HW2_solution

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Mechanical Engineering

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Dec 6, 2023

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Homework 2 In this assignment you will be analyzing the temperature distribution and heat flux through a 3-part com- posite wall under steady conditions using both hand calculations and using ANSYS. The wall dimensions, boundary conditions, and material properties are defined in the following diagram: 1 Pre-analysis (hand calculations) For this part you will solve for the rate of heat transfer and temperature distribution in the wall by hand 1. What assumptions can you make based on the problem statement that will help simplify the equations? 2. Define the mathematical model for this problem. What assumptions can you make based on the problem statement that will help simplify the equations? What is the differential equation that describes the temperature distribution for this situation (use the simplified form, accounting for your assumptions). 3. List the boundary equations in equation form. What kind of boundary conditions are these? 4. Find the temperature at the interfaces between each of the metal slabs, i.e., at x = 1 m and x = 2 m. 5. Derive the full temperature distribution as a function of position within the composite wall and plot using Matlab, Excel, or another software tool (do not plot by hand). 6. What is the heat flux at x = 1 . 5 m? How does the heat flux vary as a function of position within the wall? Geometry, Mesh, Mathematical model setup, Numerical solution Use the instructions and files for ANSYS in the homework package for Homework 2 on Blackboard to set up and solve this problem in ANSYS Mechanical. A couple of helpful notes when following the video and/or tutorial: • To select a particular face you may have to rotate the body to where you can see that face (the easiest method to do this is click and drag your mouse wheel). • The video was recorded in an older version of ANSYS, so portions of the interface may look different, but the steps and labels are the same. • Pay attention to units when meshing or defining other properties (video is in mm, the ANSYS default is m) 2 Post-processing After completing the tutorial, please submit the following results 1. Create a contour plot of the temperature distribution within the wall. As discussed in class, you must use the grayscale color map. Be sure to also include the legend in your plot. Please either save the 1

image from ANSYS directly or take a screenshot. Do not take a photo of your computer screen with your phone. 2. Copy the temperature data into Matlab, Excel, or another software plotting tool, then plot the tem- perature as a function of position within the wall. This plot should include both the hand-calculated temperature distribution from part 1 and the ANSYS predicted temperatures from this part. 3. Find the heat flux (per unit area) at x = 1 . 5 m from the simulation. 3 Verification & Validation 1. What is the maximum difference between your hand calculation of the ANSYS results? What might be the cause of this difference? (Note: if the difference is due to your own error in either your hand calculations or your ANSYS model you should go back and fix the mistake then return to this question.) 4 Critical Thinking 1. Modify this scenario by changing the boundary conditions and material properties. What did you change and what real life scenario are you modeling? 2. Create a contour plot of your new temperature distribution. 3. Explain some benefits you see to using ANSYS or other software to simulate heat transfer? What are some drawbacks or limitations that you see? Your answer to this question should be 100–250 words. 2

Homework 1 Solution 1 Pre-analysis (hand calculations) Based on the problem statement the following assumptions can be made 1. Stead state 2. No thermal energy generation 3. Constant thermal properties within each slab 4. 1 dimensional* 5. There is perfect matching between the slabs (i.e. the temperature and heat flux at the internal bound- aries are equal in each slab) *The 1-dimensional assumption is valid for hand-calculations, but ANSYS needs to use 3D geometry, thus can also say that the wall is large in the y and z directions and there are no y or z gradients (i.e. adiabatic in y and z). The governing equation is the heat diffusion equation ∂ ∂x k ∂T ∂x + ∂ ∂y k ∂T ∂y + ∂ ∂z k ∂T ∂z + ˙ q ′′′ = ρc ∂T ∂t after simplifying using the listed assumptions, the resulting ODE becomes d 2 T d x 2 = 0 The boundary conditions are specified temperatures, also known as dirichlet boundary conditions, on the left-most and right-most surfaces. In equation form these boundary conditions are T ( x = 0) = 28 ° C T ( x = 3) = − 5 ° C In order to solve for the temperature at the boundaries of the slabs we can start by solving this ODE within a single slab using the generic boundary conditions of T ( x = x 0 ) = T L and T ( x = x 0 + L ) = T R on the left and right surfaces of a particular slab, with x 0 as the x-location of the left boundary of the slab and L as the width of the slab. T = C 1 x + C 2 T = ( T R − T L ) x − x 0 L + T L Writing this out explicitly for all 3 slabs we get T ( x ) =      ( T 2 − T 1 )( x − 0) + T 1 if 0 ≤ x ≤ 1 ( T 3 − T 2 )( x − 1) + T 2 if 1 ≤ x ≤ 2 ( T 4 − T 3 )( x − 2) + T 3 if 2 ≤ x ≤ 3 To find the values of the intermediate temperatures T 2 and T 3 we can use the constant heat flux condition at each interslab boundary. − k ss ( T 2 − T 1 ) = − k al ( T 3 − T 2 ) ⇒ T 2 = k ss T 1 + k al T 3 k ss + k al − k al ( T 3 − T 2 ) = − k ss ( T 4 − T 3 ) ⇒ T 3 = k al T 2 + k ss T 4 k ss + k al 3

