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You are describing a device otherwise known as an eductor. This device converts a relatively high head, low velocity stream, into a high velocity, low head stream, and uses the kinetic energy from the high velocity to entrain another stream. I have no doubt that a low head, low velocity stream, cannot effectively be used for the application that you ...


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Peter Solo, it looks like you are posting an engineering question related to the equipment that you see on your job. Most employers (all that I have ever encountered) consider such information to be proprietary, and in my opinion, you are "walking on thin ice" regarding the engineering ethics and legality of the situation, because such information should ...


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I guess one reason why pressure and flow rate entering the compressor may vary is gas liquid contents vary, liquid is separated in the suction scrubber, so when more liquid is separated the less is pressure and flow rate at the compressor inlet.


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The work involved in accelerating the ping-pong ball is the integral of P(dV). If you further use the work-kinetic energy theorem, you arrive at the velocity of the ping-pong ball. Obviously, for the long barrel, this method doesn't work. In my opinion, this is because there is a hidden assumption that the ping-pong ball fits tightly against the tube that ...


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Well, most importantly, this question is fundamentally wrong... The units for acceleration are meters/(second**2). Either you made a typo, or you don't fully understand what you are trying to ask. I will assume it was just a typo. The first detail that must be addressed is the type of system we are analyzing. If it is something like a tube attached to an ...


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For now, let's ignore the expansion of the container due to the heating and just focus on the stress in the wall of the pressure vessel. I will also examine the case of a thin-walled, spherical vessel, but the same procedure may be applied for other geometries. Compute Stress State First, you must compute the pressure in the walls. To do this, imagine a ...


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Neutron degenerate matter can undergo a phase transition to a superfluid state. The process is thought to be analogous to Cooper-pairing, but the coupling interaction is of order 1 MeV, so can occur at temperatures below about $10^{9}$ K in neutron star interiors. The neutrons in the deep interior (which dominate the interior 100:1) can form a superfluid; ...


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This is the pressure-gradient term integrated over all volume, converted to a surface integral and using Gauss' theorem. Note that physicists prefer the differential form of such equations (see also this Wikipedia article), when the corresponding equation becomes $$ \frac{\text{d}\boldsymbol{u}}{\text{d} t} = \frac{\partial\boldsymbol{u}}{\partial t} + ...


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At least in the context of ultracold atomic fermions, the answer is no. The creation of a degenerate fermi gas is, unlike a BEC, not a phase transition. One major caveat: if there is an attractive force between the fermions, one can get a BCS-like phase transition to condensation of paired fermions. This is, of course, the case for electrons in metals, as ...


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The question comes down to buoyancy: is the mass of the flatus inside the body divided by its volume less than or greater than the mass of an equal volume of atmospheric air? To answer this we need to look at the density of atmospheric air, as well as the density of flatus. Let's start with the atmosphere. We usually see the density of air quoted as 1.2 ...


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I assume you ask this because you want to know if a fart's mass contribution to body weight can be offset by its lighter-than-air property, if any. According to About.com the typical composition of human flatulence is as follows. I added the molar mass of each component, as the density of a gas determines whether it's heavier or lighter than air: ...


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It has to be at a higher pressure than atmosphere to escape. The anus could not hold a large pressure differential so it could only be small pressure difference. Flatulence is mainly methane that is lighter than the average components of air on earth so I think heavier.


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I think as the radius increases, the velocity decreases(flow rate is constant) and pressure gets increased (fom street to house) and then due to sudden decresase in radius of pipe to 1'' at house pressure gets decreased and vel increases..


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If the flow rate is zero then the pressure is the same everywhere in the pipe. If there is some flow rate then the pressure drop per unit length of pipe is given by the Hagen-Poiseuille equation if the flow is laminar, or the Darcy-Weisbach equation if the flow is turbulent. The pressure gradient along the pipe is going to look something like: so the ...


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The pressure is clearly going to drop. Water would not flow without a pressure drop. Going from 1" to 1.5" alone does not change the pressure. The linear flow rate drops by 1 / (1.5)^2. Then when it goes back to 1" the the linear flow rate increases. At the front of the water line the end of the line you have the same volume and linear flow. ...


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A few things happen. One, the paper would bend, but lets pretend it's rigid, what happens when you lift. Your assumption is correct, air moves very fast, reducing but not completely eliminating the difference in air pressure between above and below. Air (molecules) moves at about 1,000 miles per hour. http://www.phy.mtu.edu/~suits/SpeedofSound.html ...


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Atmospheric pressure has nothing to do with it as you have not created a seal. You just have more surface area and more resistance. Karate chop versus lift is just speed. If you suspended the paper (assuming it would hold its shape) on fulcrum (no surface below) you would have the same effect. Ruler alone and enough seed and you could snap it. In a pure ...


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I hope that you don't seriously intend to try this project in the real world. There are several issues that WILL lead to failure and/or serious bodily harm: 1) The water will expand as it is heated. If you start with a container that is too full of water, it will burst the sides of the container before you get to 300 C. 2) Pressure vessels lose some ...


