anonymous
  • anonymous
Find the potential distribution in a pipe for the semi-infinite rectangular pipe if the all sides to be grounded except in at the end of the base at Vo (y,z) (look at the picture). Using Laplace equation for 3-dimention in Cartesian coordinate.
Trigonometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
@oldrin.bataku can you help me ???
anonymous
  • anonymous
have you idea about "Heat conduction in three dimensions" ???

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anonymous
  • anonymous
Laplace's equation is as follows:$$\nabla^2 V=0$$Here we're given boundary conditions:$$V(0,y,z)=V_0\\V(x,0,z)=0,V(x,a,z)=0\\V(x,y,0)=0,V(x,y,b)=0$$
anonymous
  • anonymous
@gerryliyana is \(V_0\) constant... ?
anonymous
  • anonymous
Assume \(V\) is completely separable so:$$V(x,y,z)=X(x)Y(y)Z(z)\\\frac{\partial^2 V}{\partial x^2}=X''(x)Y(y)Z(z),\frac{\partial^2 V}{\partial y^2}=X(x)Y''(y)Z(z),\frac{\partial^2 V}{\partial z^2}=X(x)Y(y)Z''(z)$$... now substitute into our equation:$$\nabla^2 V=0\\\frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial y^2}+\frac{\partial^2 V}{\partial z^2}=0\\X''(x)Y(y)Z(z)+X(x)Y''(y)Z(z)+X(x)Y(y)Z''(z)=0\\\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}+\frac{Z''(z)}{Z(z)}=0\text{ by dividing by }V\\\frac{X''(x)}{X(x)}=-\left(\frac{Y''(y)}{Y(y)}+\frac{Z''(z)}{Z(z)}\right)=-\alpha^2\\\frac{Y''(y)}{Y(y)}+\frac{Z''(z)}{Z(z)}=\alpha^2\\\frac{Y''(y)}{Y(y)}=\alpha^2-\frac{Z''(z)}{Z(z)}=-\beta^2$$Rearranging and letting \(\gamma^2=\alpha^2+\beta^2\) we have:$$X''(x)+\alpha^2 X(x)=0\\Y''(y)+\beta^2 Y(y)=0\\Z''(z)+\gamma^2 Z(z)=0$$... all of which produce trivial solutions:$$X(x)=A_1\cos\alpha x+A_2\sin\alpha x\\Y(y)=B_1\cos\beta x+B_2\sin\beta x\\Z(z)=C_1\cos\gamma z+C_2\sin\gamma z$$
anonymous
  • anonymous
Oops wait, I solved for \(Z(z)\) incorrectly... we also have yet to impose any boundary conditions.$$\alpha^2-\frac{Z''(z)}{Z(z)}=-\beta^2\\\frac{Z''(z)}{Z(z)}=\alpha^2+\beta^2=\gamma^2\\Z''(z)-\gamma^2Z(z)=0\\\implies Z(z)=C_1\cosh \gamma z+C_2\sinh\gamma z$$
anonymous
  • anonymous
Recall our boundary conditions:$$V(0,y,z)=V_0\\V(x,0,z)=0,V(x,y_1,z)=0\\V(x,y,0)=0,V(x,y,z_1)=0$$Let's start off with our Dirichlet conditions:$$V(x,0,z)=V(x,y_1,z)=0\implies Y(0)=Y(y_1)=0\implies B_1=0,\beta=n\pi/y_1\\V(x,y,0)=V(x,y,z_1)=0\implies Z(0)=Z(z_1)=0\implies C_1=0$$dangit maybe \(X(x)\) should be the one in \(\cosh,\sinh\)
anonymous
  • anonymous
