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Two cylindrical tanks, connecting pipe and outlet pipe - Differential Equation |
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| Jul13-07, 01:17 AM | #1 |
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Two cylindrical tanks, connecting pipe and outlet pipe - Differential Equation
1. The problem statement, all variables and given/known data
Two vertical cylindrical tanks, each 10 meters high, are installed side-by-side. Their bottoms are at the same level. The tanks are connected at their bottoms by a horizontal pipe 2 meters long which has an internal diameter of 0.03 meters. The first tank is full of oil and the second tank is empty. Tank 1 has a cross-sectional area twice that of tank 2. Tank one has an outlet pipe (to the environment) at it's bottom as well. It is of the same dimension as the other pipe. Both of the valves for the horizontal pipes are opened simultaneously. What is the maximum oil level reached for tank 2 before the oil drains out of both tanks? Assume laminar flow in the pipes and neglect kinetic losses and pipe entrances and exits. 2. Relevant equations The volume balance equations are as follows. TANK 1: [tex]2A\,\frac{dy}{dt}\,=\,-q_1\,-\,q_2[/tex] TANK 2: [tex]A\,\frac{dx}{dt}\,=\,q_2[/tex] [tex]q_1\,=\,\frac{k\,y}{L}[/tex] [tex]q_2\,=\,\frac{k\,\left(y\,-\,x\right)}{L}[/tex] 3. The attempt at a solution [tex]\frac{dy}{dt}\,=\,\frac{1}{2A}\,\left(-q_1\,-\,q_2\right)[/tex] [tex]\frac{dx}{dt}\,=\,\frac{1}{A}\,\left(q_2\right)[/tex] [tex]\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\,=\,\frac{\frac{1}{2A}\,\left(-q_1\,-\,q_2\right)}{\frac{1}{A}\,\left(q_2\right)}[/tex] [tex]\frac{dy}{dx}\,=\,\frac{-\left(q_1\,+\,q_2\right)}{2y}[/tex] Substituting in for [itex]q_1[/itex] and [itex]q_2[/itex]. [tex]\frac{dy}{dx}\,=\,\frac{k\,\left(x\,-\,2y\right)}{4y}[/tex] [tex]4\,\frac{dy}{dx}\,=\,k\,\frac{x}{y}\,-\,2\,k[/tex] I don't know how to proceed, please help! |
| Jul14-07, 01:52 AM | #2 |
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Recognitions:
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background: solving systems of 1st order ODE. for the homogeneous NxN linear system:
[tex]\dot{\textbf{x}}(t)=\mathbb{A} \textbf{x}(t)[/tex] where [tex]\dot{\textbf{x}}(t), \textbf{x}(t)[/tex] are N-vectors while [tex]\mathbb{A}[/tex] is a NxN matrix (with constant coefficients assumed) then the general solution takes the form: [tex]\textbf{x}(t) = c_1\textbf{V}_1\,e^{\lambda_1 t} + c_2\textbf{V}_2\,e^{\lambda_2 t}+\cdots+c_N\textbf{V}_N\,e^{\lambda_N t}[/tex] where [tex]\lambda_i[/tex] are the eigenvalues of [tex]\mathbb{A}[/tex] and [tex]c_i[/tex] are arbitrary integration constants, [tex]\textbf{V}_i[/tex] are linearly independent eigenvectors of [tex]\mathbb{A}[/tex]. For your system, it would look something like this: [tex]\begin{pmatrix}\dot{y} \\ \dot{x}\end{pmatrix}= \begin{pmatrix}-k/(AL) & k/(2AL)\\ k/(AL) & -k/(AL)\end{pmatrix}\; \begin{pmatrix}y \\ x\end{pmatrix}[/tex] |
| Jul14-07, 06:56 AM | #3 |
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can you explain what q, k, x, y and L are? I kinda can't follow what you were doing up there.
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| Jul26-07, 12:17 PM | #4 |
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Two cylindrical tanks, connecting pipe and outlet pipe - Differential Equation
yes, i wanna know too
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| Jul26-07, 02:39 PM | #5 |
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The problem doesn't say what those values are. I know some are constant and some are variables.
Here is the picture that came with the problem
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| Jul26-07, 11:35 PM | #6 |
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VinnyCee, why 2q changed to 2y?
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