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gerryliyana

  • 3 years ago

Find the steady-state temperature distribution for the semi-infinite plate problem if the temperature of the bottom edge is T = f(x) = x (in degrees; that is, the temperature at x cm is x degrees), the temperature of the other sides is 0o, and the width of the plate is 10 cm.

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  1. gerryliyana
    • 3 years ago
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    @UnkleRhaukus @oldrin.bataku @sirm3d @niksva @ParthKohli and others, would you kindly help me?

  2. mukushla
    • 3 years ago
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    would u give us a geometry of problem? :)

  3. gerryliyana
    • 3 years ago
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    i would give a partial differential equation...,

  4. mukushla
    • 3 years ago
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    ok, and boundary conditions

  5. mukushla
    • 3 years ago
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    \[\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=0\]???

  6. gerryliyana
    • 3 years ago
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    Hint: Answer:\[T = \frac{ 20 }{ \pi } \sum_{n=1}^{\infty} \frac{ (-1)^{n+1} }{ n } e^{-n \pi y/10} \sin (n \pi x/10)\] How to get that.., ??

  7. mukushla
    • 3 years ago
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    ok, u must use "separation of variables"

  8. mukushla
    • 3 years ago
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    suppose that steady state solution can be written as\[T(x,y)=X(x).Y(y)\]plug in the original equation, u will get\[X''Y+XY''=0\]\[\frac{X''}{X}=-\frac{Y''}{Y}=\color\red{\pm}\lambda^2\]now what do u think positive sign or positive?

  9. mukushla
    • 3 years ago
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    sorry positive or negative?

  10. gerryliyana
    • 3 years ago
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    what's lambda ??

  11. mukushla
    • 3 years ago
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    lambda is a positive number, an arbitrary constant...but why we put it there? see this equation\[\frac{X''}{X}=-\frac{Y''}{Y}\]LHS is a function of only \(x\) and RHS is a function of only \(y\) its not possible unless both sides are equal to a constant number

  12. gerryliyana
    • 3 years ago
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    sorry i've no idea about positive and negative, but i just know that it's separation constant

  13. mukushla
    • 3 years ago
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    what do u think? what are boundary conditions?

  14. gerryliyana
    • 3 years ago
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    why \(\pm ??\)

  15. mukushla
    • 3 years ago
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    there are several methods to verify the sign of constant of separation

  16. gerryliyana
    • 3 years ago
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    the boundary conditions is l from 0 up to 10, right ?

  17. mukushla
    • 3 years ago
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    what math u are in? Advanced Engineering Mathematics?

  18. gerryliyana
    • 3 years ago
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    mathematical physics ..,

  19. mukushla
    • 3 years ago
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    ok, boundary conditions\[T=f(x) \ \ @ \ \ y=0\]\[T=0 \ \ @ \ \ y \rightarrow \infty\]\[T=0 \ \ @ \ \ x=0\]\[T=0 \ \ @ \ \ x=10\]what do u think?

  20. gerryliyana
    • 3 years ago
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    i think |dw:1366283650608:dw|

  21. gerryliyana
    • 3 years ago
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    |dw:1366283758632:dw|

  22. mukushla
    • 3 years ago
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    exactly, and \(f(x)=x\) of course

  23. mukushla
    • 3 years ago
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    so lets think what will be the sign of \(\lambda^2\)

  24. gerryliyana
    • 3 years ago
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    because of solution may be either positive or negative ??

  25. gerryliyana
    • 3 years ago
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    ah..., i don't know

  26. mukushla
    • 3 years ago
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    first suppose positive\[\frac{X''}{X}=\lambda^2\]\[X''-\lambda^2 X=0\]\[X=A\sinh \lambda x+B\cosh \lambda x\]set boundary conditions\[T=0 \ \ @ \ \ x=0\]will result in \(X(0)=0\) and it gives \(B=0\)\[T=0 \ \ @ \ \ x=10\]will result in \(X(10)=0\) and it gives \(A=0\) so \(X\) will be equal to zero which is not acceptable so we must choose negative sign

  27. mukushla
    • 3 years ago
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    makes sense?

  28. gerryliyana
    • 3 years ago
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    ah yea..,

  29. mukushla
    • 3 years ago
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    ok, so we have\[\frac{X''}{X}=-\frac{Y''}{Y}=\color\green{-}\lambda^2\]\[X''+\lambda^2 X=0\]\[Y''-\lambda^2 Y=0\]\[X=A\sin \lambda x+B\cos \lambda x\]\[Y=A\sinh \lambda y+B\cosh \lambda y\]and finally setting boundary conditions

  30. mukushla
    • 3 years ago
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    \[T=0 \ \ @ \ \ x=0\]will result in \(X(0)=0\) and it gives \(B=0\)\[T=0 \ \ @ \ \ x=0\]will result in \(X(10)=0\) and it gives \(10\lambda=n\pi\) and\[\lambda=\frac{n\pi}{10} \ \ \ \color\green{\text{eigenvalues}}\]

  31. mukushla
    • 3 years ago
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    or if u want get the exact form wich is given in solution u can write\[Y=Ce^{\lambda y} +De^{-\lambda y} \]

  32. mukushla
    • 3 years ago
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    do u know why?

  33. mukushla
    • 3 years ago
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    general solution of\[Y''-\lambda^2 Y=0\]can be written as\[Y=C\sinh \lambda y+D\cosh \lambda y\]or\[Y=Ce^{\lambda y} +De^{-\lambda y}\]thats for u...find out why :)

  34. mukushla
    • 3 years ago
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    now we know as \(y \rightarrow \infty\) temperature approches zero so \(C=0\)

  35. mukushla
    • 3 years ago
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    its a simple limit calculation, right?

  36. mukushla
    • 3 years ago
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    so summing the answers for \(n=1,2,3,...\) we will have\[T = \sum_{n=1}^{\infty} a_ne^{-\frac{n \pi}{10} y} \sin (\frac{n \pi}{10} x)\]for calculating \(a_n\) final boundary condition will be useful\[T=f(x)=x \ \ @ \ \ y=0\]so\[x = \sum_{n=1}^{\infty} a_n\sin (\frac{n \pi}{10} x)\]from fourier series\[a_n=\frac{2}{10} \int_{0}^{10} x\sin (\frac{n \pi}{10} x) \ \text{d}x=\frac{20}{n \pi} (-1)^{n+1}\]We are done,I hope its helpful

  37. mukushla
    • 3 years ago
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    let me know if it is unclear.

  38. gerryliyana
    • 3 years ago
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    ah thnak you so much :)

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