10

I've been doing large pressure vessel design for a while, and ultimately it comes down to what is negligible v. what is not. Especially in the realm of atmospheric storage vessels, scrubbers, or vessels within 10 atmospheres of pressure, there isn't a lot of radial stress compared to the other conservative factors. In the realistic engineering world, by ...


6

The idea that I was describing in the comments earlier is pretty similar to that described by Mart - I imagined the tube (red) flaring out underneath the O ring, and the opposing taper being either pushed down by hand or driven by a standard nut (black) depending on the sealing forces required. The key difference is that instead of relying on the squashing ...


5

Vacuum pressure on a vessel is quite different than positive pressure. Under positive pressure, the shell of the vessel is essentially under a uniform membrane tension due to the shell wanting to expand outward from the pressure. Under vacuum (negative) pressure, the opposite is true, where the shell is in compression. While a cylinder is stronger in this ...


5

In the molecular flow region of pressure, the thermal conductivity of an ideal, monatomic gas is obtained by this equation. $$ k = \frac{1}{\pi^{3/2} d^2}\sqrt{k_B^3T/m} $$ where $d$ is the collision or molecular diameter and m is the molar mass divided by Avogadro's number. It is independent of pressure. Approximations have been derived for non-monatomic ...


5

NOTE ON OBROUND VESSELS While not a direct answer to your question, it should be quickly noted that ASME BPVC Section VIII, Mandatory Appendix 13, Section 10 has a section to design obround vessels (circular half sections with flat plate walls, see image below). While not the same, these can typically optimize the available area compared to elliptical ...


5

Don't do it!! Just buy a pressure regulator to put between the two tanks that will decrease the pressure to below 125. You can get these for \$17 at home depot (https://www.homedepot.com/p/Husky-1-4-in-NPT-Regulator-with-Gauge-41155HOM/205331831) and maybe even cheaper elsewhere. I'm terrified of the tank exploding and covering me with shrapnel. This is a ...


4

Both Axial stress, (parallel to the axis of cylindrical symmetry) and Hoop Stress (in the tangential direction) are "Normal Stresses". They are "Normal", because, if you consider an infinitesimal unit volume, the stresses can be considered to be entirely axial in the X and Y directions respectively. There is no stress in the Z direction because of the 'thin ...


4

Use two strap wrenches or an equivalent - one on the tank and one on the cap. You won't crush the tank :) If you need to make one, then a length of 4*2 and some old car seat belt bolted to one end works well enough - it did when I had to do something similar.


4

You could have a barely undersized O-ring around the nozzle, then use a thread around the nozzle to press on the O-ring with a nut. This would require to machining a custom nozzle and nut: The part of the nozzle where the O-ring rests and the machined part of the nut should probably be larger than in the drawing, 4.5mm or so. I would also consider a ...


3

Check Valves are available with adjustable cracking pressures, using a preload spring to control the poppet force, for example: http://www.my-ssp-usa.com/products/valves/check-valves/adjustable-cpa It's important to note, however, that there is a degree of histeresis with these valves, and once open, they will not close if the pressure difference drops a ...


3

The mathematically determined formula for a sphere with a thin wall under external pressure (which is very inaccurate - keep reading!) is simply: $$ \frac{2Et^2}{r^2\sqrt{3(1-\nu^2)}} $$ Where $E$ is the modulus of elasticity, $t$ is the thickness, $\nu$ is the Poisson's Ratio and $r$ is the radius of the sphere (to the midplane). This is a result of ...


3

Contact an engineering firm with experience designing and building such systems and vessels to code. Otherwise, you're just building a bomb.


3

It won't. Solid objects don't break in hydrostatic compression. The pressure is pushing all parts of the rod together, so there's nowhere for them to go except stay together. You can imagine an object on the ocean floor. Pressure could easily exceed shear strength but even soft animals remain intact, as do rocks. The von Mises failure criterion says a ...


3

I would say that it is true and for that case there will be many relevant codes / regulations that need to be respected. Make sure you read them all before designing / altering the system.


3

This is because of units. Getting the right answer with your units and their approach What the people have done in your question is converted everything to metres. The division of 39.93 is because that's how you convert inches into metres. L's value is actually 1m. $$ \frac{1000mm}{25.4mm} = 39.37 $$ thus: $$ 39.37\frac{inches}{metre} $$ So the equation ...


