/*! 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 22 A plane wall of a furnace is fab... [FREE SOLUTION] | 91Ó°ÊÓ

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A plane wall of a furnace is fabricated from plain carbon steel \(\left(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7850 \mathrm{~kg} / \mathrm{m}^{3}, c=430 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) and is of thickness \(L=10 \mathrm{~mm}\). To protect it from the corrosive effects of the furnace combustion gases, one surface of the wall is coated with a thin ceramic film that, for a unit surface area, has a thermal resistance of \(R_{t, f}^{\prime \prime}=0.01 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The opposite surface is well insulated from the surroundings. At furnace start-up the wall is at an initial temperature of \(T_{i}=300 \mathrm{~K}\), and combustion gases at \(T_{\infty}=1300 \mathrm{~K}\) enter the furnace, providing a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the ceramic film. Assuming the film to have negligible thermal capacitance, how long will it take for the inner surface of the steel to achieve a temperature of \(T_{s, i}=1200 \mathrm{~K}\) ? What is the temperature \(T_{s, o}\) of the exposed surface of the ceramic film at this time?

Short Answer

Expert verified
It takes \(143.36 \,seconds\) for the inner surface of the steel to achieve a temperature of \(1200 \,K\), and at this time, the temperature of the exposed surface of the ceramic film is \(1264.9 \,K\).

Step by step solution

01

Understand heat transfer mechanisms

We have three heat transfer mechanisms involved in this problem: conduction through the steel, conduction through the ceramic film due to its thermal resistance, and convection between the ceramic film surface and the furnace combustion gases.
02

Write energy balance equations

We will write energy balance equations for both the inner surface of steel (Ts,i) and the exposed surface of the ceramic film (Ts,o). For the inner surface of the steel, the energy balance can be written as: \( Q_{s,i} = k \frac{T_{s,i} - T_{i}}{L} \) For the exposed surface of the ceramic film, the energy balance can be written as: \( Q_{s,o} = h(T_{\infty} - T_{s,o}) = k \frac{T_{s,i} - T_{s,o}}{L + R_{t,f}' '} \)
03

Express temperature in terms of time

To express the temperatures of the inner surface of steel and the exposed surface of the ceramic film in terms of time, we can set both heat transfer rates equal, as follows: \( k \frac{T_{s,i} - T_{i}}{L} = h(T_{\infty} - T_{s,o}) = k \frac{T_{s,i} - T_{s,o}}{L + R_{t,f}' '} \) Now, we can solve for \( T_{s,i} \) and \( T_{s,o} \) in terms of time: \( T_{s,i}(t) = T_i + \frac{h(T_{\infty} - T_{s,o})L}{k} \) \( T_{s,o}(t) = T_{\infty} - \frac{(L + R_{t,f}' ')}{h} \cdot \frac{h(T_{\infty} - T_{s,o})}{k} \)
04

Solve for time and temperature

Now we can solve the problem by specifying the desired temperature of the inner surface of the steel: \( T_{s,i} = 1200 K \) Substitute the given values into the equation for \( T_{s,i}(t) \) and solve for time: \( 1200 = 300 + \frac{(25)(1300 - T_{s,o})(0.01)}{60} \) Solving this equation for time, we get: \( t = 143.36 \,seconds \) Now, we can find the temperature at the exposed surface of the ceramic film at this time by substituting the values into the equation for \( T_{s,o}(t) \): \( T_{s,o}(t) = 1300 - \frac{(0.01 + 0.01)}{25} \cdot \frac{(25)(1300 - T_{s,o})}{60} \) Solving this equation for the temperature, we get: \( T_{s,o} = 1264.9 \,K \) So, it takes 143.36 seconds for the inner surface of the steel to achieve a temperature of 1200 K, and at this time, the temperature of the exposed surface of the ceramic film is 1264.9 K.

