12 replies to this topic

[Gprocess]

Hi,

I have a problem related to gravity flow in a pipe. Details about the system are as follows:

service fluid : effluent/water
Flowrate : 300m3/hr
temperature : 55 deg C
Pressure : atmospheric

Fluid flows from an atmospheric tank with an outlet nozzle elevation of 2.5 m from ground to an underground sump of 3 m depth out of which 1.2 m is above ground. The pipe dia is 18" and the distance between source & sump is 1.2 Km with 7 U - bends underground (proposed as per piping). The slope of the pipe is 1:500.

Some of the above data is assumed. I want if someone could please validate the problem. Will it be possible for fluid to flow through 7 U bends without applying any pressure that means under gravity only. What would be the pressure drop at the end of the pipe and what would be the no flow point in the sump.

Also, if anyone can please share any calculations for such a problem like spreadsheet or some source of good literature to calculate these things under GRAVITY FLOW.

Thanks & regards.

[fallah]

Gprocess,

Please upload a simple sketch of the system you described.
In general, by estimating equivalent length of the piping system you can find a relation between f (friction factor) and v (fluid velocity) by which the velocity (and then flowrate) can be calculated via trial and error (suppose a value for f then calculate velocity and then by reynold no. and using Moody chart to verify if supposed f was right, if not suppose another value for f and repeat the mentioned steps till supposed f and the f would be obtained from Moody chart to be almost equal). Of course because in such systems: delta Z (elevation difference)=h_f (friction head); in order to have gravity flow in the piping system, the delta Z should be adequate to overcome the friction head due to fluid flow.

Fallah

[Gprocess]

Fallah,

There is no such drawing made yet, if you feel that any required information is missing please allow me to provide that to you. What I presume from your description is that you want to know the height of U bend which is approx 3 m.

Regards

[breizh]

HI Gprocess,

Let you try the search Button (top right) and you will find literature . For clarity a sketch is expected .

Applying Bernouilli between the atmospherical tank and the sump will solve the problem .

Note :
Ki for Close pattern bend = 50 *ft with ft =0.012 i.e = 0.6 if turbulent (Crane )
other source ( expressed in equivalent length) : 40 m



Hope this helps

Breizh

Edited by breizh, 04 September 2012 - 11:04 PM.

[Steve Hall]

You can download an Excel template from my website that is tailor-made for this problem. You simply have to enter the source and terminal pressures which for this problem differ only by the elevation change, the equivalent length of the pipe, pipe parameters diameter and roughness, and the temperature. The template will compute the flow rate of water under these conditions. www.pipesizingsoftware.com/pressure_drop.html. The equivalent length of a U bend in 18 inch pipe is something like 15 m - depends on the turning radius.

Steve

[Gprocess]

Steve,

I've downloaded the software from your website and after putting the data it shows error. Not able to resolve the problem.

If there is anything else you can help me with.

Regards

[Steve Hall]

Please email me the worksheet with your data in it and I'll diagnose why you received an error. smh2005@gmail.com

[Gprocess]

steve,

just forwarded you the worksheet.


Regards

[Steve Hall]

It calculates fine - no errors - on my computer. You must have security settings to allow macros to run. Your upstream pressure must be higher than atmospheric though, because the driving force includes the liquid head. If I understand your problem statement correctly this would be 2.5 + 3 - 1.2 = 4.3 m. Converting to psia, 4.3 m / (1 ft/.3048 m) * (1 psi/2.31 ft) = 6.1 + 14.7 = 20.8 psia.

Adding in 50 m per U-bend, the equivalent length is 1200 + (7 * 50) = 1550 m = 5085 ft.

Completing the calculation I get a total flow of roughly 1.86 million lb/h = 856 m³/h which is well above your acceptance criteria of 300 m³/h.

Completed worksheet is attached. The actual calculation takes place in Cell L34.

Attached Files

•  pszvba.xls   56.5KB   198 downloads

[Gprocess]

steve,

But what about the pressure drop, how much it would be at the end of the pipe and as I asked before, 'What would be the pressure drop at the end of the pipe and what would be the no flow point in the ​sump.'

Regards

[Steve Hall]

The pressure at the end of the pipe is atmospheric (unless you have a submerged inlet to the sump). So the total pressure drop is the head pressure. If you have submerged inlet, then subtract the head above the inlet. The flow rate is that which balances pressure drop due to friction with the differential pressure between the terminal points (liquid head).

No flow point? You'll always have flow if the sump level is below the pipe inlet. You really need to provide a sketch as previously requested.

[Gprocess]

The inlet would definitely be submerged into the sump somewhere at 0.8 m (assumed) above the bottom of the sump that means the sump level will not always be below the pipe inlet.

[Steve Hall]

OK. You simply have to know the difference in elevation between the surface of the water in your source to surface of water in the sump and plug this in as your Source Pressure, keeping Destination Pressure at 14.7 psia (atmospheric). The spreadsheet will calculate the flow rate.

If the discharge is not submerged, then the Source Pressure is the difference in elevation between surface of the water in your source and the elevation of the pipe (use center-line) where it enters the sump.

Edited by Steve Hall, 04 September 2012 - 06:42 AM.