Physics, 22.02.2020 06:11 mckinley2006

A two-dimensional rectangular plate is subjected to prescribed boundary conditions. Using the results of the exact solution for the heat equation presented in Section 4.2, calculate the temperatures along the mid-plane of the plate (x = 1 m) at y = 0.25, 0.5, and 0.75 m by considering the first five nonzero terms of the infinite series. Assess the error resulting from using only the first three terms of the infinite series.

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Answer from: jaaja

The temperature calculated with 5 terms at (1,0.25),(1,0.5) and (1,0.75) are 71.24 C, 94.53 C and 121.14 C respectively.

The temperature calculated with 3 terms at (1,0.25),(1,0.5) and (1,0.75) are 71.24 C, 94.59 C and 121.89 C respectively.

The error percentage between two temperatures for (1,0.25),(1,0.5) and (1,0.75) is 0%, 0.06% and 0.62% respectively.

Explanation:

From the given section, the equation of Temperature distribution is given as

$\theta(x,y) =\frac{T-T_1}{T_2-T_1}$

This is also given by the exact solution as

$\theta(x,y) =\frac{2}{\pi}\sum_{n=1}^{\theta}\frac{(-1)^{n+1}+1}{n}sin(\frac{n\pi x}{L})\frac{sinh({n\pi y}/{L})}{sinh({n\pi W}/{L})}$

For point 1 x=1, y=0.25

From the attached diagram

L=2

W=1

x/L=1/2

y/L=1/8

W/L=1/2

So the equation becomes

$\theta(1,0.25) =\frac{2}{\pi}\sum_{n=1}^{\theta}\frac{(-1)^{n+1}+1}{n}sin(\frac{n\pi }{2})\frac{sinh({n\pi}/{8})}{sinh({n\pi}/{2})}$

Now as seen the value of terms for n=2,4,6,8.... is zero so thus only first five odd terms (n=1,n=3,n=5,n=7,n=9) will be considered as

$\theta(1,0.25) =\frac{2}{\pi}[\frac{(-1)^{1+1}+1}{1}sin(\frac{\pi }{2})\frac{sinh({\pi}/{8})}{sinh({\pi}/{2})}+\frac{(-1)^{3+1}+1}{3}sin(\frac{3\pi }{2})\frac{sinh({3\pi}/{8})}{sinh({3\pi}/{2})}+$ $+\frac{(-1)^{5+1}+1}{5}sin(\frac{5\pi }{2})\frac{sinh({5\pi}/{8})}{sinh({5\pi}/{2})}+\frac{(-1)^{7+1}+1}{7}sin(\frac{7\pi }{2})\frac{sinh({7\pi}/{8})}{sinh({7\pi}/{2})}+\frac{(-1)^{9+1}+1}{9}sin(\frac{9\pi }{2})\frac{sinh({9\pi}/{8})}{sinh({9\pi}/{2})}]$ $\theta(1,0.25) =\frac{2}{3.14}[0.35012+(-0.01761)+0.00108+(-0.00007)+0.00000551)\\\theta(1,0.25) =0.2124\\$

From the diagram

T_2=150

T_1=50

Now the equation becomes

$\theta(1,0.25) =\frac{T-50}{150-50}\\T(1,0.25) =\theta(1,0.25)({150-50})+50\\T(1,0.25) =71.24 \, C$

Using the 1st 3 terms the solution is given as

$\theta(1,0.25) =\frac{2}{3.14}[0.35012+(-0.01761)+0.00108)]\\\theta(1,0.25) =0.2124\\$

The error is 0%.

Similarly for point (1,0.5)

x/L=1/2

y/L=1/4

W/L=1/2

The solution is given as

$\theta(1,0.5) =\frac{2}{3.14}[0.69921]\\\theta(1,0.5) =0.4453\\$

$\theta(1,0.5) =\frac{T-50}{150-50}\\T(1,0.5) =\theta(1,0.5)({150-50})+50\\T(1,0.5) =94.53\, C$

With using 1st 3 terms the value is given as

$\theta(1,0.5) =\frac{2}{3.14}[0.70019]\\\theta(1,0.5) =0.4459\\$

$\theta(1,0.5) =\frac{T-50}{150-50}\\T(1,0.5) =\theta(1,0.5)({150-50})+50\\T(1,0.5) =94.59\, C$

Ther error is given as

$error=\frac{94.59-94.53}{94.53}*100=0.06\%$

Similarly for point (1,0.75)

x/L=1/2

y/L=3/8

W/L=1/2

The solution is given as

$\theta(1,0.75) =\frac{2}{3.14}[1.11693]\\\theta(1,0.75) =0.7114\\$

$\theta(1,0.75) =\frac{T-50}{150-50}\\T(1,0.75) =\theta(1,0.75)({150-50})+50\\T(1,0.75) =121.14\, C$

With using 1st 3 terms the value is given as

$\theta(1,0.75) =\frac{2}{3.14}[1.12874]\\\theta(1,0.75) =0.7189\\$

$\theta(1,0.75) =\frac{T-50}{150-50}\\T(1,0.75) =\theta(1,0.75)({150-50})+50\\T(1,0.75) =121.89\, C$

Ther error is given as

$error=\frac{121.89-121.14}{121.14}*100=0.62\%$

The temperature calculated with 5 terms at (1,0.25),(1,0.5) and (1,0.75) are 71.24 C, 94.53 C and 121.14 C respectively.

The temperature calculated with 3 terms at (1,0.25),(1,0.5) and (1,0.75) are 71.24 C, 94.59 C and 121.89 C respectively.

The error percentage between two temperatures for (1,0.25),(1,0.5) and (1,0.75) is 0%, 0.06% and 0.62% respectively.

A two-dimensional rectangular plate is subjected to prescribed boundary conditions. Using the result

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A two-dimensional rectangular plate is subjected to prescribed boundary conditions. Using the result...