1

Designing a Temperature Distribution Profile

Jai-Kwan Bae

12/11/2014

Prof. Gracewski

2

Abstract

In order to design an appropriate temperature distribution for fish, a profile T(φ) needs to

be determined for the temperature controller. Once the profile is decided, I could use the separation

of variables and geometric symmetry on the Laplace’s equation to derive an ordinary differential

equation. Then, the initial conditions, boundary conditions and orthogonality are used to determine

the coefficients of the temperature distribution. In the last section, we use our result to calculate

the heat flow that can be used as a sanity check.

Problem Statement

Figure 1. Spherical Coordinates Figure 2. Temperature Distribution of a Pond

PDE: 𝛻2𝑢 = 0 for 0 < 𝜌 < 𝑏 and 0 < 𝛷 <𝜋

2

BC’s: 𝑢 (𝜌,𝜋

2) = 0

𝑢(𝑎, 𝛷) = 𝑇(𝛷)

The heat flow into the pudding is equal to K∯ ∇ 𝑢 ∙ 𝑛|𝜌=𝑎𝑑𝑆

Parameters: a=5m, 𝐾0 = 0.6𝑊/(𝑚 𝑜𝐶), 𝑢 (𝑟,𝜋

2) = 0, 𝑇(0) = 8℃, 𝑇(𝛷) > 0.

3

Derivation

PDE: 𝛻2𝑢 = 0 for 0 < 𝜌 < 𝑏 and 0 < 𝛷 <𝜋

2

BC’s: 𝑢 (𝜌,𝜋

2) = 0 𝑢(𝑎, 𝛷) = 𝑇(𝛷)

In spherical coordinates, Laplace’s equation has the form of:

∇2𝑢 =1

𝜌2

𝜕

𝜕𝜌(𝜌2

𝜕𝑢

𝜕𝜌) +

1

𝜌2𝑠𝑖𝑛2𝛷

𝜕2𝑢

𝜕𝜃2+

1

𝜌2𝑠𝑖𝑛𝛷

𝜕

𝜕𝛷(sin 𝛷

𝜕𝑢

𝜕𝛷) = 0

Since the temperature distribution is axisymmetric, we know:

𝜕2𝑢

𝜕𝜃2= 0

Then PDE becomes:

∇2𝑢 =1

𝜌2

𝜕

𝜕𝜌(𝜌2

𝜕𝑢

𝜕𝜌) +

1

𝜌2𝑠𝑖𝑛𝛷

𝜕

𝜕𝛷(sin 𝛷

𝜕𝑢

𝜕𝛷) = 0

- Separation of Variables

Assume

𝑢(𝜌, 𝛷) = 𝐹(𝜌)𝐺(𝛷)

Obtain:

𝐺

𝜌2

𝑑

𝑑𝜌(𝜌2

𝑑𝐹

𝑑𝜌) +

𝐹

𝜌2 sin 𝛷

𝑑

𝑑𝛷(sin 𝛷

𝑑𝐺

𝑑𝛷) = 0

4

Multiply ρ2

FG and rearrange it:

1

𝐹

𝑑

𝑑𝜌(𝜌2

𝑑𝐹

𝑑𝜌) = −

1

𝐺 sin 𝛷

𝑑

𝑑𝛷(sin 𝛷

𝑑𝐺

𝑑𝛷) = 𝜆

Broke up the above equation into two ordinary differential equations:

ODE#1: 𝑑

𝑑Φ(sin 𝛷

𝑑𝐺

𝑑Φ) + 𝜆 sin 𝛷 𝐺 = 0 0 < 𝛷 <

𝜋

2

BC #1: 𝐺 (𝜋

2) = 0, |𝐺(0)| < ∞

ODE#2: 𝜌2 𝑑2𝐹

𝑑𝜌2 + 2𝜌𝑑𝐹

𝑑𝜌− 𝐹𝜆 = 0 0 < 𝜌 < 𝑎

BC #2: |𝐹(0)| < ∞

IC: 𝑢(𝑎, 𝛷) = 𝑇(𝛷)

