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The circles are the experimental data. The line is the model calculation as I described in my most recent posts.
What was the time interval over which the pressure dropped?whiteskulleton said:I just ran another test in the same exact geometry conditions as before. The starting pressure was at -510 mmHg or 33330 Pa(abs). However I stopped the flow suddenly at 57462 Pa(abs). It then dropped to a final pressure of 54929 Pa(abs). Since the change in pressure occurs in a closed system, it is an isochoric process. Thus:$$\frac{T_1}{P_1}=\frac{T_2}{P_2}$$ Solving gets me ##T_1=309 \,K##. Based on the adiabatic assumption, the temperature of the air when I stopped it should be approximately 336 K. That's an error of 8.7%. Do you think that this is cause for concern for the models we made?
There is also significant viscous heating occurring in the air flow through the entrance hole, essentially the same as that present with the Joule Thomson effect. So even though the gas enters the tank at a lower pressure than atmospheric, its expansion cooling is fully offset by the viscous heating in flow through the hole, so that its temperature essentially does not change in passing through the hole. It then compresses within the tank, and its temperature rises as a result. This is the essence of the first law of thermodynamics analysis presented early in this thread. Either one accepts the first law of thermodynamics or one doesn't.JBA said:If using a adiabatic solution when filling from the 14.7 psia pressure air into the lower pressure tank there is a expansion cooling for that gas mass relative to the transitive pressure ratio during the fill. Only the air mass that is heated by the compression by using the initial 4.85 and final 14.7 pressures is the air mass in the vessel before filling can be this equation. The relative percentage of each gas mass and its BTU contribution must be used to determine the actual vessel air temperature of the mixture after filling.
| Pressure (psig) | Pressure (psia) | Temp (°F) |
| 0 | 14.7 | 54 |
| 500 | 514.7 | 61 |
| 1000 | 1014.7 | 67.2 |
| 1500 | 1514.7 | 74 |
| 2000 | 2014.7 | 79.9 |
| 2500 | 2514.7 | 83.4 |
| 3000 | 3014.7 | 87.7 |
| 3500 | 3514.7 | 92.3 |
| 4000 | 4014.7 | 96.4 |
| 4500 | 4514.7 | 101.9 |
I'm not familiar with the details of the experiment you describe, so I can't comment intelligently. But, if you can identify something wrong with how the first law of thermodynamics was applied (by two different members using two different equivalent versions of the first law) in the first several posts of the current thread to the problem at hand, please point it out.JBA said:This is not consistent with the bottle temperatures in the prior project where the air was being injected into 14.7 psia & 54°F atmospheric pressure bottles and being increased to 4514.7 psia with only a consistent 102°F measured final temperature in testing and verified by an equally limited pressure drop upon cooling to ambient. See the below test data example which is consistent over multiple such fillings.
Pressure (psig) Pressure (psia) Temp (°F) 0 14.7 54 500 514.7 61 1000 1014.7 67.2 1500 1514.7 74 2000 2014.7 79.9 2500 2514.7 83.4 3000 3014.7 87.7 3500 3514.7 92.3 4000 4014.7 96.4 4500 4514.7 101.9
I am really interested in being able to read any reference material that you may recommend that discusses this issue. I have not heard or read this before and all of my prior references have only made vague statements about the Joule Thomson effect and have never covered it to any depth.Chestermiller said:There is also significant viscous heating occurring in the air flow through the entrance hole, essentially the same as that present with the Joule Thomson effect. So even though the gas enters the tank at a lower pressure than atmospheric, its expansion cooling is fully offset by the viscous heating in flow through the hole, so that its temperature essentially does not change in passing through the hole.
Thank you for your kind thoughts. A few years ago, another PF member and I were very puzzled about the Joule Thomson effect, and were trying to get an understanding of why, when an ideal gas experiences an adiabatic pressure drop in passing through a valve or porous plug, its enthalpy per unit mass and temperature do not change as a result of expansion cooling. So we collaborated on this and analyzed the problem in private conversations for several weeks. We finally came to the (correct) conclusion that the viscous heating was exactly cancelling out the expansion cooling.JBA said:@Chestermiller, I continue to be astounded by the depth and width of your knowledge of the deep technical physics of many processes; and, in the event that we diverge on a subject, if ask by someone which path to take I would generally recommend that your's, by far, has the highest probability of being correct.
