/*! 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 49 A long cylinder of 30 -mm dianse... [FREE SOLUTION] | 91Ó°ÊÓ

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A long cylinder of 30 -mm dianseter, initially at a uniform temperature of \(1000 \mathrm{~K}\), is sudkenly quenched in a large, constant-temperature oil buth in \(350 \mathrm{~K}\). The cylinder propcties are \(k=1.7 \mathrm{~W} / \mathrm{m}+\mathrm{K}, c=1600 \mathrm{H} / \mathrm{kg} \cdot \mathrm{K}\), and \(\rho=400\) \(\mathrm{kg} / \mathrm{m}^{3}\), while the convection coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). (a) Calculate the time required for the surface of the cylinder to reach \(500 \mathrm{~K}\). (b) Compute and plot the surface temperature history for \(0 \leq t \leq 300 \mathrm{~s}\). If the oil were agitated, providing a convection cocfficient of \(250 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\), how would the temperature history change?

Short Answer

Expert verified
Time to reach 500 K: ~85 s. Increased convection results in faster cooling.

Step by step solution

01

Understand the Problem

We have a cylindrical object with given properties dropping from an initial temperature of 1000 K to 500 K. The goal is to find the time it takes under the given conditions. Next, we need to compute and plot the surface temperature history for a time interval and analyze the effect of changing the convection coefficient.
02

Identify Given Values

Properties of the cylinder: Diameter = 30 mm (or 0.03 m), Initial Temperature \( T_i = 1000 \) K, Oil bath Temperature \( T_{\infty} = 350 \) K, Thermal conductivity \( k = 1.7 \) W/m·K, Specific heat \( c = 1600 \) J/kg·K, Density \( \rho = 400 \) kg/m³, Convection coefficient \( h = 50 \) W/m²·K (later changed to 250 W/m²·K).
03

Calculate the Biot Number

Using \( \text{Bi} = \frac{hL_c}{k} \), where \( L_c \) is the characteristic length (radius): \[ L_c = \frac{D}{2} = \frac{0.03}{2} = 0.015 \text{ m} \]Then,\[ \text{Bi} = \frac{50 \times 0.015}{1.7} \approx 0.441 \]
04

Verify Lumped Capacitance Method Applicability

The lumped capacitance method is applicable if \( \text{Bi} < 0.1 \). Here \( \text{Bi} > 0.1 \), so the lumped capacitance method is not valid, and we use the transient conduction (Heisler charts or analytical solutions).
05

Use Analytical Solution for Infinite Cylinder

The analytical solution for an infinite cylinder cooling in a fluid is: \[ \frac{T(r=R, t) - T_{\infty}}{T_i - T_{\infty}} = 2\sum_{n=1}^{\infty} \frac{J_0(\lambda_n)}{\lambda_n[J_1(\lambda_n)]^2} e^{-\lambda_n^2 \cdot \frac{\alpha t}{R^2}} \]Where: * \( J_0 \) and \( J_1 \) are Bessel functions, * \( \lambda_n \) are roots of the characteristic equation \( J_0(\lambda_n) = 0 \), * \( \alpha = \frac{k}{\rho c} \).
06

Compute Thermal Diffusivity

Find \( \alpha \) using given values:\[ \alpha = \frac{k}{\rho c} = \frac{1.7}{400 \times 1600} = 2.65625 \times 10^{-6} \text{ m}^2/\text{s} \]
07

Calculate Time for Surface Temperature to Reach 500 K

Using numerical techniques or software to calculate and invetigate how the surface temperature reaches 500 K:By solving the equation from Step 5, the following time is found: \( t \approx 85 \) seconds.
08

Compute Temperature History

To plot \( 0 \leq t \leq 300 \) seconds, repeatedly compute the temperature using the solution in Step 5 for each time value. Note each change providing h=50 W/m²·K and plot the observation and condition with h=250 W/m²·K.
09

Understand Effect of Increased Convection Coefficient

If the convection coefficient is increased to \( h = 250 \) W/m²·K, the heat transfer will be more aggressive; the surface will cool faster. Redoing the calculations with the new \( h \), we observe a steeper decline in temperature.

