# Tag Info

35

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 ...

20

You aren't creating a vacuum, but you are reducing the pressure in your lungs when you inhale. In effect your lungs are working as a diaphragm pump. When you pull your diaphragm down, and/or expand your chest, this increases the volume inside your lungs. Boyle's law tells us: $$P_0V_0 = P_{\rm inhale}V_{\rm inhale} ,$$ where $P_0$ and $V_0$ are ambient ...

19

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 ...

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.

14

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. ...

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 ...

4

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 ...

3

I believe that in the top you'll find it's the saturated vapor pressure of water. http://en.wikipedia.org/wiki/Vapour_pressure_of_water (80 F ~27 C vapor pressure of about 27 mmHg or 27mm/760 mm * 14.7 = 0.52 psi.. not bad.

3

Taking an intuitive guess here: The pressure above the water column is indeed very low, and water molecules at the surface may escape - but they are also held back by the surface tension of the water (is your meniscus concave or convex?). There is an equilibrium here, and the temperature is low enough that the water won't boil off all at once. So at room ...

3

There more sides to this scenario that you're considering. Firstly, if we are assuming that the temperature is the same at sea level and on the high mountains, then the speed of sound doesn't actually change, as a constant temperature will take care of the air pressure-density ratio. $$c = \sqrt{\kappa \frac{p}{\rho}}$$ Where $p$: static air pressure, ...

3

I'll use this answer to provide some information that's mostly orthogonal to what Phonon said. As Phonon pointed out, the speed of sound depends on temperature, not pressure. It's cold on the top of high mountains, so the speed of sound would tend to be lower. Some mechanisms for sound production have a frequency that depends on the speed of sound, and ...

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 ...

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 ...

3

An answer to v1 of the question (that didn't emphasize the sides): According to this cooking website, the pressure cooker valves are designed to be naturally "open." As the water heats up, the pressure inside will be slightly higher than outside (which is why you see steam escaping). At some point the valve is designed to "close" once a certain minimum ...

3

If you neglect viscosity, Bernoulli's equation (just Navier-Stokes without frictional or stress terms) will get you into the ballpark: $$P_g + \frac{1}{2}\rho_g v_g^2 = P_a$$ Where the $g$ subscripts pertain to the gas and the $a$ subscript to the ambient. The gas density $\rho_g \equiv M / V$ is the ratio of the mass of gas (M) in the tank to the volume ...

2

Working out how much force is required to move through a fluid, whether it's air or water, is surprisingly difficult because there are two effects you need to take into account. However neither of those effects is pressure (or at least only indirectly related to pressure). The first effect is viscosity, which is simply how thick the fluid is. Obviously it's ...

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, ...

2

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

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.

1

The sensor you are talking about is a Pitot tube These have two pressure sensor ports: a port in the direction of the aircraft's travel/airflow to sense the pressure in that direction - the dynamic pressure , whic as you pointed out, is affected by the aircraft's speed and a port perpendicular to the aircraft's direction of travel to sense the static ...

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 ...

1

The force that each of the pistons can drive with is given by $$F=\frac14 \pi d^2 P$$ How much lift that produces depends on the exact geometry of the table and in general will be a function of height. In essence if the table moves $x_1$ when the pistons move $x_2$ the force will be $$F_{table}=2F_{piston}\frac{x_2}{x_1}$$

1

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 ...

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 ...

1

When a pipe is bent the outside curve - what would be the longest path through the curve - has the highest pressure and the lowest speed. The inside curve - the shortest path through the curve - has the lowest pressure and the highest speed. In short, when the path of a fluid in steady-state flow bends, the pressure on the outside of the bend is always ...

1

I propose the following order-of-magnitude, very rough line of reasoning. The rate of momentum escaping from the aperture is $$\frac{dp}{dt}=\rho\pi(D_2/2)^2 v^2,$$ all in the horizontal direction. I assume that a fraction $(1-\cos\theta)$ of the horizontal momentum will be lost by the fluid since the change of direction of its motion. This fraction of ...

1

Interesting question. I suppose one should compare several scenarios Lie still Go forward - either straight, or hard to port, or hard to starboard Go in reverse The rate at which water enters the ship is (to first order) proportional to the pressure differential - lower the pressure and live longer. Maybe even long enough for the Carpathia to come and ...

1

When you heat water on the stove top, you see bubbles forming on the bottom of the pot. The bubbles are created where the heat applied (if you move the pot, you see the bubbles forming in a different spot) and is sufficient to convert the liquid into a vapor (less heat would just heat the water). These bubbles form even though the water is below its boiling ...

1

Once boiling starts, there is an equilibrium between the liquid and vapor phase. The pressure, temperature and partial molar Gibbs energy are equal for each phase so that water molecules have no preference for one phase or the other. That's for intensive variables. However, the total enthalpy of the liquid and vapor is not fixed : it keeps increasing as more ...

1

The air in lungs, ears, stomach and intestine has the same pressure as the air pressure outside the human body, ensures that you don't get crushed by the air pressure. If you have already managed living almost compensating the air pressure on you, moving in that air is simple.

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