$$\nabla^2 V=0\\\frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial y^2}+\frac{\partial^2 V}{\partial z^2}=0\\X''(x)Y(y)Z(z)+X(x)Y''(y)Z(z)+X(x)Y(y)Z''(z)=0\\\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}+\frac{Z''(z)}{Z(z)}=0\text{ by dividing by }V\\\frac{Y''(y)}{Y(y)}=-\left(\frac{Z''(z)}{Z(z)}+\frac{X''(x)}{X(x)}\right)=-\beta^2\\\frac{Z''(z)}{Z(z)}+\frac{X''(x)}{X(x)}=\beta ^2\\\frac{Z''(z)}{Z(z)}=\beta^2-\frac{X''(x)}{X(x)}=-\gamma^2$$so we find:$$X''(x)-\alpha^2 X(x)=0\\Y''(y)+\beta^2 Y(y)=0\\Z''(z)+\gamma^2 Z(z)=0$$where \(\alpha^2=\beta^2+\gamma^2\). Solving for our functions we find:$$X(x)=A_1\cosh\alpha x+A_2\sinh\alpha x\\Y(y)=B_1\cos\beta x+B_2\sin\beta x\\Z(z)=C_1\cos\gamma z+C_2\sin\gamma z$$Observe our boundary conditions:$$V(0,y,z)=V_0\\V(x,0,z)=0,V(x,y_1,z)=0\\V(x,y,0)=0,V(x,y,z_1)=0$$Now let's impose our grounded boundary conditions:$$V(x,0,z)=V(x,y_1,z)=0\implies Y(0)=Y(y_1)=0\implies B_1=0,\beta=n\pi/y_1\\V(x,y,0)=V(x,y,z_1)=0\implies Z(0)=Z(z_1)=0\implies C_1=0,\gamma=n\pi/z_1$$for any integer \(n\). Lastly:$$V(0,y,z)=V_0\implies Z(0)=V_0\implies C_2=0,C_1=V_0$$
anonymous
  • anonymous
\(\gamma=m\pi/z_1\)** \(X(0)=V_0\)** As a product, we have: $$V(x,y,z)=\sum_{n,m=1}^\infty A_{nm} \cosh \alpha x\sin \frac{n\pi y}{y_1}\sin\frac{m\pi z}{z_1}$$
anonymous
  • anonymous
Can you find the Fourier coefficient \(A_{nm}\) now... ?
anonymous
  • anonymous
\[A_{nm} = \frac{ 16 V_{o} }{ \pi^{2} } \frac{ 1 }{ nm } e^{-\pi \sqrt{(n/y_{1})^{2}+(m/z_{1})^{2}}x}\] am i wrong ??
anonymous
  • anonymous
$$V(x,y,z)=\sum_{n,m=1}^\infty A_{nm} \cosh \alpha x\sin \frac{n\pi y}{y_1}\sin\frac{m\pi z}{z_1}\\V\sin\frac{j\pi y}y_1\sin\frac{k\pi z}{z_1}=\sum_{n,m=1}^\infty A_{nm} \cosh \alpha x\sin \frac{n\pi y}{y_1}\sin\frac{j\pi y}y_1\sin\frac{m\pi z}{z_1}\sin\frac{k\pi z}{z_1}$$Integrate over \(y\) with period \(y_1\) and do similar for \(z\):$$\int_0^{y_1}\sin\frac{n\pi y}{y_1}\sin\frac{j\pi y}{y_1}\,dy=\frac{y_1}2\text{ if }j=n\\\int_0^{z_1}\sin\frac{m\pi y}{z_1}\sin\frac{k\pi y}{z_1}\,dz=\frac{z_1}2\text{ if }k=m$$and so I believe we are left with $$A_nm=\frac{16V_0}{nm\pi^2\cosh\alpha x}$$
anonymous
  • anonymous
what do you say... ?
anonymous
  • anonymous
then \[V (x,y,z) = \sum_{n,m=1}^{\infty} \frac{ 16 V_{o} }{ nm \pi^{2} } \sin \frac{ n \pi y }{ y_{1} } \sin (\frac{ m \pi z }{ z_{1} })\] ???
anonymous
  • anonymous
oops, should've been \(\cosh\alpha y_1\) I think.. try finding \(A_{nm}\) again but this time use \(\cosh\alpha x\)

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