/*! 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 54 The one-dimensional system of ma... [FREE SOLUTION] | 91Ó°ÊÓ

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The one-dimensional system of mass \(M\) with constant properties and no internal heat generation shown in the figure is initially at a uniform temperature \(T_{i^{*}}\). The electrical heater is suddenly energized, providing a uniform heat flux \(q_{o}^{\prime \prime}\) at the surface \(x=0\). The boundaries at \(x=L\) and elsewhere are perfectly insulated. (a) Write the differential equation, and identify the boundary and initial conditions that could be used to determine the temperature as a function of position and time in the system. (b) On \(T-x\) coordinates, sketch the temperature distributions for the initial condition \((t \leq 0)\) and for several times after the heater is energized. Will a steady-state temperature distribution ever be reached? (c) On \(q_{x}^{\prime \prime}-t\) coordinates, sketch the heat flux \(q_{x}^{\prime \prime}(x, t)\) at the planes \(x=0, x=L / 2\), and \(x=L\) as a function of time. (d) After a period of time \(t_{e}\) has elapsed, the heater power is switched off. Assuming that the insulation is perfect, the system will eventually reach a final uniform temperature \(T_{f}\) Derive an expression that can be used to determine \(T_{f}\) as a function of the parameters \(q_{o}^{\prime \prime}, t_{e}, T_{i}\), and the system characteristics \(M, c_{p}\), and \(A_{s}\) (the heater surface area).

Short Answer

Expert verified
The temperature distribution T(x, t) can be determined by solving the differential equation \(\frac{\partial{T}}{\partial{t}} = \alpha \frac{\partial^2{T}}{\partial{x^2}}\) using the boundary conditions: \(-k \frac{\partial{T}}{\partial{x}}\Big|_{x=0} = q_o^{\prime \prime}\) and \(\frac{\partial{T}}{\partial{x}}\Big|_{x=L} = 0\), along with the initial condition \(T(x,0) = T_{i^*}\). The heat flux distribution can be deduced with \(q_x^{\prime \prime}(0, t) = q_o^{\prime \prime}\) and \(q_x^{\prime \prime}(L/2, t) = q_x^{\prime \prime}(L, t) = 0\). After switching off the heater, the system will reach a final uniform temperature \(T_f\), which can be found using an energy balance equation: \(T_f = T_i + \frac{q_o^{\prime \prime} A_s t_e}{M c_p}\).

Step by step solution

01

Differential equation

In this problem, since the system is one-dimensional, we can apply the One-Dimensional Unsteady-State Heat Conduction equation to find the temperature distribution. \( \frac{\partial{T}}{\partial{t}} = \alpha \frac{\partial^2{T}}{\partial{x^2}} \) where - T is the temperature - t is time - α is the thermal diffusivity
02

Boundary conditions

1. At \(x = 0\), the electrical heater provides a uniform heat flux \(q_o^{\prime \prime}\), which can be expressed as: \( -k \frac{\partial{T}}{\partial{x}}\Big|_{x=0} = q_o^{\prime \prime} \) 2. At \(x = L\) and elsewhere, the boundaries are perfectly insulated, which means no heat flux across the boundaries, therefore: \( \frac{\partial{T}}{\partial{x}}\Big|_{x=L} = 0 \)
03

Initial condition

Initially, the system is at a uniform temperature \(T_{i^*}\), which implies: \( T(x,0) = T_{i^*} \) #b) Temperature distribution# Since we have found the necessary differential equation, boundary, and initial conditions, we cannot directly sketch the temperature distribution for various time instances without numerically or analytically solving the governing equation. However, we can discuss qualitatively the nature of temperature distribution based on the given conditions.
04

Initial condition (t ≤ 0)

Before energizing the heater, the system is at a uniform temperature \(T_{i^*}\). So, the temperature distribution will be a horizontal line at \(T_{i^*}\) in the T-x plot.
05

After energizing the heater (t > 0)

After the heater is energized, the temperature at x = 0 will increase due to the heat flux from the heater. As time progresses, the temperature will also rise at other locations within the system. The temperature distribution will have a decreasing slope as we move away from x = 0. Finally, a steady-state temperature profile will be achieved when the temperature does not change with time at any location within the system. #c) Heat flux distribution#
06

At x = 0

At x = 0, there is a uniform heat flux \(q_o^{\prime \prime}\) provided by the electrical heater. This heat flux remains constant as a function of time, making a horizontal line in the \(q_x^{\prime \prime} - t\) plot.
07

At x = L/2 and x = L

At the insulated boundaries (x = L/2 and x = L), there will be no heat flux passing through. Thus, the heat flux will be 0 at these locations for all times, forming horizontal lines at \(q_x^{\prime \prime} = 0\) in the \(q_x^{\prime \prime} - t\) plot. #d) Final uniform temperature#
08

Energy balance

The key here is to perform an energy balance for the entire system. Suppose the heater power is switched off after the elapsed time \(t_e\), all the heat input will be entirely stored in the system, and the energy balance equation can be written as: \( \int_0^{t_e} q_o^{\prime \prime} A_s dt = M c_p (T_f - T_i) \)
09

Solve for final temperature Tf

By rearranging the above equation, we can find the expression for the final uniform temperature T_f: \(T_f = T_i + \frac{q_o^{\prime \prime} A_s t_e}{M c_p}\)

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Key Concepts

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

Differential Equation for Heat Transfer
Understanding the differential equation for heat transfer is crucial when dealing with unsteady-state heat conduction problems. The equation describes how heat diffuses through a material over time, and it forms the backbone for analyzing temperature distribution in various systems.

