Did you ask something similar a few hours ago?

1_ Yes otherwise it won't go anywhere assuming the pipe is truly horizontal and there is no gas velocity above the liquid.
2 - Look up open channel flow calculations or search open channel flow on this site

Flow in a pipe like that is a bit odd as it keeps changing cross sectional area as the liquid level goes up and down for the same mass flow rate and hence the velocity of the water keeps changing and changes the hydraulic profile.

Have a look at the attached for the start section which I think is close to what you are looking for, but you really need to deifne a start and end point as otherwise the flow is not steady state.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

If the circular channel is level, then there would be a reduction of water level height along the length.

Use Mannings equation to determine the height.

LittleInch thanks for your reply, you have right, politeness is a must, will strive to improve.
I deleted my previeus post, prefered a shorter message.

Very interesting what you image heppen: "flow ... keeps changing cross section area". It's like a wave-flow regime, even if the feeded flowrate is costant.
In the first range of time there is a very dynamic step with generation of waves, and the time needed to have the stady state could be very long. But when achieved, do you think it could be imagened a regular decreasing level profile?

Bimr, the Chezy/Manning's equation [Vavg = (1/n) x R^2/3 x S^1/2] is not directly applicable because the slope of pipe S is 0%, for this reason i'm looking for a deepening, without bother finite difference method. Circular geometry is a complication.

In my previous reply I meant to attach something which might apply, but you really need to describe what your situation is and why you're trying to find out.

Also this will only work if you have a steady state and a pipe probably no more than 10-15m long before it either empties into a pond or a vertical down pipe.

Otherwise if the pipe is infinitely long then the level over time will simply be flat with no flow.

I think velocity versus distance along will be a line with a constant slope for stead state flow.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

• https://files.engineering.com/getfile.aspx?folder=34941b3a-1ee8-4c53-8814-8

There is no simple answer.

Mannings slope is the slope of the HGL, not the slope of the pipe. When the HGL and invert at the pipe end are not parallel, the hydraulic radius continuously varies. The method for solving is complicated, so I would recommend the use of software to solve.

You haven't provided much information on your problem. Is the conduit inlet controlled or outlet controlled.

Link

Hi,
Consider this resource to support your work : https://pdhstar.com/wp-content/uploads/2019/06/CE-...

Pierre

In attached a simplified scheme, my real system is more complicated because of more feed and more diameter sections.
My goal is estimate the max discharge flow to the river.

bimr
many literature articles and tools say that mannings slope is the bottom of pipe but I think you are right, can you share a good source? Tx

• https://files.engineering.com/getfile.aspx?folder=92ecb13f-02ef-4631-885c-b

gelsi - well it won't be very high if the water doesn't reach the top of the pipe and it's a 100m long....

I think whilst setting the hydraulic slope as the pipe slope isn't super accurate, especially in a circular pipe, it's close enough as the "average" height of liquid would be somewhere close to half full if you started with a nearly full pipe. Try a few different numbers and see if it affects the flow rate.

For a long pipe the velocity is going to be pretty slow and this thing will act more than an open ditch than a pipe. The end is more or less the same as my picture above when you get to the sloped down section if it is really at 45 degrees.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

Under the assumption of uniform flow conditions the bottom slope is the same as the slope of the energy grade line and the water surface slope.

Link