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Chapter 5: Problem 110

A one-dimensional slab of thickness \(2 L\) is initially at a uniform temperature \(T_{i}\). Suddenly, electric current is passed through the slab causing uniform volumetric heating \(\dot{q}\left(\mathrm{~W} / \mathrm{m}^{3}\right)\). At the same time, both outer surfaces \((x=\pm L)\) are subjected to a convection process at \(T_{\infty}\) with a heat transfer coefficient \(h\). Write the finite-difference equation expressing conservation of energy for node 0 located on the outer surface at \(x=-L\). Rearrange your equation and identify any important dimensionless coefficients.

Short Answer

Expert verified
The finite-difference equation expressing the conservation of energy for node 0 located on the outer surface at \(x=-L\) is derived as: $$T_0 = \frac{\frac{T_1}{1+Bi} - Q\Delta x + Bi \cdot T_\infty}{1+Bi}$$ where, \(Bi\) - Biot number (\(h\Delta x/k\)), \(Q\) - Dimensionless heat generation (\(\dot{q}\Delta x^2/k\)), \(T_\infty\) - Ambient temperature.

Step by step solution

01

Understand the problem

In this problem, a one-dimensional slab is heated with a uniform volumetric heating and subjected to a convection process at both the surfaces. The goal is to apply the conservation of energy principle and derive the finite-difference equation for node 0 at x=-L.
02

Apply conservation of energy principle

The conservation of energy principle states that the net energy entering the control volume equals the net energy leaving it plus the accumulation. In this case, the control volume is node 0, which has heat being transferred into it due to convection, heat being transferred out from it into the neighboring node 1, and heat being generated inside it due to volumetric heating. Let \(T_0\) be the temperature of node 0, and \(T_1\) be the temperature of the neighboring node 1. In steady-state, there is no accumulation, and the conservation of energy principle can be written as: $$hA_s(T_\infty-T_0) + \dot{q}\Delta x = kA_c\frac{T_1-T_0}{\Delta x}$$ Here, \(h\) - Heat transfer coefficient, \(A_s\) - Surface area of the control volume, \(T_\infty\) - Ambient temperature, \(T_0\) - Temperature of node 0, \(\dot{q}\) - Uniform volumetric heating, \(\Delta x\) - Distance between node 0 and node 1, \(k\) - Thermal conductivity of the slab, \(A_c\) - Cross-sectional area of the control volume.
03

Rearrange the equation

Now, we rearrange the equation to express \(T_0\): $$hA_s(T_\infty-T_0) = kA_c\frac{T_1-T_0}{\Delta x} - \dot{q}\Delta x$$ $$T_0(hA_s + kA_c/\Delta x) = kA_c\frac{T_1}{\Delta x} - \dot{q}\Delta x + hA_sT_\infty$$ Finally, we will express \(T_0\) as: $$T_0 = \frac{kA_c\frac{T_1}{\Delta x} - \dot{q}\Delta x + hA_sT_\infty}{hA_s + kA_c/\Delta x}$$
04

Identify dimensionless coefficients

Now, we identify the dimensionless coefficients and express the equation in terms of those: Define the following dimensionless coefficients: 1. Biot number, \(Bi = h\Delta x/k\) 2. Fourier number, \(Fo = k\Delta t/(\rho c \Delta x^2)\) 3. Dimensionless heat generation, \(Q = \dot{q}\Delta x^2/k\) Now express the equation in terms of these dimensionless coefficients: $$T_0 = \frac{\frac{T_1}{1+Bi} - Q\Delta x + Bi \cdot T_\infty}{1+Bi}$$ Here, we have derived the finite-difference equation expressing the conservation of energy for node 0 located on the outer surface at \(x=-L\), considered dimensionless coefficients, and rearranged the equation.

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Key Concepts

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

Heat Transfer
Heat transfer is a crucial process in which thermal energy moves from one object to another. This movement happens due to a temperature difference between the objects. When considering a slab subjected to external environments, two main types of heat transfer occur: conduction and convection.