The key differential equation governing one-dimensional unsteady-state heat conduction is given by Fourier's second law of heat conduction:
\[\[\begin{align*}\frac{\partial{T}}{\partial{t}} = \alpha \frac{\partial^{2}{T}}{\partial{x^{2}}}\end{align*}\]\]In this equation,
  • \(T\) represents temperature,
  • \(t\) stands for time,
  • \(\alpha\) refers to the thermal diffusivity of the material—a measure of how quickly it can conduct heat relative to its ability to store heat.
Thermal diffusivity,\(\alpha\), is a property that combines the material's thermal conductivity, density, and specific heat capacity. This equation is a partial differential equation (PDE) because it relates changes in temperature to both time and space.When solving a heat transfer problem, it's important to work with the appropriate form of this equation specific to the given conditions. For instance, in the textbook problem provided, no internal heat generation is considered, and the system properties are assumed to be constant, which simplifies the equation to the form presented. If internal heat sources or varying material properties were at play, additional terms would be included in the PDE.
Boundary Conditions in Heat Transfer
Boundary conditions in heat transfer are essential components when solving the differential equation for temperature distribution. They describe the thermal behavior at the limits of the domain in which heat transfer is taking place. In the context of the textbook exercise, there are two key boundary conditions to consider:

1. At the heated surface, \(x = 0\), a given heat flux \(q_o^{\prime \prime}\) is applied, reflected mathematically as:\[\[\begin{align*}-k \frac{\partial{T}}{\partial{x}}\Big|_{x=0} = q_o^{\prime \prime}\end{align*}\]\]Here, \(-k\) is the negative of the thermal conductivity, which, when multiplied by the temperature gradient \(\frac{\partial{T}}{\partial{x}}\) at the surface, equals the known heat flux. This condition is known as a 'Neumann' or 'second type' boundary condition.

2. At the insulated boundaries \(x = L\) and beyond, there is no heat flow across the surface, yielding a zero heat flux condition:\[\[\begin{align*}\frac{\partial{T}}{\partial{x}}\Big|_{x=L} = 0\end{align*}\]\]This is also a form of the Neumann boundary condition. The zero-gradient condition reflects that an insulated boundary will not allow heat to escape, so any change in temperature across it must be zero.

Understanding and correctly applying boundary conditions are crucial for accurately determining the temperature field within the system. Incorrect boundary conditions can lead to incorrect solutions, so it is important to evaluate the physical situation carefully.
Initial Conditions in Heat Transfer
Initial conditions in heat transfer specify the temperature distribution within an object or system at the beginning of observation, before any heat transfer due to external influences has occurred. They are a snapshot of the thermal state at the starting point of analysis and play a vital role in solving time-dependent heat conduction problems.

In the example from the textbook, the entire system is initially at a uniform temperature \(T_{i^*}\). The mathematical expression of the initial condition is:\[\[\begin{align*}T(x,0) = T_{i^*}\end{align*}\]\]This condition is known as a 'Dirichlet' or 'first type' initial condition. It clearly defines the temperature at every point in the domain \(x\) at time \(t = 0\).

Initial conditions ensure that the solution to the heat conduction differential equation not only satisfies the spatial boundary conditions but also conforms to the known starting thermal state. When the heater in our example is turned on, the initial condition serves as the baseline from which the temperature changes over time are calculated. As the system evolves, this condition influences the transient behavior before reaching any potential steady state.

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Most popular questions from this chapter

One-dimensional, steady-state conduction with uniform internal energy generation occurs in a plane wall with a thickness of \(50 \mathrm{~mm}\) and a constant thermal conductivity of \(5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). For these conditions, the temperature distribution has the form \(T(x)=a+b x+c x^{2}\). The surface at \(x=0\) has a temperature of \(T(0) \equiv T_{o}=120^{\circ} \mathrm{C}\) and experiences convection with a fluid for which \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface at \(x=L\) is well insulated. (a) Applying an overall energy balance to the wall, calculate the volumetric energy generation rate \(\dot{q}\). (b) Determine the coefficients \(a, b\), and \(c\) by applying the boundary conditions to the prescribed temperature distribution. Use the results to calculate and plot the temperature distribution. (c) Consider conditions for which the convection coefficient is halved, but the volumetric energy generation rate remains unchanged. Determine the new values of \(a, b\), and \(c\), and use the results to plot the temperature distribution. Hint: recognize that \(T(0)\) is no longer \(120^{\circ} \mathrm{C}\). (d) Under conditions for which the volumetric energy generation rate is doubled, and the convection coefficient remains unchanged \(\left(h=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\), determine the new values of \(a, b\), and \(c\) and plot the corresponding temperature distribution. Referring to the results of parts (b), (c), and (d) as Cases 1, 2 , and 3, respectively, compare the temperature distributions for the three cases and discuss the effects of \(h\) and \(\dot{q}\) on the distributions.

