The Perfect Brownie Pan
Riley Dulin
Jun Seok Yeo Nick Fuerstenberg
March 2015
1
1
Abstract
This paper examines different shapes for a brownie pan such that it maximizes
the area covered in an oven, and the temperature distribution across the brownies. We also examine the relationship between the perimeter of the pan and the
temperature distribution. We used a combination of analytical and numerical
solutions to find these values, and wrote MATLAB code to simulate the temperature distribution. Through these methods, we hope to discover the optimal
shape for a brownie pan.
2
Introduction
When baking in a rectangular pan, heat is concentrated in the 4 corners and
the product gets overcooked at the corners (and to a lesser extent at the edges).
In a round pan the heat is distributed evenly over the entire outer edge and the
product is not overcooked at the edges, and there are no corners at which the
product will be extremely overcooked. However, since most ovens are rectangular in shape using round pans is not efficient with respect to using the space in
an oven. Develop a model to show the distribution of heat across the outer edge
of a pan for pans of different shapes - rectangular to circular and other shapes
in between.
Assume:
1. A width to length ratio of W/L for the oven which is rectangular in shape
2. Each pan must have an area of A
3. Pans in the oven must have a space d between each other and the sides of
the oven
4. Initially two racks in the oven, evenly spaced
Develop a model that can be used to select the best type of pan (shape)
under the following conditions:
1. Maximize number of pans that can fit in the oven (N)
2. Maximize even distribution of heat (H) for the pan
3. Optimize a combination of conditions (1) and (2) where weights p and
(1 − p) are assigned to illustrate how the results vary with different values
of W/L and p
[1]
2
3
Assumptions
First
The ideal shape of the brownie pan is a member of the family of Superellipses (also known as Lamé Curves)
x n y n
(1)
+ =1
a
b
n ∈ [2, ∞)
(2)
Where a and b are the positive half-width and half-length respectively
of the shape. Below is a picture of the ”squircle”, a superellipse with
n = 4, a = 1, b = 1:
Figure 1: A ”Squircle”: x4 + y 4 = 1
This is justified because the family of superellipses go from perfect ellipse
to perfect rectangle (n = 10 is sufficiently close to infinity to make a nearperfect rectangle), and in between they form rounded rectangles, which
would be the perfect compromise between the two extremes.
3.1
Area Assumptions
Second
The optimal layout of the pans in the oven is in a rectangular spacing
pattern
Third
a and b have the same ratio to each other as the oven’s width to length
ratio. This is justified because in order to maximize the rectangular spacing pattern, an equal amount of rows and columns of brownie pans are
necessary because a square has a higher number of pans than rectangles,
and in order to get a square pattern each pan must match the dimensions
of the oven
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Fourth
The second rack is exactly the same in alignment and heat distribution
as the first rack, so we will only analyze the situation for one rack. This
is justified because an oven heats the top and bottom rack in approximately the same way, and we assume the brownies on the top rack will
not interfere with the radiation received by the bottom rack
3.2
Heat Assumptions
Fifth
Heat flux only flows from high temperature to low temperature and does
not flow back
Sixth
The pan and the brownies are perfect black-bodies
Seventh
The temperature and the radiation in the oven are equally distributed
Eigth
The temperature of the pan is equally distributed
Ninth
All energy that the pan and the brownies receive from radiation converts
to heat with 100% transfer (ideal transfer rate)
Tenth
Thermal conductivity is constant even when there is a change in temperature
Eleventh
Oven remains the same temperature and provides unlimited heat until
equilibrium
Twelfth
The oven has a vacuum inside, and thus there are no convection currents,
and only radiative heat transfer. We assume this because it makes calculations significantly easier to make
3.3
Strategies
We had several strategies of how to model this problem. The problem was split
into two halves, the maximization of the ”Area Score”, and the maximization
of the ”Heat Score”. These are termed as scores because in order to compare
things as different as area and temperature distributions, we needed to normalize
the values. So we give each field a score between 0 and 1 by subtracting the
minimum and dividing by the range of the area and temperature values, so that
the minimum scoring shape is assigned 0, the maximum shape is assigned 1,
4
and the rest are scaled to be between those two numbers while still retaining
the shape of their plot. We then use those values to compare against each other
and decide what is the optimal shape.
During our research we found that an alternative exists for calculating the
Heat Score, namely, the Perimeter Score. More details are found in the perimeter section.
