The document is a project proposal report for analyzing heat transfer through a fire clay brick. It includes: 1) An introduction describing the furnace made of fire clay brick and the three analysis methods that will be used: one-dimensional heat transfer by conduction, flux plot analysis, and finite difference method. 2) Details of each analysis method and the assumptions made, including developing the heat transfer equation, constructing flux plots, and using a grid of 25 nodes for the finite difference method. 3) Results of applying the methods, including temperature distributions, heat flux, and heat rate calculations from the one-dimensional conduction analysis and Autodesk simulations. 4) A summary stating the different analysis methods were

1. Project Proposal Report – Andres Ramos, Michael Frazier,Shawn Robinson/Fire Clay Brick Spring2014
Texas A&M University Corpus Christi – Heat Transfer Analysisof Fire Clay Brick Page 1 of20
Heat Transfer Project Report
Heat Transfer Analysis of Fire Clay Brick
Presented to
Dr. A. Conkey
Prepared by Team Bricksquad
Prepared by: Andres Ramos, Michael Frazier, Shawn Robinson
Texas A&M University Corpus Christi
Mechanical Engineering
MEEN 3345
04-25-14

2. Project Proposal Report – Andres Ramos, Michael Frazier,Shawn Robinson/Fire Clay Brick Spring2014
Texas A&M University Corpus Christi – Heat Transfer Analysisof Fire Clay Brick Page 2 of20
Heat Analysis Project
INTRODUCTION .........................................................ERROR! BOOKMARKNOT DEFINED.
METHODS OF ANALYSIS TO APPLY.......................... ERROR! BOOKMARKNOT DEFINED.,5
APPLICATION OF METHODS................................ ERROR! BOOKMARKNOT DEFINED.,6,7,8
SUMMARY………………………………………………………………………………….8
APPENDIX…………………………………………………………………………………..9
A: Listof Abbreviations……………………………………………………………………………………9
B: Annotated Bibliography/References……………………………………………………………………9
C: Calculations…………………………………………………………………………………………….10

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Introduction
The device under study is an industrial furnace, and it is used for heating. The furnace
is made of fireclay brick having a thermal conductivity of 1.7 W/m*K. There are many
applications with industrial furnaces such as boilers, refineries, heaters, and chemical plants.
The three methods of analysis to apply for the heat transfer of the fire clay brick will be one-
dimensional heat transfer by conduction, flux plot, and finite difference. Also, an Auto
Inventor program will show a 3d model of the part, a dimensional drawing, and relevant
figures from the FE analysis (See Figure 2). The analysis will show the temperature
distribution, heat flux, and the heat rate at given lengths. The dimensions of the slab are
shown below in Figure 1, the calculations are shown in Appendix C-Calculations,and the
results are displayed in Table 1. All of the calculations for a, b, and F(x) are shown in
Appendix C-Calculations.
Table 1: Parameters that were calculated fromthe slab
Figure 1 : Dimensions to be determined of slab

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Figure 2: 3-d Model with dimensions
Present Analysis to be Conducted:
The team will conduct a one-dimensional conduction exact analysis, flux plot, and a finite
difference to determine the temperature, heat rate, and heat flux along a reference line of the
element. The one dimensional conduction analysis will be applied to develop the differential
equation to determine the temperature across the slab. Assumptions associated with the one
dimensional conduction analysis for the fire clay brick is steady-state conditions, one-
dimensional conduction through wall, and constant thermal conductivity. The equation used
to determine the heat flux, heat rate, and temperature at different points is located in Part C of
the Appendix under Fourier’s law.
Flux plot is another technique used. Flux plots are used as a graphical approach rather
than analytical or numerical. Flux plots require isotherms and heat flow line, and provides an
estimate of the rate of heat flow. The technique constructs perpendicular isotherms (same
temperature lines) to the heat flow lines to produce a network of squares. The temperature
distribution will be determined by solving the heat equation which is shown in the Appendix
table part C. Constructing the flux plot consist of drawing sets of isothermal lines
perpendicular to the adiabatic (by symmetry) top and bottom surfaces,and once the

