/*! 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 134 A support rod \(\left(k=15 \math... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(315 \mathrm{~K}\).

Short Answer

Expert verified
In this problem, we analyzed the thermal response of the rod during both heating and cooling processes. For the heating phase, we discretized the heat diffusion equation considering the convection heat transfer term and proposed a numerical solution. We compared the steady-state temperature distribution obtained numerically to that obtained analytically. For the cooling phase, we updated the convection coefficient expression, included the radiation heat transfer term, and implemented the updated numerical solution. We plotted the midspan temperature as a function of time and determined the time required for the rod to reach a safe-to-touch temperature of 315 K.

Step by step solution

01

Identify Governing Equations

The governing equation for heat transfer in a solid is the heat diffusion equation: \[\rho c_{p}\frac{\partial T}{\partial t} = k\frac{\partial^2 T}{\partial x^2}\] We will discretize the heat diffusion equation using the finite-difference method, taking into account the convection term at the exposed surface of the rod: \(Convection \thickspace term = h (T_\infty - T_surface)\) And propose a numerical solution.
02

Discretize and Implement Numerical Solution

To discretize the heat diffusion equation, we will subdivide the rod into small segments and develop a numerical solution. We will implement the numerical solution using an explicit scheme, in which the temperature at each time step is known and can be used to calculate the temperature at the next time step. The discretized equation will look like this: \[T_{i}^{n+1} = T_{i}^{n} + \frac{\Delta t}{\rho c_{p}}\left[k\frac{T_{i+1}^{n}-2 T_{i}^{n}+T_{i-1}^{n}}{(\Delta x)^{2}}-h_{i}(T_{i}^{n}-T_\infty)\right]\] The initial temperature of the rod is \(T_b\), so the temperatures at the first time step will be set to this value. Implement the numerical solution using a programming language or appropriate software.
03

Solve for Steady-State Analytical Model

To obtain the steady-state temperature distribution of the rod, we can solve the heat diffusion equation with zero time derivatives: \[\frac{d^2 T}{d x^2} - \frac{h}{k}(T_\infty - T) = 0\] We can solve this differential equation and compare the results with the steady-state temperatures obtained by our numerical solution. Now let's move to part (b):
04

Update Convection Coefficient and Implement Cooling Phenomena

After the rod reaches steady-state, the flow of hot gases is terminated, and the rod cools by free convection and radiation exchange. Update the convection coefficient expression provided in the problem: \(h = C \Delta T^n\) Also, calculate the radiation heat transfer term: \(q_r = \sigma \epsilon (T_{sur}^4 - T^4)\) Now, the discretized equation will look like this: \[T_{i}^{n+1} = T_{i}^{n} + \frac{\Delta t}{\rho c_{p}}\left[k\frac{T_{i+1}^{n}-2 T_{i}^{n}+T_{i-1}^{n}}{(\Delta x)^{2}}-h_{i}(T_{i}^{n}-T_\infty)-\sigma \epsilon (T_{i}^{n}^4-T_\infty^4)\right]\] Implement the updated numerical solution including the cooling phenomena.
05

Find the Time Required for the Rod to Reach Safe-to-Touch Temperature

Using the updated numerical solution, simulate the cooling process and monitor the midspan temperature as a function of time. To find the time required for the rod's midspan temperature to reach a safe-to-touch temperature of 315 K, record the time at which the midspan temperature drops to 315 K.
06

Analyze the Results

Compare the results obtained from both the heating and cooling processes. Plot the midspan temperature as a function of elapsed time for heating, compare the steady-state analytical and numerical models, and find the time required for the rod to reach a safe-to-touch temperature of 315 K.

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Key Concepts

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

Convection
Convection is a key mechanism of heat transfer where energy is transferred through the movement of fluids such as gases or liquids. In this exercise, the rod is exposed to a flow of hot gases, which transfers heat to the rod's surface. This process is characterized by the convection coefficient, denoted by \(h\), which quantifies the heat transferred per unit area per unit temperature difference between the fluid and the surface.
Understanding convection involves considering the type and speed of the fluid flow, as well as the properties of the fluid and the surface. For forced convection, like in this scenario, where the rod is subjected to moving gases, the heat transfer is influenced significantly by the velocity and nature of the gas flow.
This type of convection helps in rapidly increasing the rod's temperature from its initial state to approach the temperature of the flowing gas, making it essential to model this accurately in heat transfer problems.
Numerical Methods
Numerical methods are invaluable in solving complex heat transfer problems that are too arduous for analytical solutions. In this exercise, numerical methods were used to predict the temperature distribution in the rod over time. The finite-difference method, a popular numerical approach, discretizes the heat diffusion equation, allowing us to approximate and solve for temperature changes in small increments over time.
This involves dividing the rod into small segments and iteratively solving the temperature at each point and time step. In our solution, the explicit scheme was applied, where the new temperatures are calculated based on previously computed values.
Such numerical techniques offer flexibility in modeling various boundary conditions and are powerful in cases where the geometry or conditions complicate direct analytical approaches. Using programming or specialized software, these numerical simulations provide insights into transient and steady-state thermal behaviors.
Steady-State Analysis
Steady-state analysis examines the conditions where variables like temperature stop changing with time, achieving equilibrium. In the context of this exercise, the steady-state analysis involves the rod reaching a constant temperature distribution despite the continuous convective heating.
We used the differential form of the heat diffusion equation, simplified under the assumption that time-dependent changes become zero. This translates to solving a second-order differential equation without temporal factors. By comparing the results of this analytical approach with numerical simulations over long elapsed times, one can verify the numerical model's validity.
Steady-state conditions are critical in engineering practice as they represent predictable and controlled scenarios, making them particularly useful when designing systems subjected to continuous or prolonged thermal loads.