An apparatus for measuring thermal conductivity employs an electrical heater sandwiched between two identical samples of diameter \(30 \mathrm{~mm}\) and length \(60 \mathrm{~mm}\), which are pressed between plates maintained at a uniform temperature \(T_{o}=77^{\circ} \mathrm{C}\) by a circulating fluid. A conducting grease is placed between all the surfaces to ensure good thermal contact. Differential thermocouples are imbedded in the samples with a spacing of \(15 \mathrm{~mm}\). The lateral sides of the samples are insulated to ensure onedimensional heat transfer through the samples. (a) With two samples of SS 316 in the apparatus, the heater draws \(0.353 \mathrm{~A}\) at \(100 \mathrm{~V}\), and the differential thermocouples indicate \(\Delta T_{1}=\Delta T_{2}=25.0^{\circ} \mathrm{C}\). What is the thermal conductivity of the stainless steel sample material? What is the average temperature of the samples? Compare your result with the thermal conductivity value reported for this material in Table A.1. (b) By mistake, an Armco iron sample is placed in the lower position of the apparatus with one of the SS316 samples from part (a) in the upper portion. For this situation, the heater draws \(0.601 \mathrm{~A}\) at \(100 \mathrm{~V}\), and the differential thermocouples indicate \(\Delta T_{1}=\Delta T_{2}=\) \(15.0^{\circ} \mathrm{C}\). What are the thermal conductivity and average temperature of the Armco iron sample? (c) What is the advantage in constructing the apparatus with two identical samples sandwiching the heater rather than with a single heater-sample combination? When would heat leakage out of the lateral surfaces of the samples become significant? Under what conditions would you expect \(\Delta T_{1} \neq \Delta T_{2} ?\)

Uniform internal heat generation at \(\dot{q}=5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\) is occurring in a cylindrical nuclear reactor fuel rod of 50 -mm diameter, and under steady-state conditions the temperature distribution is of the form \(T(r)=a+b r^{2}\), where \(T\) is in degrees Celsius and \(r\) is in meters, while \(a=800^{\circ} \mathrm{C}\) and \(b=-4.167 \times 10^{5}{ }^{\circ} \mathrm{C} / \mathrm{m}^{2}\). The fuel rod properties are \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (a) What is the rate of heat transfer per unit length of the rod at \(r=0\) (the centerline) and at \(r=25 \mathrm{~mm}\) (the surface)? (b) If the reactor power level is suddenly increased to \(\dot{q}_{2}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\), what is the initial time rate of temperature change at \(r=0\) and \(r=25 \mathrm{~mm}\) ?

A composite one-dimensional plane wall is of overall thickness \(2 L\). Material A spans the domain \(-L \leq x<0\) and experiences an exothermic chemical reaction leading to a uniform volumetric generation rate of \(\dot{q}_{\mathrm{A}}\). Material B spans the domain \(0 \leq x \leq L\) and undergoes an endothermic chemical reaction corresponding to a uniform volumetric generation rate of \(\dot{q}_{\mathrm{B}}=-\dot{q}_{\mathrm{A}}\). The surfaces at \(x=\pm L\) are insulated. Sketch the steady-state temperature and heat flux distributions \(T(x)\) and \(q_{\mathrm{x}}^{\prime \prime}(x)\), respectively, over the domain \(-L \leq x \leq L\) for \(k_{\mathrm{A}}=k_{\mathrm{B}}, k_{\mathrm{A}}=0.5 k_{\mathrm{B}}\), and \(k_{\mathrm{A}}=2 k_{\mathrm{B}}\). Point out the important features of the distributions you have drawn. If \(\dot{q}_{\mathrm{B}}=-2 \dot{q}_{\mathrm{A}}\), can you sketch the steady-state temperature distribution?

Assume steady-state, one-dimensional heat conduction through the symmetric shape shown. Assuming that there is no internal heat generation, derive an expression for the thermal conductivity \(k(x)\) for these conditions: \(A(x)=(1-x), \quad T(x)=300\) \(\left(1-2 x-x^{3}\right)\), and \(q=6000 \mathrm{~W}\), where \(A\) is in square meters, \(T\) in kelvins, and \(x\) in meters.

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