular ends. The lateral surface area \(S_{lat} = 2\pi rh\), where \(h\) is the height (2 m). The area of the circular ends is \(2 \pi r^2\). \[ S = 2\pi rh + 2\pi r^2 = 2\pi \cdot 0.015 \cdot 2 + 2\pi \cdot (0.015)^2 = 0.1884 \text{ m}^2 \]* **Volume:** The volume \(V = \pi r^2 h\). \[ V = \pi \cdot (0.015)^2 \cdot 2 = 0.00141372 \text{ m}^3 \]
02

Calculate Biot Number

The Biot number (\(Bi\)) is calculated to determine if the system can be approximated as a lumped thermal capacity system. \(Bi = \frac{hL_c}{k}\), where \(L_c\) is the characteristic length and \(L_c = V/S\).* Calculate \(L_c\): \[ L_c = \frac{0.00141372}{0.1884} = 0.0075 \text{ m} \]* Calculate Biot number: \[ Bi = \frac{80 \times 0.0075}{110} = 0.00545 \]Since \(Bi < 0.1\), lumped system analysis can be applied.
03

Lumped System Analysis to Find Time Constant

Using the lumped system analysis, apply the formula \[ \theta = \theta_0 e^{-\frac{hA}{\rho V c}t} \] to calculate the time to reach a target temperature.Initial temperature: \(\theta_0 = 200^\circ \text{C}\)Final temperature: \(\theta = 100^\circ \text{C}\)Ambient temperature: \(\theta_{\infty} = 20^\circ \text{C}\)Rearrange the formula to solve for time \(t\):\[ \ln \left( \frac{\theta - \theta_{\infty}}{\theta_0 - \theta_{\infty}} \right) = -\frac{hA}{\rho V c}t \]Where:\[ A = 0.1884 \text{ m}^2 \]\[ V = 0.00141372 \text{ m}^3 \]\[ h = 80 \text{ W/m}^2\,K \]\[ \rho = 8600 \text{ kg/m}^3 \]\[ c = 340 \text{ J/kg\,K} \]
04

Calculate Time for Cooling

Now, calculate the time to reach 100°C using the rearranged lumped capacitance formula.Substitute values:- Initial term: \(\frac{\theta - \theta_{\infty}}{\theta_0 - \theta_{\infty}} = \frac{100 - 20}{200 - 20} = \frac{80}{180} = \frac{4}{9}\)Apply:\[ t = -\frac{\ln \left( \frac{4}{9} \right)}{\frac{80 \times 0.1884}{8600 \times 0.00141372 \times 340}} \]Calculate:\[ t \approx -\frac{\ln \left( \frac{4}{9} \right)}{1.1011} \approx 317.18 \text{ seconds} \]
05

Convert Time to Minutes

Convert time in seconds to minutes.\[ t \approx \frac{317.18}{60} \approx 5.29 \text{ minutes} \]

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Key Concepts

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

Lumped System Analysis
Lumped System Analysis is a method used in heat transfer where the temperature within a solid body is assumed to be uniform at any given time. This means the body behaves like a lump of material with uniform temperature changes, which simplifies the calculations needed to estimate heat transfer.
To apply this analysis, the material should have a substantial conductivity compared to the heat transfer from its surface. This assumption allows engineers to equate the change in internal energy to the heat lost or gained through its surface.
In the given problem, Lumped System Analysis helps us determine how long it takes for the alloy cylinder to cool from 200°C to 100°C. Using the formula \( \theta = \theta_0 e^{-\frac{hA}{\rho V c}t} \), we can calculate the time needed by considering the heat transfer coefficient, the surface area, and the specific heat capacity of the material.
This approach is efficient for systems with a Biot number less than 0.1, indicating minimal internal thermal resistance compared to external convection resistance.
Biot Number
The Biot Number is a dimensionless quantity that indicates the ratio of internal thermal resistance to external thermal resistance due to convection. It plays a critical role in determining whether lumped system analysis is applicable.
The formula for the Biot number is \( Bi = \frac{hL_c}{k} \), where:
  • \( h \) is the convective heat transfer coefficient.
  • \( L_c \) is the characteristic length, found from the volume to surface area relationship, \( L_c = \frac{V}{S} \).
  • \( k \) is the thermal conductivity of the material.
In the context of the problem, calculating the Biot number for the alloy cylinder gives us \( Bi = 0.00545 \), which is much less than 0.1.
This small Biot number confirms that lumped system analysis is a valid method here, implying the cylinder's internal temperature variations are negligible compared to the heat convection at the surface.
Convective Heat Transfer
Convective Heat Transfer is the transfer of thermal energy between a surface and a fluid flowing over the surface. It happens due to the movement of fluid, either by force (like a fan) or naturally (due to density differences in temperature gradients).
The effectiveness of convective heat transfer is quantified by the convective heat transfer coefficient \( h \). Higher values of \( h \) indicate more efficient heat transfer between the surface and the surrounding fluid.
In this exercise, the alloy cylinder cools down in air, and the cooling rate is influenced by the convective heat transfer coefficient, given as 80 W/m²K. The temperature change illustrates how effectively the surface loses heat to the surrounding air.
The equation \( q = hA(T_{surface} - T_{fluid}) \) models convective heat transfer, where \( q \) represents the heat transfer rate and \( A \) the surface area. This fundamental mechanism of heat loss governs how quickly the cylinder cools to the desired temperature.