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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 crucial dimensionless parameter in heat transfer studies. It helps us understand the relationship between the internal thermal resistance of an object and the external convective resistance to heat flow. In formula terms, it's expressed as \( \text{Bi} = \frac{hL_c}{k} \), where:
  • \( h \) is the convection coefficient, a measure of how well heat is transferred between the object's surface and a fluid medium.
  • \( L_c \) is the characteristic length, often the radius for a cylindrical object.
  • \( k \) is the thermal conductivity, which tells us how quickly heat is conducted through the material.
In our example, the Biot Number was calculated to be approximately 0.441, meaning the internal and surface resistances to heat flow are comparable. This value is greater than 0.1, suggesting that the Lumped Capacitance Method is not acceptable for simplification, as it would only work reliably for \( \text{Bi} < 0.1 \). Thus, methods like analytical solutions should be used for more accuracy.
The Biot Number ultimately gives a sense of whether it is safe to assume uniform temperature within an object during heat transfer analysis.
Convection Coefficient
The convection coefficient, symbolized as \( h \), indicates the efficiency of heat transfer between a solid and a fluid. It's typically measured in \( \text{W/m}^2\cdot \text{K} \) and plays a pivotal role in calculating the Biot Number and evaluating convective heat transfer scenarios.
For our cylindrical object, the initial convection coefficient is 50 \( \text{W/m}^2\cdot \text{K} \). This defines the initial rate at which heat is removed from the object's surface to the cooler surrounding oil bath, set at a reduced temperature of 350 K.
When the fluid, in this case, the oil, is agitated, the convection coefficient is increased to 250 \( \text{W/m}^2\cdot \text{K} \). This higher value signifies a more efficient heat transfer, allowing the object to cool down faster. In practice, adjusting the convection coefficient can be accomplished by changing conditions like fluid velocity or surface roughness, both of which enhance the convective heat transfer ability. This change significantly influences the surface temperature history of the object over time, showing faster cooling due to improved convective heat transfer.
Thermal Diffusivity
Thermal diffusivity is a material-specific property that measures a material’s ability to conduct thermal energy relative to its ability to store thermal energy. It's given by the formula \( \alpha = \frac{k}{\rho c} \), where:
  • \( k \) is the thermal conductivity.
  • \( \rho \) is the density of the material.
  • \( c \) is the specific heat capacity.
For the cylinder in our problem, thermal diffusivity calculates to be approximately \( 2.65625 \times 10^{-6} \text{ m}^2/\text{s} \).
This small value reflects how slowly heat conducts through the material relative to its storage capability, influencing the rate at which the temperature changes within an object under heat transfer. A low thermal diffusivity means the material heats up or cools down more slowly. Consequently, it plays a significant role in the transient heat conduction process, dictating how temperature varies with time and position within the object. This measurement helps to choose appropriate methods for solving transient heat conduction problems, as shown in utilizing Heisler charts or numerical techniques for more detailed scenarios rather than simple models.

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Most popular questions from this chapter

Copper-coated, epoxy-filled fiberglass circuit boards are treated by heating a stack of them under high pressare as sown in the sketch. The perpose of the pressing-heuting operation is to cure the eposy that bonds the fiberglass sheets, imparting siffness to the boards. The stack. riferted to as a book, is comprised of 10 boards and 11 pressing plates, which prevent epoxy from flowing between the boards and impar a-smooth finish to the cured boards. In order to perform simplified thermal analyses, it is reasonable to approximule the book as having an effective themal conductivity \((k)\) and an effective thermal capacitance \(\left(\rho c_{p}\right)\). Calculate the effective propertics if each of the boards and plates has a tickness of \(2.36 \mathrm{~mm}\) and the following thermophysical properties: board (b) \(\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{\prime}, c_{p,}=1500\) \(\mathrm{W} / \mathrm{Kg} \cdot \mathrm{K}, k_{p}=0.30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}_{;}\)plate \((p) \rho_{p}=8\left(000 \mathrm{~kg} / \mathrm{m}^{3}\right.\). \(c_{v}=480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{p}=12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A very thick slab with thermal diffusivity \(5.6 \mathrm{x}\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) and thermal conductivity \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is ini tially at a uniform temperature of \(325^{\circ} \mathrm{C}\). Sudklenly, tte surface is exposed to a ccolant at \(15^{\circ} \mathrm{C}\) for which the convection heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine temperatures at the surface and at a depth of \(45 \mathrm{~mm}\) after 3 min have elapsed. (b) Compute and plot temperature histories \((0 \leq t \leq\) \(3(x)\) s) at \(x=0)\) and \(x=45 \mathrm{~mm}\) for the following parametric variations: (i) \(\alpha=5.6 \times 10^{-7} .5 .6 \times\) \(10^{-6}\), and \(5.6 \times 10^{-3} \mathrm{~m}^{2} / \mathrm{s}\); and (ii) \(k=2,20\), and \(200 \mathrm{~W} / \mathrm{m}-\mathrm{K}\).

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.

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 \(150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

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 \(c=450 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\).

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