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Try imagine a thin wall placed in the center of horizontal tube to prevent the flow between two sides. You agreed that $P_{right}>P_{left}$, then try to calculate the total force acting on this imaginary wall. The surface of this wall on two side is the same but the pressure on the right is larger than the left. Total force will point toward left side, ...


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In an ideal fluid, assuming the diameter of the pipe after the contraction is the same as the diameter of the pipe before the contraction, $P_2^\prime = P_2$. There is no effect of the contraction downstream from the contraction itself. If we consider an inviscid, isentropic, incompressible flow, the total pressure in the flow is constant along streamlines ...


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If we consider temperature to be due to translational motion of the molecules and we assume the system has reached equilibrium, then the velocity distribution of the molecules is given by the Maxwell distribution: $$ f(v) = \sqrt{\left(\frac{m}{2\pi k T}\right)^3} 4 \pi v^2 \exp\left(\frac{m v^2}{2 k T}\right)$$ which will give you the velocity ...


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There's no uniform density of heated air. It depends on the temperature (higher T -> lower density) but also on the ambient air pressure. In Denver, cold air is less dense, because the ambient pressure is lower. But this same effect also increases the density of hot air, by the same percentage. So, the result is that the lift of a balloon decreases with ...


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So I found the answer....I was not taking the final step of multiplying The difference in density between the surrounding air and the heated air and then multiplying by the envelope volume. I was just focusing on the difference and getting stuck there. I chose Denver because it's a mile above sea level with a known air density. I could have chosen Mt Everest ...


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Given the imprecision in these numbers, that means that you can lift anywhere between 0 and 0.1 kg per m^3 of air. Per Wikipedia, a typical hot air balloon holds 2,800 m^3 of air in the envelope, so it can suspend something between 0 and 280 kg in the basket. A typical human weighs under 100kg, so you could probably lift between one and three people with a ...


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The air in the room must be sufficiently laden (ideally, saturated) with water vapor (or another substance, such as alcohol, that can be absorbed by the air). Then, if you drop the pressure suddenly, so that the air can expand adiabatically, the temperature can drop below the dew point of the vapor, and the vapor has an opportunity to condense into a cloud. ...


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You have to know what method are they using, they are using a tehnique called hydrostatic pressure, here is a good article about that. http://www.eolss.net/sample-chapters/c10/e5-10-04-11.pdf As you can see the pressure is applied uniformly and there are not pressure diferences as you can see from this video. That means that the object is in hydrostatic ...


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Obviously the pressure increases with load. If it didn't then you could have an infinite load! If the load is increased, eventually the pressure will increase to the point where the tire bursts (or the load is resting on the wheels, not the tires).


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When you reduce the pressure in a vessel by pumping air out you reduce the temperature of the air. If the temperature falls below the dew point then water will condense out of the air and a mist will form. The trouble is that the dew point depends on the pressure, and it falls as the pressure reduces. So whether a mist forms or not is dependant of how close ...


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It works because the plastic bottles are filled with water and water is almost incompressible. When subjected to high pressure, the volume inside the bottle stays almost the same (there is a small air bubble that will be compressed), thus the bottle is not crushed. I believe that a plastic bag would work just as well. The hydrostatic pressure is uniform so ...


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You're not getting enough Make-Up Air. I strongly encourage you to install a CO detector in you home and reassess your HVAC situation (especially the 'V' part, ventilation). Lack of sufficient makeup-air in a house with gas fired equipment is dangerous. This is usually more of a problem with new-construction homes that are built to be nearly hermetic. ...


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This is a typical problem when starting up a fireplace, wood stove, or what-have-you. The air pressure inside your house at 75 degrees is lower than the outside air pressure at 40 degrees because warm air is less dense than cold air. Result? The higher-pressure air outside the house wants to flow down the chimney and into the lower-pressure area inside your ...


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There seems to be a slightly lowered pressure inside the room the fireplace is in, which leads to air actually streaming down the chimney to equalize it, and this blows the flames into the room. When you open the door, you allow air draught to stream into the room and up the chimney, as it is supposed to, and that sucks the flames into the chimney. ...


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There's two possibilities that are immediately obvious. The first is that the pressure inside the house is slightly lower than the pressure outside the house before you light the fireplace. This would cause air to flow down the chimney into the house which would keep push the flame into the house instead of up the chimney. This would be easily testable if ...


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You need a "draft". When you light the fireplace, initially the heated combustion products want to rise directly up the chimney, but after that occurs for a few seconds it results in a partial vacuum in the house, attempting to suck air back down the chimney. This cause the fire to be blown/pulled outwards into the room. (With a conventional wood-burning ...


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Why do we boil water to cook food? It's not actually because there's anything magic about the boiling of water, or that the physical process of boiling in particular does anything. Usually it's because we want a constant-temperature heat bath. Say you are boiling vegetables. You boil water, and you know that water is at 100 degrees. Water actually cannot get ...


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This is really a comment to the two existing answers from Phoenix87 and Pranav. I agree entirely with their comments on acceleration. However if the car is not air tight, i.e. if air can flow in and out of the car, then there will be an effect of velocity as well. For example if there are small apertures in the car we'd expect the pressure inside to fall ...


