project presentation

1D and 2D heat transfer project

Project on 1 dimensional heat transfer Project Statement Create a simulation tool/calculation tool that will allow us to simulate a floor fire test and the behavior of the material in the test. See sketch below for test setup. - The tool shall be able to calculate the temperature at the top of the flooring material inside the vehicle floor - The tool shall have the ability to simulate the temperature rise at the top of the flooring material when the under said floor is subjected to a temperature curve per ASTM E-119. The tool has to be able to calculate the "top of floor" temperature at least every minute for at least 30 minutes.

Input Data Enter the number of slabs: 3 Enter material of plate 1: aluminium Enter thickness of plate 1: 0.005 Enter material of plate 2: brass Enter thickness of plate 2: 0.005 Enter material of plate 3: stainless steel Enter thickness of plate 3: 0.005 Enter value of coefficient h for given problem: 10

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The graphs obtained from the program have been plotted into EXCEL and shown above. The graphs show the variation of temperature versus time at the different interfaces. As we can see, the trend followed is similar to the source curve but the graphs are not easily distinguishable. The code takes about a minute to show up the solution.

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1D numerical approach In Simple Words

Assumptions

Heat flow is considered to be transient in nature The interfaces of the blocks are assumed to be in perfect

thermal contact so that temperature at the interface is equal for both the blocks and thus no convection takes place at the interface

Rate of heat conduction at the interfaces is assumed to be the same for both the blocks

Radiation and convection effects within the blocks are neglected (convection on the top surface is however considered)

The properties of the blocks like thermal conductivity, density etc. are assumed to be constant with temperature. The heat transfer coefficient at the convection surface is also assumed to be constant.

No heat generation in any of the blocks The block is given 5 partitions and the time step is taken

to be 0.1 minute although these values can be changed any time before the program run.

Formulation of the differential equations

𝛼1 𝜕2𝑇1𝜕𝑥2 = 𝜕𝑇1𝜕𝑡 0 < 𝑥< 𝑎1,𝑡 > 0 𝛼2 𝜕2𝑇2𝜕𝑥2 = 𝜕𝑇2𝜕𝑡 𝑎1 < 𝑥< 𝑎2,𝑡 > 0 ∶ ∙ 𝛼𝑛 𝜕2𝑇𝑛𝜕𝑥2 = 𝜕𝑇𝑛𝜕𝑡 𝑎𝑛−1 < 𝑥< 𝑎𝑛,𝑡 > 0

Boundary Conditions 𝑇1 = 𝑓1ሺ𝑡ሻ 𝑎𝑡 𝑥= 0,𝑡 > 0 𝑇1 = 𝑇2 𝑎𝑡 𝑥= 𝑎1,𝑡 > 0 ∶ 𝑇𝑛−1 = 𝑇𝑛 𝑎𝑡 𝑥= 𝑎𝑛−1,𝑡 > 0 𝑘1 𝜕𝑇1𝜕𝑥 = 𝑘2 𝜕𝑇2𝜕𝑥 𝑎𝑡 𝑥= 𝑎1,𝑡 > 0 ∶ 𝑘𝑛−1 𝜕𝑇𝑛−1𝜕𝑥 = 𝑘𝑛 𝜕𝑇𝑛𝜕𝑥 𝑎𝑡 𝑥= 𝑎𝑛−1,𝑡 > 0 𝑘𝑛 𝜕𝑇𝑛−1𝜕𝑥 + ℎ𝑇𝑛 = ℎ× 293 𝑎𝑡 𝑥= 𝑎𝑛,𝑡 > 0 Initial Conditions 𝑇1 = 𝑇2 = ⋯ = 𝑇𝑛 = 293 𝑎𝑡 𝑡 = 0

Solution Background

Fully implicit finite difference scheme of solving partial differential equations

Solving a matrix using Gauss-Seidel iterative methods The given set of equations is solved using the fully implicit finite difference. The fully implicit scheme is chosen due to its ease of implementation and because it is unconditionally convergent and stable, thereby warranting any time step and number of divisions to be chosen. In the figure shown above, each of the blocks is partitioned into a given number of segments. The finite difference equations would be applied at each of the segments. For the position derivative, the central difference approximation is used and for time derivative, the backward difference approximation. The following substitutions were made: 𝜕2𝑇𝑖𝜕𝑥2 = 𝑇𝑖+1𝑛+1 − 2𝑇𝑖𝑛+1 + 𝑇𝑖−1𝑛+1(∆𝑥)2

