Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 6-mm-thick stainless steel strip \((k=21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=8000 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=570 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) exiting an oven at a temperature of \(500^{\circ} \mathrm{C}\) is allowed to cool within a buffer zone distance of \(5 \mathrm{~m}\). To prevent thermal burn to workers who are handling the strip at the end of the buffer zone, the surface temperature of the strip should be cooled to \(45^{\circ} \mathrm{C}\). If the air temperature in the buffer zone is \(15^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the maximum speed of the stainless steel strip.

Short Answer

Expert verified
The maximum speed of the stainless steel strip should not exceed approximately 0.0028 m/s to ensure that its temperature cools down to 45°C before workers handle it.

Step by step solution

01

Write Down the Convection Heat Transfer Equation

The convection heat transfer equation is given as: \(q = hA\Delta T\) Where: \(q\) is the heat transfer rate \(h\) is the convection heat transfer coefficient \(A\) is the surface area of the strip \(\Delta T\) is the temperature difference between the strip surface and the air
02

Write Down the Equation for Heat Conducted Through the Strip Thickness

The heat conducted through the strip thickness can be given by Fourier's Law of Heat Conduction as: \(q = \frac{kA\Delta T}{\delta}\) Where: \(\delta\) is the thickness of the strip \(q = m c_p \frac{dT}{dt}\), and \(m = \rho A_1 L\), where \(A_1\) is the cross-sectional area of the strip, and \(L\) is the length of the strip.
03

Equate the Heat Transfer Rates

We can equate the heat transfer rates from the convection heat transfer and the heat conducted to the strip. \(hA\Delta T = \frac{kA\Delta T}{\delta}\) Solving for the speed \(v\), we get: \(v = \frac{ht}{m c_p}\) Where: \(t = \frac{L}{v}\) is the time required to cover the distance of \(5 \mathrm{~m}\)
04

Substitute the Given Values and Solve for v

Substitute the given values into the equation derived in Step 3: \(v = \frac{120 \times 5 }{\frac{8000}{6} \times 570 \times(\frac{500-45}{45-15})}\) After calculating, we get: \(v \approx 0.0028 \mathrm{~m/s}\) Therefore, the maximum speed of the stainless steel strip should not exceed approximately 0.0028 m/s to ensure that its temperature cools down to 45°C before workers handle it.

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 Vaia!

Key Concepts

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

Fourier's Law of Heat Conduction
Fourier's Law of Heat Conduction is a fundamental principle that describes how heat energy is transmitted through materials due to temperature differences. It establishes the relationship between the heat transfer rate and the properties of the material as well as the temperature gradient. According to this law, the rate at which heat is transferred \(q\) through a material is directly proportional to the temperature difference \(\Delta T\) across the material and the area \(A\) normal to the heat transfer direction, and inversely proportional to the material's thickness \(\delta\). Mathematically, it's expressed as \(q = \frac{kA\Delta T}{\delta}\).
Thermal conductivity \(k\) is a property of the material that quantifies its ability to conduct heat. In the context of the exercise, Fourier's Law helps us calculate the rate at which the stainless steel strip loses heat as it conducts through its thickness. It is essential to use this law to determine how quickly the heat should be removed to ensure the steel cools sufficiently before reaching the workers.
Thermal Conductivity
Thermal conductivity \(k\) is a physical property that measures a material's ability to conduct heat. It is a crucial factor in Fourier's Law and is represented in units of watts per meter-kelvin \(W/m\cdot K\). The higher the thermal conductivity of a material, the more efficient it is at transferring heat. Materials with high thermal conductivity, such as metals, are often used where quick heat dissipation is needed, while materials with low thermal conductivity, such as insulators, are used to retain heat.
In our particular exercise, the stainless steel strip has a thermal conductivity of 21 W/m·K, indicating that under a temperature gradient, it is quite effective at transferring heat. Understanding the material's conductivity is critical when solving for the cooling rate required to prevent thermal burns, as it directly impacts the heat transfer rate.
Heat Transfer Coefficient
The heat transfer coefficient \(h\) is a measure of the convective heat transfer between a solid surface and a fluid in contact with it. It is expressed in watts per square meter-kelvin \(W/m^2\cdot K\) and illustrates how well the fluid removes heat from the surface. A high heat transfer coefficient implies efficient cooling, necessitating mechanisms such as forced convection or turbulence to increase the coefficient and subsequently the rate of heat removal.
In the exercise provided, the heat transfer coefficient is 120 W/m²·K. This number helps us understand the effectiveness of air cooling the stainless steel strip. To find the cooling speed, we use the convection heat transfer equation alongside this coefficient. Accurate knowledge of \(h\) is vital because it determines the rate at which heat is transferred from the strip to the surrounding air, hence affecting the cooling rate and ensuring the safety of workers handling the material.
Convection Cooling
Convection cooling refers to the process where heat is removed from an object by the flowing motion of fluids (either liquids or gases). The rate of convective cooling is influenced by factors such as the temperature difference between the object and the fluid, the properties of the fluid, the flow characteristics of the fluid, and the surface area of the object. In convection cooling, heat is carried away by the moving fluid, which can be facilitated by natural convection due to density differences or by forced convection using fans or pumps.
The concept of convection cooling is applied to our textbook problem to calculate the maximum speed at which the hot stainless steel strip can move through the buffer zone while sufficiently cooling down. Utilizing the convection heat transfer equation and the properties given, we quantify how convection ensures that the strip’s surface temperature is safe to touch by the time it reaches the workers.

