Determine the temperature distribution in a thick, long rectangular fin by formulating and solvi

step 1 So in question 33, you created these four equations. And in question 34, which is this problem r

Determine the temperature distribution in a thick, long...

Key Concepts

Comparison Between Thick and Thin Fin Models

Analyzing the differences between thick and thin fin models highlights the limitations of the simplified one-dimensional approach typically used for thin fins. The two-dimensional analysis for thick fins reveals additional thermal resistance due to lateral conduction, leading to potentially different temperature profiles and heat flow characteristics. This comparison helps in understanding when the simplified models can be used and when a more rigorous analysis is required.

Temperature Distribution in Fins

This concept involves determining how the temperature varies throughout the fin's geometry. For a thick, long, rectangular fin, the temperature distribution is affected by conduction in two directions and is impacted by both the physical dimensions of the fin and the applied boundary conditions. Accurately predicting this distribution is crucial in designing fins to effectively dissipate heat.

Heat Flow Analysis at the Fin Base

The base heat flow refers to the total heat transferred from the fin at its attachment to the heated surface. Calculating this involves integrating the temperature gradient at the base, as determined by the conduction solution, and is an important measure of fin performance. Comparing the result with that for a long thin fin provides insight into how geometric factors modify the efficiency of thermal management devices.

Boundary Conditions in Conduction Problems

Boundary conditions are critical in determining the solution to a heat conduction problem. They specify the temperature, heat flux, or convective conditions at the edges of the domain. In this problem, conditions such as those given by Eq. (2.33c) at the fin tip and the conditions at the base and sides of the fin establish the necessary criteria to solve the governing equations, influencing both the temperature distribution and the base heat flow.

Two-Dimensional Heat Conduction

This concept refers to the analysis of temperature fields within solids where heat transfer occurs in two spatial dimensions. In the context of fins, it requires setting up and solving a partial differential equation (typically Laplace’s or Poisson’s equation) that accounts for thermal gradients not only along the fin’s length but also across its thickness. This multidimensional approach is essential when the fin's geometry does not permit a simplifying one-dimensional approximation.

Transcript

00:01 So in question 33, you created these four equations.

00:06 And in question 34, which is this problem right here, we're being asked to solve this system.

00:11 And we're going to use matrices to do this.

00:13 So the first thing i'm going to do is create my matrix.

00:16 I have four, negative one, zero, negative one, and 30.

00:29 I have negative one.

00:31 I've got zero.

00:32 I've got negative one.

00:35 I've got four, negative one.

00:37 0, taking all the coefficients, negative 1, 060, 4, negative 1, 70, negative 1, 4, and 40.

00:50 Okay.

00:51 So the first thing that we're going to do is we are going to eliminate all of the terms that are not ones here.

01:04 And so what we're going to do here is i'm going to multiply this first equation.

01:11 Here.

01:13 I'm going to multiply this by four.

01:15 I'm going to multiply this by four.

01:17 So this way, we can eliminate the first terms.

01:21 So i have four, negative one, zero, negative one, and 30.

01:27 I've got negative four, 16, negative four, zero, and two, 40.

01:36 I have, i'm going to leave this guy alone, and i'm going to end up with negative 4, 0, negative 4, 16, and 40.

01:50 And so i'm going to combine the two new equations with that first one to be able to get rid of these first terms right up here.

02:03 And so what i have right now is i'm going to leave the first one alone.

02:06 So negative, so 4, negative 1, 0, negative 1, and 30.

02:13 I'm going to leave that third row.

02:15 Alone, 0, negative 1, 4, negative 1, and 70.

02:21 And when i add the second equation to the first equation, i'm going to get 0, i'm going to get 15, i'm going to get negative 4, i'm going to get negative 1, and i'm going to get 240.

02:37 Oh, sorry, no, i'm going to get 270.

02:46 When i add this fourth row to the first row, i get 0.

02:51 I get negative 1.

02:54 I get negative 4.

02:59 I get 15.

03:02 And i get 70.

03:03 Okay.

03:04 So we're getting pretty good.

03:05 So right now this first row is in good shape.

03:07 First column is in good shape.

03:09 So let's take care of the second column.

03:11 And so in this case, what i'm going to do is i am going to take care of everything.

03:19 I'm going to multiply each of the.

03:22 You know, the first, third, and the fourth by 15 to help me to get rid of that second term.

03:32 So in this case, i'm going to end up with, let's do that down here.

03:38 So i'm going to have four times 15 is 60 minus 15, zero, minus 15, and 15, and 15, times 30 is 450.

04:04 Again, i'm going to leave that second row alone.

04:12 I'm going to do the same thing with the third row.

04:15 So i've got 0, negative 15.

04:20 I have 60, negative 15.

04:25 And then 70 times 15 is 1 ,050.

04:34 And then finally, 0, negative 15.

04:41 And then i've got negative 60.

04:44 I've got 225, and then i've got that 1050 again.

04:51 So now i'm going to combine everything with that second row.

04:59 So when i combine everything with the second row, when i add the first two rows together, i get 60, i get zero, i get negative 4, i get negative 16, and i get 720, and leave that second row alone.

05:31 Okay, and then i'm going to add the second and the third rows together.

05:34 I'm going to get zero.

05:36 I'm going to get 56.

05:39 I'm going to get negative 16.

05:46 And i'm going to get 1 ,320.

05:55 And i'm going to get zero.

05:57 I'm going to add the second in the fourth row together.

05:59 Which gives me 0, which gives me negative 64, which gives me 224, and which gives me that 1320 again.

06:12 Well, the numbers are starting to get a little bit big, so let's clean that up a little bit and work with some common factors here.

06:25 Okay, so in this case, if i divide everything by 4 here, i'm going to have 60 divided by 4 is back to 15, 0, negative 1, negative 4, and then 720 divided by 4 is 180.

06:46 So that's a little bit better.

06:49 In the second row, there's not much i can do because i've got 15 and a negative 4.

06:56 I'll leave that guy alone right now.

06:59 I'm going to take 56.

07:02 I'm going to divide that by 4, which is going to give me.

07:08 14.

07:10 It's going to give me negative 4.

07:13 And then 13, 20 divided by 4 is 330.

07:18 So that's getting better.

07:19 And then same thing down here, 0, 0, negative 16, 224 divided by 4 is 56 and 1320 divided by 4 is 330.

07:43 And again i'm going to just clean that up just one more time.

07:47 So i've got 15 -0 -9 -1 -9 -4 -180.

07:54 I've got 0 -15 -9 -4 -1 -270, and i've got 0 -0 -7 -2 - and 165 when i divide those all by 2.

08:15 And 0 -0 -0 -9 -8 -28 when i divide by 2 and 165 over here.

08:29 Okay, so we're getting closer.

08:35 So now you just have to get rid of this third row here.

08:39 And so we've got this negative, this 7 right here.

08:48 And that 7 is kind of gross right now.

08:51 So i'm going to work with this last row over here.

08:54 And i'm going to end up getting rid of the final column first before i deal with the seven.

09:01 So i am going to notice in this case right here, i've got this 28 down over here...

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