Time to Drain a Full Pipe 1

[asaldes]

Dear Team,

I need to calculate how much it takes to drain a 36" full pipe, in three main sections with different slopes. We have a 16" full open valve to drain it.


I do the calculation dividing the sections in several parts, estimate the velocity and then the volume flow trough the orifice. But this gave me a time of 11 minutes! i am not considering friction losses because I would need to estimate lambda in each part.

Can anybody give me a clue?

Best Regards to all the team

 

[asaldes]

Thanks for your interest. I give my answers:

OK,

First things.

Do you know that the pipe is full when pumping? This makes a big difference to what is happening when the pump stops. I DON'T KNOW EXACTLY HOW TO MEASURE IT, I ASSUMED FULL PIPE.

So first you need to define and make sure of the start conditions of the pipe when the pump stops before it is worth doing anything.

If there is no air vent but an open drain at the end then this line will pull a vacuum at the high point. I NEED TO CHECK IF THERE IS A VENT, I'M CONSIDERING BOTH CASES

The friction losses in the pipe cannot be ignored over 800m. At 16" there will be some resistance ot flow but not too much I think. OK, SO I NEED TO ESTIMATE FRICTION LOSSES, ANYWAY IS A 36" INCH DIAMETER, FRICTION HEAD LOSSES ARE LOW.

Until the last 70m you seem to have a fairly steady slope. YES.

So you have a head pressure of 29m minus lets say 5m for the vacuum at that altitude so 24m in 800m or 3m/100m of pipe.

For a 32" line this gives you a flowrate of about 9,000m3/hr or about 4.8m/sec. so that would be about 2 minutes.

Now that's moving so I currently don't think your pipe is running full uness you've got a back pressure valve on it. Do you? I DON'T HAVE A BACK PRESSURE VALVE.

HOWEVER your last 70m is only an elevation difference of 1m. This will take a long time to drain completly and will go into open channel flow. FOR THIS DISTANCE I SHOULD USE MANNING?

Also as the head of water gets less than 5m, air will start to enter the pipe and start to go in reverse creating a glugging two phase flow ( a bit like emptying a bottle of water vertically). So throw all the calculations out of the window at that point.

If you really really want to find out, you need a transient simulation tool. I DON'T HAVE A SOFTWARE, I NEED ONLY A ESTIMATION.


THANKS FOR YOUR TIME, REGARDS

 

[cvg]

you need a vent at the high point, that will prevent the air from coming in through the outlet valve and glugging

the flow rate will vary continuously from the beginning to the end. it will gradually decrease as the water level drops and when the water level drops to the top of the 16 inch orifice, the flow regime will change to open channel flow which will be much slower.

obtaining a precise estimate will be difficult. however there are software to estimate these things. this is fairly common in hydrology studies for dam reservoir outlet drain times. a reasonable estimate could be done relatively quickly in HEC-HMS or SWMM.

 

[asaldes]

"the flow rate will vary continuously from the beginning to the end. it will gradually decrease as the water level drops and when the water level drops to the top of the 16 inch orifice, the flow regime will change to open channel flow which will be much slower."

Thanks you CVG, I'm trying to calculte the drainage time when the water level drops at the top 16 inch orifice. I'm looking for some example

 

[asaldes]

CVG, can you suggest a way to estimate the open channel drainage when the water level drops at the top 16 inch orifice?

 

[LittleInch]

Asaldes,

The reasons for all these questions is that you're not giving us a good description of your current design or what paramters you're using or indeed what or why you're trying to estimate this.

So a proper digram , schematic, P&ID, something to tell us what is the "normal" operation?
What is the normal pumping flowrate?
At least this will give you a start point as if you turn the pump off the flow rate won't go up. Unless your normal flow is through a smaller valve than the 16".

I'm finding it difficult to believe a 36" (sorry I had 32" in my head for some reason) water pipe just empties into a pond with an open end. There is usually some form of pressure or flow control.

Or is this mysterious 16" valve an offtake part way along?
Is it normally closed? Normally wide open?

Flow in the pipe will balance out between the available head difference and the sum of all the resistances. This includes pipe friction and any exit losses. The flow will increase or decrease) until the system is in balance. You can't just assume one or other of those factors is negligible until you've done some calculations. As head decrease, so does pipe friction, but if the end losses are significnat this could vary by a different amount.

Yes, once you've got down to you last 70m and there is now air at atmospheric pressure on top of the water, you're into open channel flow.

So what is your cut off for flow to "empty". The last bit takes longer and longer. If you wait for the last drop you'll be there for hours. If you want to drain down 90%, my best guess from the information at the moment is between 5 and 10 minutes, then maybe 30 mins to get to 99% for that last 70m.

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

 

[asaldes]

Sorry LittleInch, i will show a diagram as soon as i can. When the pump stops, it leaves a volume in the line,i just assumed full pipe, from 3270 mts to 3299 mts at the top. The 16 inch valve is close to the pumps (at 3270 mts, where the blue line starts), so when I open the valve it returns the flow, from the 3299 mts to 3270 meters height.

In normal conditions the pump, wich is close to the pond, where the blue line starts, delivers flow to a point that is beyond the red arrow, passed the 3299 until a point that is at 3270 mts.

So, when the pump shuts down, the flow from the upper hill and the remaining flow on the line, returns to the pond level, wich is at 3270 meters, using a 16 inch valve. The last 70 meters, the flow is almost horizontal (1 mts height difference), so maybe that's is the bottle neck in the draining process.

I need the time estimation, because we need to do maintenance in the line.

Thanks you all for yor replies.

 

[LittleInch]

Ah, OK. I was thinking the pond was the end of the line, not the start.

Now it makes more much more sense. So this is reverse flow from the pipe high point back to the pumps?
Yes the pipe will always be full from the pump to the high point.

