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1

Okay, so this is a mediocre answer because I don't own the book nor have I read it, so I cannot assess it clearly. I am pretty curious about it so I may try to pick up a copy. However, doing a really quick literature search, the book Computational River Dynamics by CRC Press seems like it has all the ingredients to be a good introduction based on the ...


0

enter link description here Hope this image help you!!! The thing here assume the pipe only one side, gas is moving similar to given image. If it is a cone then you can see that almost all the gas entering leaves out. Hence the power is also increased.


2

The divergence of the transpose of a vector gradient is equal to the gradient of the divergence of the vector. You can prove it quickly using some index notation: $$\nabla \cdot (\nabla \vec{u})^T = \frac{\partial^2 u_j}{\partial x_i \partial x_j} = \frac{\partial}{\partial x_i}\frac{\partial u_j}{\partial x_j} = \nabla(\nabla \cdot \vec{u})$$ I assumed ...


1

I'm going to guess that the RHS really means $$ \nabla^j ( \nabla_i u_j + \nabla_j u_i ) = \nabla_j \nabla_i u^j + \nabla^2 u_i . $$ Now, $$ \nabla_j \nabla_i u^j = [ \nabla_j , \nabla_i ] u^j + \nabla_i (\nabla_j u^j) = R_{ij} u^j + \nabla_i (\nabla_j u^j) . $$ You're probably working in a Ricci flat background and the velocity vector is divergenceless ...


1

There is a simple intuitive explanation for the factor of two difference. The thrust has two contributions, as follows. 1) The air pressure in the rocket imposes a force down the nozzle of F=PA exactly as you surmised in your second method. 2) The exhaust water at the instant it leaves the nozzle is still pressurised, and imposes a back-pressure on the ...


5

The pressure drop from tank pressure to atmospheric pressure does not occur instantly at the nozzle but rather is spread out according to the area of the flow channel. This reduced pressure results in additional thrust that was not accounted for in your second solution. Here is one way to calculate this additional missing thrust: Bernoulli's equation (your ...


1

When a body moves through space filled with air, then higher pressure is created in front of it, while lower pressure/depressure behind it. The higher pressure is plus, the lower pressure is minus. I use to call this a "principle of an arrow" (− >———> +). The greater the velocity of the body is, the stronger is the plus in front of it as well as the minus ...


8

Often these rockets are launched off a vertical section of pipe or rod that extends up inside the bottle, acting as a piston until the rocket has moved far enough to clear the end of the pipe. During this phase, your second method is correct: the thrust is simply the nozzle area times the pressure. Why should the thrust increase when the rocket separates ...


20

The first method is correct. In the second you have assumed that the pressure at the nozzle is still $P$ despite the water exiting with some velocity. i.e. You have neglected the dynamic pressure. You need to use Bernoulli's principle $$ P + \frac{\rho v^2}{2} + \rho h g = {\rm constant}$$ Your first method assumes that the top surface of the water is ...


1

As you have neglected atmospheric pressure, let's assume that you are doing the experiment in a vacuum(I know that there is a spherical cow analogy somewhere :) ). You are assuming that the pressure is exerted by air inside the ballon at the surface. The air will exert the pressure in the empty part as shown by your arrows as well as on the horizontal ...


0

I think it is possible that you are neglecting compressible effects. Air that is flowing from a reservoir/compressor at 4 bar and exhausting through a pipe to atmospheric pressure will almost certainly be choked. I.e. it will be at sonic velocity (~330 m/s) at the outlet of the pipe. Appearances can be deceptive - it is very possible for an air flow to be ...


0

Those scale-estimations are an art in itself, and more so a dangerous one. The volume or scale of arbitrary fluid elements should have nothing to do with the scale you plug in. Only lazy, unreflected fluid dynamics teachers will give you those explanations. One finds in many books nonsensical values plugged into the formulas, by many physicists justified ...


0

There will almost always be atoms or molecules moving around in a fluid, so in that sense you could say that it is an average velocity. Continuity holds true as the flow rate is the same, even though there can be some turbulence in the fluid.


0

You said it yourself , when there is no viscosity.. also the velocity you talk about is about the flux of fluid through the cross section area not of individual particles.


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