𝜕𝑇𝑖𝜕𝑡 = 𝑇𝑖𝑛+1 − 𝑇𝑖𝑛∆𝑡

𝜕𝑇𝑖𝜕𝑥 = 𝑇𝑖+1𝑛+1−𝑇𝑖−1𝑛+12∆𝑥

I f the number of slabs is taken to be 3 and the number of nodes per slab to be 5 as shown in the figure, then the following system of equations will hold starting from n=0:

ሺ1+ 2𝑟1ሻ𝑇11 − 𝑟1𝑇21 = 𝑇10 + 𝑟1𝑇01 ሺ1+ 2𝑟1ሻ𝑇21 − 𝑟1𝑇11 − 𝑟1𝑇31 = 𝑇20 ሺ1+ 2𝑟1ሻ𝑇31 − 𝑟1𝑇21 − 𝑟1𝑇41 = 𝑇30 ሺ1+ 2𝑟1ሻ𝑇41 − 𝑟1𝑇31 − 𝑟1𝑇4′1 = 𝑇40

An imaginary node 𝑇4′1 is taken here to satisfy the conditions. Similarly, other such nodes will be taken when required to form the given matrix although the values obtained for these variables will hold no meaning. Continuing with the system of equations: ሺ1+ 2𝑟2ሻ𝑇41 − 𝑟2𝑇4′′1 − 𝑟2𝑇51 = 𝑇40 ሺ1+ 2𝑟2ሻ𝑇51 − 𝑟2𝑇41 − 𝑟2𝑇61 = 𝑇50 ሺ1+ 2𝑟2ሻ𝑇61 − 𝑟2𝑇51 − 𝑟2𝑇71 = 𝑇60

ሺ1+ 2𝑟2ሻ𝑇71 − 𝑟2𝑇61 − 𝑟2𝑇81 = 𝑇70 ሺ1+ 2𝑟2ሻ𝑇81 − 𝑟2𝑇71 − 𝑟2𝑇8′1 = 𝑇80

ሺ1+ 2𝑟3ሻ𝑇81 − 𝑟3𝑇8′′1 − 𝑟3𝑇91 = 𝑇80

ሺ1+ 2𝑟3ሻ𝑇91 − 𝑟3𝑇81 − 𝑟3𝑇101 = 𝑇90 ሺ1+ 2𝑟3ሻ𝑇101 − 𝑟3𝑇91 − 𝑟3𝑇111 = 𝑇100 ሺ1+ 2𝑟3ሻ𝑇111 − 𝑟3𝑇101 − 𝑟3𝑇121 = 𝑇110 ሺ1+ 2𝑟3ሻ𝑇121 − 𝑟3𝑇111 − 𝑟3𝑇12′1 = 𝑇120

𝑘1 𝑇4′1 −𝑇312∆𝑥1 = 𝑘2 𝑇51 −𝑇4′′12∆𝑥2

𝑘2 𝑇8′1 −𝑇712∆𝑥2 = 𝑘3 𝑇91−𝑇8′′12∆𝑥3

𝑘3 𝑇12′1 − 𝑇1112∆𝑥3 + ℎሺ𝑇121 − 293ሻ= 0

The given system of equations has 17 variables and 17 equations to be solved which is then encoded into a matrix and solved by MATLAB using the command A\ b. But in very high dimensional codes, the non-iterative codes used by MATLAB may not succeed and thus an iterative technique (the Gauss-Seidel method) has been worked out which gives just the same solutions but takes a lot more time inside the for loops. In the Gauss-Seidel procedure, the above system of equations would be written in the following manner:

𝑇11 = 𝑇10 + 𝑟1𝑇01 + 𝑟1𝑇21ሺ1+ 2𝑟1ሻ

𝑇21 = 𝑇20 + 𝑟1𝑇11 + 𝑟1𝑇31ሺ1+ 2𝑟1ሻ

𝑇31 = 𝑇30 + 𝑟1𝑇21 + 𝑟1𝑇41ሺ1+ 2𝑟1ሻ

𝑇41 = 𝑇40 + 𝑟1𝑇31 + 𝑟1𝑇4′1ሺ1+ 2𝑟1ሻ

𝑇4′1 = −𝑇40 − 𝑟2𝑇31 + (1+ 2𝑟2)𝑇41ሺ𝑟2ሻ

𝑇51 = 𝑇50+𝑟2𝑇61+𝑟2𝑇41ሺ1+2𝑟2ሻ

and so on till each node is explicitly written in terms of other nodes. A set of values for all 17 nodes is then assumed and the 1st node is then calculated from the above equation. The modified 1st node and the remaining nodes are then put into the 2nd node equation. The procedure keeps repeating till the consecutive values of the same node converge. For the Gauss-Seidel method, this convergence mostly happens if the given matrix is tri-diagonal or diagonally dominant. In this case, even though the formed matrix is neither, it is very close to being a tri-diagonal matrix and thus the solution does converge. After the 17 values for n=0 are obtained either by direct matrix solutions or iterative procedures, the matrix is solved again for n=1 and so on till all required values for all times are not obtained. The code for the above solution has however been developed for any number of slabs and partitions in MATLAB. The code developed is printed in Appendix B.

