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Non-turbulent flow through a pipe of radius $r$ is proportional to the cross section $\pi r^2$ times the average flow velocity. The latter is proportional to the time $t$ it takes for the momentum in the fluid to diffuse to the wall: $t \approx (1/\nu) r^2$. Here $\nu$ is the kinematic viscosity ($\eta/\rho$). It follows that the flux (area times velocity) ...

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The usual integral for the divergence of the velocity field is over a volume. Since $u$ does not depend on $y$ and $v$ does not depend on $x$, we have \begin{align} \int_V \left(\nabla\cdot \vec{U}\right) \mathrm{d}V & = \iint \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}\right) \mathrm{d} x \mathrm{d} y \\ & = \iint ... 3 Before the discussion of a sphere, I would like to mention how the flow across a long cylinder (i.e. a circle in 2 dimensions) progresses (and why so) with an increase in Reynolds number (Re). Consider a flow across the cylinder in the creeping flow regime (Re\leq 1). This means that the inertial forces are low compared to the viscous forces. Consider ... 1 Interesting question. I suppose one should compare several scenarios Lie still Go forward - either straight, or hard to port, or hard to starboard Go in reverse The rate at which water enters the ship is (to first order) proportional to the pressure differential - lower the pressure and live longer. Maybe even long enough for the Carpathia to come and ... 2 The following picture (from http://hyperphysics.phy-astr.gsu.edu/hbase/waves/imgwav/circonwave.gif) gives you a better sense of how to reconcile your observation with "circular motion": As you can see - there is circular motion for particles at the surface: they don't have to go under water to do it though. Incidentally this also shows that in the trough ... 3 If you neglect viscosity, Bernoulli's equation (just Navier-Stokes without frictional or stress terms) will get you into the ballpark:P_g + \frac{1}{2}\rho_g v_g^2 = P_a$$Where the g subscripts pertain to the gas and the a subscript to the ambient. The gas density \rho_g \equiv M / V is the ratio of the mass of gas (M) in the tank to the volume ... 0 If wanted to pump water then it would be better to "push" and not "pull", since otherwise the pressure in the pump could drop to low such that cavitation could occur. But you are dealing with "pumping" a gas this will not be a problem. The an effect which could alter the performance would be the angle of attack of the blades of the fan. If the fan would be ... 1 Watch out for the choked flow situation. The flow is only purely proportional to \sqrt{P} if density is constant. In practice the working gas often returns to ambient temperature as it passes from compressor to the nozzle. So that density will increase. The end result is that you'll have \dot m \sim P. This effect is terribly convenient for control ... 0 If you project the velocity of a water parcel traveling up the channel into radial coordinates in both the inertial frame (space frame) and the rotating frame of reference (earth frame) then I think the necessary effects will become apparent. In the earth's frame, the velocity is (use a ' to symbolize the rotating frame):$$\vec{v} = 0\cdot\hat{r}' + ...

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By continuity (assuming steady state) the outlet flow must equal the inlet flow. If you have constant mass flow rate into the tube and block one hole, the flow through the others will increase. Whether the flow is sonic or not, that will force an increase in the pressure in front of the nozzle. Well below sonic speed the flow rate is proportional to the ...

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