5. Project Proposal Report – Andres Ramos, Michael Frazier,Shawn Robinson/Fire Clay Brick Spring2014
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isothermal lines are drawn the heat lines are drawn perpendicular showing the heat transfer
from the hotter surface to the cooler surface.
The final method the team used to determine the temperature, heat flux, and heat rate was
the finite difference method. The team broke the slab into small rectangles with a grid of 25
nodes to get specific temperatures at specific points. The finite difference method is a
numerical technique, and the benefit of a numerical technique is it can be extended into a
three dimensional problem. The team will determine the temperature distribution through
different nodes both by hand calculations and by simulating steady-state one dimensional
conduction using Autodesk Inventor and Autodesk Multiphysics Simulation.
By conducting the different methods of analysis, the team should be able to obtain the
temperature distribution through the fireclay brick successfully, and the optimal results
should be consistent and the values would have as little variance as possible between the
different methods of analysis.
Application of Methods
Solutions applied:
The first method applied to the slab was the one dimensional conduction exact
analysis. The figure of the model is shown in the Appendix C-Calculations (one dimensional
conduction analysis). The thermal conductivity of the fire clay brick was 1.7 W/m*K. The
one dimensional conduction analysis was found by using Equation 1: Fourier’s law.
Equation 1: Fourier’s law - (d dx)
The heat flux was calculated by obtaining the temperatures at the beginning of the slab and at
the end of the slab, and dividing by the length of the slab shown in Appendix C-Calculations
(one dimensional conduction analysis) to get the overall heat transfer through the brick by
conduction. The heat flux represents the rate of heat transfer through a section of the area.
The heat loss of the brick is obtained by the area under the curve multiplied by the width of
the slab and the heat flux. The temperature across the element was found by obtaining the
differential equation and solving for the temperature at a specific length. The results of the
one dimensional conduction analysis is very close in temperature at specific lengths to the
simulations sketch in autocad inventor (See Graph 1). All of the calculations for one
dimensional conduction exact analysis are shown in Appendix C-Calculations.

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Graph 1: Nodal Temperatures plotted for the exact method
A flux plot was by creating four isothermal lines and three heat flow lines. The
isotherms are vertical and heat lines horizontal, and they are perpendicular to one another.
The symmetry means the top and bottom surface are adiabatic, thus a flux plot can be created
(Figure3)
Figure 3: Flux plot with isotherms and heat flow lines

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The finite difference was done by hand by first graphing the top curve and scaling it
down by a factor of two on graph paper. To determine the location of the nodes a dx of 18mm
was set as constant throughout the geometry; a vertical line was drawn from each dx step to
an intersection at the top of the curve, over the domain 0<x<108mm, give a total of 5 vertical
lines (see A3). The dy step was dynamically changing over the rage of 0<y<37, determined
by finding the height of each dx intersection with the curve, and drawing a line from that
point, leftwards to side A. This gave a total of 6 horizontal lines. The changes in dy could be
measured form the line spacing’s that hit the y axis (see A3). After the lines were drawn, we
had a total of 25 nodes, labeled from the in sequential from the top to bottom, then moving
left the next dx step, starting with the node located at (18mm, 26.4mm). Assumptions made
were that the top and bottom sides of the graph were adiabatic; steady state finite difference
will be applied. Starting with node 1, we use equation (4.31), with k = .0017 W/mm*K,
temperature at side A being 200C, giving a q-in from node 2 and wall A. This method was
applied to nodes 1-25. When working out the geometry the dynamic dy of the overhang of
the top nodes was ignored. When applying the finite difference a pattern was observed, that
the coefficients in front of the temperature differences of a node, were the same as the
coefficients of the node to the left of it, except when a new “arc node” was introduced. This
was due to the similarities in geometry caused by the constant dx step (see A4). After the
energy balance for the nodal network was applied, the temperature’s coefficients were pulled
out and formed into a matrix, and set equal to the constants. This matrix was solved by
taking the inverse of it and multiplying it to the constants column, which would yield a
column of node temperatures (see A5). The values we received after evaluating the matrix
were not consistent with the values obtained by the other methods. This is most likely due to
false assumptions in the geometry of the top nodes, or other calculation errors. Further
investigation may help obtain the values of finite difference that are consistent with other
solutions.
CAD modeling and simulation was completed using Autodesk Inventor and Autodesk
Simulation 2013, and the temperature distribution about 40 nodes along the center reference
line were acquired. The fireclay brick is modeled in Autodesk Inventor, and then in Autodesk
Simulation we are able to perform a steady-state conduction by creating a mesh grid (See
Figure 4) and setting the constant temperatures at each end of the model. Once the mesh grid
was created and nodes were chosen, temperatures at these nodes were acquired and plotted
(See Graph 2).

