Instant pressure change gas in pipeline

[eltjo]

Hi All,

This question is a bit frustrating since I am thinking about it all day!

The problem: I have a pipeline of 5000 meter, on one side I have a compressor and on the other side I have a consumer. In one scenario I have an equilibrium and for instance the pressure at the consumer is 4,7 bar(g) and pressure at compressor is 5,5 bar(g) (Pressure of compressor is controlled by pressure controller at consumer).

If the consumer consumes more, the pressure at the consumer will decrease, to speak in numbers, for instance to 4,5 bar(g). The compressor will need to correct this and will compress more kg/s, which will result in an pressure increase by the consumer till it will reach the 4,7 bar(g) again (New equilibrium).

My question: How fast will this pressure change be?

Is is the speed of sound of the gas? For instance speed of sound is 300 m/s pressure change will take 5000/300 = 17 seconds?

Hope someone can give me an answer with good formula's or text!

Thanks in advance,
sincerely,
Eltjo

 

[quark]

Pressure builds up when the rate of generation is more than consumption. For a fixed rate of generation, pressure drops with increase in rate of consumption.

Say, your air compressor capacity is 100 cfm and initially you are consuming at the rate of 60 cfm and then your consumption increased to 80 cfm. So, it should take 20*60/100 = 12 seconds.

 

[eltjo]

@ quark,

Thanks for the fast response.

I understand your explanation, but you don't give an answer of the speed of pressure change over the 5000 meter pipeline. Maybe, I did not made my question correct.

What is the time, when at one end of a 5000 meter gas pipeline the pressure instantly increases, for the change is noticeable at the other end of the pipeline?

 

[quark]

Calculate the volume of the pipe. Check the volume of air required to increase the pipe pressure from initial to final condition, using isothermal conditions (approximately). Volume required/(compressor capacity-consumption rate) gives you the time.

 

[eltjo]

So you say that there is no "delay" in pressure change and a sudden increase in pressure in one end of the pipe is without any delay seen at the other end?

My feeling says this is not correct, since it is a gas and not a liquid. I think it takes some time for the pressure change to be seen at the end of the pipe.

If the pressure increase for instance is in a sphere I think you are correct, but this is a 5000 meter pipeline.

 

[katmar]

I agree that the pressure sensor at the consumer cannot "see" any change in the pressure in less than 17 seconds because the velocity of the gas is limited to the speed of sound. In reality, you gas velocity is probably closer to 1/10th of this , ie 30 m/s and the full increase in pressure will more likely take 170 seconds to be seen at the consumer.

It would be a difficult calculation to do exactly by hand because you have a varying pressure profile all the way along the pipeline and you would need to perform a finite element analysis with many iterations to get an accurate answer. Depending on why you want this information, it might be more expensive to do the calculation than the value it gives. Self tuning PID controllers are so cheap now it is probably the most practical answer to just install one and let it sort itself out. But it is an interesting question and I can see why it has troubled you all day.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[quark]

Katmar/Eltjo,

The procedure I gave seems to give much lesser time than that based upon sonic flow calculation (because of my example). However, it takes much more time. If I consider a 1" sch40 pipe for this application, which has an ID of 0.0266m, the total volume will be 5000*0.00056 = 2.8 cu.mtr.
If we have a difference of 36 cfm, which is about 0.0168cu.mtr/s(corresponding to 30m/s velocity), the time would be 2.8/0.0168 = 166 sec.

My approach and velocity approach, both are same. I said it as Q/V (where Q is difference between rate of generation and rate of consumption) and V is pipe volume.

Q = 3.142xd²xv/4 and V = 3.142xd²xL/4, where v is velocity and L is pipe length.

So, Q/V = v/L (which speaks of dividing the pipe velocity with pipe length)

 

[katmar]

Hi Quark - yes you are right that the volume and velocity approaches are really the same thing for a pipe of constant diameter. Both are approximations of the correct result which would have to be obtained via FEA.

Harvey

 

[eltjo]

@ Katmar

Thanks for your answer.
I think however there is a flaw in your answer. Let us say that we have a 5000 meter pipeline without any flow and just a static pressure. If in one end we put some more gas molecules, i.e. the pressure will get higher, there is no gas velocity (dynamic pressure), but the pressure will level out in the whole pipe. My colleagues think the pressure increase at the end of the pipe will be seen calculated with the average speed of gas molecules (approx. 350 m/s) in 15 seconds.

Your 30 m/s of the gas velocity is the actual gas speed through the pipe? (Pipe size dependent?) I ask this because this value will change significantly since consumers consummation will change significantly? Velocity will be around 0 - 15 m/s and with little to none consuming your answer will become very large.