Then we solve the Eigenvalue problems:

- Suppose λ=0,

Observe,

𝜌2𝑑𝐹

𝑑𝜌= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝑘1, sin Φ ∗

𝑑𝐺

𝑑Φ= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝑘2

Then,

F =𝑘1

𝜌+ 𝑘1

′, 𝐺 = 𝑘2 ∫ csc Φ 𝑑Φ = 𝑘2 (log (tanΦ

2)) + 𝑘2

′

By the boundary conditions, |F(0)|, |G(0)| < ∞, 𝐺 (𝜋

2) = 0

∴ F = 𝑘1′, 𝐺 = 𝑘2

′ = 0, 𝑢0 = 0, 𝑡𝑟𝑖𝑣𝑖𝑎𝑙

5

- Suppose λ>0,

For ODE#1, in Sturm Louiville Form, we can see:

𝑝(𝛷) = 𝑠𝑖𝑛 𝛷 , 𝑞 = 0, 𝜎(𝛷) = 𝑠𝑖𝑛 𝛷

∴ 𝐺𝑚(𝛷)=𝑃𝑚(𝑐𝑜𝑠𝛷), 𝜆𝑚 = 𝑚(𝑚 + 1)

However, since G(𝜋/2)=0, m is always odd.

∴ 𝐺𝑛(𝛷)=𝑃2𝑛−1(𝑐𝑜𝑠𝛷), 𝜆𝑛 = (2𝑛)(2𝑛 − 1), for n=1,2,3, ….

For ODE#2, Assume 𝐹(𝜌) = 𝐶𝜌𝑠

2𝜌(𝐶𝑠𝜌𝑠−1) + 𝜌2(𝐶𝑠(𝑠 − 1)𝜌𝑠−2) − 𝜆𝐶𝜌𝑠 = 0

𝐶𝜌𝑆(2𝑠 + 𝑠(𝑠 − 1) − 𝜆) = 0

𝑠2 + 𝑠 − 𝜆 = 0

𝑠 = 2𝑛 − 1 𝑜𝑟 − 2𝑛

∴ 𝐹𝑛(𝜌) = Anρ2n−1 +Bn

ρ2n,

By the boundary condition #2:

∴ 𝐹𝑛(𝜌) = Anρ2n−1,

∴ 𝑢𝑛 = 𝐹𝑛𝐺𝑛,

𝑢 = ∑ 𝑢𝑛

∞

𝑛=1

= ∑ Anρ2n−1𝑃2𝑛−1(𝑐𝑜𝑠𝛷)

∞

𝑛=1

Note that since the differential equations have the Strum Louiville Form, λ ≥ 0.

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- Orthogonality

∫ 𝑃𝑛(𝑐𝑜𝑠𝛷)𝑃𝑚(𝑐𝑜𝑠𝛷) 𝑠𝑖𝑛 𝛷 𝑑𝛷

𝜋2

0

= 0 when 𝑚 ≠ 𝑛

∫(𝑃2𝑛−1(𝑐𝑜𝑠𝛷))2 𝑠𝑖𝑛 𝛷 𝑑𝛷

𝜋2

0

=1

4𝑛 − 1

The initial conditions is :

𝑢(𝑎, 𝛷) = 𝑇(𝛷) = ∑ 𝐴𝑛𝑎2𝑛−1𝑃2𝑛−1(𝑐𝑜𝑠𝛷)

∞

𝑛=1

∴ ∫ 𝑇(𝛷)𝑃2𝑚−1(𝑐𝑜𝑠𝛷) 𝑠𝑖𝑛 𝛷 𝑑𝛷

𝜋2

0

= ∑ 𝐴𝑛𝑎2𝑛−1 ∫ 𝑃2𝑛−1(𝑐𝑜𝑠𝛷)𝑃2𝑚−1(𝑐𝑜𝑠𝛷) 𝑠𝑖𝑛 𝛷 𝑑𝛷

𝜋2

0

∞

𝑛=1

=𝐴𝑚𝑎2𝑚−1

4𝑚 − 1

∴ 𝐴𝑚 =4𝑚 − 1

𝑎2𝑚−1∫ 𝑇(𝛷)𝑃2𝑚−1(𝑐𝑜𝑠𝛷) 𝑠𝑖𝑛 𝛷 𝑑𝛷

𝜋2

0

Thus, the solution is:

