Gas temperature increase during filling 5

[CostasV]

I am trying to explain (or even better to find a relationship describing) the gas temperature increase in the scraper-trap of a natural gas pipeline system, when it is filled with gas (from 0 bar to 50 bar) wich is supllied from a 50 bar underground (at steady temperature of aprox. 8 deg C) pipeline.

At the end of the filling proccess, the wall of the above-ground scraper-trap was aprox 35 deg C (aprox. 40 on the upper section of the diameter, and aprox. 20 on the lower section of the diameter)

Length of scraper-trap : 20 m
Diameter of scraper-trap : aprox. 1 m
Ambient air temperature : 6 – 8 deg C (= initial wall temperature)

Method of filling : through a 4 inch plug-valve and a (partially open) 4 inch gate-valve, which are connected to the trap with a 100 m long, 4 inch underground steel pipe. The filling, from 0 to 10 bar took about 5 minutes, then for 5 minutes no increase in the pressure (time to check for leaks), and then, about 5-10 minutes to go from 10 bar to 50 bar.

It can not be the P1/T1 = P2/T2, since this law is for steady substance, which is not the case here.

If I remember well, during the filling (from the 50 bar pipeline, through the valves, into the initial “empty” trap), there should be a drop in the gas temperature, not an increase.
What is the main phenomenon in this proccess? How would an increase or decrease in filling time, effect the final temperature?

Costas

 

[rzrbk]

At a guess, there is air or some other gas already in the trap that you are compressing as it is filled.

 

[sailoday28]

A quasi-steady approach by solution of the following differential equation will yield the gas temperature in the space to be filled:

dE =Ho*dm + dQ

Where E is the internal energy of the space E=me
m= mass and e specific internal energy

Q heat transfer to the gas
Ho specific stagnation enthalpy of the filling gas.

Regards

 

[dcasto]

rzrbk is on track, you compressed the gas already in the trap and the trap/temperture measuring device didn't come to equalibrium for several minutes.

 

[zdas04]

Yep, you compressed the gas in the pipe. The heat of compression is:

T(f) = T(i) * )(Pf/Pi)^((k-1)/k)

If the trap was air-filled then k=1.4, R=Pf/Pi=11 and your temp will increase by about 1.98 times (so, trapped air would increase from 281K to 562k or 289C, the autoignition temperature of methane is 704C).

Since the steel only increased to 40C then the air temperature probably got very hot, but there weren't enough BTU's in the gas to heat the steel very far, most likely it didn't catch fire.

It could be that you were really lucky. About 1 person per month dies somewhere in the world doing things a lot like what you described. Had you gone all the way to 50 bar in one step you most likely would have blown the end off the trap.



David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

The harder I work, the luckier I seem

 

[sailoday28]

zdas04 (Mechanical)Your formulation is for an isentropic compression of a perfect gas. What is the basis for your approach?
There is no way that source gas throttled into a chamber can come close to a constant entropu process.
The formulation in my post has an analytic and practical basis. It follows the basic laws of thermo. All that is needed is integration of the equations. For simple cases, the solutions are in basic thermo texts.

Regards

 

[BigInch]

There is a way. Many gas pipelines operations closely approximate the isentropic process.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[zdas04]

Saloday28,
When you look at actual suction in a compressor (inlet pressure minus valve losses), and actual discharge pressure (outlet pressure minus valve losses), then the equation above will be within a couple of percent of measured values (generally within the accuracy of the field instruments). I've run that calculation hundreds of times and it is always darn close. It may not "follow the basic laws of thermo", but it works.

I'm not going to argue that air is close enough to an ideal gas that it is very common for working engineers with real-world deadlines and forces to simply apply ideal-gas arithmetic to air. Natural gas is a bit farther from ideal than air, but we still find useful answers from using ideal-gas arithmetic. I'll accept your argument that it isn't an ideal gas.

The process of introducing a sonic stream into a confined static stream has been very well documented for over 100 years. The temperature increase in the static stream closely follows the isentropic equation above in every study I've seen. The company I used to work for had a fatality that was directly traced to a line "dieseling" and my boss asked me to prepare a class on the subject. I did that and have since taught the class several hundred times to engineers, field techs, and (once) to a group of geologists. Every single time I've put it on, someone in the class has had a personal example of a line burning from this phenomenon.

I frequently hear engineers say "I haven't solved a differential equation since I was a Sophomore, and I'm so glad". Your "analytical and practical basis" would be rejected out of hand by every engineer I know.