Background info:
I am designing a network where a compressor needs to maintain the pressure at the end of the pipe. If the consumer at the end of the pipe consumes more, the pressure there will decrease and the compressor needs to do some more work and needs to put some more kg/s into the pipeline. In order to let it work I need to know this value so I can estimate the ramp up/down of the consumer (=ramp up/down compressor + "answer of this question" + safety margin)

 

[eltjo]

Edit to background info:
Pipe length is approx. 5000 meter with internal diameter of approx. 250 mm --> Volume will be around 230 m3

 

[quark]

eltjo,

Where do you put the gas molecules from? If the gas molecules don't flow then how will they enter the pipe?

What I am trying to reiterate is that the approach you considered gives the minimum time possible (based on sonic velocity) to get the pressure to an equilibrium state.

When there is no consumption, the entire compressor capacity is used to fill the pipe (i.e pressurise it) and so time taken would be less.

 

[eltjo]

Of course the gas molecules will flow because it will pressurize the whole pipeline, but I need to give a decent ramp up/down to the consumers and therefore I need to know how much time it will take for the pressure to be stabilized after pressure drop.

And since the compressor will rise the pressure in one end of the pipe, even when there is no consummation, there will be a pressure wave to stabilize the pressure in the whole pipeline.

ramp up/down of the consumer =ramp up/down compressor + "answer of this question" + safety margin

 

[quark]

The only way is to calculate the the difference between the capacity and consumption rate as I said above. However, generation rate is constant (if you have a PD compressor) but rate of consumption may vary every time. So, it is difficult to speed up and down the compressor based on one fixed value.

Probably, you can install a pressure transmitter at the end and use the pressure signal to control the VFD. The inbuilt PID controller can check the rate of pressure drop (with respect to the set value) and run the compressor accordingly. Life becomes easy if you have auto tuning type PID controller.

Harvey,

I think FEA is useful if you have to calculate the time with respect to the pressure build up at various points along the length of a pipeline, at any instant of time. It may be redundant, to check at the end.

 

[katmar]

My guess of 30 m/s was simply a typical gas velocity for flow in pipelines. Now that you have given more information it is possible to calculate that for air with a supply pressure of 5,5 barg and a consumer pressure of 4,7 barg in a 250 mm pipe the velocity will be closer to 10 m/s and flow is about 5500 SCFM.

If this is the velocity under these steady state conditions then I think this is the scenario you should be considering as your base case. The scenario of somehow injecting moelcules into one end while there is no flow is an artificial one and I cannot see how it adds to the solution.

If your consumer is an on-off application you will be controlling the supply end down to 4,7 barg during the no flow periods and there will be little capacity for the pipeline to suddenly deliver when the consumer requires flow. You should consider putting an air accumulator tank at the consumer end and controlling its pressure to a value a bit above that required by the consumer. You could then regulate the pressure to the consumer while controlling the accumulator pressure independently.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[zdas04]

You can think of the system as a huge spring. When you are adding more mass than you are removing, the spring compresses. When you are removing more mass than you are adding, the spring relaxes. This has a significant dampening effect of the system (i.e., a change in consumption will be supplied/stored by a change in the pressure gradient long before the compressor will even know that something has changed).

This pipeline storage/dampening effect is very well known in pipeline operations, and the pressure changes at both ends are so cyclical that you often see pressure at the delivery point increasing while consumption is increasing due to the position of the spring in its oscillation cycle.

In real flows of real gases you never see steady state conditions for more than a few milliseconds at a time, and the time lag between a change in consumption and a change in the average pressure at the supply can vary widely (I've seen it appear to be faster than the speed of sound and I've seen it to be slower than the mean velocity in the line). FEA will give you an average of averages and can be pretty close on 900 tests out of 1,000, but no particular data set will necessarily match any particular FEA run on the scale that this discussion covers.

David

 

[Compositepro]

The pressure disturbance will propagate much slower than sonic velocity. I don't know the exact answer but the analogy is the speed of surface waves on water or, as Zdas04 says, a spring-mass system.

 

[MortenA]

If its really important get a simluation done of the system (IMHO).

If not use the rough estimates given here and then tune the system.

Best regards

Morten

 

[eltjo]

Thank you all for the fast responses!

I did a little workaround for my problem:
- I know now that the ramp up/down for the compressor unit is 5 minutes (0 --> 100%).
- I have calculated the time the molecules will stay in the pipeline @ maximum consummation. (Which is also around 5 minutes)

As a safety margin I will set the maximum ramp up/down of the consumers @ 10 minutes. (The sum of both) Of course this is conservative and the real number is lower, but for me this safety measure is fine.

Again thanks for your responses!