  • Conduction involves the transfer of heat through a material. It happens when molecules collide and pass energy to one another. In our slab example, heat moves from the inner part of the slab towards the surface.
  • Convection occurs when heat is transferred to a fluid (like air) flowing over a surface. For our slab, convection happens on the outer surfaces exposed to the surrounding air, which is cooler.

By understanding these two mechanisms, you can better analyze how energy flows within and from the slab.
Biot Number
The Biot number (\(Bi\)) is a dimensionless coefficient that helps assess the relative effectiveness of conduction and convection in a system. It is defined as the ratio of the convective heat transfer at the surface to the conductive heat transfer within the material.

By using the formula:\[Bi = \frac{h \Delta x}{k}\<\]

  • Where:

    • \(h\) is the heat transfer coefficient, showing how efficiently heat is transferred convectively.
    • \(\Delta x\) is the characteristic length, typically the thickness of the material.
    • \(k\) is the thermal conductivity, indicating how well the material conducts heat.

A low Biot number (\(< 0.1\)) implies efficient internal conduction compared to the convection outside. A high Biot number (> 1) suggests a significant temperature difference between the surface and the core, often requiring simplifications in energy assessments.
Volumetric Heating
Volumetric heating occurs when heat is generated uniformly throughout a material due to an internal energy source. In our example, this might happen from electrical energy being converted to heat inside the slab.

  • Effects of Volumetric Heating: It can raise the overall temperature of the slab, impacting both the internal and surface temperatures.
  • Mathematical Representation: Volumetric heating can be expressed as \( \dot{q} \) in W/m3, representing the rate of heat generated per unit volume.

This heating needs to be balanced with the heat loss through conduction and convection to maintain energy conservation.
Energy Conservation
Energy conservation in thermodynamics involves ensuring that energy entering, leaving, and stored in a system is balanced. For the slab example, this principle is applied to derive the finite-difference equation.

  • Key Components:

    • Energy In: Through convection from the surroundings when external temperature (\(T_\infty\)) is higher than slab temperature (\(T_0 \)
    • Energy Out: Conducted outwards to adjacent nodes, and convectively lost if the surface is warmer than its environment.
    • Energy Generated: Within the slab from volumetric heating (\( \dot{q} \Delta x\)

In a steady state, these processes are balanced, meaning no net energy accumulation occurs. Thus, the energy conservation equation tells us how temperature distribution stabilizes over time.

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Most popular questions from this chapter

A chip that is of length \(L=5 \mathrm{~mm}\) on a side and thickness \(t=1 \mathrm{~mm}\) is encased in a ceramic substrate, and its exposed surface is convectively cooled by a dielectric liquid for which \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=20^{\circ} \mathrm{C}\). In the off-mode the chip is in thermal equilibrium with the coolant \(\left(T_{i}=T_{\infty}\right)\). When the chip is energized, however, its temperature increases until a new steady state is established. For purposes of analysis, the energized chip is characterized by uniform volumetric heating with \(\dot{q}=9 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Assuming an infinite contact resistance between the chip and substrate and negligible conduction resistance within the chip, determine the steady-state chip temperature \(T_{f}\). Following activation of the chip, how long does it take to come within \(1^{\circ} \mathrm{C}\) of this temperature? The chip density and specific heat are \(\rho=2000 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=700 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

A microwave oven operates on the principle that application of a high- frequency field causes electrically polarized molecules in food to oscillate. The net effect is a nearly uniform generation of thermal energy within the food. Consider the process of cooking a slab of beef of thickness \(2 L\) in a microwave oven and compare it with cooking in a conventional oven, where each side of the slab is heated by radiation. In each case the meat is to be heated from \(0^{\circ} \mathrm{C}\) to a minimum temperature of \(90^{\circ} \mathrm{C}\). Base your comparison on a sketch of the temperature distribution at selected times for each of the cooking processes. In particular, consider the time \(t_{0}\) at which heating is initiated, a time \(t_{1}\) during the heating process, the time \(t_{2}\) corresponding to the conclusion of heating, and a time \(t_{3}\) well into the subsequent cooling process.