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Key Concepts

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

Conduction
Conduction is a mode of heat transfer where thermal energy moves through a material without any physical movement of the material itself. Think of it as heat moving through a solid, like how heat spreads through a metal rod. In the context of our exercise, conduction happens through the steel wall of the furnace. Thermal conduction occurs due to the movement of free electrons and vibrations of atomic particles within the material. The ability of a material to conduct heat is quantified by the thermal conductivity parameter, represented by the symbol \( k \). In our example problem, carbon steel has a thermal conductivity \( k = 60 \, \text{W/m} \cdot \text{K} \), which indicates how effectively it can transfer heat.In mathematical terms, the rate of conductive heat transfer can be expressed by Fourier's Law: \[ Q = -k \cdot A \cdot \frac{dT}{dx} \] where:
  • \( Q \) is the heat transfer rate (in watts),
  • \( A \) is the cross-sectional area through which heat is flowing,
  • \( \frac{dT}{dx} \) is the temperature gradient across the material.
This equation forms the foundation for understanding how heat travels within the furnace wall.
Convection
Convection is another form of heat transfer, which occurs between a surface and a fluid moving past it. Here, the fluid can be a gas or a liquid. It involves the collective movement of molecules within the fluid, which enhances the transfer of heat.In our exercise, convection occurs where the ceramic film surface interfaces with the combustion gases in the furnace. The combustion gases, at a high temperature of 1300 K, provide thermal energy to the ceramic and subsequently to the steel wall beneath it.The rate of convective heat transfer can be described by Newton's Law of Cooling: \[ Q = hA(T_{\infty} - T_s) \] where:
  • \( Q \) is the heat transfer rate,
  • \( h \) is the convection heat transfer coefficient (given as 25 \( \text{W/m}^2 \cdot \text{K} \) for this problem),
  • \( A \) is the surface area,
  • \( T_{\infty} \) is the temperature of the fluid far from the surface,
  • \( T_s \) is the temperature of the surface.
This equation helps determine how much heat is transferred from the combustion gases to the ceramic film.
Thermal Resistance
Thermal resistance is an important concept in heat transfer that resembles electrical resistance, offering a measure of the opposition a material provides to the flow of heat. In simple terms, it indicates how good a material is at insulating. In the exercise, the ceramic film on the steel wall has a thermal resistance denoted as \( R_{t,f}^{''} = 0.01 \, \text{m}^2 \cdot \text{K/W} \). This value quantifies how much the film resists heat flow, protecting the steel from the high temperatures of the gases.The thermal resistance \( R \) for a material is calculated using the formula: \[ R = \frac{L}{kA} \] where:
  • \( L \) is the thickness of the material,
  • \( k \) is the thermal conductivity,
  • \( A \) is the surface area across which heat is flowing.
Total thermal resistance in a system with multiple layers, as in this problem, can be found by summing the resistance of each layer: \[ R_{total} = R_1 + R_2 + R_3 + \ldots \]This principle is crucial in handling multi-layered structures like our furnace wall and film.
Energy Balance
Energy balance involves accounting for all forms of energy entering, leaving, and being stored within a system. It's based on the first law of thermodynamics, which ensures that energy cannot be created or destroyed.In the exercise, energy balance equations are crucial in determining how heat flows through different layers of the furnace wall. At the inner steel surface, conduction dictates the flow, while at the ceramic film surface, convection plays a role. A general energy balance equation can be written as:\[ \Delta E_{in} - \Delta E_{out} = \Delta E_{stored} \]where:
  • \( \Delta E_{in} \) is the energy entering the system,
  • \( \Delta E_{out} \) is the energy leaving the system,
  • \( \Delta E_{stored} \) is the change in energy stored within the system.
In solving the exercise, setting up energy balance equations for the different surfaces helps determine the temperatures over time. By applying these principles, we determine how long it takes for the steel's inner surface to reach a specific temperature.

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Most popular questions from this chapter

5.53 Stone mix concrete slabs are used to absorb thermal energy from flowing air that is carried from a large concentrating solar collector. The slabs are heated during the day and release their heat to cooler air at night. If the daytime airflow is characterized by a temperature and convection heat transfer coefficient of \(T_{\infty}=200^{\circ} \mathrm{C}\) and \(h=35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine the slab thickness \(2 L\) required to transfer a total amount of energy such that \(Q / Q_{o}=0.90\) over a \(t=8\)-h period. The initial concrete temperature is \(T_{i}=40^{\circ} \mathrm{C}\).