3.4
3.4.1
Constants and Variables
Constants
W
L
r
A
d
p
h
pt
Kb
cb
cp
σ
ρb
ρp
TR
TO
∆t
t0
tf
∆x
ccube
Qtop
Qbottom
Qside,j
Width of the oven (m)
Length of the oven (m)
Ratio of W/L
Area of the pan (m 2 )
Distance between each pan (m)
Weight of Area Score (Heat Score is 1 − p)
Height of the brownie pan (m)
Thickness of the brownie pan (m)
Thermal conductivity of brownies (W/(m*K))
Specific heat of brownies (J/(kg*K))
Specific heat of the pan (J/(kg*K))
Stefan-Boltzmann’s Constant (W/(m2 ∗ K 4 ))
Density of the brownies (kg/m3 )
Density of the pan (kg/m3 )
Room temperature (K)
Oven temperature (K)
Time step (s)
Start time (s)
End time (s)
Spatial step (m)
Specific heat for one cube (J/(kg*K))
Heat from the oven radiation (J/kg)
Heat from the pan (J/kg)
Heat from sides of a cube (J/kg)
5
3.4.2
Variables
n
a
b
x
y
t
N
P
Tp
Tm
Tside,j
T (x, y, t)
SA (W, L, a, b, d)
SH (a, b, n)
SP (a, b, n)
4
Parameter of the superellipse
Half-width of the superellipse (m)
Half-length of the superellipse (m)
Spatial x coordinate (m)
Spatial y coordinate (m)
Time value (s)
Number of pans
Perimeter of the superellipse
Temperature of the pan (K)
Temperature for one cube on the mth second (K)
Temperature of the jth side of a brownie (K)
Temperature function (K)
Area Score
Heat Score
Perimeter Score
Model
The formal solution to this problem is the value of n that maximizes the following expression
p ∗ SA (W, L, a, b, d) + (1 − p) ∗ SH (a, b, n)
(3)
where n is the parameter of a superellipse given by (1) and (2) and each S is
normalized in the following way
SA :
V − min(V)
S=
(4)
max(V) − min(V)
SH , SP :
S−1 =
min(V) − V
+1
max(V) − min(V)
(5)
where V is a vector of values that are determined by each respective score and
S is the normalized score vector. SH and SP use the ”inverse” normalization
function because their values are opposite to the desired measurement. See
Sections 4.1.3, 4.2.5, and 4.3.1 for individual explanations of why the scores
have their formulae.
4.1
Area Score
To find the Area Score we followed this outline:
1. Find a and b
2. Calculate N , the number of pans
3. Normalize the values
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4.1.1
Find a and b
To find a and b, we use a formula for finding the area of a superellipse:
2
Γ 1 + n1
A = 4ab ∗
Γ 1 + n2
a = br
(6)
(7)
We then solve these equations, and thus we have found the a and b for this value
of n. We will use this for all of the scores we compute.
4.1.2
Find N
To find N , we must form equations for how many pans can fit in the oven. We
know a and b, along with the constants that represent the oven size, and we
form these two linear equations:
(Ncol + 1) ∗ d + Ncol ∗ 2a = W
4.1.3
(8)
(Nrow + 1) ∗ d + Nrow ∗ 2b = L
(9)
N = Nrow ∗ Ncol
(10)
Normalize
Once we finish calculating N for each n, we normalize them using (4), because
the number of pans that can fit in the oven is proportional to our desired behavior. In other words, as the number of pans increases, so should the score, so
we use the first normalization function.
4.2
Heat Score
The Heat Score is the most complicated of all the scores to calculate, and
requires the creation of a simulation of the cooking brownies. To simulate the
event, we break the brownie mixture down into very small cubes and find the
energy transfer through each cube. To accomplish this we wrote some code in
MATLAB that simulated the transfer of heat across the brownie mixture. See
Appendix A for more details on the simulation, and see Appendix B for our
consideration of an analytical solution
4.2.1
Heat Transfer for the pan
As shown in the diagram, the pan accepts heat from the oven in the form of
radiation according to Stefan–Boltzmann’s Law:
∂Qpan
= AσTO4
∂t
Qpan = cp ∗ ∆Tp
∆Tp =
AσTO4
cp
7
∆t
(11)
(12)
(13)
Figure 2: Diagram of Heat Transfer
4.2.2
Heat Transfer for a Single Brownie Cube
Figure 3: Segment of brownie into cubes
The total energy delivered to a single brownie cube of width and length ∆x
is given by the sum of the heat transferred by all of its neighboring brownies, or
the pan if it is touching the pan, plus the heat transfer from the top of the oven
and the heat transfer from the bottom of the pan. To determine if the cube is
touching the pan, we use (1) with the x and y of the brownie and see if that
is > 1. If so, then that side will be set to the Tp . The heat transfer equations
from radiation (from the oven on the top of the cube) and conduction (from the
pan on the bottom of the cube) are as follows:
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4.2.3
Heat Transfer for a Single Brownie Cube
