Volume in a sloped Pipe

[Blu1913]

Am I have a friggin brain fart or is this more complicated than im thinking it should be.

Say you have a 24" pipe, at 1%, 100' long, beginning elevation is 1000.

If the pipe is capped at both ends and you fill the pipe to 1002.5', what is the volume in the pipe??!?!?!?!!?!!? :brick:

Similarly, what is the volume at 1000.7?????

Is this a integral problem or am i missing the easy solution?

 

[ferryg]

I have a stage/storage curve for the condition you are talking about....email me at [email protected] I will send it to you.

 

[RIP - VTEnviro]

If the pipe is capped at both ends and you fill the pipe to 1002.5', what is the volume in the pipe??!?!?!?!!?!!? 

If the pipe is 100 feet long, at 1%, you will only gain/lose 1 foot of elevation over the run.

The elevation will never be 1002.5'.

Even if you started at 1001 and filled the pipe at that elevation to 1002.5, that water will flow towards the bottom of the pipe. The pipe will be full from the bottom to some point, then empty above that point.

 

[ferryg]

If the pipe is capped at both ends and you fill the pipe to 1002.5', what is the volume in the pipe??!?!?!?!!?!!??

If the pipe is 100 feet long, at 1%, you will only gain/lose 1 foot of elevation over the run.

The elevation will never be 1002.5'.

Even if you started at 1001 and filled the pipe at that elevation to 1002.5, that water will flow towards the bottom of the pipe. The pipe will be full from the bottom to some point, then empty above that point.

This is true...but the one foot is added to the pipe invert...not the crown. The invert at the upstream end will be 1001....but the crown will be at 1003 (24" pipe).

This is a simple underground detention pipe problem. Assuming there is no outlet. I have the stage/storage curve for it...but I don't know how to post an image here. To answer the initial question....the storage at WSE 1002.5 is about 305+/- CF and at WSE 1000.7 is about 29+/- CF

 

[RIP - VTEnviro]

^ Read my second note above. That's what I was getting at.

I guess you could go to one of those circular pipe tables, figure out what the ratio of V/Vfull is for d/dfull = 0.75. Then multiple by the volume of the pipe assuming it's full.

 

[ferryg]

See my modified post.

 

[Blu1913]

^ Read my second note above. That's what I was getting at.
I guess you could go to one of those circular pipe tables, figure out what the ratio of V/Vfull is for d/dfull = 0.75. Then multiple by the volume of the pipe assuming it's full.

But what would you use for d (depth of water in the pipe)? It varies all along the length of the pipe...

 

[ferryg]

Think of it like a detention basin...if you have experience with those. Create a stage/storage curve...you can easily figure out the storage at even increments along the pipe...then plot a curve along this points (roughly mind you)...then use the curve to determine the storage at any elevation. With a pipe like this...it would be more advantageous to use a software that can model a pipe with zero outflow....basically an underground tank that slopes.

 

[Blu1913]

Right right, thats what I thought. Were basically talking about using integrals if you want to solve it quickly...or you can use a program to do it for you.

I was just trying to figure out how to do it by hand. I was pretty sure it was going to be an integral...but i just wanted to make sure i wasnt missing anything...

Sapper thats also how i did it, but if you keep getting into different area, it seems to just be a pain in the ass. Im just going to let programs do it for me - less chance for human error....(ie fat finger you calc

 

[ferryg]

This is not easily done by hand...I didn't mean to make it sound that way. There are plenty of commercial softwares and freeware from underground detention system manufacturers that can solve a problem like this.

Your volume is pretty close to my volume...and I am simply reading it from a curve...so I might be off a little as well. It seems like your formula is pretty close to the way it would have to be done by hand.

 

[Blu1913]

^---- BINGO thats more what i was looking for. Damn it, see i knew it was there somewhere

Sapper -> :bow: I bow to you....

 

[Blu1913]

No doubt...damn it, Im so pissed I worked most of the day and couldnt get that...(then again I did spend most of that time on this site)

 

[Blu1913]

here you go -> $8 to CASH

 

[tmckeon_PE]

Figure out where the halfway mark is along the surface of the water (if there is surface water inside the pipe). This is also going to be at the halfway (diameter) of the pipe. Figure volume based on a full pipe with this distance. Problem solved.

 

[tmckeon_PE]

I was the one wondering.

 

[Sapper PE LS]

I'm kind of wondering what the hell my solution was that got me a bow... but more importantly, why am I able to type "sapper engineerboards" in google and get this from 7 years ago (and 13 other pages of google results), clearly my plan to eradicate my internet footprint back in April failed miserably.... Jeeeeeez, nothing dies on the interwebs.

 

[ptatohed]

Am I have a friggin brain fart or is this more complicated than im thinking it should be.

Say you have a 24" pipe, at 1%, 100' long, beginning elevation is 1000.

If the pipe is capped at both ends and you fill the pipe to 1002.5', what is the volume in the pipe??!?!?!?!!?!!? :brick:

Similarly, what is the volume at 1000.7?????

Is this a integral problem or am i missing the easy solution?

How does water enter if the pipe is capped at both ends? Where is the 'beginning elevation'? Upstream or downstream end?