CMME Assignment 2

8/11/2019 CMME Assignment 2

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When i=1,j=0:

When i=2,j=0:

When i=3,j=0:

When i=1,j=1:

When i=2,j=1:

When i=3,j=1:

When i=1,j=2:

When i=2,j=2:

When i=3,j=2:

When i=1,j=3:

When i=2,j=3:

When i=3,j=3:

When i=1,j=4:

When i=2,j=4:

When i=3,j=4:

When i=1,j=5:

When i=2,j=5:

When i=3,j=5:

Hence, we can form the matrix by using the equation that we got.

3. Method A: Microsoft Excel

After the governing equations are formed and boundary conditions are applied into the

formula, a matrix, which is name as A, is formed by assembling all governing equation to

form a matrix equation. Next, those values which are at the right hand side of the equation

will be assembled into another matrix which is B.


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A Matrix B Matrix

-4 1 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 -4 1 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0

0 1 -4 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0

1 0 0 -4 1 0 1 0 0 0 0 0 0 0 0 0 0 0

0 1 0 1 -4 1 0 1 0 0 0 0 0 0 0 0 0 0

0 0 1 0 1 -4 0 0 1 0 0 0 0 0 0 0 0 0

0 0 0 1 0 0 -4 1 0 1 0 0 0 0 0 0 0 0

0 0 0 0 1 0 1 -4 1 0 1 0 0 0 0 0 0 0

0 0 0 0 0 1 0 1 -4 0 0 1 0 0 0 0 0 0

0 0 0 0 0 0 1 0 0 -4 1 0 1 0 0 0 0 0

0 0 0 0 0 0 0 1 0 1 -4 1 0 1 0 0 0 0

0 0 0 0 0 0 0 0 1 0 1 -4 0 0 1 0 0 0

0 0 0 0 0 0 0 0 0 1 0 0 -4 1 0 1 0 0

0 0 0 0 0 0 0 0 0 0 1 0 1 -4 1 0 1 0

0 0 0 0 0 0 0 0 0 0 0 1 0 1 -4 0 0 1

0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 -4 1 0

0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 1 -4 1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 1 -4

=

-650

0

-150

-650

0

-150

-650

0

-150

-650

0

-150

-650

0

-150

-650

0

-150

T2

T3

T4

T7

T8

T9

T12

T13

T14

T17

T18

T19

T22

T23

T24

T27

T28

T29


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(P.S: To form matrix A and B in the faster way, we define the parameters of matrix on

variable editor rather than on SciNotes directly. First, we input our data into Microsoft excel,

and then declare a new variable in the variable editor. Lastly, the raw data at Excel file is

copied and pasted on the new variable in variable editor. Same steps are used to form matrix

B)

Matrix C is the inverse matrix of A. This can be done by using Scilab with the following

command:

C=inv(A);

Data is shown in excel form in softcopy

In order to obtain the final answer, the matrix C must be multiplied with matrix B to

form a new matrix which is D. To perform this, following commands are used.

D=C*B;

disp(D)

T2

=

525

T3 400

T4 275

T7 525

T8 400

T9 275

T12 525

T13 400

T14 275

T17 525

T18 400

T19 275

T22 525

T23 400

T24 275

T27 525

T28 400

T29 275

T1 = T6 = T11 = T16 = T21 = T26 = 650oC

T5 = T10 = T15 = T20 = T25 = T30 = 150oC

(Boundary condition)


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4. Method B:IHT Software

//Stainless Steel-AISI 302 Material Property is defined

Rho=rho_300k(Stainless Steel-AISI 302) //Density, kg.m^3alpha=k/(rho*cp) //Thermal Diffusivity, m^2/s

k=k_T(Stainless Steel-AISI 302,300) //Thermal Conductivity, W/m.K

cp=cp_T(Stainless Steel-AISI 302,300) //Specific Heat, J/kg.K

qdot=0

deltax=0.003

deltay=0.003

Thermal Diffusivity, = 3.90E-6 m2

/s

Specific Heat, Cp = 480 J/kgK

Thermal Conductivity, k =15.1 W/mK

Density, = 8055 kg/m3

//Boundary Conditions are specified

T1=650

T6=650

T11=650

T16=650

T21=650

T26=650

T5=150

T10=150

T15=150

T20=150

T25=150

T30=150

//Commands used

/* Node 2: interior node;e, w, n, s labeled 3, 1, 7, 7. */

0.0 = fd_2d_int(T2,T3,T1,T7,T7,k,qdot,deltax,deltay)