Input Data Enter the number of slabs: 3 Enter material of plate 1: aluminium Enter thickness of plate 1: 0.005 Enter material of plate 2: brass Enter thickness of plate 2: 0.005 Enter material of plate 3: stainless steel Enter thickness of plate 3: 0.005 Enter value of coefficent h for given problem: 10

Without Gauss-Seidel iteration

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Interface 2

Interface 3

With Gauss-Seidel iteration

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Interface 1 gs

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Comparison (with and without Gauss-Seidel)

As we can see, both the non-iterative and the iterative programs yield exactly the same results with the respective graphs in both the methods overlapping. The iterative procedure however takes up slightly more time and memory space although it guarantees solution for very large matrices too.

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Interface 3

Convergence of Gauss-Seidel procedure

The above graph is shown for the convergence of temperatures of the top-most surface of all the slabs at different time intervals for the Gauss-Seidel method. This confirms that even though the formed matrix in this case does not rigorously satisfy convergence criteria, but the implemented code does happen to converge.

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25min

Improvements converting the program into GUI increasing options for user to input values like

surrounding temperature etc. creating separate functions for some part of the code for

better reusability and optimizing code to work faster validating the answer obtained through experimental

techniques making the problem more and more generalized by

reducing number of assumptions

Conclusion The fully implicit finite difference method, both by with and without Gauss-Seidel iteration, gives a seemingly good and accurate solution which can now be verified by experimental techniques. Now that a seemingly right methodology has been accomplished, the scope of expanding the project to higher dimensions (2D has been included next) or including radiation conditions always remains.

2D numerical approach In Simple Words

Assumptions

Heat flow is considered to be transient in nature The interfaces of the blocks are assumed to be in perfect

thermal contact so that temperature at the interface is equal for both the blocks and thus no convection takes place at the interface.

Rate of heat conduction at the interfaces is assumed to be the same for both the blocks

Radiation and convection effects within the blocks are neglected (convection on the top surface and the sides is however considered)

The properties of the blocks like thermal conductivity, density etc. are assumed to be constant with temperature. The heat transfer coefficient at the convection surface is also assumed to be constant.

No heat generation in any of the blocks There are two values of interface temperatures found at

each time step which differ by a slight proportion. Thus, the first has been taken as the valid temperature in such a scenario.

Formulation of the differential equations 𝛼1ቆ𝜕2𝑇1𝜕𝑥2 + 𝜕2𝑇1𝜕𝑦2ቇ= 𝜕𝑇1𝜕𝑡 0 < 𝑥< 𝑎1,0 < 𝑦< 𝑎𝑦, 𝑡 > 0

𝛼2ቆ𝜕2𝑇2𝜕𝑥2 + 𝜕2𝑇2𝜕𝑦2ቇ= 𝜕𝑇2𝜕𝑡 𝑎1 < 𝑥< 𝑎2,0 < 𝑦< 𝑎𝑦, 𝑡 > 0 ⋮ 𝛼𝑛 ቆ𝜕2𝑇𝑛𝜕𝑥2 + 𝜕2𝑇𝑛𝜕𝑦2 ቇ= 𝜕𝑇𝑛𝜕𝑡 𝑎𝑛−1 < 𝑥< 𝑎𝑛,0 < 𝑦< 𝑎𝑦, 𝑡 > 0

