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Dave, Yes it is possible, but first we need to know the original temperature of the nitrogen in the bottle, and if we want the problem to be easy we have to assume that the flow coming out of the bottle is isentropic, or has a constant entropy. An isentropic flow is one that is both adiabatic and reversible. "Adiabatic" means that there is no heat transfer from the flow to the surroundings or vice versa. "Reversible" in this context means that you could reverse the process and recover the fluid's original state, i.e. there are no losses such as friction which would take some of the usable energy in the flow and turn it into unusable heat. It also means that there are no large gradients in the flow, i.e. that there are no large changes in pressure, temperature, velocity, etc. such as shock waves. Strictly speaking, there are no isentropic flows in the real world since there is always some kind of friction or gradient in the flow, however in this case the flow will be very close to isentropic so we could expect to see the real result be off by only a few percent from the isentropic result. The extra work to do a more detailed calculation would be immense by comparison, and in a real-world application there is probably more error in your pressure regulator than there is in assuming the flow is isentropic. All of that being said, just keep in mind that this is a key assumption that we've made to solve the problem, so if you were to do something quirky like put lots of drag inducers or vortex generators in the flow, or expand the flow so slowly that a significant amount of heat is transferred in from the surroundings, or something like that, you will see a different resulting temperature. For a simple expansion through an orifice or nozzle, this should be a pretty good approximation. I'll give references below, but the isentropic flow equation we're interested in for this problem is: P/Pt = (T/Tt)^(k/(k-1)) P = Static pressure, i.e. the pressure that you've expanded the flow to. In this case this is atmospheric pressure. Pt = Total pressure or stagnation pressure. This is the pressure the flow would have if you isentropically slowed the flow to zero velocity. Since we are assuming an isentropic process that starts from rest, this is simply the pressure at the exit of the regulator, 50 psi. T = Static temperature. This is the answer you're looking for. It's the temperature of the moving flow after it's been expanded. Tt = Total temperature or stagnation temperature. This is similar to the total pressure and again it will merely be the temperature of the flow at the exit of the regulator, which will be the temperature of the gas in the tank. k = Ratio of specific heats of the gas. This is a physical property of the gas and for most gasses it is relatively constant for a wide range of temperatures and pressures. We can simply look this up in a textbook or some other reference. For nitrogen it is about 1.4. Interestingly, if you ask a mechanical engineer what the symbol for the ratio of specific heats is, they will likely say "k" while an aerospace engineer will typically recognize the lowercase greek gamma as the ratio of specific heats. Don't look at me, I didn't decide to teach it differently to different people. So we simply plug in the known values of P, Pt, Tt, and k and then solve for T. If we go from 50 psi to atmospheric pressure (about 14.7 psi) and if the temperature in the tank starts out at 540 Rankine (about 80 degrees Fahrenheit), we'd get a final temperature of about 380.6 Rankine, or about -70 degrees Fahrenheit. A couple of other notes while I'm at it: As you allow nitrogen to escape from the bottle and the remaining nitrogen expands, its temperature will be reduced according to the Ideal Gas Law (Pv=nRT). Keep that in mind if you're very concerned about the resulting temperature being uniform, as this will make a noticable difference if you leave the orifice open for very long. Note that I have not even mentioned the rate at which nitrogen escapes from the bottle. For given values of upstream and downstream pressure, the flow rate of gas through the orifice will depend only on the size of the orifice. You can either pick the size and calculate the resulting flow rate, or pick the flow rate and calculate the required orifice size. You can do these calculations using the same isentropic flow equations found on the NASA webpage I've listed in the references, but it might not be as straightforward as the calculation above. Keep in mind that the mass flow rate will be the product of the density, velocity, and cross-sectional area at a given location, and the volume flow rate will be the product of the velocity and cross-sectional area. If you run into problems, feel free to follow up with another question and I'd be glad to go into more detail. I hope this helps, David Coit References: http://www.grc.nasa.gov/WWW/K-12/airplane/isentrop.html Introduction to Flight, 3rd ed., John D. Anderson Jr.

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