/*! 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 67 As part of a heat treatment proc... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 67

As part of a heat treatment process, cylindrical, 304 stainless steel rods of \(100-\mathrm{mm}\) diameter are cooled from an initial temperature of \(500^{\circ} \mathrm{C}\) by suspending them in an oil bath at \(30^{\circ} \mathrm{C}\). If a convection coefficient of \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of \(50^{\circ} \mathrm{C}\), at which point it is withdrawn from the bath? If 10 rods of length \(L=1 \mathrm{~m}\) are processed per hour, what is the nominal rate at which energy must be extracted from the bath (the cooling load)?

Short Answer

Expert verified
It takes approximately 553.6 seconds or 9.23 minutes for the centerline of a rod to reach a temperature of 50掳C. The nominal rate at which energy must be extracted from the bath (the cooling load) is approximately 38.3 kW.

Step by step solution

01

Calculate the temperature difference at the centerline of the rod

The temperature difference at the centerline of the rod, \(\Delta T\), is equal to the difference between the initial temperature and the final temperature. Therefore, \(\Delta T = T_{initial} - T_{final} = 500^{\circ}C - 50^{\circ}C = 450 \mathrm{K}\)
02

Use the Biot number to find the time it takes for the rod to cool down

In this cooling process, we can use the Biot number, \(Bi\), to find the time it takes for the rod to cool down. The formula for the Biot number is: \(Bi = \frac{hL_c}{k}\) Where \(h\) is the convection coefficient, \(500 \mathrm{W/m^2 \cdot K}\), \(L_c\) is the characteristic length of the rod, which is the diameter in this case, \(100 \times 10^{-3} \mathrm{m}\), and \(k\) is the thermal conductivity of the 304 stainless steel, \(16.3 \mathrm{W/m \cdot K}\). Calculating the Biot number, \(Bi = \frac{500 \mathrm{W/m^2 \cdot K} \times 100 \times 10^{-3} \mathrm{m}}{16.3 \mathrm{W/m \cdot K}} = 3.0675\) Since the Biot number is greater than 0.1, we can proceed to use the transient heat conduction analysis by using the following formula for the time, \(t\), it takes for the rod to cool down: \(t = \frac{0.2 \rho c_p L_c^2}{3.0675 \cdot k}\) Where \(\rho\) is the density of the 304 stainless steel, \(7900 \mathrm{kg/m^3}\), and \(c_p\) is the specific heat capacity of the 304 stainless steel, \(500 \mathrm{J/kg \cdot K}\). Calculating the time, \(t\): \(t = \frac{0.2 \times 7900 \mathrm{kg/m^3} \times 500 \mathrm{J/kg \cdot K} \times (100 \times 10^{-3} \mathrm{m})^2}{3.0675 \times 16.3 \mathrm{W/m \cdot K}} = 553.6 \mathrm{s}\) So, it takes approximately 553.6 seconds or 9.23 minutes for the centerline of a rod to reach a temperature of 50掳C.
03

Determine the energy exchanged between the rod and the oil bath

The energy exchanged between the rod and the oil bath is given by: \(Q = mc_p \Delta T\) Where \(m\) is the mass of the rod, which can be found using the density and volume: \(m = \rho \times V = 7900 \mathrm{kg/m^3} \times \pi (0.05 \mathrm{m})^2 \times 1 \mathrm{m} = 61.63 \mathrm{kg}\) Calculating the energy exchanged, \(Q\): \(Q = 61.63 \mathrm{kg} \times 500 \mathrm{J/kg \cdot K} \times 450 \mathrm{K} = 1.38 \times 10^7 \mathrm{J}\)
04

