# Tag Info

0

If we consider the air (outside the vacuum can) and the can as an isolated system, then in the absence of external forces, the position of the center of mass should stay the same. As the air enters the can through the hole on the right of the can, the position of the air's center of mass will move to the left. To compensate, the can will move slightly to the ...

0

The NCEES: FE Reference Handbook has some good material on fluid flow through a submerged orifice in its fluid mechanics section. You can search for it online. NCEES will provide you with one free of charge.

1

I have used the Darcy Formula together with the following formulas for a quick numeric solution (only a few iterations needed) $$h_f = \frac{\Delta P}{\rho g}$$ $$f = {\rm Darcy}(Re)$$ $$h_f = f\,\frac{L}{D}\,\left( \frac{v^2}{2 g} \right)$$ Solve above for $v$ $$Re = \frac{\rho D\,v}{\mu}$$ Go to step 2 until $f$ converges to a value.

0

From Wolfram article we get the simplified Bernouilli equation: $Q = a c \sqrt { 2gh }$ Where: $Q$: the flow rate (${m^3}/s$). $a$: the area of the hole ($m^2$). $c$: flow coefficient. $g$: the gravity acceleration ($m/s^2$). $h$: the depth of the hole ($m$). That is valid for a small enough hole, but since you hole can be big, we have to use ...

1

First assume that $h$ doesn't change very much because you have a large body of water (we can relax this condition later). Let's also assume that the hole is small compared to the depth ($d \ll h$) - we'll relax this too. For this case, the answer is straightforward, you'd use Bernoulli's equations and simply set the static pressure ($\rho g h$) equal to ...

1

The surface formed by the bubble is such that its energy is minimized. Since increasing the interface between a liquid and air increases its energy due to surface tension, the bubble tends to reduce its radius, which implies that the pressure inside it must be higher than the pressure outside, and following this reasoning you may also get a quantitative ...

0

You can do it without raising the internal pressure. First you have your pool, at 1 atm or ambient inside the vessel. Then you have a wet chamber which can be sealed at both ends. Enter the pool, then only the inner door of the wet lock opens, enter this area, and close the door to the moon pool behind you. When the internal door is shut you can open the ...

3

The implications are that you'd have to pass through an airlock to get to the room, and that it would only work to a certain depth. Correct x 2. You have 2 choices with an underwater habitat - build it really strong to take the pressure, or just pressurize it and you can make the whole thing out of plastic. Submarines chose the former, as they may ...

0

Below the yield stress your fluid is behaving like an elastic solid. Imagine putting your tank in zero g, so there are no forces, and then removing the base. The result would look like the left hand figure in the diagram below: Now turn on gravity, or apply an external force and the result will be the middle diagram. As long as the stress is below the ...

3

It is like a balloon. The pressures of the inner and outer air tend to equilibrate, creating a force over the balloon surface from the higher pressure to the lower one, trying to make them equal (the force goes from inside to outside, when you inflate it, from outside to inside when you deflate it). That's why it changes its size, because the gas pressures ...

2

If the flow is laminar, i.e. not turbulent, then the relationship between flow rate and pressure is given by the Hagen–Poiseuille equation: $$\text{Flow rate} = \frac{\pi r^4 (P - P_0)}{8 \eta l}$$ where $r$ is the radius of the pipe or tube, $P_0$ is the fluid pressure at one end of the pipe, $P$ is the fluid pressure at the other end of the pipe, ...

1

I will vote for Water filling the sewer so quickly, that it made the air pressure this strong? Would this really be the easiest way for the air to leave the sewer? In the two videos John Rennie linked to it is evident that the air is coming out with pressure aerating the water which is high around the manhole. In your video not enough water is seen ...

13

Is there a particular way that you think this scheme will fail? Rather than have an airlock with that particular portion of ship, you can simply pressurize the entire vessel. There are practical reasons why you would not want to do this at great depths (related to how much gas you use and toxicity), but the problems are not related to how the access works. ...

31

I drew an image to illustrate the forces at play. For any curved surface of the bubble, the tension pulls parallel to the surface. These forces mostly cancel out, but create a net force inward. This compresses the gas inside the bubble, until the pressure inside is large enough to counteract both the outside pressure, as well as this additional force from ...

5

This can be computed for small changes in the pressure by considering the partial derivative of the temperature w.r.t. pressure at constant entropy. If we suddenly raise the pressure a bit, then this is to a good approximation an isentropic process as no heat is exchanged and it is not a violent process causing large irreversible effects. So, we want to ...

18

The increased pressure is caused by the surface tension between the soap and the surrounding air. This can be seen by a simple equilibrium energy argument. The total energy of the system reads $$E = E_i + E_o + E_s \;,$$ where $E_i$ is the energy associated with the air inside the bubble, $E_s$ is the interfacial energy, and $E_o$ denotes the energy ...

