Quote:
Originally Posted by fovea1959
Why is running air at 70PSI more dangerous than running water at 70PSI?
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There are those who will tell you that 70psi air pressure is more likely to cause a rupture than 70PSI of water pressure at the time of switching. I don't disbelieve their arguments, but the much more compelling argument to me is this: Air at 70psi has a
lot more energy than water at 70psi. That is, should the thing rupture, there's a whole lot more potential for damage.
For scale purposes, I shall use a single FRC Clippard tank (574 ml). I shall also simplify the problem to assume that everything is at the same temperature, 0C. It's a bit colder than the typical FRC playing field, but if you increase the temperature or recognize that air compressors heat the air up as they go, the situation only gets more dangerous.
Acknowledging the hazards of info from Wikipedia, and promising that I will verify them against the CRC Handbook on Monday when I get back to the office:
Quote:
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Originally Posted by wikipedia page for "Properties_of_water"
The compressibility of water is a function of pressure and temperature. At 0 °C, at the limit of zero pressure, the compressibility is 5.1×10−10 Pa−1.[31] At the zero-pressure limit, the compressibility reaches a minimum of 4.4×10−10 Pa−1 around 45 °C before increasing again with increasing temperature. As the pressure is increased, the compressibility decreases, being 3.9×10−10 Pa−1 at 0 °C and 100 MPa.
...The low compressibility of water means that even in the deep oceans at 4 km depth, where pressures are 40 MPa, there is only a 1.8% decrease in volume.[32]
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70 psi is equal to 483 kPa (from google units, but agrees well with my understanding that 1 atm ~ 14.2 psi and also ~ 10
5 Pa).
Finally, recall that Energy is the ability to do work, so let's caculuate how much work each can do:
Let's assume that compressibility scales linearly with pressure over the range given for water. This means that at 70 psi, the compressibility at the end point is: (5.1 - (5.1 - 3.9) * 0.483 / 100) * 10
-10Pa
-1 ~ 5.095 x 10
-10Pa
-1. This is very little change (much less than the implied resolution), so I'll use a constant 5.1 * 10
-10Pa. This means that a clippard tank full of water at 70 psig which is allowed to expand to ambient will increase by about 574 ml * 5.1*10
-10Pa
-1 * 483 kPa = .141 ml (or about 1/7 of a cubic centimeter!). If it were pushing a piston it would do .141 ml * 483 kPa / 2 = 34 lPa = 34mJ, or .034 joules. For those who think in English units, a joule is about three-quarters of a foot pound, so this would lift one pound about 5/16 of an inch.
Now, let's do the same thing with air. To keep the math simpler, I'll assume that the temperature remains constant. Then, as we are talking about the same mass of air, we can use Boyle's Law, which is that PV = constant. For Boyle's Law, we must use total pressure, so that our end pressure is 1 atmosphere, or 101 kPa, and our starting pressure is 483 kPa + 101 kPa = 594 kPa. Our initial volume is again 574 ml. We have: 574 ml * 594 kPa = V * 101 kPa. Solving for V, we get 3370 ml. That is, the gas expanded by 2800 ml, or nearly three quarts. Because the pressure does not decrease linearly with volume, the average force is not just half the initial force, but a bit less. The work that can be done by this expansion is 332 l kPa, or 332 J. In English units, this will lift one pound about 244 feet!
In the actual case of an exploding tank or valve, the temperature will not remain constant. The rapidly expanding air will cool, decreasing the pressure and therefore the amount of mechanical energy released. Air is about 40 percent more compressible in the isothermal (constant temperature) than in the adiabatic (constant entropy, or very rapid) case. This means that the actual amount of energy released will be closer to 150 foot-pounds, but there's still five orders of magnitude more energy in the air than the water.