/* Node 8: interior node;e, w, n, s labeled 9, 7, 13, 3. */0.0 = fd_2d_int(T8,T9,T7,T13,T3,k,qdot,deltax,deltay)


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Result:

Node Temperature,

T1 650

T2 525

T3 400

T4 275

T5 150

T6 650T7 525


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T8 400

T9 275

T10 150

T11 650

T12 525

T13 400T14 275

T15 150

T16 650

T17 525

T18 400

T19 275

T20 150

T21 650

T22 525

T23 400T24 275

T25 150

T26 650

T27 525

T28 400

T29 275

T30 150

5. Comparison between Method A & Method B:

Data T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11 T12 T13 Ti

Excel 650 525 400 275 150 650 525 400 275 150 650 525 400 All

SameIHT 650 525 400 275 150 650 525 400 275 150 650 525 400

i=14,15,16.....30

By comparing both methods which are Scilab and IHT software, we found that IHT software

will be more accurate compare to scilab. The main reason is both software use different

method in solving the question. For Scilab, it uses numerical method in solving the question.

On the other hand, IHT it uses analytical method to solve the problem. Numerical method

does not include all the external factors such as density , thermal conductivity K, Specific

heat capacity Cp, and alpha in the calculation. However, analytical method does include all

the external factors in calculating the temperature.


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When the effects of external factors are considered, any changes in the external factors will

lead to a different in calculation and eventually leads to different answers. If there is heat

generation, scilab and IHT will give a different answer for different material. However, due

to no generation in our question, both scilab and IHT will give us the same answer. Thus, the

values remain the same.

6. Discussion:

a) Does the thermal conductivity, k play a role in the solution?

From the results that we have obtained, we found that thermal conductivity does not

play a role in the solution. It is adequate to assume the thermal conductivity is the

same in all directions, named isotropic materials. Hence, the equation of 2D, steady

state conduction heat transfer can be simplified to

+=0. Since, our

question does not include any heat generation; the thermal conductivity has no effect

in the calculation both methods are applicable to get the true result (

=0). So,

method B which is IHT will be the same as method A which is scilab because there is

no external factors like heat generation in this question is involve.

Supposing if heat generation is involved, the formula in IHT should be changed to

q= fd_2d_int(T2,T3,T1,T7,T7,k,qdot,deltax,deltay)

where q=rho*cp*der(T2,t) for node 2

b)

Extra Question: We have fully understood how to use IHT and we would try it on

another question. Since this is a discussion, we will only include the code and answer

towards this extra question. We also fulfil the condition asked :

irregular boundary, rather than the square boundaries in the example

less nodes(We prefer less nodes)

insulated, (refer to CMME lab 3 )

Our imagination

Since the question mentioned that step size x must equal to y, or vice versa,so we can

only think of x=3mm, of y=3mm


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Code:

rho=8055 //Density, kg.m^3

alpha=3.9E-6 //Thermal Diffusivity, m^2/s

k=15.1 //Thermal Conductivity, W/m.K

cp=480 //Specific Heat, J/kg.K

qdot=0deltax=0.003

deltay=0.003

T1=650

T7=650

T13=650

T19=650

T23=650

T6=150

T12=150

T18=150T2=100

T3=100

T4=100

T5=100







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Result:

Node Temperature,

T1 650

T2 100

T3 100T4 100

T5 100

T6 150

T7 650

T8 372.6

T9 260.9

T10 203.1

T11 160.9

T12 150

T13 650T14 479.6

T15 367.8

T16 290.5

T17 190.6

T18 150

T19 650

T20 528

T21 440

T22 400.7

T23 650

T24 542.4

T25 463.7

T26 432.2

7. Conclusion:

In conclusion, without heat generation, Microsoft Excel and IHT software can get similar

results. We can say that both of the methods in this case is quite accurate. In 1D steady state

without heat generation, we can also form the equation by using finite difference method.

This shows that finite difference method is sufficient enough to analyze the heat transfer

problem.

IHT would still be preferred to be used though because of its simplicity and ability to solve

real engineering problem applicable in real life. Excel can also be used but the result is not

accurate in reality as there is surely heat generation in between the conduction of the material.

When there is heat generation, external factors should also be involved in the calculation and

therefore IHT can provide that solution.

CMME Assignment 2

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