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Although this may be a small effect, I think pushing is better than pulling. If you pull air, the you are pumping air that starts a at a pressure (and density) less than atmospheric. This means that the pressure differential across the length of the duct cAn never be more than one atmosphere. On the other hand if you push air you can easily generate a ...


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When you open the tube, yes, air will be drawn into it. This is indirectly due to gravity, in the sense that what will drive the flow is a pressure gradient. The pressure in the tube is lower than the pressure outside (after you pump air out and let things settle down, this is true no matter where you open the hole), so air is drawn in. Gravity has a role in ...


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The volume of the shape drawn there is $dS\,v\,dt\cos(\theta)$. $dS$ is not a cross section, it is at an angle to the axis $v$ is aligned with.


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I don't understand... Those two cylinders have the same volume $S.L$: So all particles exactly on (or beyond) surface $S$ at point $A$ and time $t$ will be exactly on (or beyond) surface $S$ shifted at point $B$ by time $(t+dt)$ assuming they all move at constant velocity $\vec{v}$. I don't see how the orientation of the surface changes anything. I can't ...


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You can't simply multiply $dS$ with $\vec{v}$ to obtain the volume. If we assume a coordinate system $x$ perpendicular to $dS$, and $y$ in the plane of $dS$. The volume is defined as: $$V=\int^{s_{begin}}_{s_{end}} \: S \vec{dx}_{\perp S}$$ However, if we want to express it as a function of $v dt$ we get: $$V=\int^{t_{begin}}_{t_{end}} \: S \: \vec{(v ...


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The way I see it, with larger $\theta$ (as long as it's not $\pi/2$) it is still cylinder of volume $dS.v.dt$ that will pass through the surface $dS$


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Look at large $\theta$ : fewer particles per unit time will reach the surface because their perpendicular velocity is much less. When $\theta = \pi/2 $ zero particles cross the surface.


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I think that it helps to define appropriate control volumes. See the image below where I define surfaces A and B. Here, we can say that the pressure at A is given by $\rho g h_A$ and the pressure at B is given by $\rho g h_B$, recognizing that $h_a$ and $h_b$ are functions of time. If the tank is open to atmosphere the $P_A$ and $P_B$ terms will be equal ...


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The intuitive way to think about this is to consider a gas inside a glass container (that cannot expand). If the gas expands, then what must happen as a result? The gas leaks out of the container. Similarly, if try I put more gas into the container, then the gas compresses. The vector field $\mathbf F$ is what we use to describe the flow of a fluid. The ...


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I answer my own question and give a good thanks to DavidPh, who has not really gave the answer, but in fact, it was impossible for him to give it. Here is "why": I'm French, so I've many fire hydrant data but from France. And when applying them to the formulas, the result was wrong... In fact, the problem is not the formula but the way we measure the ...


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The discrepancy is that the pressure as measured by the Pitot tube is not just the kinetic energy term of the pressure, but instead is a combinaiton of static pressure and the kinetic energy term. See if pages 16-34 of the following reference are helpful, though not metric: http://www.southsaltlakecity.com/uploads/documents/%5E_Fire_Flow_Calculations.pdf ...


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If the tips of the propeller blades are moving near the speed of sound a shock wave can form. Supersonic flow has very different character than subsonic. A propeller designed to operate at subsonic speeds will be inefficient at supersonic ones due to shock waves. In general, shock waves cause a loss of efficiency. You might have noticed that subsonic ...



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