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By the equivalence principle you can take into considerations the effects of gravity on pressure for instance. Then it is known that pressure depends on altitude (although this dependence is not simple due to many other factors). The general relation for the pressure gradient is $$\nabla p=\rho\mathbf g$$ where $\mathbf g$ is the gravitational acceleration, ...


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Yes. Air has mass, and anything with mass has inertia. Due to these inertial effects, the air in a car will be redistributed as you would expect any other object with mass to move in a non-inertial frame. See Why does a helium filled ballon move forward in a car when the car is accelerating?


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The answer to your question is that based on the dimensions, halving the width from 1 to 2 feet will halve the force. Were the width larger than the depth this would be untrue, as the formula F=0.5 x depth squared times width applies to dimensions of width=depth & larger. So the math breaks down when width is less than height in which case the formula F= ...


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The relationship between flow rate and pressure is given by the Darcy-Weisbach equation. The problem is that using it requires you to know the geometry of the pipe underground. I suspect you'll struggle to do an accurate calculation. Your best option is probably to try and borrow a pressure gauge.


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Under common assumptions and ignoring potential energy, static pressure is the expression of the fluid's temperature (internal energy) and dynamic pressure is the expression off the fluid's velocity, so if the fluid is brought to a rest adiabatically, their sum is equal to the stagnation pressure. The stagnation pressure represents the total energy of the ...


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Yes, $F_b \approx 2 F_a$. The force needed to advance the screw into the water is $F = PA$ where $P$ is the pressure on the end of the screw and $A$ is the cross-sectional area of the screw. When the height of the water is doubled, the pressure is also doubled. The relation is not exact because your scenario doubles the height of the water from the bottom ...


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Both cases are true. Basically the force the object will feel is due to the net differences in pressure and viscous forces acting on its upstream and downstream faces. How the pressure field and viscous stress are distributed around the object is as function of a number of factors: the shape of the object, its size relative to the pipe, the wall roughness, ...


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To fluid dynamicists, Bernoulli's equation is better known as the 'Energy Equation' since it does indeed account for the energy changes that occur along a fluid path. The energy equation says that the energy is constant along any given streamline. Static or stagnation pressure can exist in the absence of fluid velocity creating a potential energy component. ...


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The quantity $\frac{1}{2}\rho v^2$ is called dynamic pressure for two reasons: because it arises from the motion of the fluid, and because it has the dimensions of a pressure. It is not really a pressure at all: it is simply a convenient name for the quantity (half the density times the velocity squared), which represents the decrease in the pressure due ...


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Using dimensional analysis, and the relations $p=mv$ and $E=mgh$, we can write: $$[P]=\dfrac{kg^2\cdot m^2\cdot s^{-2}}{kg\cdot m^2\cdot s^{-2}}=kg$$ And kilogrammes are obviously not a unit of pressure, therefore your relation doesn't have the correct dimensions.


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IF all of the water above the leak point has indeed completely drained away, and the water is completely stationary throughout the pipe's entire length (those are important "if"s), then your problem is quite straightforward! With the density of water, we can easily come up with a direct relation between water height (regardless of the horizontal path taken ...


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Based on your responses to the previous answers, I understand that there are actually two distinct bodies of gas involved in the setup you describe: The gas which, at the start of your experiment, is already in the sample chamber (gas A) The gas which, at the start of your experiment, is stored in a separate vessel at a higher pressure (gas B) These two ...


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Consider an extreme version. Ammonia gas in one, and water in another. This is a classic chemistry experiments called an Ammonia Fountain Pressure decreases substantially


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Tall buildings are subject to the stack effect. Because the interior is held at a different temperature than outside (due to indoors temperature control), the vertical pressure gradient is different. Typically it is a reasonable assumption that the pressure inside the building is continuous from one floor to the next via the stairwell (if nothing else). ...


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Best answer (besides everything said about pressure) I can give you is this: Any isolated physical system evolves not only towards the state of lowest energy, as you said, but also highest entropy. Now entropy is a delicate topic in it self but for answering your question it will suffise to think of it as a measure of order. The lower the entropy the higher ...


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If your objective is to obtain precise and equal flow rates through the three holes however constrained by serial arrangement you sketched then you need to progressively increase the radii of each duct that are downstream of one another.


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I'm guessing the flow rates in your system are slow so the flow is laminar. In that case the flow is described by the Hagen-Poiseuille equation: $$ \Delta P = \frac{8\mu\ell V}{\pi r^4} $$ where $\Delta P$ is the pressure drop, $\mu$ is the viscosity of your saline solution, $\ell$ is distance along the pipe, $r$ is the pipe radius and $V$ is the volume ...


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As mentioned in DumpsterDoofus's answer, those potentials are likely relative to atmospheric pressures but at magnitudes of 1 GPa that's pretty much irrelevant. It's the pressure of the gas vapor that determines equilibrium. The pressure of the liquid is nearly irrelevant, so in this case even if the capillary forces lowered the pressure below absolute ...



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