𝑢 = ∑ 𝑢𝑛

∞

𝑛=1

= ∑ Anρ2n−1𝑃2𝑛−1(𝑐𝑜𝑠𝛷)

∞

𝑛=1

Where

𝐴𝑛 =4𝑛 − 1

𝑎2𝑛−1∫ 𝑇(𝛷)𝑃2𝑛−1(𝑐𝑜𝑠𝛷) 𝑠𝑖𝑛 𝛷 𝑑𝛷

𝜋2

0

7

- Heat flow of the hemisphere surface

K∯ ∇ 𝑢 ∙ 𝑛|𝜌=𝑎𝑑𝑆= K∯∂u

∂ρ�̂� ∙ �̂� 𝜌 sin Φ 𝑑𝜃𝜌𝑑Φ = 2πK𝑎2 ∫

∂u

∂ρ

𝜋

20

sin Φ 𝑑Φ

= 2πK𝑎2 ∫ ∑(2𝑛 − 1)𝐴𝑛𝑎2𝑛−2𝑃2𝑛−1(𝑐𝑜𝑠𝛷) sin Φ 𝑑Φ

∞

𝑛=1

𝜋2

0

= 2𝜋𝐾 ∑(2𝑛 − 1)𝐴𝑛𝑎2𝑛 ∫ 𝑃2𝑛−1(𝑐𝑜𝑠𝛷) 𝑠𝑖𝑛 𝛷 𝑑𝛷

𝜋2

0

∞

𝑛=1

= 2𝜋𝐾 ∑(2𝑛 − 1)𝐴𝑛𝑎2𝑛 ∫ 𝑃2𝑛−1(𝑥)𝑑𝑥

1

0

∞

𝑛=1

- Heat flow of the upper surface

K∯ ∇ 𝑢 ∙ 𝑛𝑑𝑆 = 𝐾 ∯1

ρ

𝜕𝑢

𝜕ΦΦ̂ ∙ Φ̂𝜌𝑑𝜃𝑑𝜌 = 2𝜋𝐾 ∫

𝜕𝑢

𝜕Φ𝑑𝜌

𝑎

0

= 2πK ∫ ∑ 𝐴𝑛𝜌2𝑛−1𝜕𝑃2𝑛−1(𝑐𝑜𝑠𝛷)

𝜕Φ𝑑𝜌

∞

𝑛=1

𝑎

0

= 2πK ∑ 𝐴𝑛

𝑎2𝑛

2𝑛

𝑑𝑃2𝑛−1(𝑐𝑜𝑠𝛷)

𝑑Φ

∞

𝑛=0

|Φ=

π2

∴ The total Heat flux = 2πK ∑ 𝐴𝑛𝑎2𝑛 [(2𝑛 − 1) ∫ 𝑃2𝑛−1(𝑥)𝑑𝑥

1

0

−1

2𝑛

𝑑𝑃2𝑛−1(𝑥)

𝑑𝑥|𝑥=0]

∞

𝑛=1

Note that this value should be 0 as a sanity check.

8

Results

(a) 𝑇1(𝛷) = 𝑇0 cos Φ

Figure 3. Plot of 𝑇0 cos Φ

𝐴1 =3

𝑎∫ 8 cos2 𝜃 sin 𝜃 𝑑𝜃

𝜋/2

0=

24

a∗

1

3=

8

𝑎, 𝐴𝑛 = 0 𝑤ℎ𝑒𝑛 𝑛 ≠ 1

∴ 𝑢1 =8

𝑎𝜌 𝑐𝑜𝑠 𝛷

In the previous section, we derived the heat flux: K∯ ∇ 𝑢 ∙ 𝑛|𝜌=𝑎𝑑𝑆

=2𝜋𝐾 ∑ (2𝑛 − 1)𝐴𝑛𝑎2𝑛 ∫ 𝑃2𝑛−1(𝑥)𝑑𝑥1

0∞𝑛=1 = 2𝜋𝐾 ∗ 𝐴1𝑎2 ∗

1

2= 24𝜋

Figure 4. Plot of 𝑢1(𝜌, 0)