For each of the following cases, determine an appropriate characteristic length \(L_{c}\) and the corresponding Biot number \(B i\) that is associated with the transient thermal response of the solid object. State whether the lumped capacitance approximation is valid. If temperature information is not provided, evaluate properties at \(T=300 \mathrm{~K}\). (a) A toroidal shape of diameter \(D=50 \mathrm{~mm}\) and cross-sectional area \(A_{c}=5 \mathrm{~mm}^{2}\) is of thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The surface of the torus is exposed to a coolant corresponding to a convection coefficient of \(h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) A long, hot AISI 304 stainless steel bar of rectangular cross section has dimensions \(w=3 \mathrm{~mm}\), \(W=5 \mathrm{~mm}\), and \(L=100 \mathrm{~mm}\). The bar is subjected to a coolant that provides a heat transfer coefficient of \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at all exposed surfaces. (c) A long extruded aluminum (Alloy 2024) tube of inner and outer dimensions \(w=20 \mathrm{~mm}\) and \(W=24 \mathrm{~mm}\), respectively, is suddenly submerged in water, resulting in a convection coefficient of \(h=37 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the four exterior tube surfaces. The tube is plugged at both ends, trapping stagnant air inside the tube. (d) An \(L=300-m m\)-long solid stainless steel rod of diameter \(D=13 \mathrm{~mm}\) and mass \(M=0.328 \mathrm{~kg}\) is exposed to a convection coefficient of \(h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (e) A solid sphere of diameter \(D=12 \mathrm{~mm}\) and thermal conductivity \(k=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is suspended in a large vacuum oven with internal wall temperatures of \(T_{\text {sur }}=20^{\circ} \mathrm{C}\). The initial sphere temperature is \(T_{i}=100^{\circ} \mathrm{C}\), and its emissivity is \(\varepsilon=0.73\). (f) A long cylindrical rod of diameter \(D=20 \mathrm{~mm}\), density \(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\), specific heat \(c_{p}=1750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and thermal conductivity \(k=16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is suddenly exposed to convective conditions with \(T_{\infty}=20^{\circ} \mathrm{C}\). The rod is initially at a uniform temperature of \(T_{i}=200^{\circ} \mathrm{C}\) and reaches a spatially averaged temperature of \(T=100^{\circ} \mathrm{C}\) at \(t=225 \mathrm{~s}\). (g) Repeat part (f) but now consider a rod diameter of \(D=200 \mathrm{~mm}\).

The heat transfer coefficient for air flowing over a sphere is to be determined by observing the temperature-time history of a sphere fabricated from pure copper. The sphere, which is \(12.7 \mathrm{~mm}\) in diameter, is at \(66^{\circ} \mathrm{C}\) before it is inserted into an airstream having a temperature of \(27^{\circ} \mathrm{C}\). A thermocouple on the outer surface of the sphere indicates \(55^{\circ} \mathrm{C} 69 \mathrm{~s}\) after the sphere is inserted into the airstream. Assume and then justify that the sphere behaves as a spacewise isothermal object and calculate the heat transfer coefficient.

A thick steel slab \(\left(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, c=480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is initially at \(300^{\circ} \mathrm{C}\) and is cooled by water jets impinging on one of its surfaces. The temperature of the water is \(25^{\circ} \mathrm{C}\), and the jets maintain an extremely large, approximately uniform convection coefficient at the surface. Assuming that the surface is maintained at the temperature of the water throughout the cooling, how long will it take for the temperature to reach \(50^{\circ} \mathrm{C}\) at a distance of \(25 \mathrm{~mm}\) from the surface?
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