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Most popular questions from this chapter

The thermal resistance per unit area of clothing is often expressed in the unit clo, where \(1 \mathrm{clo}=0.88 \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} \mathrm{hr} / \mathrm{B}\) tu. (i) What is 1 clo in SI units? (ii) If a wool sweater is \(2 \mathrm{~mm}\) thick and has a thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m}\) \(\mathrm{K}\), what is its thermal resistance in clos? (iii) A cotton shirt has a thermal resistance of \(0.5 \mathrm{clo}\). If the inner and outer surfaces are at \(31^{\circ} \mathrm{C}\) and \(28^{\circ} \mathrm{C}\), respectively, what is the rate of heat loss per unit area?

Two small blackened spheres of identical size-one of aluminum, the other of an unknown alloy of high conductivity-are suspended by thin wires inside a large cavity in a block of melting ice. It is found that it takes \(4.8\) minutes for the temperature of the aluminum sphere to drop from \(3^{\circ} \mathrm{C}\) to \(1^{\circ} \mathrm{C}\), and \(9.6\) minutes for the alloy sphere to undergo the same change. If the specific gravities of the aluminum and alloy are \(2.7\) and \(5.4\), respectively, and the specific heat of the aluminum is \(900 \mathrm{~J} / \mathrm{kg} \mathrm{K}\), what is the specific heat of the alloy?

A reactor vessel's contents are initially at \(290 \mathrm{~K}\) when a reactant is added, leading to an exothermic chemical reaction that releases heat at a rate of \(4 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). The volume and exterior surface area of the vessel are \(0.008 \mathrm{~m}^{3}\) and \(0.24 \mathrm{~m}^{2}\), respectively, and the overall heat transfer coefficient between the vessel contents and the ambient air at \(300 \mathrm{~K}\) is \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If the reactants are well stirred, estimate their temperature after (i) I minute. (ii) 10 minutes. Take \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=3000 \mathrm{~J} / \mathrm{kg} \mathrm{K}\) for the reactants.

A hot-water cylinder contains 150 liters of water. It is insulated, and its outer surface has an area of \(3.5 \mathrm{~m}^{2}\). It is located in an area where the ambient air is \(25^{\circ} \mathrm{C}\), and the overall heat transfer coefficient between the water and the surroundings is \(1.0 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), based on outer surface area. If there is a power failure, how long will it take the water to cool from \(65^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) ? Take the density of water as \(980 \mathrm{~kg} / \mathrm{m}^{3}\) and its specific heat as \(4180 \mathrm{~J} / \mathrm{kg} \mathrm{K}\).

A \(1 \mathrm{~cm}\)-diameter sphere is maintained at \(60^{\circ} \mathrm{C}\) in an enclosure with walls at \(35^{\circ} \mathrm{C}\) through which air at \(40^{\circ} \mathrm{C}\) circulates. If the convective heat transfer coefficient is \(11 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), estimate the rate of heat loss from the sphere when its emittance is (i) \(0.05\). (ii) \(0.85 .\)

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