Flow in Gas Pipeline

[dingelde]

Hello,

I have a quick question that I cant seem to get an answer to. The velocity for gas in a pipe is calculated taking into account temperature and pressure:

v=14.7349(Q_b/d_s)(P_b/T_b)(ZT/P)

I understand this to be the velocity of a gas particle on a streamline travelling from point a to point b.

My question is, what is Flow/Area for gas in a pipeline? It is a velocity, but what is this velocity? Is it a gas particle on a streamline as well? For the same system, the two equations give two very different velocities. Thanks!

 

[LittleInch]

Without knowing where this comes from or what the constant is trying to do, Q seems to be the standard volumetric flowrate (scfh for example) as it is divided by the pressure, but without knowing what the b stands for might actually be the actual flow in volume per second. If you take b as the "base" then this seems to work as assuming you are at standard conditions you divide by the pressure. ds is area?

Can't work out what the 14.73 is, but assume this is some sort of conversion factor from whatever the units are on the RHS to a different set on the LHS.

I would just work out the actual volumetric flowrate at whatever pressure and temp you are running at is as a single line first to be clearer what you are doing and then divide by the internal surface area to get your velocity

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[dingelde]

Hi LittleInch,

My apologies, I made a mistake in the formula. It is:

v=14.7349(Q_b/D²) * (P_b/T_b) * (ZT/P)

v=velocity (m/s)
Qb=flow rate (m³/hr) at standard conditions
P_b= base pressure kPa
T_b=base temperature K (273 + degrees celcius)
P=pressure kPa
T=average gas flowing temperature K(273 + degrees celcius)
Z=gas compressibility factor at the flowing temperature

The 14.7349 is just a factor for the conversion into SI units, it is dimensionless.

The formula comes from the book "Gas Pipeline Hydraulics" by E. Shashi Menon

Using modelling software, the flow is solved for, and so is the velocity in a pipeline. Yet when I take the flow that the model give me, and divide it by the cross sectional area of the pipeline, I get a much faster velocity then the actual velocity of the gas molecules, as predicted by the above equation.

My question then remains, what is the difference between these two velocities? They are substantially different.

 

[EnergyMix]

That formula now makes a bit more sense. I couldn't see how units worked out otherwise. Now as to why it differs from your modeling software -- it's pretty hard to tell. Especially since it sounds like you're doing additional calculations.
Somewhere, hopefully, you have the manual for your modelling software. What equation is it using? Otherwise what results are you getting (include units please) from your modeling program and what is your cross-sectional area? Actual numbers might help.

Want to know the do's and don'ts of Eng-Tips? Read FAQ731-376.
English not your native language? Looking for some help in getting your question across to others or understanding their answers? Go to forum1529.

 

[katmar]

The original book was written in US Customary units. I suspect it went wrong when they tried to convert to SI Units. For the formula to work (assuming D is in mm) the constant 14.7349 should be replaced with 353.6

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[ione]

Katmar,

Nothing wrong there with the conversion factor. Dingelde has just missed that in the formulation reported in the book of E. Shashi Menon (SI) the flow rate is expressed in m3/day and not in m3/hr. In fact 14.7349*24 = 353,6376.

 

[katmar]

Well spotted Ione. It's reassuring to know that the value of such a good book is not diminished by horrible typos.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[zdas04]

Now that we've not answered his question, maybe we should.

The velocity that is calculated that way is "bulk velocity" (the equation is a crutch, all you are doing is converting from SCM to ACM, you do that with a ratio of density, and then dividing the ACM by flow area [pi/4 is buried somewhere in the constant]). We all know that there is a velocity profile across the pipe that ranges from zero at the pipe wall to maximum at some point within the pipe (hopefully fairly close to the center). Plus in turbulent flow there are velocity vectors of randomly varying magnitude in every possible direction.

Volume flow rate over area does not give you that. It "resolves" that chaos into a plug the size of the pipe moving at a "bulk velocity". The number is slower than max velocity and faster than zero, but it is not necessarily the average of all of the velocity vectors. It is a convenient number to use in other calculations and it is as close to "right" as you are going to get in most real flows.