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Combining the two T 2 = k ss T 1 + k al k al T 2 + k ss T 4 k ss + k al k ss + k al = k ss k ss + k al T 1 + k 2 al ( k ss + k al ) 2 T 2 + k al k ss ( k ss + k al ) 2 T 4 T 2 − k 2 al ( k ss + k al ) 2 T 2 = k ss k ss + k al T 1 + k al k ss ( k ss + k al ) 2 T 4 T 2 = k ss k ss + k al T 1 + k ss k al ( k ss + k al ) 2 T 4 1 − k 2 al ( k ss + k al ) 2 Doing the same to solve for T 3 , or alternatively recognizing the symmetry in the problem to inverse the apparent boundary conditions, you can find T 3 = k ss k ss + k al T 4 + k ss k al ( k ss + k al ) 2 T 1 1 − k 2 al ( k ss + k al ) 2 You could substitute the above into the piece-wise temperature profile, but it’s easier to leave in this form. Using the given values for surface temperatures and thermal conductivities results in intermediate temper- atures of T 2 = 13 . 37 ° C T 3 = 9 . 636 ° C To calculate the heat flux at the center point, we can use Fourer’s Law for the second slab (which includes the x = 1 . 5 m location) q ′′ = − k d T d x = − k ( T 3 − T 2 ) = − 237 . 5 W/m-K · (9 . 636 − 13 . 37) ° C = 885 . 5 W/m 2 While we calculated the heat flux at the center point, we could’ve calculated it at any point as the heat flux is constant in the entire wall. 2 Post-processing The final solution will look like this: 4

Copying the temperature values given by ANSYS into Excel and plotting together with the temperature profile from the hand calculations we can see they are nearly identical The heat flux calculated by ANSYS is 885.17 W/m 2 . 3 Verification & Validation In my simulations, the maximum difference between an analytical solution and the numerical solution is − 0 . 0049 K at x = 2 m, or about 0.015% of the full temperature range. This difference may be caused by differences in rounding between the hand calculations and ANSYS. Another possible source is that ANSYS is solving a discretized form of the differential equation, while our hand calculation solved the exact differential equation. Finally, ANSYS is solving the problem in 3D while the hand calculations use 1D. This problem was a 1D problem and we only solved it in 3D in ANSYS because we had to. This will be a much smaller problem than the other two issues, but it may still affect the solution. 4 Critical Thinking Some advantages and disadvantages of simulation tools such as ANSYS are that they are able to simulate complicated geometries and boundary conditions that it may be either very difficult or impossible to solve exactly using hand calculations. Once an initial case is defined, further modifications to material properties or certain boundary conditions are also relatively easy compared to experiments. Simulations also provide 5

much more detail of a solution than an experiment of the same conditions is able to provide. Simulations are limited by having to solve a discrete version of the governing equations, thus their solutions are only approximations of the true solution and care is needed to make sure the approximation of what would happen in an experiment is a good one, which takes extra time and effort. Simulations also take time to complete, and for very complicated models this can be much longer than an experiment and/or require access to specialized high performance computing resources to complete the simulation. 6

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