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Most popular questions from this chapter

Two large blocks of different materials, such as copper and concrete, have been sitting in a room \(\left(23^{\circ} \mathrm{C}\right)\) for a very long time. Which of the two blocks, if either, will feel colder to the touch? Assume the blocks to be semi-infinite solids and your hand to be at a temperature of \(37^{\circ} \mathrm{C}\).

Standards for firewalls may be based on their thermal response to a prescribed radiant heat flux. Consider a \(0.25\)-m-thick concrete wall \(\left(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is at an initial temperature of \(T_{i}=25^{\circ} \mathrm{C}\) and irradiated at one surface by lamps that provide a uniform heat flux of \(q_{s}^{\prime \prime}=10^{4} \mathrm{~W} / \mathrm{m}^{2}\). The absorptivity of the surface to the irradiation is \(\alpha_{s}=1.0\). If building code requirements dictate that the temperatures of the irradiated and back surfaces must not exceed \(325^{\circ} \mathrm{C}\) and \(25^{\circ} \mathrm{C}\), respectively, after \(30 \mathrm{~min}\) of heating, will the requirements be met?

A spherical vessel used as a reactor for producing pharmaceuticals has a 5 -mm-thick stainless steel wall \((k=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(D_{i}=1.0 \mathrm{~m}\). During production, the vessel is filled with reactants for which \(\rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), while exothermic reactions release energy at a volumetric rate of \(\dot{q}=10^{4} \mathrm{~W} / \mathrm{m}^{3}\). As first approximations, the reactants may be assumed to be well stirred and the thermal capacitance of the vessel may be neglected. (a) The exterior surface of the vessel is exposed to ambient air \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) for which a convection coefficient of \(h=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) may be assumed. If the initial temperature of the reactants is \(25^{\circ} \mathrm{C}\), what is the temperature of the reactants after \(5 \mathrm{~h}\) of process time? What is the corresponding temperature at the outer surface of the vessel? (b) Explore the effect of varying the convection coefficient on transient thermal conditions within the reactor. A spherical vessel used as a reactor for producing pharmaceuticals has a 5 -mm-thick stainless steel wall \((k=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(D_{i}=1.0 \mathrm{~m}\). During production, the vessel is filled with reactants for which \(\rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), while exothermic reactions release energy at a volumetric rate of \(\dot{q}=10^{4} \mathrm{~W} / \mathrm{m}^{3}\). As first approximations, the reactants may be assumed to be well stirred and the thermal capacitance of the vessel may be neglected. (a) The exterior surface of the vessel is exposed to ambient air \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) for which a convection coefficient of \(h=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) may be assumed. If the initial temperature of the reactants is \(25^{\circ} \mathrm{C}\), what is the temperature of the reactants after \(5 \mathrm{~h}\) of process time? What is the corresponding temperature at the outer surface of the vessel? (b) Explore the effect of varying the convection coefficient on transient thermal conditions within the reactor.

The density and specific heat of a particular material are known \(\left(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1250 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\), but its thermal conductivity is unknown. To determine the thermal conductivity, a long cylindrical specimen of diameter \(D=40 \mathrm{~mm}\) is machined, and a thermocouple is inserted through a small hole drilled along the centerline. The thermal conductivity is determined by performing an experiment in which the specimen is heated to a uniform temperature of \(T_{i}=100^{\circ} \mathrm{C}\) and then cooled by passing air at \(T_{\infty}=25^{\circ} \mathrm{C}\) in cross flow over the cylinder. For the prescribed air velocity, the convection coefficient is \(h=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If a centerline temperature of \(T(0, t)=40^{\circ} \mathrm{C}\) is recorded after \(t=1136 \mathrm{~s}\) of cooling, verify that the material has a thermal conductivity of \(k=0.30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (b) For air in cross flow over the cylinder, the prescribed value of \(h=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) corresponds to a velocity of \(V=6.8 \mathrm{~m} / \mathrm{s}\). If \(h=C V^{0.618}\), where the constant \(C\) has units of \(\mathrm{W} \cdot \mathrm{s}^{0.618} / \mathrm{m}^{2.618} \cdot \mathrm{K}\), how does the centerline temperature at \(t=1136 \mathrm{~s}\) vary with velocity for \(3 \leq V \leq 20 \mathrm{~m} / \mathrm{s}\) ? Determine the centerline temperature histories for \(0 \leq t \leq 1500 \mathrm{~s}\) and velocities of 3,10 , and \(20 \mathrm{~m} / \mathrm{s}\).

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 thermal diffusivity of \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), at what time will a surface temperature 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 period \(0 \leq t \leq 300 \mathrm{~s}\). In a similar manner, assess the effect of thermal conductivity by considering values of \(k=1.5,15\), and \(150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

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