2d heat conduction using matlab

7/28/2019 2D heat conduction using Matlab

1/13


2/13

Assumptions

Use Finite Difference Equations shown intable 5.2

2D transient conduction with heat transferin all directions (i.e. no internal corners asshown in the second condition in table5.2)

Uniform temperature gradient in object

Only rectangular geometry will beanalyzed


3/13

Program Inputs The calculator asks for

Length of sides (a, b) (m)

Outside Temperatures (T_inf 1-T_inf 4) (K) Temperature of object (T_0) (K)

Thermal Convection Coefficient (h1-h4)(W/m^2*K)

Thermal Conduction Coefficient (k) (W/m*K)

Density () (kg/m^3) Specific Heat (Cp) (J/kg*K)

Desired Time Interval (t) (s)


4/13

Transient Conduction Example problem

suppose we have an object with rectangular cross-

section with these boundary conditions:

T_inf 1, h1

T_inf 2, h2

T_inf 3, h3

T_inf 4, h4

a

b

Origin


5/13

Conditions%Userdefined h values

h(1) = 10;h(2) = .1;

h(3) = 10;

h(4) = .1;

%Boundary conditions

%Userdefined T infinity values in kelvin

T_inf(1) = 293;

T_inf(2) = 293;

T_inf(3) = 353;T_inf(4) = 353;

%Initial condition (assume uniform initial temperature)

%Userdefined initial temperature value

T_0 = 573;

%Material properties

%Userdefined material values

k = .08;

rho = 7480;

c_p = .460;

%Userdefined physical variables

a = 1; %height of cross section

b = 1.3; %width of cross section

t = 3600; %time at which results are given


6/13

Time Step (t)

We assumed a value of x = y = gcd(a,b)

Using each of the conditions (except thesecond) in the table 5.2, we calculate thet and choose the smallest value

Using that t we calculate Fo

Our outputs for delta_x, delta_t, Forespectively

0.0500, 3.7078, 0.0345


7/13

Method

Using the Finite Difference Method,matlab generates a matrix of

temperature values that are representedin the graph shown on the next slide

This method allows for the calculation ofevery node in any 2D direction


8/13

Results

0

0.5

10 0.2 0.4 0.6 0.8 1 1.2 1.4

250

300

350

400

450

500

550

a (m)

Transient conduction (the origin of the plot is the top left corner of the cross section)

Temperature(K)

00.20.40.60.810

0.5

1

1.5250

300

350

400

450

500

550

a (m)


b (m)

Temperature(K)


9/13

Solution to different Problem%Userdefined h values

h(1) = 0;

h(2) = 1000;

h(3) = 1000;

h(4) = 100;

%Boundary conditions

%Userdefined T infinity values in kelvin

T_inf(1) = 273;

T_inf(2) = 150;T_inf(3) = 590;

T_inf(4) = 273;

%Initial condition (assume uniform

initial temperature)

%Userdefined initial temperature value

T_0 = 250;

%Material properties%Userdefined material values

k = .8;

rho = 1000;

c_p = .460;

%Userdefined physical variables

a = 1; %height of cross section

b = 1.3; %width of cross section

t = 20; %time at which results are given

00.2

0.40.6

0.8

1

0

0.5

1

1.5100

200

300

400

500

600

a (m)


b (m)

Temperature(K)


10/13

Conclusion and

Recommendations

Works only in rectangular geometry

High values of h and t>1 causes errors tooccur due to lack of memory

Use a better method to find x and t


11/13

Appendix-References

Incropera, Frank P. DeWitt, DaviD P.Fundamentals of Heat and Mass Transfer

Fifth Edition, R. R. Donnelley & SonsCompany. 2002 John Wiley & Sons, Inc


12/13

Appendix-hand work


13/13

Appendix-hand work

2d heat conduction using matlab

Documents