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Filling a rigid tire from a pressure regulated compressed air line
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[QUOTE="erobz, post: 6636284, member: 700856"] I would like to try and cast some doubt by showing that for viscous, incompressible flow between two tanks, equilibrium is archived in a finite amount of time. [ATTACH type="full" alt="Two tanks.jpg"]301976[/ATTACH] For incompressible viscous flow between two points 1 & 2 we have the following result derived from Conservation of Energy applied to Reynolds Transport Theorem: $$ \frac{P_1}{\gamma} + z + \frac{v_1^2}{2g} = \frac{P_2}{\gamma} + h + \frac{v_2^2}{2g} + \sum_{1 \to 2 } h_L \tag{1} $$ For the system above we have that: ## v_1 = \dot z ## ## v_2 = \dot h ## ## P_1 = P_2 = 0 \rm{gauge} ## ## \sum_{1 \to 2 } h_L = k \frac{v_p^2}{2g} ## Plugging that into (1): $$ z + \frac{ \dot z^2}{2g} = h + \frac{ \dot h^2}{2g} + k \frac{v_p^2}{2g} \tag{2}$$ Next, put everything in (2) in terms of ##h## the height of the liquid surface in tank 2 Since the flow is incompressible the total volume of the system must be preserved: $$ A_1 z + A_2 h + {V\llap{-}}_p = V\llap{-} \implies z = \frac{1}{A_1} \left( V\llap{-} - {V\llap{-}}_p - A_2 h \right) \tag{3} $$ Furthermore, by taking the derivative of (3) with respect to time ##t##: $$ \dot z = - \frac{A_2}{A_1} \dot h \tag{4} $$ By continuity we also have that: $$ A_p v_p = A_2 \dot h \implies v_p = \frac{A_2}{A_p} \dot h \tag{5} $$ Substituting back into (2): $$ \frac{1}{A_1} \left( V\llap{-} - {V\llap{-}}_p - A_2 h \right) + \left( \frac{A_2}{A_1} \right)^2 \frac{ \dot h^2}{2g} = h + \frac{ \dot h^2}{2g} + k \left( \frac{A_2}{A_p} \right)^2 \frac{\dot h^2}{2g} \tag{6} $$ Rearranging (6) and combining constants we end up with a first order ODE of the form: $$ \dot h = \beta \sqrt{ \lambda - \varphi h } \tag{7}$$ Where: ## \beta = \sqrt{ \frac{2g}{ 1 + k \left( \frac{A_2}{A_p} \right)^2 - \left( \frac{A_2}{A_1} \right)^2 } } ## ## \lambda = \frac{1}{A_1} \left( V\llap{-} - {V\llap{-}}_p \right) ## and ## \varphi = 1 + \frac{A_2}{A_1} ## Now, the steady state height of fluid in tank 2 can be found by setting eq (7) ## \dot h = 0 ## $$ h_{ss} = \frac{\lambda}{ \varphi} $$ From (3) it follows that: $$ z = \lambda - \left( 1 - \varphi \right)h $$ $$ z_{ss} = \lambda - \left( 1 - \varphi \right) \frac{\lambda}{ \varphi} = \frac{\lambda}{ \varphi} \implies z_{ss} = h_{ss} $$ Now, all that is left to do is show that equilibrium must be reached in a finite time by solving (7) by method of separation of variables: $$ \int \frac{dh}{ \sqrt{ \lambda - \varphi h } } = \beta \int dt $$ Performing the integration: $$ - \frac{2}{\varphi} \left( \sqrt{ \lambda - \varphi h} - \sqrt{ \lambda - \varphi h_o} \right) = \beta t \tag{8} $$ If we substitute ## h = h_{ss} = \frac{\lambda}{ \varphi} ## into (8) we are left with: $$ t = \frac{2}{ \varphi \beta} \sqrt{ \lambda - \varphi h_o} \tag{9} $$ From this analysis it is apparent that the heights of the fluid in each tank are equivalent after a finite amount of time has passed. Now, what ( if anything) about compressible flow destroys the "finite" result? [/QUOTE]
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