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I want to build a thermal model of a pin fin:

enter image description here

On the left side of the cylinder (red line), a temperature of 140°C heats up the fin. Both on the surface and on the right side of the cylinder (blue lines), a convective heat transer is cooling down the fin. Inside the pin, heat conduction is present.

For such a setup, analytical solutions of the temperature distribution over the length of the pin fin T(x) can be found in literature, e.g. here (formula 18.12):

enter image description here

with:

enter image description here

where:

  • h_conv = Convective heat transfer in W/m²K
  • h_cond = Conductive heat transfer in W/mK
  • S = Surface area of the pin in m²
  • A = Cross sectional area of the pin fin in m²
  • T_amb = Ambient temperature in °C
  • T_base = Temperature on the left end of the fin tip in °C
  • T_x = Temperature of the pin fin at location x

I put all the equations to a Matlab script to evaluate the temperature distribution over the length of the rod:

% Variables
r       = 0.01;             % Radius of the pin fin in m
l       = 0.2;              % Length of the pin fin in m
h_conv  = 500;              % Convective heat transfer in W/m²K
h_cond  = 500;              % Conductive heat transfer in W/mK
T_base  = 140;              % Temperature on the left end of the fin tip in °C
T_amb   = 40;               % Ambient temperature in °C    
n_elem  = 20;               % Number of division
x       = [0:l/n_elem:l]';  % Vector of division, necessary for evaluation

A       = r^2 * pi;         % Cross sectional area of the pin fin in m²
S       = 2 * pi * r * l;   % Surface area of the pin in m²

% Analytical solution
beta    = sqrt((h_conv*S)/(h_cond*A));
f       = cosh(beta*(l-x))/cosh(beta*l);

temp    = f*(T_base-T_amb)+T_amb;

% Plot of the temperature distribution
plot (x,temp);
axis([0 0.2 40 140]);

The resulting temperature distribution looks like this:

I tried to build a Simscape model of that setup based on the example Heat Conduction through Iron Rod. After solving the problem with Simscape, I did a comparison between the analytical and the Simscape solution:

x_ss     = [0:0.05:0.2]';                    % Vector of division, necessary for evaluation of the Simscape results
temp_ss  = [T_base,temp_simscape(end,:)]';   % Steady state results of Simscape model at 1/4, 2/4, 3/4 and 4/4 of the length

% Plot analytical vs. Simscape solution
plot (x,temp);
hold on;
plot (x_ss,temp_ss,'-o');
axis([0 0.2 40 140]);

The corresponding plots of the analytical and the Simscape solution looks like this:

enter image description here

As you can see from the graph, the Simscape model (blue curve) predicts much lower temperatures compared to the analytical solution (orange curve). As I was not able to find out the reason for that difference, I appreciate any help!

You can download the model here. The filehoster (www.xup.in) converts the name of the model from "PinFin.mdl" to "PINFIN.MDL", therefore you need to modify the file-extension back to ".mdl" in order to open it in Matlab.

Regards, Phil

Phil U.
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1 Answers1

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you're using the right approach. The results aren't matching because of the computation of the beta parameter. You have beta = sqrt((h_convS)/(h_condA));

The dimensions are wrong, you want: beta = sqrt((h_convP/(h_condA));

where P is the perimeter, here P = 2*pi*r.

With this corrections you'll find both outputs (analytical & simulation) match nicely given a fine enough discretization (I used n_elem = 20).

ngautier
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