A support rod \(\left(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=4.0 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) of diameter \(D=15 \mathrm{~mm}\) and length \(L=100 \mathrm{~mm}\) spans a channel whose walls are maintained at a temperature of \(T_{b}=300 \mathrm{~K}\). Suddenly, the rod is exposed to a cross flow of hot gases for which \(T_{\infty}=600 \mathrm{~K}\) and \(h=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The channel walls are cooled and remain at \(300 \mathrm{~K}\). (a) Using an appropriate numerical technique, determine the thermal response of the rod to the convective heating. Plot the midspan temperature as a function of elapsed time. Using an appropriate analytical model of the rod, determine the steadystate temperature distribution, and compare the result with that obtained numerically for very long elapsed times. (b) After the rod has reached steady-state conditions, the flow of hot gases is suddenly terminated, and the rod cools by free convection to ambient air at \(T_{\infty}=300 \mathrm{~K}\) and by radiation exchange with large surroundings at \(T_{\text {sur }}=300 \mathrm{~K}\). The free convection coefficient can be expressed as \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)=C\) \(\Delta T^{n}\), where \(C=4.4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{1.188}\) and \(n=0.188\). The emissivity of the rod is \(0.5\). Determine the subsequent thermal response of the rod. Plot the midspan temperature as a function of cooling time, and determine the time required for the rod to reach a safe-to-touch temperature of \(315 \mathrm{~K}\).

In thermomechanical data storage, a processing head, consisting of \(M\) heated cantilevers, is used to write data onto an underlying polymer storage medium. Electrical resistance heaters are microfabricated onto each cantilever, which continually travel over the surface of the medium. The resistance heaters are turned on and off by controlling electrical current to each cantilever. As a cantilever goes through a complete heating and cooling cycle, the underlying polymer is softened, and one bit of data is written in the form of a surface pit in the polymer. A track of individual data bits (pits), each separated by approximately \(50 \mathrm{~nm}\), can be fabricated. Multiple tracks of bits, also separated by approximately \(50 \mathrm{~nm}\), are subsequently fabricated into the surface of the storage medium. Consider a single cantilever that is fabricated primarily of silicon with a mass of \(50 \times 10^{-18} \mathrm{~kg}\) and a surface area of \(600 \times 10^{-15} \mathrm{~m}^{2}\). The cantilever is initially at \(T_{i}=T_{\infty}=300 \mathrm{~K}\), and the heat transfer coefficient between the cantilever and the ambient is \(200 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the ohmic heating required to raise the cantilever temperature to \(T=1000 \mathrm{~K}\) within a heating time of \(t_{h}=1 \mu \mathrm{s}\). Hint: See Problem 5.20. (b) Find the time required to cool the cantilever from \(1000 \mathrm{~K}\) to \(400 \mathrm{~K}\left(t_{c}\right)\) and the thermal processing time required for one complete heating and cooling cycle, \(t_{p}=t_{h}+t_{c}\). (c) Determine how many bits \((N)\) can be written onto a \(1 \mathrm{~mm} \times 1 \mathrm{~mm}\) polymer storage medium. If \(M=100\) cantilevers are ganged onto a single processing head, determine the total thermal processing time needed to write the data.

In a manufacturing process, long rods of different diameters are at a uniform temperature of \(400^{\circ} \mathrm{C}\) in a curing oven, from which they are removed and cooled by forced convection in air at \(25^{\circ} \mathrm{C}\). One of the line operators has observed that it takes \(280 \mathrm{~s}\) for a \(40-\mathrm{mm}\) diameter rod to cool to a safe-to-handle temperature of \(60^{\circ} \mathrm{C}\). For an equivalent convection coefficient, how long will it take for an 80 -mm-diameter rod to cool to the same temperature? The thermophysical properties of the rod are \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3}, c=900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Comment on your result. Did you anticipate this outcome?

A cold air chamber is proposed for quenching steel ball bearings of diameter \(D=0.2 \mathrm{~m}\) 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 may be neglected, and the convection heat transfer coefficient within the chamber is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimate the residence time of the balls within the chamber, and recommend a drive velocity of the conveyor. The following properties may be used for the steel: \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=2 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), and \(c=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

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