# Tag Info

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For commercial uses, usually these gases are in liquefied form .You can find more details, for LPG lets say , here: https://en.m.wikipedia.org/wiki/Liquefied_petroleum_gas In general you are right.

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The Ideal Gas Law PV=nRT explains what happens. It can be written as PV/T=nR. For our purposes nR can be considered a constant, PV&T are all expressed in absolute units. Thus if the container volume is constant, P/T is a constant.

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Apart from volume and temperature, there is one more parameter affecting pressure, that is number of molecules. At filling up the container you added more molecules in, consequntly you get higher pressure.

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Yes and No! The vessel is open to the surrounding atmosphere so that it will always be in equilibrium with it. However, if the opening is very small and the supply of high pressure gas of a high enough volume, then the pressure in the vessel would rise. This is analogous to a car tyre having a blowout or a slow puncture.

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It depends on the particular membranes that are employed. Example, if conversion rate is greater than 60 percent, I believe the savings will be small. However, if the membranes conversion rate is say only 30 percent, the savings can be large. Here is why: A million gallons per day plant uses membranes with a 30 percent flux rate, than to produce the ...

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When you say "pressure", I think you are probably referring to something like "the height to which the water shoots when you operate the fountain". There are actually two things that affect this: The pressure of the water supply. Any obstruction to the pipe (such as corrosion). As a thought experiment, imagine that you have a completely clear pipe except ...

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You're wondering why pressure nodes form at an open end of a tube. The answer is, they don't! It's just a reasonably good approximation. Physically, consider the air molecules at the center of the tube. Since they're far away from the edges, there's no way for them to "know" exactly when the tube ends, so the sound wave must "leak out" slightly. The ...

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Just visualize the upper beaker as a tube connected to the siphon tube(both tubes having different area of cross section)

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In this case, it is allowed. As a thought experiment, you can replace the top bucket with the siphon hose by an S-shaped tube that has a gradual change in diameter, like this: _ / \ |www| | | \www| | | \ww| | | \w| | | \|__/ | | | With this type of analysis, it is more ...

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Blood pressure is measured in mm-Hg. If we take a typical blood-pressure 120-80, that can be adjusted to height above or below the heart provided we make a very bad assumption that the human body is a bag of water. In reality, the human body is nowhere near that simple. 120-80 mm-Hg where Hg is a density of 13.59 g/cc and blood is about 1.06, then every ...

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There is a pump somewhere upstream in the system to provide water pressure. Apparently it cannot maintain exactly needed pressure but alternatively goes start-stop depending on actual pressure. Once the pump started it delivers pressure as 'high' adjusted, then stops. Then pressure slowly gets down as the water is taken out so when pressure reaches certain ...

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Old pipes get rusted or otherwise clogged by different stuff, such as salts. This will increase the friction of the water flow, resulting in a reduced pressure. The longer the pipes the larger the effect. I do not see any other solutions than to change the clogged pipes, assuming first that you can find which one are the clogged ones (iron pipes instead of ...

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From the Wikipedia article on sound: In physics, sound is a vibration that propagates as a typically audible mechanical wave of pressure and displacement. To fully understand how is air vibrating in an open pipe, you have to consider not only the acoustic pressure wave, $$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial ... 2 (b) is correct. Your equation applies only if all the fluid is the same fluid. Suppose that, as in your problem, E = gz, where z is the elevation above the base. So, the component of your equation in the z direction becomes$$\frac{dp}{dz}=-\rho g$$Because points 3 and 4 involve the same fluid, the pressure at these two points is the same, say ... 0 if the negative pressure is due to high velocity, there is possibility of cavitation to occur but if the negative pressure is due to height, it will just try to suck water, may be form other side or from the same side. 0 No, it isn't. Consider to the below container. There is water and air inside the container. The pressures of the points A and B are different. What that is common for every point of the water (like A and B) is the pressure of the air (P_{air}) that is transmitted to any point of the water (like points are on the walls) without changing (Pascal's ... -1 Looking at natural aspects of the world we can see first that a lack of pressure in no way means water won't and some point coalesce amongst themselves given the ability (eg clouds). So does; by creating a lack of pressure do we also decrease the relative amount of water molecules? NO, but pulling out the molecules via vacuum pump will decrease the overall ... 2 Gage pressure readings are taken assuming that 0 gage pressure = 1 atmosphere. By specifying a vacuum of 25~\textrm {kPa}, that suggests that the gage pressure is 25~\textrm {kPa} lower than local atmospheric. When you want to find absolute pressure, then P_{abs} = P_{atm} + P_{vac} = 70.7 - 25 = 45.7~\textrm {kPa}. 0 For a bit more empirical approach, try the compressibility factor. With this factor, the ideal gas equation becomes PV = znRT, where z is described in very great detail by https://en.wikipedia.org/wiki/Compressibility_factor. For places on the "z plot" that differ substantially from a value of z=1, you will find that argon starts behaving more and more ... 0 The result is problematic that the compressibility factor is greater than 1. The well-known Van der Waals equation is,$$(P+\frac {n^2a}{V^2})(V-nb)=nRT$$And ideal gas EOS is,$$P_{ideal}V=nRT$$where n is molar number and R is universal gas constant. You can use particle number if you prefer that way. From VDW equation, we get$$ P+\frac ...

