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The quantity Lighthill referred to as the 'virtual mass' is usually called 'added mass' in ocean engineering and naval architecture. It is basically the force on the body due to its acceleration in the fluid. The term 'added' is a bit misleading, as in certain cases, it can be negative. Note that the added mass depends also on the mode of motion. So, for ...


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The higher the water level in the tank, the faster the flow will exit from the orifice in the bottom. How fast will it exit? Viscous effects are of course present in the flow, but they can be neglected for the streamtube which exits the container, so we can employ Bernoulli's equation. Mathematically, the flow variables will conform to the following ...


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A stationary state, usually referred to as the steady state, is the particular case when the partial time-derivative of the variable in question is zero. For your question, $$ \frac{\partial v}{\partial t}=0 $$ Or, in macroscopic terms, $$ \frac{\Delta v}{\Delta t}=0\to v_{in}=v_{out} $$ which is what jhobbie said in the comments. The water level will ...


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Starting from the conservation of mass: $$ \dot m_{1}=\dot m_{2} $$ This translates to $$ \rho_{1} S_{1} V_{1}=\rho_{2} S_{2} V_{2} $$ Assuming incompressible flow, thus $\rho_{1}=\rho_{2}$ gives: $$S_{1} V_{1}= S_{2} V_{2} $$ With $S_{1} V_{1} = Q_{1}$ , the formula you are using. This formula follows directly from the mass balance, with only the ...



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