0.5 1.0 1.5

2

4

6

8

1 2 3 4 5

2

4

6

8

9

Figure 5. Contour plot of 𝑢1(𝜌, Φ)

(b) 𝑇2(𝛷) = 3 𝑐𝑜𝑠 𝛷 + 5 𝑐𝑜𝑠3 𝛷

Figure 6. Plot of 𝑇2(𝛷) = 3 𝑐𝑜𝑠 𝛷 + 5 𝑐𝑜𝑠3 𝛷

𝐴1 =3

𝑎∫ (3 𝑐𝑜𝑠 𝛷 + 5 cos3 𝛷) cos 𝛷 sin 𝛷 𝑑𝛷

𝜋/2

0=

6

a= 1.2,

𝐴2 =7

𝑎3∫ (3 𝑐𝑜𝑠 𝛷 + 5 𝑐𝑜𝑠3 𝛷)𝑃3(cos 𝛷) sin 𝛷 𝑑𝛷

𝜋/2

0

=2

𝑎3=

2

125

0.5 1.0 1.5

2

4

6

8

10

𝐴𝑛 = 0 𝑤ℎ𝑒𝑛 𝑛 ≠ 1, 2

∴ 𝑢2(𝜌, 𝛷) = 𝐴1𝜌 𝑐𝑜𝑠 𝛷 + 𝐴2𝜌3𝑃3(𝑐𝑜𝑠 𝛷)

=6

5𝜌 𝑐𝑜𝑠 𝛷 +

1

125𝜌3(5 cos3 Φ − 3 𝑐𝑜𝑠 𝛷)

In the previous section, we derived the heat flux: K∯ ∇ 𝑢 ∙ 𝑛|𝜌=𝑎𝑑𝑆

=2πK𝑎2 ∫∂u

∂ρ

𝜋

20

sin Φ 𝑑Φ = 2𝜋𝐾 ∑ (2𝑛 − 1)𝐴𝑛𝑎2𝑛 ∫ 𝑃2𝑛−1(𝑥)𝑑𝑥1

0∞𝑛=1

= 2πK (𝐴1𝑎2 ∗1

2+ 3𝐴2𝑎4 ∗ (−

1

8)) = 13.5𝜋

Figure 7. Plot of 𝑢2(𝜌, 0)

Figure 8. Contour Plot of 𝑢2(𝜌, Φ)

1 2 3 4 5

2

4

6

8

11

Discussion

In order to design the temperature profile for fish, I had to choose a specific species of fish

that I am targeting for. Lake Trout is one of the most common fish that you can see in a pond and

the optimal temperature for the species is around 40℉ which is about 4℃. Compare to the original

solution, you can observe that the second profile has more gentle temperature change. I also tried

to make the profile as simple as possible such that the fish inside the pond won’t get confused by

the artificial manipulation. The sanity checks were achieved by checking all kinds of various

conditions, such as T(Φ) > 0, u (ρ,π

2) = 0, T(0) = 8. The most characteristic sanity check was to

see if the total heat flux is 0. We proved above that the total heat flux becomes 0 when

(2𝑛 − 1) ∫ 𝑃2𝑛−1(𝑥)𝑑𝑥1

0=

1

2𝑛

𝑑𝑃2𝑛−1(𝑥)

𝑑𝑥|𝑥=0. It turns out the equation is satisfied in case n=1 and

n=2. From the conservation energy, we can deduce that it is true for all eigenfunctions.

Summary

This project focuses on solving the Laplace’s equation in an axisymmetric situation. We

used the separation of variables to obtain the ordinary differential equation and could observe the

eigenfunctions are Legendre polynomials. By setting a temporary initial condition with fixed

boundary conditions, all the coefficients were obtained and the graphs were drawn with

Mathematica as figures.