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 5-mm-thick stainless steel strip \((k=21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=\) \(8000 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=570 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is being heat treated as it moves through a furnace at a speed of \(1 \mathrm{~cm} / \mathrm{s}\). The air temperature in the furnace is maintained at \(900^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the furnace length is \(3 \mathrm{~m}\) and the stainless steel strip enters it at \(20^{\circ} \mathrm{C}\), determine the temperature of the strip as it exits the furnace.

Short Answer

Expert verified
In the given problem, an equation for energy balance in the form of an ordinary differential equation (ODE) was derived: \(\frac{dT}{dt} = \frac{k}{\rho W v c_p}\frac{dT}{dx} - \frac{80}{\rho W v c_p}(900 - T(x))\) Here, the temperature, \(T(x)\), varies along the length of the furnace and is subjected to the initial condition \(T(0) = 20^{\circ}\text{C}\). Calculate the temperature of the stainless steel strip as it exits the furnace (\(T(L)\)) when the length of the furnace, \(L\), is 3 meters.

Step by step solution

01

Calculate the heat transfer rate to the strip

We can calculate the heat transfer rate to the strip per unit area using Newton's Law of Cooling: \(q = h\times (T_\text{air} - T)\) where \(T\) is the strip temperature. We are given \(h = 80 \, \text{W/m}^2 \cdot \text{K}\) and \(T_\text{air} = 900^{\circ}\text{C}\). The strip temperature \(T\) is a function of the position along the furnace, so we can rewrite the equation as: \(q(x) = 80 \times (900 - T(x))\)
02

Calculate the heat transfer rate within the strip

We can calculate the heat transfer rate within the steel strip using the heat conduction equation: \(q(x) = -k \frac{dT}{dx}\) We are given \(k = 21\,\text{W/m}\cdot\text{K}\), and we need to find the derivative \(dT/dx\).
03

Apply energy conservation

The strip is gaining heat from the surrounding air (at the rate \(q(x)\)) and it's transferring the heat internally. As the strip is moving at a constant speed, the net heat transfer rate to the strip must equal the rate at which its internal energy increases. The internal energy increase rate can be calculated using the strip mass flow rate, specific heat, and the temperature change. The mass flow rate of the strip is given by: \(\dot{m} = \rho \times A \times v\), where \(\rho = 8000 \, \text{kg/m}^3\) is the density, \(A\) is the cross-sectional area, and \(v = 0.01 \, \text{m/s}\) is the speed. The internal energy increase rate per unit width of the strip is given by: \(\dot{m}c_p\frac{dT}{dt} = \rho Wv c_p \frac{dT}{dt}\), where \(W\) is the width of the strip. Considering energy conservation, we can equate the heat transfer rate to the internal energy increase rate per unit width as follows: \(\rho Wv c_p \frac{dT}{dt} = -k \frac{dT}{dx} + 80 \times (900 - T(x))\)
04

Solve the energy balance for the temperature

The above energy balance equation is a first-order ordinary differential equation (ODE) in the form: \(\frac{dT}{dt} = \frac{k}{\rho W v c_p}\frac{dT}{dx} - \frac{80}{\rho W v c_p}(900 - T(x))\) Now, we can solve this ODE subject to the initial condition \(T(0)=20^{\circ}\text{C}\). Finally, we can find the value of \(T(x)\) at \(x=L = 3\,\text{m}\) to get the temperature of the strip as it exits the furnace. This equation can be solved either analytically or using numerical methods (e.g., using Python or MATLAB). In either case, you will find the exit temperature of the strip.

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.

Newton's Law of Cooling
Newton's Law of Cooling is fundamental to understanding how an object exchanges heat with its surroundings. Newton's Law asserts that the rate at which an object coalesces or dissipates heat is proportional to the difference in temperature between the object and its environment. In mathematical terms, the heat transfer rate, denoted by q, can be expressed as:

\[ q = h \times (T_{\text{air}} - T) \]

where h is the convection heat transfer coefficient, T_{\text{air}} is the ambient temperature, and T is the temperature of the object—in this case, the stainless steel strip. This principle is crucial when analyzing heat treatment processes, such as the exercise scenario where a stainless steel strip is heated as it moves through a furnace. By estimating the heat transfer rate with Newton's Law, we can further delve into the specifics of the material's temperature change during the process. This law is particularly useful when the system reaches a quasi-steady state, where the temperature gradient is stable.
Heat Conduction Equation
The heat conduction equation is central to studying how heat moves within a solid material. This equation, derived from Fourier's law of heat conduction, tells us how thermal energy is transported due to temperature gradients within an object. For a one-dimensional case, the heat conduction equation is usually expressed as:

\[ q(x) = -k \frac{dT}{dx} \]

where q(x) is the local heat flux, k is the thermal conductivity of the material, and \( \frac{dT}{dx} \) represents the temperature gradient in the material. It's important to note that a negative sign is present because heat flows from high to low temperature, which means that heat flux is in the opposite direction of the increasing temperature. For the stainless steel strip in our exercise, we calculate this heat transfer rate within the strip itself to establish how heat travels from the surface toward the strip's center. Understanding this concept allows us to model the temperature distribution across the thickness of the strip accurately.
Energy Conservation in Heat Transfer
The principle of energy conservation in heat transfer is an encompassing concept that ensures the energy balance within a system. It implies that all the heat entering a system must be accounted for either through storage within the system or by heat leaving the system. Applied to our stainless steel strip, it means that the heat absorbed from the furnace air must equal the increase in internal energy of the strip as it travels through the furnace. Mathematically, we can illustrate this principle with the equation:

\[ \rho Wv c_p \frac{dT}{dt} = -k \frac{dT}{dx} + h \times (T_{\text{air}} - T(x)) \]

The left-hand side of the equation details the energy rate increase within the strip, incorporating the mass flow rate, the specific heat (\(c_p\)), and the temperature change rate of the strip. The right-hand side shows the sum of heat conducted within the strip and the heat transferred from the air. This balance must be maintained for the strip's temperature to change predictably, enabling the calculation of the strip's exit temperature from the furnace. Maintaining energy conservation allows us to define the relationship between the temperature distribution, the properties of the material, and the furnace's environmental conditions.

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 6-cm-diameter 13-cm-high canned drink ( \(\rho=\) \(\left.977 \mathrm{~kg} / \mathrm{m}^{3}, k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) 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 \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long it will take for the drink to \(\operatorname{cool}\) 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) \(6.7 \mathrm{~min}\)

Carbon steel balls \(\left(\rho=7833 \mathrm{~kg} / \mathrm{m}^{3}, k=54 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\), \(c_{p}=0.465 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and \(\left.\alpha=1.474 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right) 8 \mathrm{~mm}\) in diameter are annealed by heating them first to \(900^{\circ} \mathrm{C}\) in a furnace and then allowing them to cool slowly to \(100^{\circ} \mathrm{C}\) in ambient air at \(35^{\circ} \mathrm{C}\). If the average heat transfer coefficient is \(75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long the annealing process will take. If 2500 balls are to be annealed per hour, determine the total rate of heat transfer from the balls to the ambient air.

Consider the freezing of packaged meat in boxes with refrigerated air. How do \((a)\) the temperature of air, (b) the velocity of air, \((c)\) the capacity of the refrigeration system, and \((d)\) the size of the meat boxes affect the freezing time?

Why are the transient temperature charts prepared using nondimensionalized quantities such as the Biot and Fourier numbers instead of the actual variables such as thermal conductivity and time?

How can we use the transient temperature charts when the surface temperature of the geometry is specified instead of the temperature of the surrounding medium and the convection heat transfer coefficient?

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