Figure 4: Diagram of Heat Transfer
Figure 4 shows how the heat is transferred onto one cube from multiple
sources
Qtop = ∆t∆x2 σTO4
(14)
Tp − Tm
h
= ∆thkb (Tside,j − Tm )
Qbottom = ∆t∆x2 2kb
(15)
Qside,j
(16)
Since
ccube = cb ρb ∆x2 h
(17)
Q = ccube ∗ ∆T
(18)
the equation for the change in temperature for one cube from the top and bottom
is:
∆t∆x2
Tp − Tm
∆T1 =
(19)
σTO4 + 2kb
ccube
h
and the equation for the change in temperature for one cube from its neighboring
cubes is:


4
∆tkb h X
∆T2 =
(Tside,j ) − 4Tm 
(20)
ccube
j=1
Combine them together to get the net change in temperature for one brownie
cube:
Tm+1 = Tm + ∆T1 + ∆T2
(21)
T0 = TR
(22)
We split the mixture into a 200x200 matrix of these cubes, and calculate this
change in temperature for each cube for m = 1200, or 20 minutes (the average
baking time for a brownie recipe). [2] [3]
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4.2.4
Average Variance
In order to assign a particular number for a heat distribution to be described
by, we decide to use the average standard deviation of the temperature over the
whole time period. The average standard deviation is defined as the square root
of the average variance:
v
u P
u tf 2
u
σ
t m=0 Tm
σT =
(23)
tf
This number increases approximately linearly as the n value increases, then
plateaus at around n > 8.
4.2.5
Normalize
Similarly to the other scores, we then normalize this score using (5), because an
increase in average variance means that the heat is not as well distributed, which
goes against our desired characteristic. Thus we use the ”inverse” normalization
function to make low variances higher valued.
4.3
Perimeter Score
We developed a third score, the Perimeter score, as a response to the length of
time it took to run the heat simulation. We assume that the strength of the
Heat Score is dependent only on one factor, the perimeter of the superellipse.
This is a logical assumption because brownies typically only burn at the edges
and corners, and so if the area is fixed, then the minimal perimeter would have
the best Heat Score. We will verify if this is true by comparing it to the Heat
Score and seeing if it is a decent approximation. Finding the Perimeter Score
is fairly straightforward, after finding a and b during the Area Score part, we
simply need to find the arc length of the superellipse to get the perimeter:
v
2(n−1)
Z au
u
n
b2
xn
t
1+ 2
4∗
dx
(24)
n
n
a
a −x
0
We can solve this using numerical integration software.
4.3.1
Normalize
This perimeter value is then normalized using (5), because as perimeter increases, the heat is not as well distributed, which goes against our desired
characteristic. Thus we use the ”inverse” normalization function to make low
variances higher valued.
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5
Verification
After running the simulation and obtaining our results, we find that the perfect
shape of the brownie is given by n = 3.3. The shape of this pan and the plot
of the scores are shown in Figures 6 and 5, respectively. Figure 7 shows the
simulation of the heat distribution over time for n = 3.3 in a 3D plot and a 2D
contour plot.
Figure 5: Area Scores(Blue), Heat Scores(Orange), and Perimeter Scores(Red)
Figure 6: The Shape of the Pan
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(a) t = 8
(b) t = 8
(c) t = 32
(d) t = 32
(e) t = 56
(f) t = 56
(g) t = 80
(h) t = 80
Figure 7: Temperature Graphs (t in minutes)
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5.1
Perimeter Perspective
As you can see in Figure 5, the perimeter and the heat score are somewhat
similar in their distribution, and the heat score moves closer to the perimeter
score as ∆x is decreased. And if the perfect n is recalculated using perimeter
instead, we find n = 3.3 again. Therefore we conclude that the perimeter score
is a suitable approximation for the heat score. For future tests then, we would
use the perimeter instead of our simulation because finding the perimeter is
orders of magnitude faster than finding the heat distributions.
6
Discussion
The true goal of modelling some situation is that in the future, the model
could be used instead of doing any actual physical experiment. In order to
assess whether our model meets these goals, we look at the predictive power
of the model, and the model’s behavior under changes in the initial conditions
or constants. After coming to a conclusion about these characteristics, we will
then assess what are the weaknesses and strengths of our model, and what we
would change if we were to do this again.