Boundary Conditions 𝑇1 = 𝑓1ሺ𝑡ሻ 𝑎𝑡 𝑥= 0,0 < 𝑦< 𝑎𝑦,𝑡 > 0 𝑇1 = 𝑇2 𝑎𝑡 𝑥= 𝑎1,0 < 𝑦< 𝑎𝑦,𝑡 > 0 ∶ 𝑇𝑛−1 = 𝑇𝑛 𝑎𝑡 𝑥= 𝑎𝑛−1,0 < 𝑦< 𝑎𝑦,𝑡 > 0 𝑘1 𝜕𝑇1𝜕𝑥 = 𝑘2 𝜕𝑇2𝜕𝑥 𝑎𝑡 𝑥= 𝑎1,0 < 𝑦< 𝑎𝑦,𝑡 > 0 ∶ 𝑘𝑛−1 𝜕𝑇𝑛−1𝜕𝑥 = 𝑘𝑛 𝜕𝑇𝑛𝜕𝑥 𝑎𝑡 𝑥= 𝑎𝑛−1,0 < 𝑦< 𝑎𝑦,𝑡 > 0 𝑘𝑛 𝜕𝑇𝑛−1𝜕𝑥 + ℎ𝑇𝑛 = ℎ× 293 𝑎𝑡 𝑥= 𝑎𝑛,0 < 𝑦< 𝑎𝑦,𝑡 > 0 𝑘1 𝜕𝑇1𝜕𝑦 = ℎ(𝑇1 − 293) 𝑎𝑡 𝑦= 0,0 < 𝑥< 𝑎1 𝑡 > 0 ⋮ 𝑘𝑛 𝜕𝑇𝑛𝜕𝑦 = ℎ(𝑇𝑛 − 293) 𝑎𝑡 𝑦= 0,𝑎𝑛−1 < 𝑥< 𝑎𝑛 𝑡 > 0

𝑘1 𝜕𝑇1𝜕𝑦 + ℎ(𝑇1 − 293) = 0 𝑎𝑡 𝑦= 𝑎𝑦,0 < 𝑥< 𝑎1 𝑡 > 0

Initial Conditions 𝑇1 = 𝑇2 = ⋯ = 𝑇𝑛 = 293 𝑎𝑡 𝑡 = 0 Solution Background

The alternating direction implicit (ADI) finite difference scheme of solving partial differential equations which is well suited for a 2D heat transfer problem The above set of equations is solved using a slight variant of the fully implicit FDM scheme called the alternating direction implicit method. A usual implementation of a fully implicit FDM in this case would result in a near penta-diagonal matrix which would have to be coded all over again. Instead, the ADI scheme breaks a time step into two halves. In the first half, the x-derivative is written as an implicit central-difference approximation and the y-derivative as an explicit central-difference approximation. The reverse holds true for the second half of the time step. Each of these halves result in matrices very similar to tri-diagonal matrix, the results of which are then combined to obtain the final solution.

This time the blocks were divided into segments in both x and y directions such that it resulted in creation of point nodes. At each of these nodes, the equations and boundary conditions stated above are valid. The following substitutions were made: For (n+12)th time step

𝜕2𝑇𝑖,𝑗𝜕𝑥2 = 𝑇𝑖+1,𝑗𝑛+12 − 2𝑇𝑖,𝑗𝑛+12 + 𝑇𝑖−1,𝑗𝑛+12(∆𝑥)2 𝜕2𝑇𝑖,𝑗𝜕𝑦2 = 𝑇𝑖,𝑗+1𝑛 − 2𝑇𝑖,𝑗𝑛 + 𝑇𝑖,𝑗−1𝑛(∆𝑦)2

𝜕𝑇𝑖,𝑗𝜕𝑡 = 𝑇𝑖,𝑗𝑛+12 − 𝑇𝑖,𝑗𝑛∆𝑡

𝜕𝑇𝑖,𝑗𝜕𝑥 = 𝑇𝑖+1,𝑗𝑛+12 −𝑇𝑖−1,𝑗𝑛+122∆𝑥

For (n+1)th time step

𝜕2𝑇𝑖,𝑗𝜕𝑥2 = 𝑇𝑖+1,𝑗𝑛+12 − 2𝑇𝑖,𝑗𝑛+12 + 𝑇𝑖−1,𝑗𝑛+12(∆𝑥)2 𝜕2𝑇𝑖,𝑗𝜕𝑦2 = 𝑇𝑖,𝑗+1𝑛+1 − 2𝑇𝑖,𝑗𝑛+1 + 𝑇𝑖,𝑗−1𝑛+1(∆𝑦)2

𝜕𝑇𝑖,𝑗𝜕𝑡 = 𝑇𝑖,𝑗𝑛+1 − 𝑇𝑖,𝑗𝑛+12∆𝑡

I f the number of slabs is taken to be 3 and the number of segments per slab to be 5 in both x and y direction as shown in the figure, then the following system of equations will hold starting from n=0:

