New answers tagged

3

The force the fluid does on the bottom piece does not depend on the height of the water column of the reservoir. It does depend on the height $h$ of the water column in the plate. This can be easily seen by the fact the water is static so the pressure at any horizontal plane is the same. The pressures in $a$ and $b$ are the same. The force of the water on ...


2

The critical pressure is given by $$P_c=\frac{a}{27b^2},$$ while the critical temperature is $$T_c=\frac{8a}{27bR}=\frac{8bP_c}{R}.$$ The parameter $b$ is related to to the effective volume occupied by the molecules, $$b=4N_0V_0,$$ where $V_0$ is the volume of the molecule and $N_0$ is the Avogadro number. So at least theoretically you can chose $P_c=1\, \...


3

Whether an answer exists depends on your definition of "near" compared to STP. There are a few fluids that have their critical point at a temperature close to STP, but higher pressure. For example, (see http://www.engineeringtoolbox.com/critical-point-d_997.html) material Tc(K) Pc(atm) acetylene 309.5 61.6 ethylene 283.1 50.5 ethane ...


1

I tried to look around but I couldn't find anything. It does seem like supercritical $\mathrm{CO_2}$ is very popular in applications because the critical temperature is just a little over $30^{\circ}\mathrm{C}$, but it still requires 73 atmospheres of pressure. An interesting thing mentioned on the Wiki page is that Venus may have had supercritical $\mathrm{...


0

If you cannot understand, you are not alone; even teachers find difficulty to grab. The difficulty of understanding is due to the rubbish ways of explanation in academic/ textbook. Let's see the actual concept. Cv is heat capacity at constant volume. Constant volume has no direct meaning in heat retaining capacity. Constant volume indirectly means work ...


0

You need some flowmeters. The type depends on the quantity of the gas that you use, and the type. You would then need to come up with a full description of a program to use those values. For example, you might just want to store in a file the quantity of gas supplied to furnaces as a function of time, updated every second (or minute). Or you might want to ...


1

Damping The density of atmospheric air is approximately 1.225 kg/m$^2$. At 3000 atm, the density would be 3675 kg/m$^2$ compared to the density of water of 1000 kg/m$^2$. I don't know the viscosity of high density air, but as @DirkBruere answered, it could be a significant factor in the damping of the spring decompression. The relative difference in ...


2

In 3000atm it's speed of decompression will be slower because it is facing greater air density and the expanding spring has to move it. There will also be less "resonance" as the denser air damps the spring movement.


2

Gold will compress to about half of its volume at atmospheric pressure if you compress it to 2 million atmospheres at room temperature, which is something that I'm sure has been done with diamond anvil cells. For many metals, the atomic lattice will also undergo structural phase transitions from one lattice type to another at certain pressures, but I don't ...


-1

I would say that gold will remain gold and do not go into nuclear fusion. However I am not saying that it is impossible. You may have heard of neutron stars and black holes, when the matter is under immense pressure the electrons around the nucleus could not withstand it and they fall into the nucleus and convert everything in a giant neutronic mass. ...


5

Will a tennis ball go further if i hit it with the side of the racket? No. You want the racket to deform, not the ball. This means using the strings to elastically store energy and return it to the ball. The Ball The ball's deformation upon impact is undesirable because "a tennis ball is required by the rules of tennis to dissipate a fraction of ...


0

To impart certain amount of kinetic energy to tennis ball you will have to do work on it. Now, work done= Power x Time. The difference between hitting using netting and hitting sideways is that, in the former case you have more contact time with the tennis ball available, so you need smaller average power to do necessary amount of work. When you hit sideways,...


6

Here's some tennis racket physics from Rod Cross, including links to several Am. J. Phys articles (the physics educators' journal, thus excellent for learning from) and this excellent diagram: There are at least three "sweet spots": The node, at the center of the strings, is a point where the natural standing waves in a vibrating racket don't have any ...


-3

Interesting indeed... My guess is if you hit the ball on the frame it should go farther. Here's my thinking: Hitting with the frame causes all the compression energy to go into the ball. The frame does not compress at all. Hitting on the stringing causes both the ball and strings to compress, which expends more energy than compressing the ball alone. ...


0

Think of an inflated balloon. Pressure inside the balloon is greater than atmospheric pressure. Entire system is in mechanical equilibrium because of tension provided by balloon surface. If you keep inflating the balloon, pressure inside the balloon keeps increasing, and so does tension in balloon surface, until it can take no more and bursts, thus ...


2

A deodorant can contains a liquid hydrocarbon, typically a propane/butane mixture, and the pressure inside the can is due to the vapour pressure of this hydrocarbon. The pressure can be set to any desired value by varying the composition of the propellant - more propane makes a higher pressure while more butane makes a lower pressure. For a deodorant the ...


