Pipe flowing compressible fluid

[Luiz]

For pipe design flowing incompressible fluid, I use:

Darcy equation hf = f*(LV^2)/(2gD) and

Colebrook equation f = 0.25*[-log(0.27*k/D + 2.51/(Re*f^0.5))]^-2

where:
hf = friction loss
f = friction factor
L = length
V = velocity
g = gravity
D = diameter
k = equivalent roughness
Re = Reynolds number

I would like to know: how to procedure with compressible fluid?
Can anyone here help me?

Thanks.
Luiz

 

[engware]

Hi Luiz:

In general, for the incompressible flow, friction loss is either neglected or can be calculated by using some rule of tumb -- well it is isentropic efficiency for both compression and expansion. To be more specific and accurate, there is a way on how to calculate friction -- any good book on gas dynamics will have a separate chapter devoted to the flow with friction.

Gordan Visit the Engineering Software web site --

http://members.aol.com/engware

-- where you can find free on-line calculators, demos and other material/information related to energy conversion systems (power and propulsion) ...

 

[MJCronin]

I suggest the use of the method in the classic text:

Crane Technical Paper #410 - "Flow of Fluids"

Comprssible fluid flow is handled for various gasses and steam. Nomographs are also used.

Contact Crane Co of King of Prussia PA to purchase (610) 265-5050 or (800) 323-3679

A software program by ABZ Incorporated has also been written incorporating the Crane method Call (800) 747-7401

For more elaborate multi-branced systems, consider the AFT Arrow software:

http://www.aft.com

Good Luck !!

MJC

 

[dcasto]

To answer your question directly, the equation you presented will work. In order to get the velocity in the equation, the density is going to come into play. On a noncompressible, the density is constant, so velocity is constant. With compressible fluids, not true. You could literally take your equation and intergrate it manually. In other words calculate the pressure drop evey 1 foot. With the new pressure, calculate a new velocity, and so on until you end up at the end of the pipe. Practically, this is done emperically with the equations you will find. Weymouth, Panhandle, ect..

 

[GGW]

Luiz,

For non-significant changes in fluid density the equations you present in your thread can be used as "dcasto (chemical)" points out (July 8, thread response).

However, for signifcant changes in fluid density the standard fluid mechanic relationships are used with the addition of a thermodynamic equation of state to account for the change in fluid density. Depending on the requirements of the problem the equation of state can range from the ideal to very complex .

 

[Luiz]

GGW:

This is the question: I would like to use Darcy and Colebrook equations
to compute satured steam flow pipes size, say, 120 psi.
Can you tell me what is this equation of state?

Thanks.
Luiz

 

[dazma]

Dear Luiz,

You have to find out the Mach no. of your steam flow. If it happens to be less than 0.3 you can neglect the compressible effects and use the same formulas applicable for non-compressible flow. Generally any industrial compressible flow rate are chosen such that the Mach. No. does not exceed 0.3. This is avoid choking in the areas like bends, entrance and exit areas.

Regards

 

[Luiz]

Hello dazma,

Can you tell me how to get the Mach number?

Thanks.
Luiz

 

[MorelG]

The Mach number is defined as the ratio of the velocity of the fluid to the velocity of sound. Ma=U/c. For air st standart conditions a flow can be considered incompressible if the velocity is less than about 100 m/s.

 

[Luiz]

Hi MorelG

My fluid is satured steam flowing in pipes, say, at 35 m/s under 120 psi.
So Ma = 35/340 = 0.1
Is it considered incompressible?

Thanks

 

[MorelG]

Hi Luiz,
Yes, you can consider it as incompressible flow and use the relevant equations (incompressible flow eq.). Regards, Morel.

 

[Luiz]

Ok Morel.

For satured steam in steel pipe under 8 Kgf/cm2 (113.7 psi)
and
D=0.10226 m (4" schedule 40)
k=0.0001m (0.0000305 ft)
V=30 m/s (9.14 ft/s)
v=0.0000035 m2/s (0.0000011 ft2/s)
L=500 m (152.4 ft),

by Colebrook and Darcy formulas, we have:
hf=4,442.73 m (1,354.14 ft)

But, by Spirax Sarco formula:
hf=26.4 m (8.05 ft)

Anything is wrong.
Can you tell me?

Regards
Luiz