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MATLAB for Breakfast Evolution of heat distribution across bottom of frying pan on spiral-shaped MATLAB for Breakfast Evolution of heat distribution across bottom of frying pan on spiral-shaped

MATLAB for Breakfast Evolution of heat distribution across bottom of frying pan on spiral-shaped - PowerPoint Presentation

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Uploaded On 2019-06-22

MATLAB for Breakfast Evolution of heat distribution across bottom of frying pan on spiral-shaped - PPT Presentation

Tim Nickell EPS 109 Fall 2013 Prof Burkhard Militzer Context Spiral Stovetop Type of electriccoil stovetop Made from highresistance nichrome alloy 80 nickel 20 chromium High electrical resistivity allows coil to rapidly heat up as electric current passes through it ID: 759680

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Presentation Transcript

Slide1

MATLAB for Breakfast

Evolution of heat distribution across bottom of frying pan on spiral-shaped electric Coil stovetopTim Nickell – EPS 109, Fall 2013 – Prof. Burkhard Militzer

Slide2

Context: Spiral Stovetop

Type of electric-coil stovetop

Made from high-resistance

nichrome

alloy (~80% nickel, ~20% chromium)

High electrical resistivity allows coil to rapidly heat up as electric current passes through it

Goal of simulation: to model heat distribution across bottom surface of aluminum frying pan sitting on this type of burner

Slide3

Coding: methods, techniques, etc.

Main method = variant of the 2D time-dependent heat equation (includes internal heat source): Adapted this equation to 2D, using form from Lab/HW 8 (discretized form of PDE): Pan is composed of aluminum  use the density, heat conductivity, and specific heat of Al to derive kappa (heat diffusivity = k/rho*cp )

Slide4

Conditions and Approximations

Approximations:Pan is extremely thinHeat loss to air via convection = negligible on short time scales (for rough modeling purposes)Lip of pan = excluded – only looking at the bottom surface of pan (2D)Spiral burner = very thin (in the simulation, approximated to width of pixel)

Source term, Q: spiral-shape (like shape of burner) Boundary conditions (bottom of pan): insulating  shaped like a circle (see below)cooling via convection < heat transfer through pan aluminumFor both spiral and circle BC, method from Lab 2 used (if statement with radius r):

Slide5

“And now, our feature presentation:”

Slide6

Bon Appétit! (Thanks for listening!)

You can run my code by:

1. Setting up source distribution pattern (spiral)

First, use

i

, j nested loop with an n = ~201x201 matrix (e.g., for I = 1:n; for j = 1:n)

Next, use method from Lab 2 to get a radius in terms of

i

, j (same one for making circles)

Then, come up with theta in terms of x(

i

), r (two loops, one for top half, one for the other)

Finally, set matrix = 1 (or whatever source temp) for all matrix(

i,j

) with r close to the

r-value

for theta at that point (solve for

rnew

(theta), set equal to r-old)

2. Running the PDE as described above

3. Insulating boundary conditions (eight loops!)