0

The answers in the comments may be possible, but unlikely. Air would be able to escape through other places such as storm drains/gutters. Its seems more likely to me that the lid is not being pushed up by high pressure underneath so much as it is being pulled by low pressure above the hole. It has to do with something called Bernoulli's Principle. You can ...

0

Two things against: It's unlikely that the waves will actually all interfere constructively. In a realistic setting the sounds won't be constant, so points of constructive interference are very temporary and don't have time to do anything even if they could. Each of the sound sources is putting out a certain amount of power as sound (in all directions), ...

3

This page quotes the pressure inside a soap bubble as $\frac {4\gamma }R$, where $\gamma$ is the surface tension, about $25\text { dyne}/\text{cm }$ for soapy water, and $R$ is the radius of the bubble. For $R=1$ cm, the pressure is then $100 \text { dyne}/\text{cm}^2 = 10 \text{ Pa}$. This is released when the bubble pops. It doesn't seem like much with ...

16

The air pressure inside the (intact) bubble is larger than in the surrounding. This pressure difference is called Laplace pressure and is caused by the surface tension between the soap film and the air. When the bubble pops the compressed air expands, thus creating a pressure wave, which you ultimately hear as the typical popping sound.

0

Cavitation is the formation of bubbles in a liquid when a sufficiently strong negative pressure is applied. A point in the liquid experiences a “negative pressure” if the local pressure goes below the average pressure in the liquid. This can happen when water in a pipe has a very abrupt turn, near the propellers of ships and submarines, in presence of a ...

0

You're taking outside air, compressing it to more than double the pressure, ($14.96+15$, compared to atmospheric $14.96$) and putting it into a 9 cubit foot container. So the pump must suck in around 18 cubic feet of outside air and run it into the container. If we assume that the quoted figure is the through-put of the pump with no back pressure, and that ...

0

Actually, it's not as easy as that. In 3.6 minutes, you have filled the space, but there is no additional pressure. You need to run the pump as long again to get an atmosphere of pressure inside the paddle board. Also, as the pressure increases in the paddle board, the efficiently of the pump decreases, therefore taking longer to fill it to the correct ...

0

I'm not sure what our question is exactly, so I'll try to answer what I guess your question is. In general, the entropy of a fluid is a function of both $V$ and $T$. During an isentropic compression, the decrease in entropy from the reduction of volume is compensated by an increase due to the temperature rise. The net effect is zero. For an ideal gas we ...

1

By definition, $dS =\frac{dQ}{T}$, so an adiabatic process doesn't change entropy. But you can find more details at http://en.wikipedia.org/wiki/Adiabatic

2

Some of these responses are hilarious, yet some have a bit of accuracy. For the man who asked the question... never believe anything seen on TV. Ever. Have divers been to depths greater than 100m, 200m? Absolutely. Does it require a monumental amount of time, money, planning, and training? You bet. Not just the decompression, but the acclimation to the ...

0

Your derivation of the additional force on the scale for the falling rope is wrong, both cases yield the same results. If I understand correctly you are comparing the effect of a rope falling on a scale, to the similar fall of a liquid, something like this schematically: The problem of your reasoning is related to a misconception of the stagnation ...

2

No. the whole weight will not act on the base of the container 2. If the whole weight had acted on the base of container 2, then the pressure on the base of container 2 would be equal to that of container 1 i.e., mg/A, where mg is the whole weight of the fluid and A is area of the base. But as you know the pressure at a depth depends on the height of the ...

5

No, the entire weight will not directly rest on the base of the slanted container (although it does indirectly). There are a number of ways to approach this, but the easiest way is to observe that the total force acting on the bottom of the container is equal to the sum of the hydrostatic pressure force (the pressure at the bottom of the container multiplied ...