Calculate the cooling load

Since 10 rods of length 1 meter are processed per hour, the rate at which energy must be extracted from the bath is given by: \(Q_{rate} = \frac{1.38 \times 10^7 \mathrm{J} \times 10 \mathrm{rods}}{3600 \mathrm{s}} = 3.83 \times 10^4 \mathrm{W}\) So, the nominal rate at which energy must be extracted from the bath (the cooling load) is approximately 38.3 kW.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Biot number
The Biot number (Bi) is a dimensionless number important in transient heat conduction problems. It compares the resistance to heat conduction within an object to the resistance to heat transfer across the object's boundary due to convection. The Biot number is defined by the formula:

\[\begin{equation}Bi = \frac{h L_c}{k}\end{equation}\]

where h is the convection coefficient, L_c is the characteristic length of the object, and k is the thermal conductivity of the material. A small Biot number (Bi < 0.1) suggests that conduction within the object is fast compared to convection, and the temperature can be assumed uniform inside the object (lumped capacitance method). Conversely, a high Biot number indicates significant temperature gradients within the object, necessitating a more detailed transient analysis. For the rods in our example, the Biot number is greater than 0.1, thus confirming temperature variation within the rod and validating our use of a transient heat conduction analysis for accurate time prediction.
Convection coefficient
The convection coefficient (h), also known as the heat transfer coefficient, measures how well heat is transferred between a surface and a fluid moving past it via convection. It's given in units of Watts per square meter per degree Kelvin (W/m虏路K). The convection coefficient is crucial for calculating the rate of heat transfer due to convection, and it depends on a variety of factors, including the velocity of the fluid, the fluid's properties, and the type of flow (laminar or turbulent). In the provided exercise, the convection coefficient is given as 500 W/m虏路K. This value determines how quickly heat will transfer from the hot steel rods to the cooler oil bath, ultimately affecting the cooling time for the rods.
Cooling load calculation
The cooling load calculation is a technique used to determine the rate of energy removal necessary to maintain a desired temperature in a space or substance, such as during a cooling process. It is especially relevant in environmental engineering, HVAC design, and various industrial processes. In our heat treatment example, the cooling load refers to the nominal rate at which energy must be extracted from the oil bath to process the rods. It's given by the formula:

\[\begin{equation}Q_{rate} = \frac{Q_{total}}{t_{process}}\end{equation}\]

where Q_{total} is the total energy to be removed and t_{process} is the process time. Here, we found it to be approximately 38.3 kW for the hourly processing of 10 rods. The cooling load gives insights into the capacity requirements for the equipment used in the cooling process, such as heat exchangers, pumps, and cooling towers.
Energy transfer in heat treatment
Heat treatment processes involve the energy transfer in the form of heat to and from materials to alter their physical, and sometimes chemical, properties. During cooling in heat treatment, such as in quenching steel rods, energy is transferred from the hot material to the cooler fluid surrounding it. The amount of energy exchanged can be calculated using the equation:

\[\begin{equation}Q = mc_p \Delta T\end{equation}\]

where m is mass, c_p is specific heat, and \(\Delta T\) is the change in temperature. This formula is essential for determining the energy removed from the material, which is equal to the energy absorbed by the cooling medium, under the assumption of an adiabatic process. In practical applications, understanding the energy transfer helps in designing effective cooling systems and predicting the material's microstructural changes after treatment.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Small spherical particles of diameter \(D=50 \mu \mathrm{m}\) contain a fluorescent material that, when irradiated with white light, emits at a wavelength corresponding to the material's temperature. Hence the color of the particle varies with its temperature. Because the small particles are neutrally buoyant in liquid water, a researcher wishes to use them to measure instantaneous local water temperatures in a turbulent flow by observing their emitted color. If the particles are characterized by a density, specific heat, and thermal conductivity of \(\rho=999 \mathrm{~kg} / \mathrm{m}^{3}\), \(k=1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(c_{p}=1200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively, determine the time constant of the particles. Hint: Since the particles travel with the flow, heat transfer between the particle and the fluid occurs by conduction. Assume lumped capacitance behavior.

