Pressure drop of gas in a pipe line 9

[enrjdean]

Hi Guys

I'm currently looking at sizing a pipe to carry 6.55m^3/s of exhaust gas over 1609m @ 40 psig in a 12" pipe. I've gone through these set of equations to look at the pressure drop over the line:

http://www.engineersedge.com/fluid_flow/pressure_drop/pressure_drop.htm

But the answer I'm getting back (4036Pa) seems extremely low? I'm recently graduated, so perhaps its my lack of feel for this sort of problem.

Would this be a reasonable figure, are the equations I'm using appropriate or not?

I'd appreciate any help or comments you guys have

Jim

 

[Bribyk]

Do you know the density of your exhaust gas (kg/m^3 or lb/ft^3)?

If you can find/figure that your approximate pressure drop will be approximately:

P(inH20) = 21850 X Density (lb/ft^3) [engine manufacturer's reference]

I'm guessing about 400 degF for your (constant) temperature so your density works out to about 0.17 lb/ft^3.

Therefore, your pressure drop should be about 925 kPa (3714 inH2O).

Are you sure on your units? It looks like your 4036 estimate is about right if your results are supposed to be in inches of water column, not Pa.

 

[enrjdean]

Hi, I've used SI units throughout, so the pressure drop should be in Pa?

To calculate the density, I've used rho=Pabs/R*T which gives: 376kPa / ((287J/kg*K)* (333K)) = 3.94kg/m^3

The temp is actually 140F, so 333K and Ive used the universal gas constant for air. I've then had a look at the Reynolds number so I can calculate the friction coefficient which I calculate to be 4.8*10^-5.

Jim

 

[Bribyk]

Is this a (natural?) gas being vented/flared or the products of combustion (exhaust gases)?

We need some more info on the gas/application before we can help.

 

[katmar]

A flowrate of 6.55 m³/s in a 12" pipe gives a velocity of around 90 m/s. This velocity over 1609 m is definitely going to give you a pressure drop of more than 4 kPa. Your friction factor seems much too low to me. How did you calculate it?

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[enrjdean]

Hi

I don't have any other information yet on the gas other than it is a natrual gas being flared off from an oil rig.

To calculate the friction factor I used: 64/Re

I've been through my calcs over and over to make sure I'm making an error with the units.

I agree that the pressure drop should be higher through gut feeling, but ovbiously I dont have as much experinence as you guys.

Jim

 

[Bribyk]

Okay, I was initially under the assumption that this was some kind of combustion exhaust gas.

Ignore my previous numbers and see the attached document for natural gas sizing, the Canadian Gas Code (CSA B149) uses the same calculation.

 

[Bribyk]

Using the simplified natural gas formula I get a pressure drop of about 13.3 psi (40 psig at inlet, 26.7 psig at outlet).

 

[Latexman]

64/Re is for laminar flow! Your friction factor is too low. Use a friction factor for turbulent flow and be VERY accurate on pipe roughness. Using a 0.0018 inch roughness for new commercial steel, I got ~ 35 psi pressure drop. when I used a 0.004 inch roughness for "design", the flow choked.

Good luck,
Latexman

 

[katmar]

Assuming your gas is pure methane with a MW of 16 gives a density of 2.2 kg/m³ at the upsteam condition. If we treat this flow as incompressible (i.e. like a liquid) we get a pressure drop in a 12" line of 650 kPa. Your inlet pressure of 40 psig (375 kPa abs) is less than this so it is an impossible situation. Taking the gas compressibility into account makes the pressure drop higher and the situation worse.

Under the incompressible assumption the Reybolds number is 5x10⁶ and the friction factor is 0.0135 (Moody/Darcy) or 0.0034 (Fanning) with a pipe roughness of 0.05 mm. Sorry for the mixed units - you say you are working in SI but you give your data as psig and Fahrenheit?

If you have not made an error in your data then it looks to me that you need at least an 18" line. In this case Re= 3.6x10⁶ and friction factor = 0.0128 (Moody).

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Latexman]

BTW, the 35 psi drop was for adiabatic compressible flow with MW = 19.5 and c_p/c_v = 1.27.

Good luck,
Latexman

 

[quark]

I have got yet another value but mine is close to Bribyk. At 60C and 19.5MW, pressure drop is about 0.83 bar, Reynolds Number is 4780461 and f is 0.0133. I am just following Crane's method. I considered 0.05mm roughness.

 

[Latexman]

Sorry, I forgot to convert kg to lb for my software. Now I'm aligned with katmar. You either have an error in your data or you have/need an 18 inch line. Sorry quark, I think you have an error too.

Good luck,
Latexman

 

[enrjdean]

Hi Guys

Thanks for the replies, I really appreciate it! However, I'm still not able to get the values you have quoted. I've attached a spreadsheet with four different cases with different bore sizes. Maybe you might be able to point out where I'm going wrong?

Jim

 

[Latexman]

Using your numbers from Case 3.0 and row 18, I get the same answer, 15.7 bar pressure drop using incompressible methods. However, this is impossible because the pressure drop exceeds (significantly) the starting pressure!

In gas (compressible) service if the pressure drop divided by the inlet absolute pressure is < 10%, you can use incompressible methods using either the upstream or downstream conditions. If the pressure drop divided by the inlet absolute pressure is > 10% and < 40%, you can use incompressible methods using the average of upstream and downstream conditions. This is iterative. If the pressure drop divided by the inlet absolute pressure is > 40%, you need to use compressible flow methods.







Good luck,
Latexman

 

[BigInch]

At those low pressures, I wouldn't bother with compressibility factors.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[quark]

Latexman,

So far, it didn't come to my observation that either you are katmar missed a point grossly, in any previous posts. So, I give both of you benefit of doubt.

However, there are couple of issues that may require some understanding. First, why the assumption of either incompressible flow or adiabatic flow in this case?

Secondly, if the flow is choked, how 6.55 cu.mtr/s flow rate is maintained?

Panhandle's formula gives me much higher flow rate at 0.83 bar pressure drop.

Am I missing a point here?

 

[Latexman]

quark,

All good questions! I bet the inconsistancies in the units of this post have contributed to these differing opinions. I still have some doubts about that. The spreadsheet says 20 MCFD. In my industry and location, which uses King George's units, that's 20,000 CFD. In the mixed-Metric world I've seen it mean 20,000,000 CFD. I hope the OP can shed more light on this.

Good luck,
Latexman

 

[enrjdean]

Hi

I've taken 20MCFD to be: Million cubic foot per day? Which converts to 6.55m^3/s.

I must appolgise for the inconsistancies in units, I've had these past onto me, then have had to convert them into something I recognise, such as SI units.

Jim

 

[BigInch]

For natural gas work M is 1000, MM = 1E6
and flow should be discussed Standard units, unless pressure & temperature are also stated. It doesn't hurt to also specifically state the standard conditions in any case, just to be sure.

kPag is a more conventional international gas pressure unit.
Standard flowrates are most often mentioned in m3/h or m3/day.

So was that original 6.55 m3/s a flowrate at 40 psi or at STP?


**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/