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Yes, the temperature of the gas would decrease quite fast, given that the molecules in the container are still, which implies zero temperature for container. However, if the container's temperature is non-zero, it sometimes happens that gas molecules will instead gain energy because the molecule it collides with is moving fast enough in the opposite ...


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If you assume that the gas is ideal then each collision of a molecule of gas with the wall conserves the kinetic energy. Hence the temperature will stay the same.


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The temperature of the gas will eventually reach equilibrium with the walls of the container, and since a perfect insulator is not possible, the gas, walls and outside environment will, given enough time, be at the same temperature.


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You are correct in a sense: under the pressures of the Mariana Trench, the thin plastic of the bottle walls will become slightly thinner as it is squeezed between the water inside and outside. This does not mean that the bottle is "crushed" or will deform, and the effect is likely to be quite small, depending on the bottle material. The integrity of the ...


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Water is almost incompressable, unlike air. If you filled the bottle with air, then sent it into the Marianas Trench, then the bottle would crush until the pressure of the air inside matched that of the water outside. At that point, the bottle would be almost (but not quite) flat. If you fill the bottle with water, then the water inside cannot compress, ...


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If the cube is sealed, doesn't expand and the temperature doesn't change, the pressure inside the cube will remain at 14.7 psi. That pressure is applied to all 6 sides continuously, simultaneously. Pressure in a sealed container will depend only on the volume and the temperature, not the number of sides, nor the external pressure. On the other hand, there ...


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Sometimes I feel Wikipedia is a funny place... In the article you quote they provide a calculation from our patent application (see, e.g., http://akhmeteli.org/wp-content/uploads/2011/08/vacuum_balloons_cip.pdf ) proving that a homogeneous shell made of any existing material cannot be both light enough to float in air and strong enough to withstand ...


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The maximum change of pressure caused by a sound wave is its pressure amplitude. This would be the difference between high and low pressure areas in the sound wave. When sound is measured in pascals, however, for the purpose of computing decibels by comparing with other sounds, it's just the high pressure against the measuring surface, to the extent that ...


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To know the force, you'd know the area. Since Psi is a unit of pressure, you could have almost have no pressure and lift thousands of tons. On the contrary, you could have just a tiny fraction of a millionth of a gram but a very high pressure. Since pressure is defined as force per unit area, $p = \frac{F}{A} \Leftrightarrow pA = F$, the pressure needed to ...


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To lift 100 kg (220 lbs) using a pressure of 1450 psi (about 100 atm), the minimum area needed to apply this force is: $$area=\frac {220\ lbs}{1450\ psi}$$ $$area=0.152\ in^2$$


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Consider the typical powered speaker; a paper cone on a solenoid. When current is applied to the solenoid, the cone moves (forward or backward, depending on direction of current). Let's say you apply a steady DC current to the solenoid and the cone pushes forward. It will push a high-pressure wave ahead of it, and then pressure against the cone will ...


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Yes, based on your comment. There is a low pressure area which follows the high pressure but it also travels at the speed of sound, so you wouldn't catch up to it at 50 m/s.


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When the pressure increases, the flow decreases thereby providing the same "power". You can have larger flow if you naturally decrease the pressure, or larger pressure if you restrict the flow.


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If it is water with a specific gravity of 1 you will lose 0.0361 psi per inch in the container. Example- If the level of the container drops 12 inches you would take 12 x .0361 and you would have a pressure loss of 0.4335 psi. If it is a material with a different specific gravity you would have to make adjustments, such as mercury would be .491 psi per inch. ...


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There isn't a simple formula for fan noise, but the physics can be worked out from the fundamental equations of fluid mechanics and acoustics. It isn't a simple problem however. The noise created by fans is complex and from several fundamental sources, and its amplitude depends on frequency. Here is an example of a fan noise frequency spectrum for a cooling ...


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Are you interested in the relative pressure difference between two points or the absolute pressure? In a container the pressure on the walls due to a gas can be calculated using the ideal gas law (pV=NRT). However, for a column of water, or the pressure at earth's surface due to a column of air, the pressure can be calculated as P=F/A=ρgh. The first ...


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Your answer lies in the ideal gas law: $$PV = nRT$$ where $P$ is pressure, $V$ is volume, $n$ is the amount of substance (usually in moles), $R$ is the "ideal gas" constant, and $T$ is temperature. You can see from the equation that if you're adding substance (i.e. increasing $n$), $V$ must increase proportionally (i.e. the piston must be displaced) to ...


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The ideal gas law is derived without consideration of the gravitational field. The kinetic behavior of a gas is, by far, more complex than the very simple example of the ideal gas law. So, answering to your two questions: Yes, the range of application of each equation is different. Use the hydrostatic equation and not the ideal gas one. Anyway, kinetic ...


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Here it is an incomplete answer, but just to think about some issues. An ideal gas is assumed to have particles non-interacting and with no extension (point-like particles). In particular for an ideal gas you discard gravity. While the Stevin's law (the relation $\rho g h$) is a direct consequence of the fact that you have a fluid in a gravitational field.


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The answer easily follows from the definition of pressure itself. It is the force that the particles apply on the sides of the container per unit of distance. Particles do have dimensions, but for what concerns the calculations their dimensions are irrelevant when compared to the sides of the container (although little modifications can apply, if you use any ...



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