3

Here are the limitations to this approach. First of all a howitzer is specifically designed for relatively short-range use, by using a very steep (in some cases, almost vertical) launch angle which allows the howitzer projectile to be lofted high up above trees, buildings, and even hilltops and then strike the intended target by falling almost vertically ...


3

I understand your setup as follows (in flow direction): compressor with cutout set to 150 PSI tank (T1) rated for compressor, with safety valve (planned) pressure regulator (PR) additional tank (T2), rated 120 PSI, with safety valve If this is so the safety valve of the second tank will protect you unless you tinkered with or replaced this safety valve. ...


3

In industrial settings, the correct piece of equipment is referred to as either a Pressure Relief Valve (PRV) or Pressure Safety Valve (PSV). These are spring-loaded valves with a few specific characteristics The valve is designed with a spring load, and is ordered with a relief pressure in mind (in your case, 120 PSI) the valve is sized with a valve ...


2

By dividing the diameter and length dimensions, which are in inches, by 39.37 you are converting them to metres. The problem you are trying to solve has inputs that are in both English and metric units, but the answer needs to be in metric/SI, so you need to convert everything to SI units prior/during calculation. 1 inch = 2.54 cm = 0.0254 m 1/0.0254 = 39....


2

Spherical vessels have a stress in the wall equal to, $$ \sigma = \frac{p r}{2 t} $$ where $p$ is the internal gage pressure, $r$ is the inner radius of the sphere, and $t$ is the thickness of the wall, where the only constraint is $r/t \geq 10$.


2

As a reality check lets consider industrial gas tanks as the specifications are easily available. A BOC size D hydrogen tank has a gross weight of 57kg and contains 23kg of hydrogen at 230bar. At atmospheric pressure hydrogen has a density of 0.09 kg per cubic metre so you get about 255 cubic metres of gas at atmospheric pressure. This will give lift ...


2

No, liquid check valves function on density, viscosity or velocity or a combination and the mass is taken into account - gas check valves are designed to deal with a fluid 700 or more times less dense than a liquid. Some gas check valves will function with a liquid - a check valve designed for blowing up party baloons for example will work with a liquid at ...


2

Hydrostatic pressure of water changes per depth according to Pascal's law which equates to $1 \text{Pa} = 1000 \text{kg}/\text{m}^3 * 9.81 \text{m}/\text{s}^2 * h$ or $101 \text{Pa} /\text{m}$ or $1.46*10^{-2}\text{PSI}/\text{m}$ or $4.45*10^{-3}\text{PSI}$ per foot. That is not a lot of margin to measure especially when the water is actively boiling which ...


2

The pressure in the silo is determined by the temperature rise, caused by the combustion. The end temperature after combustion can be calculated like this: $$T_{end}= \dfrac{E}{(m_{air}+m_{fuel})C_V}+T_{begin}$$ Where $E$ is the added energy added per space(eg. m3), $m$ is the mass of the fuel and air in kg present in that space, and $C_V$ is the specific ...


2

For spheres,stresses in the material is same in all directions.So ,hoop stress and longitudinal stresses are the same. Using similar abbreviations, as P for pressure(gauge) inside sphere ,FS for factor of safety,S for allowable stress and additionly Ri for inner radius,Ro for outer radius P(gauge pressure)=((Ro^2-Ri^2)×S)÷((Ri^2)×FS) For ...


2

A section of a long straight pipeline can be considered as plane strain. When the internal pressure creates radial strain in the pipe, the Poisson's ratio effect can not reduce the length of the pipe so the strain in the axial direction is always zero. (Think about a pipeline that is say 100mm in diameter and 100km long - it doesn't make sense to say that if ...


2

There are a lot of variables to consider for a thermal analysis, but none of them require pressure as a boundary condition. As alephzero indicated in their comment, this is function of heat source and sink. Assuming the most basic situation, steady-state conduction, you could use the equivalent resistance equation: $$Q = (T_1-T_2)/R$$ Where $Q$ is heat ...


2

The reason that you can't find anything on Google is that it is a bad idea to make a rectangular pressure vessel. You get stress concentrations at the corners which makes the vessel much weaker. This is bad because when pressure vessels fail, they release a huge amount of energy very suddenly. It only takes a little googling to find examples of fatalities ...


2

The only way to keep pressure constant as volume is reduced is by let out the same amount of liquid/gas as the volume contracts. That is exactly what a pressure reducing valve does. As long as it has the capacity to vent at the rate you need you should be fine.


2

Have a second separate balloon that expands as the first is compressed. This will also keep the pressure constant as the volume is increased if that is part of your requirement.


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