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If the propane in the tanks are under pressure,then on connecting them, both tanks will be full of propane as it is a gas. Had it been water, or in a liquid form, then both the tanks will have the same level no matter how you arrange them.

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In a liquid (or any fluid), the molecules are in random motion (best to say is vibration). So each molecule is vibrating and hence collides with each other. Likewise the molecules in contact with the container also collides with the container walls. Assuming perfect elastic collision, the collided molecules are pushed backwards as insisted by Newton's third ...

1

The reason is because liquids don't have preferred directions. It's true that if you squeeze a solid by pushing down on it, it'll push back up on your hand but it won't push to the sides. If you model a solid as a cubic lattice of masses connected by springs, this makes sense, because only the vertical springs get compressed. A solid has enough order to ...

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Consider a sphere with no hole in the middle. It is also under stress when placed in a pressurized fluid. Your situation is no different. The material of any object which is pressurized deforms very slightly until the stresses produced exactly balance the external pressures. In essence, the material of the object is just like the fluid, and has its own ...

2

Since you have an equilibrium situation the net force on any portion of the water is zero. In region $B$ the horizontal forces acting on the water are either provided by the walls of the container in the left-hand diagram of by the water in regions $A$ and $C$ in the right-hand diagram. So as far as the water in region $B$ is concerned it makes no ...

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Because a Newtonian liquid's molecular characteristics, they tend to possess a definite volume without any specific shape. The important aspect is the height of the liquid not the shape of the container. If the height of a liquid with a specific gravity of 1 is 2.31 feet high (rounded off), The weight of that column is one pound over an area of 1^2 inch. ...

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You can indeed write $F=\rho gdA$, but the problem is that the force isn't constant throughout the entire area. You need to find an area with constant force, which would be a horizontal strip, because the depth is the same. The force on small strip (at the $w$ side) at depth $x$ with height $\Delta x$ is $\rho g x A=\rho g xw\Delta x$. The force isn't ...

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Pressure is indeed F/A however force varies upon the depth of the water therefore to find the force you need to "divide each area into small bands of areas, each having its own pressure (rho)gh, then add up those little forces you have got to gain the total force on the wall" Which this is actually the concept of integration Therefore you will get ...

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The pressure at a particular depth, $h$, is $h\rho g$.Force on a rectangular strip of infinitesimal height $\text dh$ and width $w$ is $h\rho g\text dhw$ which is when integrated for the entire surface becomes $\frac{1}{2}\rho gh^2w$.Similarly for other sides.

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At a certain depth a liquid exerts the same pressure on all sides. So if you want to calculate the force exerted by water on sides w and l then you have to put in the value of depth in the formula. Whatever the answer will come will be the pressure exerted by water on the sides w and l for that particular depth. As for the ratio F/A is concerned, it states ...

1

Many ways of looking at this: If there were any differences in pressure in horizontal direction, the pressure difference would mean a force that would induce motion of the fluid until the pressure difference would drop to 0. So in stationary state, pressure must be constant horizontally, and vertically, the difference in pressure between different heights ...

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I cannot comment (yet) - that is why I have to ask via an answer... Please let me know if I understand your question right: You want to know why the pressure on the lower, inner surface of the bucket is the same, considering that there is more water in the bucket with the inclined side surfaces... If this is your question think about that: There is more ...

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The book answer looks wrong to me. I think $k$ can only appear in combination with $x$, since the tension $T=kx$ in the spring is relevant here but $x$ on its own is not. It could be relevant as a geometrical factor, but there is no length marked $x$ in the diagram. I agree with your answer - except that you should have $k$ in there with $x$. I do not ...

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Use below free body diagrams and note that $P_2=\rho gh_2$ and $h_2=h+y-x$ and y is lower piston displacement. Note that volume of the displaced water for 2 pistons is equal. Explanation of the answer: At first, let’s review the problem description. We have a container is shown in the problem description above. There is water inside the container. The ...

2

In simple terms, the glass is being constantly shaken by weather. There isn't very much difference in mass between Oxygen and Nitrogen and the air at low altitudes is dense enough that molecules collide after a very short distance and so the atmosphere is very well mixed. At very high altitudes where the density is very low there is no weather, and little ...