1

As Rahulgarg mentioned, the pressure does not depend on the shape but on the depth. However, the direction of the force caused by pressure can be approximated as being normal to the surface, hence the total force on the sides will depend on the shape. For a fluid at rest like the one I think you are assuming, the pressure at any depth will be $p=p_0+\rho g ... 0 As that wikipedia link suggests, gauge pressure is compared to the surrounding air pressure. It really is intended to be used for "what a pressure gauge reads" in "everyday life". The all-surrounding presence of air on the Earth makes it a convenient zero point, so it is immediately obvious when things are not in equilibrium with the environment. Your ... 1 If you replace the fluid scenarios with solids with matching have the same shapes, then indeed the pressure at bottom would be the weight of each solid divided by base area, i.e., they would be different! Then what's different about the fluid case? I presume in scenario 1 you have a prism, in scenario 2 you have a bottom-heavy frustum, and in scenario 3 ... 2 Consider this diagram showing the three columns you describe all connected to the same body of water: Your question asks whether the three pressures$P_1$,$P_2$and$P_3$will be the same. The answer is obviously yes, because the columns are all connected to the same body of water. For example if$P_1 > P_2$then water would flow from the base of ... 0 It looks like the question boils down (at least in part) to the following: can a fluid have negative ABSOLUTE pressure? This question has been discussed here several times. My take is: it can (although such state is probably metastable in the best case), because the force between two molecules can be attractive. See, e.g., ... 2 Let's take your last question first. Let the stress tensor at a point (x,y,z) in the fluid be given as$\sigma$. You can pick a Cartesian basis$\{ e_1, e_2, e_3 \}$and express the components of the tensor in that basis$$\begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & ... 0 It seems like the dyson fan uses a regular fan to pull air into its base creating its air multiplying effect. If that's the case a single fan and motor could drive multiple dyson structures at once, which could result in an overall lower net weight. Not sure how well these things would perform outdoors in air turbulence, but they might make for nifty indoor ... 0 Google didn't immediately come up with anything significant for "Ludvigsen's methodology", but let me give this a shot nonetheless. Sound is a propagating pressure wave. So as it goes by, the pressure increases, then decreases, then increases again, etc. Pressure increasing means the particles in the material (typically air) are closer together for some ... 1 There is no quick answer, except if the droplet is completely non-wetting or if it is at least partly wetting. If it is completely non-wetting, it will be move towards the wide side of the funnel until it is a spherical drop touching only its wall. If it is at least partly wetting, it will move to the narrow side until it reaches its apex (if air is ... 4 Water vapor is invisible. I think you mean fog - fine water droplets condensed from water vapor. Pressurized planes fly with an 8000 foot equivalent altitude and humidity in the cabin is low. But, at 35,000 feet (called Flight Level 35) it is likely one would get a brief fog. If loss of pressure is fast, you would only get to watch for a few seconds. At ... 1 The piezo electric effect is caused by the distortion of, normally, a crystalline solid. By applying pressure, you are creating an electromechanical interaction between the crystals electric and mechanical states. It is also a reversible process, i.e. applying a current to an piezo electric material will distort it (this is how most buzzers work or the ... 0 Real bodies aren't airtight. In fact, one of the most important principles guiding the evolution of multicellular organisms is that materials (such as oxygen) need to be exchanged throughout the organism, and so many interfaces are actually conducive to equilibrating pressure. When the ambient air pressure increases, the air pressure in your lungs increases ... 0 It is due to Archimedes' principle. The total force on the 'air bubble' under the water is the force of gravity on it (downwards) plus the force of gravity on the water that the 'air bubble' displaces. Since the same volume of water that replaces the air is more massive (because of higher density) there is a net upward force on the bubble and causes it to ... -2 It was due to the slight expansion of the liquid and the slight contraction of the bottle gained when in the freeze. Normally, when liquids tend to reach their lower fixed point/freezing point they tend to expand which is the opposite of solids. If the bottle was left in the freezer for some time the bottle was going to explode. 0 You could start with a pack of cards and ask how long does it takes for the whole pack to free fall after the bottom card supporting the rest of the pack is released all cards are individually held from the sides, and then released at the same time. From this, I think the answer to your original question is that it depends upon the shape of the ... 1 As a scuba diver you know that pressure increases when you go deeper. Imagine a cylinder held vertically under water. The force on the top of the cylinder is pressure times area (by definition of pressure). On the bottom of the cylinder the area is the same but the force is greater (deeper, more pressure). The difference between the two is the buoyancy ... 0 Answer to question (1) is NO: Using the equations of the comments$(P_1 V_1=P_2 V_2$), you get:$P_2=P_1/(1+.06666)$(in this case of a small drop in pressure the answer will be approximately correct though, because$1/(1+x) \simeq 1-x$for$x<<1$) Answer to question (2) is YES: Using$n_2RT_1=(1-0.06666)n_1RT_1$you get$P_2=(1-.0666)P_1$0 Pressure in its simplest sense is just a force acting over an area. Imagine all the particles in the air in the car. The pressure of the air is really a measure of the average force these particles are pushing against each other with. When we bring in a helium balloon to float in the car, the air particles push against the helium particles, and the helium ... 2 Basic idea Picture in your mind a deep ocean of water. Imagine a column of the water, going from the surface down to a depth$d$. That column of water has some weight$W$. Therefore, there is a downward force of magnitude$W$on that column of water. However, you know the column of water is not accelerating, so there must be an upward force of magnitude$W\$ ...

0

Well, I've always thought of it as gravitational pull on a non-equilibrium state. Try to picture 2 different balls one on top of the other falling from the sky (In earth atmosphere). If the lighter ball is on top of the heavier ball, the lighter ball will separate from the heavier ball. If the heavier ball is on the top of the lighter ball, then we have 2 ...

1

Consider this container in pressurized air but zero-gravity (and ignore surface tension, which would make the liquid ball up). If your guess were right, the liquid would squirt out the small hole on the right, but that ignores the role of the wall on the right, which counters the pressure on the left. Think of the liquid as a collection of horizontal ...

Top 50 recent answers are included