Pipeline Drainage - Simplified Approach

[Jack.C]

Hi,

Kindly asking for your help on belows query.

Let's assume a 10km long DN1000 pipeline with 3 drains at the low points. The initial head is H1 (orange line).
After a time step dt a certain amount of water has left the system (I'm using the Toricelli approach) through the 3 drains.
How would you calculate the head drop / new head H2 in the system? I would like to avoid a complex numerical equation to calculate the drop but rather go with a simplified approach.

Our simplified approach is shown in the red box in the below figure.



Please note that we are aware that the head line will behave different for the individual sags once it is equal to the height of the profile. So at one point Q1 and Q2 will drain the sag 1 whereas Q3 will drain sag 2 and that there will be different heads in these section. This will be considered.

Your inputs would be highly appreciated.

Thanks and regards!

 

[HTURKAK]

Please note that we are aware that the head line will behave different for the individual sags once it is equal to the height of the profile. So at one point Q1 and Q2 will drain the sag 1 whereas Q3 will drain sag 2 and that there will be different heads in these section. This will be considered.

This assumption is not reasonable.
If the valves at both ends are open, the max flow will be at Q3 since the deepest one. The size of the PL 1000 mm and length 10000 m.
The simplest approach would be calculate the amount of discharge for each drain , find the total discharge then corresponding head drop.

 

[Jack.C]

This assumption is not reasonable.
If the valves at both ends are open, the max flow will be at Q3 since the deepest one. The size of the PL 1000 mm and length 10000 m.
The simplest approach would be calculate the amount of discharge for each drain , find the total discharge then corresponding head drop.

Hi Hturkak. The valves at both ends are closed. We are aware that Q3 is highest. But the question here is how to calculate the head drop. Regards, Jack.

 

[HTURKAK]

Hi Hturkak. The valves at both ends are closed. We are aware that Q3 is highest. But the question here is how to calculate the head drop. Regards, Jack.

If the valves at both ends are tightly closed and if there are no air valves at high points, this will be completely another story.
The pressure will rapidly drop with a small amount of discharge and vacuum will develop at high locations and if the heights of discharge points more than suction head, water will start to evaporate .The variables are not only diameter , heights but also temperature , altitude ...

 

[Jack.C]

If the valves at both ends are tightly closed and if there are no air valves at high points, this will be completely another story.
The pressure will rapidly drop with a small amount of discharge and vacuum will develop at high locations and if the heights of discharge points more than suction head, water will start to evaporate .The variables are not only diameter , heights but also temperature , altitude ...

We are aware that there must be air vents installed to avoid vacuum in the pipeline during drainge. The question is related to define a simplified approach to evaluate the head drop over time. Let's leave the temperature, altitude effects etc. away as this would not lead to a simplified approach.

 

[LittleInch]

You can't simplify something which you have shown to be quite complex.

If you place the air vents at the high point and open or have one on the valve at the left hand side would be a start, but that hump before the valve on the right messes things up quite a lot.

Actually what am I saying. The head H1 in a closed liquid system (valves closed both ends) will drop pretty much instantly to the high point to the left of the valve on the right. The amount of liquid to drop a closed in liquid head will be <0.5% of the total volume That is a relatively simple calculation about bulk modulus and expansion of the pipe. After that you really need to allow air into the system.

If the low point at pt Q3 is less than 10m below the high point then all flow will just stop until air is let into the system. If its more than that then you will start to pull a vacuum at the high point and probably also just next to the valve on the LHS.

 

[Snickster]

Could you provide the elevations and horizontal km locations of the high and low points in the piping. I might have a solution.

 

[Jack.C]

You can't simplify something which you have shown to be quite complex.

If you place the air vents at the high point and open or have one on the valve at the left hand side would be a start, but that hump before the valve on the right messes things up quite a lot.

Actually what am I saying. The head H1 in a closed liquid system (valves closed both ends) will drop pretty much instantly to the high point to the left of the valve on the right. The amount of liquid to drop a closed in liquid head will be <0.5% of the total volume That is a relatively simple calculation about bulk modulus and expansion of the pipe. After that you really need to allow air into the system.

If the low point at pt Q3 is less than 10m below the high point then all flow will just stop until air is let into the system. If its more than that then you will start to pull a vacuum at the high point and probably also just next to the valve on the LHS.

Hi LittleInch

Thanks!
Let's assume that there is sufficient air intake capacity provided. All we are trying to understand is how the HGL will drop over time when the drains are open. At some point of course the HGL will be at the level of that hump on the right. So there will be water remaining in this section between hump and right shut-off valve.
Then at some point the HGL will reach the intermediate hump. Then Q3 will only contribute to the HGL decrease between "small" and "big" hump. Q1 and Q2 will drain the section between left shut-off valve and intermeidate hump.

Kind regards

 

[Jack.C]

Could you provide the elevations and horizontal km locations of the high and low points in the piping. I might have a solution.

We could just assume something. I also randomly generated the profile and the amount and locations of the three drains.

Let's assume:
Lowpoint / Q1 at KP 02+500 / Elevation = 22 masl
Lowpoint / Q2 at KP 05+000 / Elevation = 20 masl
Lowpoint / Q3 at KP 07+500 / Elevation = 10 masl

Highpoint 'small' at KP 06+000 / Elevation = 25 masl
Highpoint 'big' at KP 08+500 / Elevation = 30 masl