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Let's take a look at this in a very simple way. At 60 F, 1 ft^3 of water weighs 62.37 lbs. that weight is resting on the base of 144 in^2. The force (weight) of the one foot high water on one square inch is: P = (62.37 lbs)/(144 in^2) = 0.433125 psi.

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1) Your question states that, until the compressor is switched off, a constant pressure of $5\; atm$ is maintained in the box. This answers the 1st part of your question. (However, perhaps you mean that the compressor delivers a certain amount of gas per second while the 2nd hole is open, and you wish to know the pressure of the gas in the box when ...

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Inflating a Rubber Balloon www.engr.uvic.ca/~struchtr/2002balloons.pdf University of Victoria by A DUVTCEV - ‎Cited by 26 - ‎Related articles

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You cannot subtract a pressure from a pressure gradient because they have different dimensions. However suppose you express the pressure gradient as: $$\frac{dP}{dx} = \frac{P_1 - P_2}{x_1 - x_2}$$ then you can subtract some constant pressure $P_0$ from both $P_1$ and $P_2$: \begin{align} P'_1 &= P_1 - P_0 \\ P'_2 &= P_2 - P_0 \end{align} ...

0

The stress-energy tensor is a stand-in for whatever forces you want to account for. For example, electromagnetic waves carry energy (as well as momentum, angular momentum, etc.) and you can choose to include these energies in the stress-energy tensor if you wish. Please see the Wikipedia article about the stress-energy tensor for E&M: ...

1

Bernouli does not explain wing lift. You can measure an older light plane with a "plank" wing, factor in the wing area, distance over the upper and lower surfaces, cruise speed, and air density, and come up with a total lift figure of about 25% of the aircraft weight. Bernouli equations were published in an aviation text decades ago and the error propagated ...

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There are two reasons why the bottom hole puts out more water. 1, it is exposed for a longer period of time than the two holes above it, add, 2, the velocity of the water coming from the lowest hole in greater than the other two holes above it. See formula below: V = sq. root of 2gh where: V = velocity in ft/s g = accel const. 32.2 ft/ s^2 h = height of ...

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The other way around is more intuitive; if the pressure is lower on the right, the fluid would feel a net positive force in that direction and accelerate toward right. hence it will have higher velocity there. So, lower pressure will result in higher velocity. you can rephrase the above in a way that it sounds as what you may like but is not scientifically ...

0

Don't let the size and length of the tube confuse you. What you described can be viewed as a big nozzle with a diameter of 20 meters with a water pressure of 500 atm. entering the back side of the nozzle. The formulas that can be used are: V = sq. root of 2gh where: V = velocity in ft/s. g = acceleration const = 32.2 ft/s^2 ...

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If the parts are perfectly solid and they are at equilibrium, then the pressure is constant (ideally). $$q(x,y) = \frac{\int p(x,y)\,{\rm d}A}{\int \,{\rm d}A} =\frac{\text{applied force}}{\text{area}}$$ But if either the floor or the body are ever so slightly elastic then the pressure distribution is given by a non-hertzian contact which concentrates all ...

3

The effect of temperature is included in determining the density as a function of temperature and pressure via the equation of state. For example, in the case of an ideal gas in hydrostatic equilibrium, the density is related at each point in the gas locally to the temperature and pressure by $$\rho=\frac{pM}{RT}$$where M is the molar mass. So the ideal gas ...

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Yes, it is independent in the case to which I believe you are referring, in which one has a container of liquid which is open to the atmosphere here on Earth, and the system is in static equilibrium. However these conditions are merely a special case in which the various effects of temperature cancel each other out perfectly. There are many possible ...

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PSI does not change with pipe size, only the surface area it is pushing on. Static head pressure is .433 per vertical foot. Water cannot flow over the top of a vessel which is gravity fed. It will only flow as high as the water level in it, because the water in the hose is also pushing back with gravity. A lot of these answers you are reading are ...

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Because hydrostatic pressure depends on depth, the problem requires integration. The pressure as a function of $y$ according to Pascal's Law is: $$p(y)=p_0+\rho g y\sin \theta$$ Where $p_0$ is the atmospheric pressure and $\theta=45\:\mathrm{degrees}$. $y\sin \theta$ is the depth. On an infinitesimal piece of door of length $dy$, at position $y$ and ...

2

Short answer: It can be compressed, due to extreme pressure, but only slightly. The Wikipedia page on the Mariana Trench says that the seawater density is 4.96% greater there than at the surface. Also refer to Hydrostatic pressure - doesn't density vary with depth?. Further explanation (and assuming everything you currently know about water is that ...

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