I can only assume therefore that the 16" valve is normally closed and this is only open for drain down.

But all my previous points apply - Depends how accurate you want to get, but I think you're in the area of up to 10 minutes for the main part of the pipe, then three times that for the last section.

Also if you high point is not very "pointy", then you will get a slow flow from the line just the far side of the highest point which will continue at slow flow for some time until the water level is exactly at the high point bottom of pipe.

Which part of the line needs maintenance?

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

 

[asaldes]

LittleInch, my answers:

Now it makes more much more sense. So this is reverse flow from the pipe high point back to the pumps? YES,full pipe from the 3299 mts and the 3270 pond
Yes the pipe will always be full from the pump to the high point.

I can only assume therefore that the 16" valve is normally closed and this is only open for drain down. YES,is normally closed, it opens to empty the line.

But all my previous points apply - Depends how accurate you want to get, but I think you're in the area of up to 10 minutes for the main part of the pipe, then three times that for the last section. OK It's very fast,but I think that i need a vent at the top, because if not, i would have a water column in equilibrium with the outside pressure.

Also if you high point is not very "pointy", then you will get a slow flow from the line just the far side of the highest point which will continue at slow flow for some time until the water level is exactly at the high point bottom of pipe.

Which part of the line needs maintenance? The discharge manifold, very close to the pond, so I need to empty all the line.

 

[LittleInch]

"but I think that i need a vent at the top, because if not, i would have a water column in equilibrium with the outside pressure."

Correct. The 70m flat bit would probably prevent air from going past the water plug for quite a long time but so long as the slope is flat or always down then air will eventually make its way up the pipe and you will get some glugging going on again for a long time.

Could you not put on a freeze plug downstream the bit you want to repair?

Might take a while to develop the ice plug on a 36" line, but overnight should do it. search "pipeline ice plug"

Or put in a pig and blow the contents down the line?

Always more than one way to skin the cat.



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

 

[GeoEnvGuy]

I took the five minutes to run an iterative Hazen Williams flow calculation and based on the information provided it will take 89.75 minutes to get to 0.01 m of flow in the pipe. It will still have 6.23 m3 of water left in the pipe and flowing at roughly 0.001 m3/s.

The math portion explained

Q = V A

A = D2 / 4 x ( θ -1/2 sin(2θ) )
θ = cos-1(1 - 2y/D)
y Depth of water
D diameter in this case 0.91 m

V = k C R^0.63 S^0.54
K = 0.849 for si
C H-W Coefficient assumed 140 for this one
R hydraulic radius
R = D / 4 x (1 - Sin(2θ) / 2θ )
S = slope m/m in this case 29m/859m

I first ran the formula assuming a full 36 inch pipe to determine how many m3/s would be coming out.
Then I figured out the total volume of 859 m of 36 inch pipe and subtracted 1 second of flow determined.
I then iterated the y depth of pipe with the new volume divided by the total volume
5385 seconds later y = 0.01 m

 

[LittleInch]

That assumes a free inflow of air into the top?

and constant slope?

It's the last 70 flat bit and the vacuum which will slow this down.

So we're in the region of one to two hours in reality.

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

 

[GeoEnvGuy]

That assumes a free inflow of air into the top?

yes

and constant slope?

yes

It's the last 70 flat bit and the vacuum which will slow this down.

Shouldn't froud number of a pipe flowing at a slope of 1/70 m still stays above 1 indicating supercritical upstream control except for a portion where the pipe is closer to full does the froud number drop below 1, but that should last less than 100 seconds.


Chart of outputs Flowrate starts just above 5

 

[cvg]

the flow is limited by the orifice, you need to use an orifice equation - not Hazen Williams

 

[asaldes]

I think that a way would be to use an orifice equation until the last 70 meters, or until we have a height of 0,4 meters (the orifice diameter, 16 in). Then use Manning to the remain flow. And Add the two times

 

[GeoEnvGuy]

I looked up the orifice equations (Flow through Partially Submerged Orifice) and found that in this case of 29 m of head a 0.40 m orifice will have 10 m3 left in about 600 seconds meanwhile the hazen williams 0.91 m pipe will have 10 m3 left in 3000 seconds. Granted the orifice equation does start with a lower flowrate at the beginning but after a specific amount of time HW will tell its slower assuming you still have a uniform partial flow condition over the entire 859 m length. You could join these methods and even recalculate the length of pipe to determine the new average slope, or you could be confident that after half an hour the majority of the volume and flow will be gone and it will continue to trickle out for hours.

 

[asaldes]

GeoEnvGuy (Geotechnical)27 Aug 20 17:08
I took the five minutes to run an iterative Hazen Williams flow calculation and based on the information provided it will take 89.75 minutes to get to 0.01 m of flow in the pipe. It will still have 6.23 m3 of water left in the pipe and flowing at roughly 0.001 m3/s.

The math portion explained

Q = V A

A = D2 / 4 x ( θ -1/2 sin(2θ) )
θ = cos-1(1 - 2y/D)
y Depth of water
D diameter in this case 0.91 m

V = k C R^0.63 S^0.54
K = 0.849 for si
C H-W Coefficient assumed 140 for this one
R hydraulic radius
R = D / 4 x (1 - Sin(2θ) / 2θ )
S = slope m/m in this case 29m/859m

I first ran the formula assuming a full 36 inch pipe to determine how many m3/s would be coming out.
Then I figured out the total volume of 859 m of 36 inch pipe and subtracted 1 second of flow determined.
I then iterated the y depth of pipe with the new volume divided by the total volume
5385 seconds later y = 0.01 m


How did you get the y depht with the new Volume?

 

[GeoEnvGuy]

I calculated the new y with 0.91 x new volume / total volume at start.