0

There are three possible contributors to the leaking: 1) a reduction in ambient pressure as you drive to a higher altitude; 2) a possible expansion of the liquid if it heats up during transport, which makes the liquid expand and decreases the vapor space (hence increasing the air pressure inside the container); 3) If the liquid has a moderate vapor pressure,...


0

(NASA lost a Mars Probe) What you can do: There is a well-known $pV=\frac{m}{M}RT$ universal gas formula. Here: $p$ is the pressure $V$ is the volume $m$ is the mass $M$ is the molar mass of the gas $R$ is the universal gas constant ($8.314$) $T$ is the absolute temperature (in Kelvin) Although your question literally interpreted asks for the weight of ...


0

I believe all these answers miss a practical detail - and that we have to be careful about how we use absolute and gage pressures too. At depth the tire would probably collapse under the squeeze of the increasing external water pressure and the bead would probably pop off the rim and all the air inside would probably escape before it reached that depth. ...


1

I think this happens because when windows are open then there is free airflow in the room and doesn't affect the shutting of the door but when the windows are closed then there is obstruction in airflow. The pressure of air inside the room become less than the pressure of air outside the room and hence obstruct the shutting of the door in the room having all ...


0

Take gas in a container with a piston connecting to its surrounding as an example. When, at the moment you release the piston, the pressure difference between the surrounding and the container is very small, the piston will move slowly because there is not a lot force to accelerate it. In this situation, the system's response (temperature, pressure change) ...


1

A thermodynamic process is called reversible if an infinitesimal change of the external condition reverses the process. Consider a gas enclosed by a freely moving piston in a cylinder. Let us say it is in mechanical equilibrium with the atmosphere, that is, the pressures on the piston match. If you increase the external pressure infinitesimally the piston ...


1

Since the pressure $p$ is intensive and the volume $V$ and enthalpy $H$ are extensive variables the function $p=p(V,H)$ is homogeneous degree $0$ so you always have $$H \frac{\partial p}{\partial H} + V \frac{\partial p}{\partial V} = 0$$


0

The percentage by which the pressure lowers on the leeward side of your finger is minuscule, so it barely affects the rate of evaporation.


0

It seems like the high speed particles would collide with the metal walls and slowly transfer their energy to the slower particles outside the container. The mechanism you describe is correct, but you have to keep in mind that average kinetic energy, $\langle K \rangle$, is only proportional to temperature: $$\langle K \rangle = \frac 3 2 N k T$$ So (...


1

re: "Why don't high pressure gases stored in containers lose energy?" They can gain & lose energy: Energy (heat) is lost from a gas as the gas is compressed (whether thru mechanical compression or thru cooling compression (e.g. passing a gas thru a tube that is immersed in a very cold liquid -- like liquid nitrogen). Energy (heat) is gained by a gas ...


6

Gases in containers at high pressures have those pressures because there are more molecules in them than in the same container at atmospheric pressure, not because there is a difference between the molecular energies. At the same temperature, two containers with different numbers of molecules in them have the same probability distribution of energies. The ...


2

There are two ways you can change the internal energy of a gas, one is macroscopic, that is, performing work on or by the gas, if the gas either expands or contracts. The other is microscopically through heat. If the compressed gas is at the same temperature than the outside gas, these microscopic collisions will not result in an exchange of energy, because ...


0

Assuming there is only one molecular in this box and assuming it is a closed system with initial P-V state is defined. The question becomes: can the system move to anywhere on the PV diagram? Well we can adjust volume to any number. Then the question becomes: can pressure reaches to any values on PV diagram? Pressure relates to impacting intensity and ...


1

In a polytropic process other than adiabatic, you are controlling the temperature in tandem with P and V in such a way that n is constant. You can certainly achieve negative values of n by controlling the temperature appropriately. From the ideal gas law, if T and P are expressed parametrically in terms of V, then:$$\frac{P}{P_0}=\left(\frac{V_0}{V}\right)^...


3

It's simply inherent to the definition of polytropic processes that they don't allow the system to increase both its pressure and volume at the same time. That doesn't mean you can't increase a system's pressure and volume. You just need a non-polytropic process to do so. For example, it could be a compound process consisting of two polytropic processes with ...


3

First, the reason why the finger becomes more wind-sensitive with some saliva isn't that the saliva evaporates but because the saliva, or water, is a good thermal conductor. The finger has to be warmer than the air so the heat flows from the finger to the air and a good thermal conductor such as saliva helps this flux to take place. Second, because it's the ...


