Question

For the slab shown, we have derived the equations of the temperature distribution on the Two dimensions to be: 2. T(x,y)-o, (sin A, xHsinh , y) Where an the coefficient of boundary conditions using Orthogonality was given by:-- 2-1 f(x)sin λ"xar L sinh λ W If: fx) - To.sin(x/a), where To is constant Find equations for temperature distribution, txy) and heat flux gtry your works in sequence, formula cited firs and then numerical values Units need for all answers. f the or formulas not listed, the answers EVENIE correct will have no points Page: 2 You shal show all

Answer 1

The given temperature distribution T ( x, y ) is
  
And so,
$\frac{\partial T}{\partial x}=\sum_{n=1}^{\infty}a_n\lambda_n\cos(\lambda_n x)\sinh(\lambda_n y)$
And
   $\frac{\partial^2 T}{\partial x^2}=\frac{\partial }{\partial x}\left ( \frac{\partial T}{\partial x} \right )=-\sum_{n=1}^{\infty}a_n\lambda_n^2\sin(\lambda_n x)\sinh(\lambda_n y)$
And similarly
      $\frac{\partial T}{\partial y}=\sum_{n=1}^{\infty}a_n\lambda_n\sin(\lambda_n x)\cosh(\lambda_n y)$
And
   $\frac{\partial^2 T}{\partial y^2}=\frac{\partial }{\partial y}\left ( \frac{\partial T}{\partial y} \right )=\sum_{n=1}^{\infty}a_n\lambda_n^2\sin(\lambda_n x)\sinh(\lambda_n y)$
And thus we get
$\frac{\partial^2 T}{\partial x^2}=-\sum_{n=1}^{\infty}a_n\lambda_n^2\sin(\lambda_n x)\sinh(\lambda_n y)=-\frac{\partial^2 T}{\partial y^2}$
   $\Rightarrow \frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=0$

And the heat flux is defined as
   $\vec{q}(x,y)=-k\nabla T(x,y)=-k\left ( \hat{x}\frac{\partial T}{\partial x}+\hat{y}\frac{\partial T}{\partial y} \right )$
where, k is the heat conductivity. And so,
   $\Rightarrow \vec{q}(x,y)=-k\left ( \hat{x}\sum_{n=1}^{\infty}a_n\lambda_n\cos(\lambda_n x)\sinh(\lambda_n y)+\hat{y}\sum_{n=1}^{\infty}a_n\lambda_n\sin(\lambda_n x)\cosh(\lambda_n y) \right )$

$\Rightarrow \vec{q}(x,y)=-k \sum_{n=1}^{\infty}a_n\lambda_n\left (\hat{x}\cos(\lambda_n x)\sinh(\lambda_n y)+\hat{y}\sin(\lambda_n x)\cosh(\lambda_n y) \right )$

The given temperature distribution T (x, y) is T (x, y) = sigma^infinite_n = 1 a_n sin (lambda_n x) sin(lambda_n x) sinh(lambda_n y) And so, partial differential T/ partial differential x = sigma^infinite_n = 1 a_n sin (lambda_n x) sin(lambda_n x) sinh(lambda_n y) And partial differential^2 T/ partial differential x^2 = partial differential/ partial differential x(partial differential T/ partial differential x) = -sigma^infinite_n = 1 a_n sin (lambda_n x) sin(lambda_n x) sinh(lambda_n y) And similarly partial differential T/ partial differential y = sigma^infinite_n = 1 a_n lambda^2_n sin (lambda_n x) sin(lambda_n x) sinh(lambda_n y) And thus we get partial differential T/ partial differential x^2 = -sigma^infinite_n = 1 a_n lambda^2_n sin (lambda_n x) sin(lambda_n x) sinh(lambda_n y) = -partial differential^2 T/ partial differential y^2 partial differential^2 T/ partial differential x^2 + partial differential^2 T/ partial differential y^2 = 0 And the heat flux is defined as q^rightarrow (x, y) = -k(x sigma^infinite_n = 1 a_n lambda_n sin (lambda_n x) sinh(lambda_n y) + y sigma