David Simpson, PE
MuleShoe Engineering

Law is the common force organized to act as an obstacle of injustice Frédéric Bastiat

 

[dingelde]

Thanks zdaso4. That was my theory as well. The modelling software I use calculates flow using your standard flow equation. I am *assuming* that it uses a similar formula for velocity as to how I posted. I could not find any info in the user guide, however I will be contacting the programmers to get a better understanding.

The way I see it, is the Q/A velocity is your average velocity of all the particles in the cross section. This velocity is actually much quicker than the velocity used to calculate the individual velocity of a gas particle. I think what it tells us is that while the plug of gas is flowing quite quickly, the individual gas molecules are traveling much slower.

At the heart of this is a way to determine the time it takes for a gas line purge. Tough to do, as really this is a non steady state problem, and we only have steady state software.

 

[katmar]

You can get an approximate idea of the velocity of the individual streamlines as a function of the distance from the centerline of the pipe using the power law (see for example

http://me.queensu.ca/People/Sellens/PowerLaw.html

), but I don't believe this is the answer to your problem. Especially now that you have redefined the problem as a purge calculation. I suspect that the answer will be more empirical than theoretical, but there is nobody better placed to answer that than zdas04 so I will retreat to the sidelines as an interested observer and leave it to David to guide us.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[ione]

I am not there this time (probably my fault). I think we all, or at least should, agree that the average velocity is, as the name implies, an average value of the values which represent the velocity profile across a pipe section (zero at the pipe wall and max value at the centreline for a laminar flow etc.). Now what the formula reported in Dingelde's second post (dated 14th of March) gives is an average velocity value. I suspect, but cannot be sure of this, that there should be a mess with standard and actual conditions. The formula gives an average velocity referred to actual conditions, although what one is asked to enter in the formula is the volumetric flow rate (m3/day) at standard condition. If one then used the same value (standard conditions) for the volumetric flow rate (just converted to m3/s) in order to calculate velocity as v=Q/A, then it wouldn't get the right value at actual conditions. Since gas is generally operated at higher than standard pressure (I don't take into account temperature here) it would get a higher value. A practical case, with real numbers involved, could dispell my doubt.

 

[katmar]

Ione, I assumed that the factor (ZTP_b)/(T_bP) in the 14 March formula was what corrected the standard volumetric flow to an actual volumetric flow, allowing the calculation of an actual velocity. But dingelde doesn't really want to know what the velocity is. He wants to know how much gas (or how much time) will be required to purge the pipeline. I think practical rules of thumb will serve better than trying to theoretically estimate the boundary layer velocity in relation to the average velocity.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[ione]

Katmar, You are obviously right when you say that the factor (ZTPb)/(TbP) allows to convert volumetric flow from standard to actual conditions, and so the formula allows to calculate an actual velocity. Nor with my previous post I mean to divert the discussion about pipeline purging, I definitely leave that to people who are certainly more skilled than I am.

What I cannot understand is how one can get different values of velocity using the formula reported in the book of Shashi E. Menon (Gas hydraulics pipeline) or in general a formula used by a reliable software from that which comes from dividing the actual volumetric flow rate by the area of the relevant pipe cross section. I’m just curious to see a practical example, on figures, which shows such a discrepancy. I have run some numbers on my PC and got the same values.

 

[dingelde]

I can give an actual problem. It will all be SI (canadian...).

Starting pressure (source node)P₁ = 500 kPa (gauge)
End pressure (flare stack outlet) P₂= 0 kPa (gauge)

Pipeline inner diameter D₁ = 137.74 mm
Pipeline length L₁ = 900 m

60 mm PE Ball Valve V₁ constant = 25 m³/hr-kPa

Purge Hose inner diameter D₂ = 25.732 mm
Purge Hose length L₂ = 10.0 m

Flare Stack inner diamter D₃ = 49.34 mm
Flare Stack length L₃ = 2.0 m

​

Friction Factor Equation - Colebrook-White
Base Temperature = 15 degrees celcius
Specific Gravity = 0.6103
Viscosity = 0.01144 cp
Friction Factor = 0.015
Roughness = 0.0015

From the attached picture, this is how I have it set up in the model. I have two known pressures, 500 kPa and 0 kPa. From that, the model can calculate the flow. Based on the diameters and legths, I get a flow of 1,784.273 standard m³/hr. The resultant upstream/downstream velocity for Pipe 1 is 5.6 m/s & 5.6 m/s. For Pipe 2 162.2 m/s & 838.9 m/s. For Pipe 3 228.2 m/s & 255.8 m/s.

