Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 57

Starting with an energy balance on a volume element, obtain the steady two- dimensional finite difference equation for a general interior node in rectangular coordinates for \(T(x, y)\) for the case of variable thermal conductivity and uniform heat generation.

Short Answer

Expert verified
Based on the derived finite difference equation for a general interior node in rectangular coordinates for the temperature T(x, y), provide a short answer discussing the contributions of various terms. The finite difference equation for a general interior node with variable thermal conductivity and uniform heat generation is: \(k_E T_{i+1,j} + k_W T_{i-1,j} + k_N T_{i,j+1} + k_S T_{i,j-1} - (k_E + k_W + k_N + k_S) T_{i,j}= -g\Delta x^2\Delta y^2\) The contributions of various terms in this equation are: 1. \(k_E T_{i+1,j}\): Heat entering from the right neighbor (east direction) due to T(x+Δx, y) 2. \(k_W T_{i-1,j}\): Heat entering from the left neighbor (west direction) due to T(x-Δx, y) 3. \(k_N T_{i,j+1}\): Heat entering from the top neighbor (north direction) due to T(x, y+Δy) 4. \(k_S T_{i,j-1}\): Heat entering from the bottom neighbor (south direction) due to T(x, y-Δy) 5. \(-(k_E + k_W + k_N + k_S) T_{i,j}\): The total heat leaving from the interior node due to various conductivities in all directions 6. \(-g\Delta x^2\Delta y^2\): Contribution from the uniform heat generation within the volume element The equation balances the heat transfer between the neighbors and the internal node, accounting for variable thermal conductivity and uniform heat generation, providing a numerical method to solve for temperatures in a two-dimensional domain.

Step by step solution

01

Apply the energy balance equation on a volume element

Considering a rectangular volume element of dimensions \(\Delta x\), \(\Delta y\), and \(\Delta z\), the energy balance equation states that the heat entering the volume minus the heat leaving the volume plus the heat generated within the volume equals the change in internal energy of the volume. But, since we are dealing with steady-state, the internal energy does not change, we then have: Heat entering - Heat leaving + Heat generated = 0
02

Determine the heat flux in x and y direction

To find the heat flux in x and y direction, we'll use Fourier's law of heat conduction. In general, the heat flux components are given as: \(q_x = -k\frac{\partial T}{\partial x}\) and \(q_y = -k\frac{\partial T}{\partial y}\) where \(k\) represents the variable thermal conductivity, \(T\) denotes Temperature, and \(q_x\) and \(q_y\) refer to the heat flux in x and y directions.
03

Calculate heat flux entering and leaving

Now, let's calculate the heat flux entering and leaving the volume element in both x and y directions. Due to variable thermal conductivity, we'll take an average value of neighboring conductivities (\(k_E\), \(k_W\), \(k_N\), and \(k_S\)) in each direction: Heat entering in x-direction: \(q_{x_{in}} = -k_W\frac{T_{i-1,j}-T_{i,j}}{\Delta x}\) Heat leaving in x-direction: \(q_{x_{out}} = -k_E\frac{T_{i+1,j}-T_{i,j}}{\Delta x}\) Heat entering in y-direction: \(q_{y_{in}} = -k_S\frac{T_{i,j-1}-T_{i,j}}{\Delta y}\) Heat leaving in y-direction: \(q_{y_{out}} = -k_N\frac{T_{i,j+1}-T_{i,j}}{\Delta y}\)
04

Determine the heat generated in the volume element

Let \(g\) be the uniform heat generation rate (energy generated per unit volume). The total heat generated in the volume element is given by the product of the heat generation rate and the volume (\(\Delta x \cdot \Delta y \cdot \Delta z\)): Heat generated = \(g\Delta x \cdot \Delta y \cdot \Delta z\)
05

Apply energy balance and obtain the finite difference equation

We are now ready to apply energy balance. By substituting the heat entering, leaving, and generated, we get: \(\left(-(- k_W\frac{T_{i-1,j}-T_{i,j}}{\Delta x} + k_E\frac{T_{i+1,j}-T_{i,j}}{\Delta x})\right)\Delta x\Delta z +\left(-(-k_S\frac{T_{i,j-1}-T_{i,j}}{\Delta y} +k_N\frac{T_{i,j+1}-T_{i,j}}{\Delta y})\right)\Delta y\Delta z + g\Delta x \cdot \Delta y \cdot \Delta z = 0\) After canceling the common terms and rearranging the terms, the resulting finite difference equation for a general interior node in rectangular coordinates for \(T(x, y)\) is: \(k_E T_{i+1,j} + k_W T_{i-1,j} + k_N T_{i,j+1} + k_S T_{i,j-1} - (k_E + k_W + k_N + k_S) T_{i,j}= -g\Delta x^2\Delta y^2\) This is the desired finite difference equation for the given problem with variable thermal conductivity and uniform heat generation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A hot brass plate is having its upper surface cooled by impinging jet of air at temperature of \(15^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(220 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The \(10-\mathrm{cm}-\) thick brass plate \(\left(\rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=\right.\) \(110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=33.9 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) 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, \((c)\) the temperature at the center plane of the brass plate after 1 minute of cooling, and \((d)\) compare the result in \((c)\) with the approximate analytical solution from Chapter 4 .

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 \(95^{\circ} \mathrm{C}\) in a kitchen at \(25^{\circ} \mathrm{C}\). 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.

Consider one-dimensional transient heat conduction in a plane wall with variable heat generation and variable thermal conductivity. The nodal network of the medium consists of nodes 0,1 , and 2 with a uniform nodal spacing of \(\Delta x\). Using the energy balance approach, obtain the explicit finite difference formulation of this problem for the case of specified heat flux \(\dot{q}_{0}\) and convection at the left boundary (node 0 ) with a convection coefficient of \(h\) and ambient temperature of \(T_{\infty}\), and radiation at the right boundary (node 2 ) with an emissivity of \(\varepsilon\) and surrounding temperature of \(T_{\text {surr }}\).

What is a practical way of checking if the round-off error has been significant in calculations?

In the energy balance formulation of the finite difference method, it is recommended that all heat transfer at the boundaries of the volume element be assumed to be into the volume element even for steady heat conduction. Is this a valid recommendation even though it seems to violate the conservation of energy principle?
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Measurements

Read Explanation

Classical Mechanics

Read Explanation

Magnetism

Read Explanation

Waves Physics

Read Explanation

Solid State Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free