Sorry, the world works like it works, and wishing there was more rigor in engineering analysis just ain't going to make it happen.

David

 

[sailoday28]

Compressing or filling a closed space from a pressurized or varying pressure source is not the same as going thru the stage of a compressor where work is done on the fluid.
The simplest comparison for the posted question might be the example of filling of a automobile tire with negligible expansion. If you or biginch belive that the process within the tire is isentropic you are greatly mistaken.
My posted equation,does not include the PdV work of the small expansion of the tire. Try solving, that problem based on a fixed source. The solution is very simple.
Biginch states that the process in many cases is close to isentropic. Clearly, that depends on time. If the process is slow, then heat transfer is involved. If very slow, the process may be represented as isothermal. If very fast, as adiabatic-not isentropic. In the many cases experienced by biginch, I can only assume the process is rapid.

Regards


However, perhaps I am misunderstading the problem. I am picturing, the process as a filling of a closed fixed volume.

 

[zdas04]

Think of a stream at sonic velocity entering a confined space. There will be no mixing until the sonic stream slows to Reynolds Numbers less than about 5,000. That is a lot of velocity change. The thing that accelerates the velocity change is the work that is done by the sonic stream on the static stream. Just like a compressor piston that evaporates at the end of its stroke.

David

 

[BigInch]

Exactly right Sailoday. Isentrophic ... due to lack of time, perhaps never exactly true in the field, but always so in the office.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[sailoday28]

zdas04 (Mechanical)
Are you stating that the entropy of the jet and the static stream are the same?
If the jet is not mixing, then you are describing a "contact surface" and you are in a new realm of fluid dynamics.
Think about the automobile tire or better yet a simple case of an evacuatied rigid can. What happens to the contents if there is a small puncture? The can problem is an elementary thermo problem and could give you more insight to the heatup problem.

Regards

 

[zdas04]

This "new realm of fluid dynamics" only works if the static stream is confined. Puncture a can in the open and it will push the air out of the way, but won't compress it.

You can't think of the sonic stream as a gas, it is a moving mass like a piston. The important issue is the temperature of the static volume when mixing begins. That typically happens when the static volume has been compressed to about half the pressure upstream of the valve (my example above was a bit dramatic on purpose). The case that kills people is an air-filled vessel that gets hit with a sonic stream of flammable gas that has enough excess pressure to raise the temperature of the static volume to the autoignition temperature before mixing begins. When mixing begins you get an explosion.

David

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

The harder I work, the luckier I seem

 

[CostasV]

Thank you for your answers.

It is my mistake that I didn't say that the scraper-trap (which is a closed fixed volume) after the depresurisation, was purged by nitrogen, and then the scraper-trap opened to the air. We then closed the scraper-trap having a mixture of nitrogen and air (say 90% N2 and 10% O2). After that, we start introducing natural gas until we measure more then 95% CH4 at the vent stack of the scraper-trap. But, due to the geometry of the pipes, we must accept that there is aprox 50-60% CH4 (supplied from the 50 bar network through the 4 inch valves and pipe) and 50-40% nitrogen+air (remained in the scraper trap) . All that at nearly 0.5 bar (g). Then we fill the trap up to 10 bar, check for leaks, and then again from 10 to 50 bar.

The ratio CH4/(air+CH4) is always kept above 15% which is the upper explosive limit, so no risk for fire even at temperature of 700 C.

Costas

 

[sailoday28]

zdas04 (Mechanical) 30 Mar 07 7:06 You state
This "new realm of fluid dynamics" only works if the static stream is confined. Puncture a can in the open and it will push the air out of the way, but won't compress it.


What happens to the pressure in the previously evacuated can? Clearly it rises. I would call that compression.

You state "You can't think of the sonic stream as a gas, it is a moving mass like a piston."

Well then you have a transient and a good example of where contact surfaces are applied. If the piston, moves fast enough, it will generate shocks. See any good book on unsteady flow. With these shocks in the compressed region the entropy changes. And the process is not isentropic. If the piston moves very slowly, then the change in entropy might be neglected.

Again, I suggest you try and analyze the punctured can, which is initially evacuated. You will see that for a perfect gas, constant specific heat, the Tfinal/Tinital =gamma. Where the temperatures that I refer to are absolute. And the initital pressure, rises from complete vacuum to atmospheric.
If initially with some mass in a partially evacuated can, the analysis is still simple, although the final temp will not be as high as the first case. And I would still call it a compression problem.