A tile-iron consists of a massive plate maintained at \(150^{\circ} \mathrm{C}\) by an embedded electrical heater. The iron is placed in contact with a tile to soften the adhesive, allowing the tile to be easily lifted from the subflooring. The adhesive will soften sufficiently if heated above \(50^{\circ} \mathrm{C}\) for at least \(2 \mathrm{~min}\), but its temperature should not exceed \(120^{\circ} \mathrm{C}\) to avoid deterioration of the adhesive. Assume the tile and subfloor to have an initial temperature of \(25^{\circ} \mathrm{C}\) and to have equivalent thermophysical properties of \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=1.5 \times 10^{6}\) \(\mathrm{J} / \mathrm{m}^{3} \cdot \mathrm{K}\) Tile, 4-mm thickness Subflooring (a) How long will it take a worker using the tile-iron to lift a tile? Will the adhesive temperature exceed \(120^{\circ} \mathrm{C} ?\) (b) If the tile-iron has a square surface area \(254 \mathrm{~mm}\) to the side, how much energy has been removed from it during the time it has taken to lift the tile?

A very thick slab with thermal diffusivity \(5.6 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) and thermal conductivity \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is initially at a uniform temperature of \(325^{\circ} \mathrm{C}\). Suddenly, the surface is exposed to a coolant 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 \(3 \mathrm{~min}\) have elapsed. (b) Compute and plot temperature histories \((0 \leq t \leq\) \(300 \mathrm{~s}\) ) 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 \(5.6 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\); and (ii) \(k=2,20\), and \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A constant-property, one-dimensional plane wall of width \(2 L\), at an initial uniform temperature \(T_{i}\), is heated convectively (both surfaces) with an ambient fluid at \(T_{\infty}=T_{\infty, 1}, h=h_{1}\). At a later instant in time, \(t=t_{1}\), heating is curtailed, and convective cooling is initiated. Cooling conditions are characterized by \(T_{\infty}=T_{\infty, 2}=T_{i}, h=h_{2}\) (a) Write the heat equation as well as the initial and boundary conditions in their dimensionless form for the heating phase (Phase 1). Express the equations in terms of the dimensionless quantities \(\theta^{*}, x^{*}, B i_{1}\), and \(F o\), where \(B i_{1}\) is expressed in terms of \(h_{1}\). (b) Write the heat equation as well as the initial and boundary conditions in their dimensionless form for the cooling phase (Phase 2). Express the equations in terms of the dimensionless quantities \(\theta^{*}, x^{*}, B i_{2}\), \(\mathrm{Fo}_{1}\), and \(\mathrm{Fo}\) where \(\mathrm{Fo}_{1}\) is the dimensionless time associated with \(t_{1}\), and \(B i_{2}\) is expressed in terms of \(h_{2}\). To be consistent with part (a), express the dimensionless temperature in terms of \(T_{\infty}=T_{\infty, 1^{*}}\) (c) Consider a case for which \(B i_{1}=10, B i_{2}=1\), and \(F o_{1}=0.1\). Using a finite-difference method with \(\Delta x^{*}=0.1\) and \(\Delta F o=0.001\), determine the transient thermal response of the surface \(\left(x^{*}=1\right)\), midplane \(\left(x^{*}=0\right)\), and quarter-plane \(\left(x^{*}=0.5\right)\) of the slab. Plot these three dimensionless temperatures as a function of dimensionless time over the range \(0 \leq F o \leq 0.5\). (d) Determine the minimum dimensionless temperature at the midplane of the wall, and the dimensionless time at which this minimum temperature is achieved.

In a tempering process, glass plate, which is initially at a uniform temperature \(T_{i}\), is cooled by suddenly reducing the temperature of both surfaces to \(T_{s}\). The plate is \(20 \mathrm{~mm}\) thick, and the glass has a thermal diffusivity of \(6 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). (a) How long will it take for the midplane temperature to achieve \(50 \%\) of its maximum possible temperature reduction? (b) If \(\left(T_{i}-T_{s}\right)=300^{\circ} \mathrm{C}\), what is the maximum temperature gradient in the glass at the time calculated in part (a)?
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Circular Motion and Gravitation

Read Explanation

Solid State Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Mechanics and Materials

Read Explanation

Physics of Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.