Answer:
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
[tex]\theta(x,y) =\frac{T-T_1}{T_2-T_1}[/tex]
This is also given by the exact solution as
[tex]\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})}[/tex]
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
[tex]\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})}[/tex]
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
[tex]\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})}+[/tex][tex]+\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})}][/tex][tex]\theta(1,0.25) =\frac{2}{3.14}[0.35012+(-0.01761)+0.00108+(-0.00007)+0.00000551)\\\theta(1,0.25) =0.2124\\[/tex]
From the diagram
T_2=150
T_1=50
Now the equation becomes
[tex]\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[/tex]
Using the 1st 3 terms the solution is given as
[tex]\theta(1,0.25) =\frac{2}{3.14}[0.35012+(-0.01761)+0.00108)]\\\theta(1,0.25) =0.2124\\[/tex]
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
[tex]\theta(1,0.5) =\frac{2}{3.14}[0.69921]\\\theta(1,0.5) =0.4453\\[/tex]
[tex]\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[/tex]
With using 1st 3 terms the value is given as
[tex]\theta(1,0.5) =\frac{2}{3.14}[0.70019]\\\theta(1,0.5) =0.4459\\[/tex]
[tex]\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[/tex]
Ther error is given as
[tex]error=\frac{94.59-94.53}{94.53}*100=0.06\%[/tex]
Similarly for point (1,0.75)
x/L=1/2
y/L=3/8
W/L=1/2
The solution is given as
[tex]\theta(1,0.75) =\frac{2}{3.14}[1.11693]\\\theta(1,0.75) =0.7114\\[/tex]
[tex]\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[/tex]
With using 1st 3 terms the value is given as
[tex]\theta(1,0.75) =\frac{2}{3.14}[1.12874]\\\theta(1,0.75) =0.7189\\[/tex]
[tex]\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[/tex]
Ther error is given as
[tex]error=\frac{121.89-121.14}{121.14}*100=0.62\%[/tex]
So
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.