2

Assuming an incompressible liquid, Bernoulli for instationary flow (neglecting friction) is $$ \int_1^2 \frac{\partial c}{\partial t} \, \mathrm{d}s + \tfrac12 (c_2^2-c_1^2) + g(z_2-z_1) + \frac1{\rho}(p_2-p_1)=0 $$ with velocity $c$, gravitation accceleration $g$, height $z$, density $\rho$ and pressure $p$ and $1$ and $2$ denoting the two positions ...


2

Suppose you do a force balance on the portion of the fluid situated between elevations z and $z +\Delta z$ in the left column. You get: $$p(z+\Delta z)S-p(z)S+\rho g S\Delta z=\rho S\Delta z \frac{dv}{dt}\tag{1} $$where $v$ is the downward velocity in the left column:$$v=-\frac{dx}{dt}\tag{2}$$ The latter equation is correct because the fluid is ...


0

There are many assumptions made in the problem. But one thing we know is that the motion is accelerated and hence the flow in unsteady. For unsteady flow you cannot use bernoullis equation either. So i guess it'll have to be experimental. I may be wrong, thete may be other methods but none within simplistic limits.


1

No, a feather dropped in a vacuum jar will not drop at the same speed as in air. There must be a thousand video's out there of this very topic but my favourite is this one. An object moving through air, or any fluid, will experience a drag force resisting its motion. This force increases as the speed of the object through the fluid increases. It also ...


2

There are two questions: "Why does vaccum crush the steel tank?" and Why the tank implode?" lemon's answered the first question perfectly - multiply the 1 atm pressure by the surface area of the tank and you will get the force, that crushed it. The second answer is not that simple. The tank walls are designed to transform the pressure forces (perpendicular ...


4

A tank is shaped for pressure from the inside, not the outside. The hull of the tank is convex. Pressure on the inside will cause the hull to assume a shape maximizing the volume per surface which leads to spherical or cylindrical shapes. This does not need much rigidity: balloons come in similar shapes. Pressure on the outside instead will maximize ...


1

Depends on the wall thickness, for example you can collapse a plastic bottle sucking with your mouth but you can't with a glass bottle. There is an Asme code to calculate the minimum wall thickness of a steel tank. The code for external pressure is diferent for internal pressure because geometry of the vessel is very important. Flat and convexe geometry ...


0

There is a difference between pressure and displacement: pressure is "absolute", displacement has a direction. With speakers on the left and right, pointing to each other, the pressure wave from the right will be the mirror image of the one from the left, the displacement wave is not only mirrored but also changes sign (mirrored by the x-axis). Taking that ...


0

I think some cargo holders generally aren't concerned about maintaining pressure, most things survive, it's only when pets and people are involved that pressure becomes an important aspect. If the air pads (I guess you're talking about large bubble wrap) were filled with enough air to be tense on the ground the reduced air pressure at 10,000ft would have a ...


3

If you look at the tank from its circular side you could see how it has to perform like an arch to support the load of atmospheric pressure. Let's imagine we cut a section 1 meter long of this cylinder and cut the bottom part off to have a nice round arch and inspect how it works. It is roughly 3 meters diameter so it has to support a load of 3 x 1 meters x ...


4

Drawing a vacuum in the tank puts the tank walls under a compressive load. The ability of a structure to take compressive load depends on its stability. For a tank car, if we ignore the end caps, compressive loads are acting in two directions - lengthwise and radial. The cylindrical tank will be very stable in lengthwise compression - any buckling forces are ...


19

First of all, as mentioned, atmospheric pressure can exert very high loads when integrated over significant areas. As an example, an overpressure of just 2psi is sufficient to destroy many houses and can kill people. That's about 13% of atmospheric pressure. Secondly there is an important scale question. You give an example of a bike tyre: a road bike ...


47

Atmospheric pressure is equivalent to supporting a weight of 10 tonnes (about 10 average cars) per metre squared. Put like that, it's not surprising that those metal tanks crumple. However, in the comments you raise the point that you pump your bike tyres to 40 psi (about 3 atm) and yet they don't explode. I think this gets to the crux of your confusion. ...


1

According to the second law of thermodynamics,entropy of an isolated system tends to increase. Considering the high pressure region and low pressure region as an isolated system, its total entropy goes up, making fluid flow from high pressure to low pressure to increase the disorder(entropy of the system).This behavior follows from statistical models of ...


2

Interactions between the molecules of the gas are not required. In fact ideal gases are modeled as if the molecules have zero interaction. They do however move and interact with the container. That is sufficient to explain the behavior. Imagine that you have a vessel with two identical halves that are connected by a small portal that can be opened and ...


4

The micron used in this way is a unit of pressure. It's short hand for "micron of mercury". It's the pressure that causes the column of mercury in a mercury manometer (pressure gauge) to rise one micro meter. One Torr is one millimeter of mercury, and atmospheric pressure is 760 Torr. 1 $\mu$ = 0.133 Pa.



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