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How does the finite difference formulation of a transient heat conduction problem differ from that of a steady heat conduction problem? What does the term \(\rho A \Delta x c_{p}\left(T_{m}^{i+1}-T_{m}^{i}\right) / \Delta t\) represent in the transient finite difference formulation?

Short Answer

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Answer: The main difference between the finite difference formulation of transient heat conduction problems and steady heat conduction problems is the presence of a time-dependent term in transient problems. This term accounts for the change in temperature within an object over time, which is essential for solving transient problems. In contrast, steady heat conduction problems do not have this time-dependency.

Step by step solution

01

Steady vs Transient Conduction Problems

Let's first understand the difference between steady and transient heat conduction problems. A steady problem refers to a situation where the temperature distribution in an object does not change over time. On the other hand, a transient problem deals with the temperature change within an object over time.
02

Finite Difference Formulation for Steady Problems

The finite difference formulation for a steady heat conduction problem derives from the steady-state heat conduction equation, which can be expressed as: \(\frac{\partial}{\partial x}\left(k\frac{\partial T}{\partial x}\right) = 0\). In this case, the focus is on solving for temperature distribution within an object, and since it is a steady-state problem, there is no time-dependency.
03

Finite Difference Formulation for Transient Problems

For a transient heat conduction problem, the finite difference formulation derives from the unsteady heat conduction equation, which is expressed as: \(\rho c_{p}\frac{\partial T}{\partial t} = \frac{\partial}{\partial x}\left(k\frac{\partial T}{\partial x}\right)\). In comparison to the steady heat conduction equation, this form includes an unsteady term, being the first term on the left hand side, which represents the change in temperature with respect to time.
04

Understanding the Given Mathematical Term

The given term, \(\rho A \Delta x c_{p}\left(T_{m}^{i+1}-T_{m}^{i}\right) / \Delta t\), is from the finite difference formulation of a transient heat conduction problem. It represents the energy storage term in the transient heat conduction equation. Here, \(\rho\) is the density, \(A\) is the area, \(\Delta x\) is the spatial step size, \(c_{p}\) is the specific heat capacity, \(T_{m}^{i+1}\) and \(T_{m}^{i}\) are the nodal temperatures at time steps \(i+1\) and \(i\), and \(\Delta t\) is the time step size. This term essentially signifies how the temperature changes within an object over time due to energy storage in the object. In conclusion, the main difference between the finite difference formulation of transient heat conduction problems and a steady heat conduction problems is the presence of a time-dependent term. This term is responsible for considering the temperature change within an object over time and is essential for solving transient problems.

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Most popular questions from this chapter

A hot brass plate is having its upper surface cooled by an impinging jet of air at a temperature of \(15^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(220 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The 10 -cm-thick brass plate $\left(\rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\(, and \)\alpha=33.9 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}$ ) had a uniform initial temperature of \(650^{\circ} \mathrm{C}\), and the lower surface of the plate is insulated. Using a uniform nodal spacing of \(\Delta x=2.5 \mathrm{~cm}\), determine \((a)\) the explicit finite difference equations, \((b)\) the maximum allowable value of the time step, and \((c)\) the temperature at the center plane of the brass plate after 1 min of cooling, and (d) compare the result in (c) with the approximate analytical solution from Chap. \(4 .\)

Consider a rectangular metal block $(k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( of dimensions \)100 \mathrm{~cm} \times 75 \mathrm{~cm}$ subjected to a sinusoidal temperature variation at its top surface while its bottom surface is insulated. The two sides of the metal block are exposed to a convective environment at \(15^{\circ} \mathrm{C}\) and have a heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\). The sinusoidal temperature distribution at the top surface is given as $100 \sin (\pi x / L)\(. Using a uniform mesh size of \)\Delta x=\Delta y=25 \mathrm{~cm}$, determine \((a)\) finite difference equations and \((b)\) the nodal temperatures.

Consider a stainless steel spoon $(k=15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\varepsilon=0.6$ ) that is partially immersed in boiling water at \(100^{\circ} \mathrm{C}\) in a kitchen at \(32^{\circ} \mathrm{C}\). The handle of the spoon has a cross section of about $0.2 \mathrm{~cm} \times 1 \mathrm{~cm}\( and extends \)18 \mathrm{~cm}$ in the air from the free surface of the water. The spoon loses heat by convection to the ambient air with an average heat transfer coefficient of $h=13 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ as well as by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}=295 \mathrm{~K}\). Assuming steady one- dimensional heat transfer along the spoon and taking the nodal spacing to be \(3 \mathrm{~cm},(a)\) obtain the finite difference formulation for all nodes, \((b)\) determine the temperature of the tip of the spoon by solving those equations, and (c) determine the rate of heat transfer from the exposed surfaces of the spoon.

Quench hardening is a mechanical process in which the ferrous metals or alloys are first heated and then quickly cooled down to improve their physical properties and avoid phase transformation. Consider a $40-\mathrm{cm} \times 20-\mathrm{cm}\( block of copper alloy \)\left(k=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)$ being heated uniformly until it reaches a temperature of \(800^{\circ} \mathrm{C}\). It is then suddenly immersed into the water bath maintained at \(15^{\circ} \mathrm{C}\) with $h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ for quenching process. However, the upper surface of the metal is not submerged in the water and is exposed to air at $15^{\circ} \mathrm{C}\( with a convective heat transfer coefficient of \)10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Using an explicit finite difference formulation, calculate the temperature distribution in the copper alloy block after 10 min have elapsed using \(\Delta t=10 \mathrm{~s}\) and a uniform mesh size of \(\Delta x=\Delta y=10 \mathrm{~cm}\).

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