Regards

 

[25362]

It could indeed be we are facing a combination of effects:

[•] During step one (up to 10 barg) that took 5 minutes, at the steady-state filling rate of ~0.3 kg/s (because of critical conditions), assuming no appreciable change in kinetic energy, the J-T isenthalpic effect could have been felt and the gas may have cooled down a bit in this process, although the pressure drop in the throttling process is constantly being reduced o/a of the increasing back-pressure in the trap.

[•] The filling rate on step two was -on the average- quicker than in step one resulting in gas heat-up by compression, with some heat lost to the surroundings.

[•] Both above processes could explain why the temperatures didn't reach the level expected from an adiabatic compression as shown by zdas04's equation.

[•] At some point in step two -and w/o changing the valves' opening- the filling rate would be steadily reduced on account of the increased back-pressure inside the trap.

[•] As for the rate (time) effect on temperature I tend to agree with sailoday28's message:

If the process is slow, then heat transfer is involved. If very slow, the process may be represented as isothermal. If very fast, as adiabatic -not isentropic.

As a result, if the trap were thermally insulated the final gas temperature would be higher.

 

[zdas04]

Varsamids,
Sounds like your procedure is well thought out that the answer to your original question is "heat of compression".

I don't usually specify the nitrogen purge because it requires the field guys to carry nitrogen with them. For pig-traps of all sorts, my procedures include a clearing purge to sweep the air from the trap and replace it with pure methane (statistically, 3 volumes of natural gas, introduced at about 1.1 bar with an adequate vent open gives you a 98% confidence that there is no unswept volume of air). Been doing it that way for 15 years without a hot pipe.

Sailoday,
OK, the can was sealed in outer space and there is NOTHING in it. Puncture the can and the air rushing in is at sonic velocity, but there is nothing to compress so it just hits the far wall and mills around, rapidly losing velocity to collisions with the can walls. In a very short time, the pressure in the can reaches the critical pressure (call it 10 inHG) and the incoming velocity falls below sonic and then rapidly the dP goes to zero. What was I supposed to have learned from this? Maybe that you can't compress nothing? I would expect the temperature to be very slightly less than ambient temperature due to Joule-Thompson effects.

We didn't start with a rigorous definition of "compression" so I guess you can call any pressure increase by that term. In general usage "gas compression" is something like "application of mechanical energy to raise the pressure of a fixed volume of gas". I don't have a problem calling the wave front of a sonic stream "mechanical energy" because gases at Mach 1 behave so differently from gases in any other state. The world-view that I work with says that a stream moving 500 m/s going 20 m in 40 ms is quick enough to be considered constant entropy (i.e. isentropic). The sonic stream is also adiabatic for the duration of the compression event. Things are very different a few dozen mili-seconds later, but if you are going to have an explosion it already happened by the time events approach equilibrium.

David

 

[dcasto]

I'll tell the story again, an ethylene plant had a huge fire because the operator didn't slowly pressure up a borbon tube pressure gauge. The gauge was purged with N2 and screwed into place. When the 1200 psig ethylene compressed the N2 in the tube, the The heat of compression cause a temperature rise and the themperature was high enough at the ethylene N2 interface to start a decomposition and fire.

 

[sailoday28]

zdas04 (Mechanical)You don't have to take the can into space. Evacuate it here on earth, get a good partial vacuum. Let the can sit around so that its initial temp, Ti, is that of the surrounding ambient. Puncture it. The contents in the can will pressurize. AND the final tempTf will be approximately Tf=gammaTi where the tempeartures are absolute. I say approximate because, there is some mass in the can prior to puncture.


25362 (Chemical) 30 Mar 07 9:23 You state

• During step one (up to 10 barg) that took 5 minutes, at the steady-state filling rate of ~0.3 kg/s (because of critical conditions), assuming no appreciable change in kinetic energy, the J-T isenthalpic effect could have been felt and the gas may have cooled down a bit in this process, although the pressure drop in the throttling process is constantly being reduced o/a of the increasing back-pressure in the trap.

The above is a misapplication to the original problem.
The upstream of the filling volume to the downstream volume is not a J-T process. See my post with the quasi steady energy balance.

Regards

 

[25362]

To sailoday28, I beg to disagree, the low linear velocity in the 4-in pipe, with the partly-closed valve and the large magnitude of the pressure drop, could result in a local "throttling" process.