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The derivation is not correct. The mass within the control volume is $\rho A\Delta x$. The rate of energy accumulation within the control volume is $\rho A\Delta xC\frac{\partial T}{\partial t}$. So the heat balance should be:$$\rho A\Delta xC\frac{\partial T}{\partial t}=Q_x-Q_{x+\Delta x}$$Dividing by $\Delta x$ and taking the limit as $\Delta x$ ...

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Here some details \begin{equation*} \partial _{t}\mathbf{w}(x,t)+\mathbf{A}\cdot \partial _{x}\mathbf{w}(x,t)=0 \end{equation*} Let $\mathbf{A}$ be an $n\times n$ matrix. Then $\mathbf{w}$ must be $n$ -dimensional. Let us assume that $\mathbf{A}$ has real entries and $\mathbf{w }$ has real components. \begin{eqnarray*} \mathbf{w}(x,t) &=&\exp ...

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I find Pulliam's notes for the Euler equations to be a pretty good introduction to this topic using the equations of fluid motion. The idea is that you start with a conservation law: $$\frac{\partial \vec{Q}}{\partial t} + \frac{\partial \vec{F}\left(\vec{Q}\right)}{\partial x} = 0$$ where $Q$ is your variable vector and $F$ is your flux function. You can ...

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Your setup of the problem isn't correct. Let V be the volume of water in the container. Then the rate of accumulation of heat in the container is equal to the heat in minus the heat out. The correct equation for this is: $$V\rho C_p\frac{dT}{dt}=q\rho C_p(T_{NEW}-T)-kV\rho C_p(T-T_a)$$This assumes that the tank is well-mixed so that the exit temperature ...

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This is more of a comment than an answer, but I can't fit this into the amount of characters; Writing a quick bit of code, it looks to me like there's not much wrong with the method: The numerical and the analytical solution go on top of one another. N = 256 T = 256*128 L = 1. dt = 0.000001 x = linspace(0., N-1, N)*L/N psix = exp(1j*2*pi*x) psik = ...

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