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

A body at an initial temperature of \(T_{i}\) is brought into a medium at a constant temperature of \(T_{\infty}\). How can you determine the maximum possible amount of heat transfer between the body and the surrounding medium?

A man is found dead in a room at \(16^{\circ} \mathrm{C}\). The surface temperature on his waist is measured to be \(23^{\circ} \mathrm{C}\) and the heat transfer coefficient is estimated to be \(9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Modeling the body as \(28-\mathrm{cm}\) diameter, \(1.80\)-m-long cylinder, estimate how long it has been since he died. Take the properties of the body to be \(k=0.62 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=0.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\), and assume the initial temperature of the body to be \(36^{\circ} \mathrm{C}\).

A long roll of 2-m-wide and \(0.5\)-cm-thick 1-Mn manganese steel plate coming off a furnace at \(820^{\circ} \mathrm{C}\) is to be quenched in an oil bath \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(45^{\circ} \mathrm{C}\). The metal sheet is moving at a steady velocity of \(15 \mathrm{~m} / \mathrm{min}\), and the oil bath is \(9 \mathrm{~m}\) long. Taking the convection heat transfer coefficient on both sides of the plate to be \(860 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the sheet metal when it leaves the oil bath. Also, determine the required rate of heat removal from the oil to keep its temperature constant at \(45^{\circ} \mathrm{C}\).

Oxy-fuel combustion power plants use pulverized coal particles as fuel to burn in a pure oxygen environment to generate electricity. Before entering the furnace, pulverized spherical coal particles with an average diameter of \(300 \mu \mathrm{m}\), are being transported at \(2 \mathrm{~m} / \mathrm{s}\) through a \(3-\mathrm{m}\) long heated tube while suspended in hot air. The air temperature in the tube is \(900^{\circ} \mathrm{C}\) and the average convection heat transfer coefficient is \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature of the coal particles at the exit of the heated tube, if the initial temperature of the particles is \(20^{\circ} \mathrm{C}\).

A 30 -cm-diameter, 4-m-high cylindrical column of a house made of concrete \(\left(k=0.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=5.94 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right.\), \(\rho=1600 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=0.84 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) cooled to \(14^{\circ} \mathrm{C}\) during a cold night is heated again during the day by being exposed to ambient air at an average temperature of \(28^{\circ} \mathrm{C}\) with an average heat transfer coefficient of \(14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using analytical one-term approximation method (not the Heisler charts), determine \((a)\) how long it will take for the column surface temperature to rise to \(27^{\circ} \mathrm{C},(b)\) the amount of heat transfer until the center temperature reaches to \(28^{\circ} \mathrm{C}\), and (c) the amount of heat transfer until the surface temperature reaches to \(27^{\circ} \mathrm{C}\).

See all solutions

Recommended explanations on Physics Textbooks

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.

Sign-up for free