Dont get hung up on the valve. This is to represent the riser coming out of the bell hole, which has a valve that can be throttled to ensure that the flare continues at a nice even pace.

The criteria for purging are:

1. Minimum pipeline velocity of 0.8 m/s to ensure displacement purge
2. Maximum flare veloicty of 150 m/s to ensure noise control
3. Minimum flare velocity of 5 m/s to ensure flashback will not occur.
4. Specify the amount of time it will take before a flare is seen

So, the question is, for 1 & 4, is it the "resultant velocity" that I use, or the standard flow over area that I use?

 

[dingelde]

Ok, the picture did not work. I will try a PDF:

 

[zdas04]

I don't think I understand the problem. The continuity equation requires the mass flow rate to be constant at every point in this system where you are only adding or removing mass at the ends of the pipe. Mass flow rate is constant. Volume flow rate at actual conditions (and therefore velocity) is not required to be constant from one meter of pipe to the next meter of pipe. It changes continuously as pressures, temperatures, and pipe diameters change.

All the velocities are "bulk velocity" I don't know what the heck "resultant velocity" would even be.

One of the posters above said that the q/A velocity would be average. I can think of a half dozen ways to calculate "average" and they are all valid and each give a different number. The only number that makes all the other equations work is q/A. Any other valid way of calculating an average give you wrong answers when you use that velocity for anything (such as calculating acceleration or force).

David Simpson, PE
MuleShoe Engineering

Law is the common force organized to act as an obstacle of injustice Frédéric Bastiat

 

[dingelde]

It sounds like I need to discuss this with the makers of the program. It makes sense to me that if your velocity is changing throughout the pipe, then so should the volumetric flow rate. Yet the model gives a standard flow rate, and it is identical in all the segments of the pipe.

Assuming for a second that I can get velocity, does it make sense to use that velocity to predict the time required to purge? From the time they release the squeezer to the time that the flare begins is all unsteady state. The pressure in the pipe is building as the gas propogates throughout the pipe, displacing the air. I thought maybe using the flow or velocity the model gives would help to approximate the time to see a flare from squeezer release.

My theory is that using the flow or velocity from the model will give you a much quicker time than actual. As soon as the squeezer is released, the initial pressure in the pipe would be near zero. The gas would propogate, displacing the air. I think that using a starting pressure of 500 kPa (system pressure) is unrealistic, and should maybe reduced to around 30 kPa to account for this effect.

 

[Latexman]

What is a "squeezer"?

Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

 

[zdas04]

You are mixing two very important terms. "Volume flow rate at standard conditions" is a very good surrogate for mass flow rate and it does follow the continuity equation (i.e. as long as no mass is added or removed volume flow rate at standard conditions will be exactly the same everywhere along a line regardless of flowing pressure, flowing temperature, or pipe size). "Volume flow rate at actual conditions" changes continuously along a pipe.

The reason for standard flow rate to act like mass flow rate is because the flowing gas is converted to imaginary conditions at every point and this imaginary temperature and pressure is the same at every point. You cannot use imaginary conditions to calculate a physical property like velocity. You have to convert to actual. v(actual)=q(std)*ρ(std)/ρ(actual)/A. If you skip the density over density step you get nonsense from imaginary land.

Yes, I use bulk velocity to estimate when gas will arrive at the end of the pipe.


David Simpson, PE
MuleShoe Engineering

Law is the common force organized to act as an obstacle of injustice Frédéric Bastiat

 

[ione]

I agree 100% with zds04. This is exactly what I was trying to explain (probably in a muddy way) in my post dated back 17 Mar 14 8:12, that is a mess between standard and actual conditions.

dingelde, in order to get an estimate of the time required to see flare, I would use an average value of the actual velocities for each segment (v1+v2)/2, being v1 the actual velocity at the beginning of a segment and v2 the actual velocity at the end of the same segment.