Calculation question

1 point for the insertion of XKCD into any one of my threads. Add 9 more for their appropriate use.

"Well-played, Mauer."

While I get, SWS, that 1PSI = 70cmH20 and change, I'm still stumped on the flow rate, considering the minimal orifice between the generator and the hose end, 6 feet away, is 15mm...



I don't ever want to see a resistance problem again. I hated that class. Although... I could just tear it up and say, "Resistance is futile."

Jules, PRECISELY. A "flush test" type scenario, if you will. (and this is all hypothetical, lest some DME decide to track my butt down)

Chunky, yes, I'm good with computers, but I still need the formula.

Roman,
(I see you are from central NY. Rome, perhaps?) I think I see what you are trying to figure out here.

If you have a flow generator generating 1 cm of pressure, and you have a certain flow through a certain length of hose, of a certain diameter, what is the pressure at the other end?

This is actually moderately hard to calculate accurately.

First of all, lets look at pressure. When pressure is given in terms of a column of liquid (be it water, mercury, etc.), the diameter of the column is irrelevant. Assume you have two U-shaped tube filled with water. One tube is 2 cm in diameter, and the other is 4 cm in diameter. Now, you seal off one end of the U, and apply a pressure of 1 cm/water to the sealed-off side each tube. The water level will go down 1 cm in the pressurized leg, and go up 1 cm in the unpressurized leg. The reason diameter is irrelevant is that air is a compressible medium, and it exerts a force that is the same per unit of area, in the vessel containing it. (A u-shaped pipe like this for measuring air or gas pressure is called a manometer.)

As mentioned by an earlier poster, the pressure at the end of the hose is sort of calculated like you would calculate voltage, current and resistance in electronics. The important thing in electronics is, that whenever a current flows, there are losses caused by resistance. So, if you have a conductor (assume a wire in this case, which is similar to a hose) with a certain resistance (all conductors have at least some resistance), and cause a certain current to flow through that conductor, you will find that the voltage at the far end of the conductor is less than the voltage at the end that is powered. I am sure you understand this. Same with the hose. If the pressure is a certain amount at the flow generator end, and you are flowing a certain quantity of air though the hose, the pressure will be lower on the other end. The diameter of the hose, the length, and the friction the hose offers to the air are the primary determinants of how much the pressure will drop. (And like in electricity, the loss is turned into heat. That is how a resistance heater, like a space heater works. In the case of the hose though, the amount of heat generated is very small, too low to practically measure.)

There are formulas available to roughly calculate the pressure loss in the hose, but I don't know them (I might research them tonight). You could rig up appropriate pressure and flow gauges and measure it (or a calibrated leak at the other end, like a blanked-off mask with a known leak vs pressure curve). This would be considerably more accurate than calculating the loss, as real-world conditions have a great effect on the loss. (Generally, they say you lose about 1 cm in a typical standard CPAP hose, with typical length, and a typical mask/patient at the other end.) The pressure gauges you can make yourself, as described above. Air flow gauges are available from places like Omega Scientific.

Roman Hokie wrote: Chunky, yes, I'm good with computers, but I still need the formula.

I'm unclear why and exactly what you are trying to calculate, but as a suggestion, take the manufacturer's design leak rate for your mask for a given pressure. That represents the flow (volume per unit of time) through the exhaust ports on the mask. Measure the total cross sectional area of those exhaust ports, then scale it up to the total cross-sectional area of the hose. This may not be precise because the manufacturer may have a fudge factor in its design leak rate numbers.

e.g. (all numbers are made up). exhaust ports total cross sectional area = 0.28 sq cm. Hose cross sectional area = 2.8 sq cm, or 10 times larger. Manufacturer's leak rate for 10 cm pressure = 30 L per minute. Therefore, flow rate through the hose (with no mask or control on the end) is 300 L per minute.

If you are trying to calculate the flow through the hose while you are wearing the mask, then just use the manufacturer's leak rate numbers.

Thanks, folks. It is a genuinely hypothetical exercise, but based on a post someone else made last week or so stating a maximum leak for his or her night at 157 l/m and change or something thereabouts.

I'm trying to figure out what possible pressure rating (cmH20) might correspond with a completely open mask (100% leak or a removed mask) that might offer 157 (or 158) l/m.

I thought that that particular number was extremely high.

So, I thought, in theory, that a hose with NO MASK (and thus, no mask leak rate to be considered) connected to a flow gen would have an associated "air thruput" to it, like water going through a sprinkler system.

(think of an automated sprinkler system. If the PSI through the city water line is X PSI, and you run the sprinkler system for an hour (or a minute) how many gallons will you get?

And yes, I'm in Rome.

I'm an engineer and, well, though I passed my EIT nearly 15 years ago, I never had to take a fluid dynamics class...

In this case, with nothing at the end of the hose, you can still use suitable formulae to determining pressure drop in a pipe, except using a pressure of zero for the open end-- the air pressure equivalent of a short circuit. In this case, the 'source impedance' of the flow generator becomes significant. It probably isn't much, and a pressure of 1 cm or less wouldn't surprise me. In this case, it is a good experiment to try this with a manometer (made from a piece of flexible plastic tubing bent in a U, with water in it) connected to the flow generator end of the hose.

BTW, used to drive by the Rome army base all the time when I used to live in Rochester (1988-2000). I would frequently see some of the albino deer that live there along the fenceline.

Thanks, Timba. Yeah, the base was a casualty of BRAC back in the 90s. The city has recovered about as well as it will, which is to say hardly at all.