6.1
Predictive Power
The model’s predictive power is relatively strong as long as the assumptions
made under section 3 are used. For any given value of n, we can see how many
brownie pans will fit in the oven, and what the temperature distribution across
the pan will be; however, beyond the realm of superellipses, we would need to
adjust our model to allow other shapes to be analyzed in the same way. The
changes would be minimal though, because regardless of the shape, the Area
Score and the Heat Score or even the Perimeter Score would still yield similar
results, and could still be used to find the maximum point of a shape. A possible
extension to this model would be to change the assumed perfect shape of the
pan to instead have the form of the superformula:
r(ϕ) =
! 1
cos mϕ n2 sin mϕ n3 − n1
4
4
+
a
b
(25)
This equation allows for many more various shapes, and would be a much more
general solution of this problem. If we used this formula, it would become
almost infeasible to test all of the shapes possible, but we would instead rely
wholly on an analytical solution, by finding the maximum of the perimeter and
the minimum of the dimensions. Due to the amount of time it takes to produce
a Heat Score and the difficulty of solving a partial differential equation for any
given boundary conditions, we would focus on perimeter instead.
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6.2
Changes in Initial Conditions
Our model does not handle changes in the initial conditions well. The Area
Score is highly sensitive to changes in W , L, A, and d. By changing those
values we saw behavior that varied from a constant function (every shape took
up the same number of pans in the oven, this happens when the size of the
oven is too small compared to the area of the pan, so that minute changes in
the dimensions of the pan don’t help as much) to a function with a large gap
(when d was large enough that a slightly smaller pan size allowed it to fit one
more pan in, and cause a large jump in the graph). To make the Area Score
more noticeable, we increased the size of a standard oven by a factor of 7, or in
other words turned it into an industrial sized oven, and this produced a much
more dynamic Area Score. The Heat Score is also sensitive to changes, the most
dramatic of which is changing ∆x, the size of the cubes. If this value gets too
large (> 0.001) the average variance actually decreases as n increases, which
produces a curve in the exact opposite shape. In order for our model to be the
most correct, the limit should be taken as lim∆x→0 SH . The same thing also
applies to ∆t, and it should also be as close to 0 as possible. In this way, our
model is sensitive to changes in the initial conditions, and thus will not adapt
well to different parameters.
6.3
Simulation Error
Another note to add is that our Heat Score simulation stops at t = 1200 seconds, or equivalently 20 minutes; however, in section 5 it is noticeable that the
equilibrium occurs after 80 minutes. We had tested this before, and the length
of time observed changed the variances (in relation to each other) very little,
so for the sake of the time it would take to simulate until equilibrium we shortened it to be until a normal brownie baking time. One possible reason that the
oven took so long was because under the Twelfth Assumption, we considered
convection currents caused in the oven to be negligible, but they could account
for a much larger percentage of the heat transfer than we realized. We would
probably include convection currents if we were to do this model again.
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A
Simulation Analysis
In our numerical simulation in MATLAB, we have to consider any possible
sources of error. One of the largest possible sources is numerical instability.
Our simulation creates 2 200x200 matrices (one for current and one for the
last temperature) and runs operations on it m = 1200 times. This means
that the simulation is O(ijm) (i is the number of rows in one matrix, j is the
number of columns in one matrix, and m is the number of times to iterate
over). Since this is a large number of calculations (due to ∆x and ∆t being
close to 0), the chance for error is non-negligible. In the calculation of the
temperature change from the side cubes (20), we are subtracting two very similar
numbers, Tside,j − Tm and summing those 4 differences. Subtraction of very
similar numbers creates numerical instability; however, we could not see a way
to mitigate these subtractions. Another possible source is when we compute the
average variance (23), we subtract the sum of squares and the square of sums
over m, which are two large numbers being subtracted from each other that are
relatively close in magnitude, so there could be some instability there as well.
B
Analytical Solution
An analytical solution exists for the brownies if we simplify the model more. If
we assume that the pan reaches the oven temperature almost instantly (plausible
given the thermal diffusivity, α2 , of aluminum pans is much greater than the
thermal diffusivity for chocolate brownies), and that the change in temperature
from the oven emitting radiation on the top of the brownies directly is negligible
(because it hits all brownies evenly, it does not affect the distribution). Under
these assumptions, this is the partial differential equation we came up with to
describe the analytic solution:
αb2 (Txx + Tyy ) = Tt
Subject to the following boundary and initial conditions:
!
r
xn
n
T x, b 1 − n , t = TO
a
T (x, y, 0) = TR
(26)
(27)
(28)
Unfortunately, we do not know of a way to solve this equation, and the underlying assumptions are questionable, so we decided not to pursue this route.
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References
[1] The ultimate brownie pan. Mathematical Contest in Modeling, 2013.
[2] R. Byron Bird, Warren E. Stewart, and Edwin N. Lightfoot. Transport
Phenomena. John Wiley Sons, Inc, 1960.
[3] H. S. Carslaw and J. C. Jaeger. Conduction of Heat in Solids. Oxford
University Press, 1959.
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