I find the above to be amazing. In all of my reviewing of reference texts available to me on the analysis and physics of fluid and gas flow the focus was on Bernoulli's equations and the standard array P/T = PT etc using k = 1.4 etc. I have never seen anything with as much depth as presented here. That is why for the majority of the thread I was satisfied to be a "fly on the wall" observer.
On the expansion cooling issue, after reviewing my prior project I realize there is a disconnect between that system and the one here with regard to expansion cooling of the source air being delivered to the tank. In this case, there is an infinite supply of ambient temperature air provided during the filling. In my tank to bottle filling project there is a finite initial high pressure supply in the supply tank. As a result, as that tank is depressurized it goes through and expansion cooling of the remaining air in the tank; and, therefore, the air delivered to the bottle is progressively colder as the filling progresses and it has nothing to do with the expansion thru the flow valve to the bottle. I also now realize, if the viscous heating you mentioned occurs, the delivered air can then be subject to compression heating from its initial feed reduced temperature, all of which is something I need to investigate.
At the same time, I still have a problem resolving the nozzle flow ΔT/ΔP first law issue simply because in years of flow testing, I have not observed the degree of temperature change the classical first law predicts; and, I am most experienced with pressures in the 1 to 10 thousand psig range where there are can be extreme ΔP changes, i.e. such as the 14.7 to 4500 psia compression as exists in the scuba filling tests or flowing valve and nozzle certification tests with the reverse condition.
I am not taking a hard stand on this issue, it is just that I am having difficulty reconciling my prior observations with the first law results.
Above you stated:
I am really interested in being able to read any reference material that you may recommend that discusses this issue. I have not heard or read this before and all of my prior references have only made vague statements about the Joule Thomson effect and have never covered it to any depth.
| P i = | 4.85 | psia |
| T I = | 70 | °F |
| T I = | 530 | °R |
| P f = | 14.7 | psia |
| k air = | 1.4 | |
| T f = | 655 | °R |
| T f = | 281 | °F |
655 R = 195 F, not 281 F. This agrees roughly with the results of my calculations and those of the OP.JBA said:Thank you for taking the time to review and reply to my extended post. I have read your summary report and while it does not provide any specific supporting analyses; and, because of my confidence in your rigorous approach to problem solving as well as the fact that I no longer have access to a required flow testing facility to investigate it in that manner I am fully prepared to accept the validity of your conclusions.
At the same time, I am afraid I am back again with my concerns on the 1st Law issue. This time it relates directly to the issue addressed in this thread as an example of my concerns.
Using this as an example, using the 1st Law equation as presented in post#8, I to calculate the final Tank temperature, ignoring tank heat transfer, to be as follows:
P i = 4.85psia T I = 70°F T I = 530°R P f = 14.7psia k air = 1.4T f = 655°R T f = 281°F
Which would indicate that even with an large amount of Tank heat loss during the 9 sec filling time, there would be an additional amount of filling time required as the Tank consumed the air required to maintain its target 14.7 psia pressure level; or, if the fill valve were to be closed, a substantial post fill pressure drop would be expected as the tank cools to ambient temperature.
If you are willing to continue consuming your time with my ongoing inquiries I would really appreciate your feedback on this issue; if not, I fully understand because it getting to one of those extended "OK, but what about" situations.
Jack
The analysis in this thread (so far) assumes that the process takes place adiabatically, and does not include heat transfer from the gas to the tank walls and surroundings. This, of course, can be included, but, for a 9 second process, its inclusion does not seem worthwhile.JBA said:My error, checking my conversion program confirms that 195°F is indeed correct; but, that still does not address my concerns regarding the issue of a pressure drop associated with the return of the tank's temperature to ambient or does your calculation method account for this as well.
If so, this type of calculation with a single equation that crosses the Pcr point and accounts for the pressure vs temperature reduction is something I would very much like to learn and understand.
OK. The calculations involve integrating an equation expressing the time derivative of pressure as a function of pressure itself. The key parameter in the calculation is the expansivity factor ##\epsilon##, which is a function of the ratio of the pressure in the tank to the outside pressure. There are two regions of functionality, as you indicated, critical and sub critical. The expansivity factor is continuous at the transition between these regions.JBA said:Adiabatic, I understand, thanks
With regard to my question regarding the critical pressure, it wasn't focused upon the critical pressure; but, on the fact that, if I understand it correctly, you appear to have a method to calculate the flow rates for the full range of pressure ratios including both the critical and sub critical pressure regions without having to calculate the two regions separately and sum the results; and, if so, I really want to understand that method.