ሺ1+ 2𝑟𝑥1ሻ𝑇1,112 − 𝑟𝑥1𝑇2,112 = ൫1− 2𝑟𝑦1൯𝑇1,10 + 𝑟𝑦1𝑇1,20 + 𝑟𝑦1𝑇1,00 + 𝑟𝑥1𝑇0,112

ሺ1+ 2𝑟𝑥1ሻ𝑇2,112 − 𝑟𝑥1𝑇3,112 − 𝑟𝑥1𝑇1,112 = ൫1− 2𝑟𝑦1൯𝑇2,10 + 𝑟𝑦1𝑇2,20 + 𝑟𝑦1𝑇2,00


ሺ1+ 2𝑟𝑥1ሻ𝑇4,112 − 𝑟𝑥1𝑇4′ ,112 − 𝑟𝑥1𝑇3,112 = ൫1− 2𝑟𝑦1൯𝑇4,10 + 𝑟𝑦1𝑇4,20 + 𝑟𝑦1𝑇4,00 ⋮

And so on for the remaining slabs. From the equations obtained a 17×17 matrix is formed and solved for corresponding temperatures. Once the temperatures for j=1 have been obtained, we move on to j=2.

ሺ1+ 2𝑟𝑥1ሻ𝑇1,212 − 𝑟𝑥1𝑇2,212 = ൫1− 2𝑟𝑦1൯𝑇1,20 + 𝑟𝑦1𝑇1,30 + 𝑟𝑦1𝑇1,10 + 𝑟𝑥1𝑇0,212



ሺ1+ 2𝑟𝑥1ሻ𝑇4,212 − 𝑟𝑥1𝑇4′ ,212 − 𝑟𝑥1𝑇3,212 = ൫1− 2𝑟𝑦1൯𝑇4,20 + 𝑟𝑦1𝑇4,30 + 𝑟𝑦1𝑇4,10 ⋮

And so on for the remaining slabs.

After having all the values, we move to the next half of the solution method for finding values at n=1. The following equations hold:

൫1+ 2𝑟𝑦1൯𝑇1,11 − 𝑟𝑦1𝑇1,21 = ሺ1− 2𝑟𝑥1ሻ𝑇1,112 + 𝑟𝑥1𝑇2,112 + 𝑟𝑦1𝑇0,112 + 𝑟𝑦1𝑇1,01

൫1+ 2𝑟𝑦1൯𝑇1,21 − 𝑟𝑦1𝑇1,31 − 𝑟𝑦1𝑇1,11 = ሺ1− 2𝑟𝑥1ሻ𝑇1,212 + 𝑟𝑥1𝑇2,212 + 𝑟𝑦1𝑇0,212



൫1+ 2𝑟𝑦1൯𝑇1,51 − 𝑟𝑦1𝑇1,5′1 − 𝑟𝑦1𝑇1,41 = ሺ1− 2𝑟𝑥1ሻ𝑇1,512 + 𝑟𝑥1𝑇2,512 + 𝑟𝑦1𝑇0,512 ⋮ and so on. Similarly, here we form the matrix at the i th value and keep solving for each i until we have the temperatures for each node in the mesh at n=1. We then move on to solving for n=2 by the same procedure described above. The difference from the previous chapters is that the temperatures at each node here can be different even varying along the x direction although only slightly.

Input Data Enter the number of slabs: 3 Enter material of plate 1: aluminium Enter thickness of plate 1: 0.005 Enter material of plate 2: brass Enter thickness of plate 2: 0.005 Enter material of plate 3: stainless steel Enter thickness of plate 3: 0.005 Enter equal width of all plates: 0.01 Enter value of coefficent h for given problem: 10

The graph above has been shown for the corner nodes of all the interfaces and follows similar trends as discussed in the previous chapters.

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Source temperature

The graph above shows temperature curves for all nodes of the top surface. As we see, the temperature variation along the y direction is not much and the temperature curves for all these nodes overlap.

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node 5

This graph shows the temperature variation along the top surface at time t=30 min. As expected, the temperature is symmetric with respect to the middle node with heat flowing from both sides of the middle node to the surroundings at ambient temperature.

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30 min

Improvements converting the program into GUI increasing options for user to input values like

surrounding temperature etc. creating separate functions for some part of the code for

better reusability and optimizing code to work faster validating the answer obtained through experimental

techniques making the problem more and more generalized by

reducing number of assumptions can be extended to the 3 dimensional case

Conclusion The ADI finite difference technique works for the 2 dimensional case and gives good results which can now be validated using experimental techniques.

References

Heat conduction by M Necati Ozisik

http://highered.mcgraw-hill.com/sites/dl/free/0073129305/314124/cen29305_ch04.pdf

en.wikepedia.org

www.mathworks.in

www.wolframalpha.org

Computational Fluid Dynamics by J ohn D. Anderson

Tata Technologies Limited 1

Thank You!!!

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