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Figure 4: Mesh grid of the thermal distribution in Autodesk Simulation 2013
Graph 2: Nodal temperature distribution of fireclay brick vs. position (mm)

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Summary
The experiment demonstrated different methods of heat transfer through a fire clay
brick at known dimensions. An understanding of the different methods to solve heat transfer
problems are important in determining unknown surface temperatures, and heat flux from
known dimensions throughout the body. The methods used were one dimensional
conduction, flux plot, finite difference, and Autodesk Inventor and Autodesk Multiphysics
Simulation. It was concluded from the results that one-dimensional conduction can be found
by beginning with the heat equation and deriving the expression resulting in values for heat
flux, heat rate, and temperature distribution. The flux plot shows the heat flow lines as well as
the perpendicular isotherms lines flowing through the brick when the top and bottom surfaces
are adiabatic. Finite difference method showed different temperatures at nodal points of the
brick, and the solution of solving was the matrix inversion. The knowledge gained aids the
team of engineers in predicting how a material will react in the presence of heat, and also
demonstrates the different methods that can be applied. It was concluded from the results that
at different dimensions throughout the brick showed similar temperatures when applying the
different methods. Overall, the experiment was carried out successfully, and that all methods
were met.
Appendix
A: List of Abbreviations
k - Thermal conductivity, units ( W/m*K)
T - Temperature, units (Kelvin or Celsius)
q = heat flux, units ( W/ )
dT = Change in temperature units, (Kelvin or Celsius)
dx = Change in lengths units (m)
L = length ( m)
w = width (m)
B: Annotated Bibliography/References
[1] Bergman, Theodore L., Adrienne S. Lavine, Frank P. Incropera, and David P.
Dewitt. Fundamentals of Heat and Mass Transfer. Hoboken,: Wiley, 2011. Print.

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C: Calculations
A1– Calculation ofparameters
A2– Calculation ofexact solution

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A3– Nodalnetworkforfinite difference calculation

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A4 – Finite difference calculations
A5 – Table of finite difference equations
T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11 T12 T13 T14 T15 T16 T17 T18 T19 T20 T21 T22 T23 T24 T25 EQUALS Constant
-0.219 0.219 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -0.266
0.218 -0.347 0.106 0 0 0 0 0.001 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -0.452
0 0.106 -0.192 0.08 0 0 0 0 0.003 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -0.634
0 0 0.08 -0.007 0.065 0 0 0 0 0.004 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -0.8
0 0 0 0 0 0.065 -0.133 0.059 0 0 0 0 0.0047 0 0 0 0 0 0 0 0 0 0 0 0 -0.932
0 0 0 0 0.059 -0.109 0.039 0 0 0 0 0.006 0 0 0 0 0 0 0 0 0 0 0 0 0 -1.2
0 0 0 0 0 0.038 -0.046 0 0 0 0 0 0.0037 0 0 0 0 0 0 0 0 0 0 0 0 -0.74
0 0.001 0 0 0 0 0 -0.107 0.106 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0.003 0 0 0 0 0.106 -0.191 0.801 0 0 0 0.0018 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0.004 0 0 0 0 0.08 -0.153 0.0653 0 0.004 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0.005 0 0 0 0 0.0653 -0.133 0.0588 0 0 0 0.0047 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0.006 0 0 0 0 0.0588 -0.109 0.0383 0 0 0 0.006 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0.004 0 0 0 0 0.0383 -0.046 0 0 0 0 0.0037 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0.002 0 0 0 0 -0.078 0.0801 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0.0653 0 0 0 0.0801 -0.152 0.0655 0 0 0.0022 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0.0047 0 0 0 0.0653 -0.133 0.0588 0 0 0.006 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0.006 0 0 0 0.0588 -0.109 0.0383 0 0 0.006 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0.0037 0 0 0 0.0383 -0.046 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0022 0 0 0 -0.068 0.0653 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0047 0 0 0.0653 -0.131 0.0588 0 0.006 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.006 0 0 0.0588 -0.109 0.0383 0 0.006 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0037 0 0 0.3825 -0.39 0 0.006 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0025 0 0 -0.056 0.0588 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.006 0 0.0588 -0.107 0.0383 -0.111
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0037 0 0.0383 -0.046 -0.111