How to determine depth of flow in pipe in open-channel flow?

[KernOily]

Hi guys. I have a combined open-channel and full-pipe problem here. I have a tank overflow piping system that flows by gravity. If the tank goes into overflow, the water exits the tank through the overflow nozzle in the tank shell and flows on through several hundred feet of overflow header piping until it reaches its destination, which is another tank, in a lower part of the plant.

The drain line is composed of horizontal and vertical segments as it makes its way to the endpoint. Some of the vertical segments are risers and some are downcomers. This configuration, plus the fact that the endpoint of the line is lower in elevation than the the starting point, means some segments of the line run full but others run in open-channel flow.

How do I determine the depth of flow for the pipe sections running in open-channel flow? I can't determine the head loss and line max capacity without knowing the hydraulic radius, and I can't calc hyd. radius without knowing how full the pipe is running. The horizontal segments are not sloped so I can't use Manning. I will be using Darcy with the hydraulic radius.

I may be missing something fundamental here. Wouldn't be the first time...

Thanks guys ! ! !

 

[BigInch]

What you probably have there is something that never runs in steady state until the system is entirely full, since at changes of slope you may have one flowrate upstream and another flowrate downstream, one being supercritical flow and another subcritical flow, and even hydraulic jumps at some changes in profile. For example flowing at a constant rate into a downcomer turning horizontally at the bottom, the flowrate in the downcomer may be a water fall towards the top, but filling the pipe at the bottom and even increasing the fluid level there and also the head on the downstream pipe, thus increasing the flowrate in the downstream pipe. At changes of profile, you will probably find that there are accumulations and reductions in the amount of fluid being stored at those points, until steady state conditions are reached, if they ever are.

In any case, you must find the hydraulic grade lines (HGL). Where the HGL goes below the top of the pipe, it will not be flowing full, but will be flowing at the depth of the HGL - bottom of pipe elev - v^2/2g. If the pipe is above the HGL, no flow there at all. So start drawing with the maximum possible HGL beginning at the fluid level at the tank overflow and going to the fluid level at the outlet. That's the maximum energy usage you can get out of the system. Now start making the HGLs for each pipe segment. As you go below the maximum HGL, pressure and velocity will interchange with one another, depending on the diameter and the elevations of the pipe segment in question. Keep track of the HGLs of each segment and plot them in relation to the maximum HGL. And remember, where the top of pipe goes above the max, its not flowing full and if the bottom goes above the max, its not flowing at all.

Actually I would be tempted to design the system for THE flowrate possible with the inlet head to that system equal to the fluid elevation at the tank overflow level and some given discharge head on the outlet end, if the outlet is submerged = the maximum basin level, or 0 head if its not submerged, since in that condition the system is effectively under maximum differential pressure. It may be possible to have some maximum flows in various segments that yield a faster flowrate during some transient condition, but as soon as the pipes become full, the system is then limited by the maximum pressure differential that you can impose on it and that's the overflow level at the inlet and (say) 0 at the outlet. That's the condition that I would find the most interesting, but true, the critical condition depends on the intent of your analysis which is not explained.

This spreadsheet recently appeared for pipes not flowing full,

Spreadsheet for pipe flow/capacity
thread770-192109

and its still available for downloading at,

http://rapidshare.com/files/42529967/TrueMannings1.xls.html

http://virtualpipeline.spaces.msn.com

 

[bduane]

I agree with BigInch's suggestion of drawing a line from the tank overflow hydraulic datum to the hydraulic datum at the receiving tank. However, I disagree with the statement that there will be no flow in the pipe segments that are above this line. In such a case, the hydraulic datum of the system becomes the centerline of the pipe that is above the grade line that was drawn and that will determine the flowrate in the system until the high pipe segment is full. If there is an air release valve at the high pipe segment, the maximum energy grade line is as first discussed. However, if there is a vacuum release valve on the high pipe segment, the maximum energy grade line becomes the overflow inlet datum and the high pipe centerline. The flowrate determined by the high segment gradeline becomes the maximum of the system.

If there is no vacuum relief valve (which I doubt) and there is a way to vent air from the high segment, the initial flowrate will be as determined by the high pipe segment. It will then increase up to the maximum determined by the difference between the inlet and outlet datums. A siphon is created and the high pipe segment will be operating at less than atmospheric pressure. If you are dealing with water, this may or may not be a problem depending on the temperature and the magnitude of the negative pressure - you can reach a condition where the contents boil. If you are working with something that has a high vapor pressure, you could have a problem in the negative pressure area.

As the system overflows, you will have various flow conditions in the overflow piping until you reach the maximum flowrate determined by the maximum energy grade line.

Bduane

 

[BigInch]

bduane,

There is no energy to drive flow above the HGL. As the HGL is the sum of both (absolute) pressure head and velocity head, any point above would be at the vapor pressure of the fluid which is insufficient to drive a siphon. If there was an air release valve, the energy available would not be sufficient to raise the fluid higher and drive out the air. Any fluid on the "other side" of the intersection point with the HGL would separate from the upstream segment and move downstream, if possible, leaving only fluid vapor at the fluid's vapor pressure behind it.

If you used gage pressure to draw the HGL, its possible that you might have enough pressure head to drive a siphon above the HGL (gage ref), but only if the gage pressure remained above the negative x atmospheric pressure plus the fluid's vapor pressure.

If in a transient condition, the HGL (abs ref) is no longer of constant slope through time. It will change through time such that its end point is always located at the first point of zero horizontal flow in the system. In other words, as flow slows, pressure rises sufficiently such that it could drive the fluid level to as high as the original tank overflow level, if it were possible to contain the fluid to that level (as in a closed pipe).

http://virtualpipeline.spaces.msn.com

 

[KernOily]

Hi guys. Thanks for the awesome replies.

I don't see how I can find the HGL because I don't know the friction loss. I don't know the friction loss because I don't know the hydraulic radius in the sections that don't flow full.

Difficult to explain these things in words, no?

Sorry - I should have stated my goal. (I was trying to keep my original question as short as possible.) My goal is to size the overflow line so that the overflow rate leaving the tank equals the inlet rate entering the tank. In other words, size the pipe so the friction loss is equal to or less than the available driving force.

Thanks guys! ! ! ! Pete

 

[BigInch]

depth of flow vs hydraulic radius and velocity vs slope is an iterative calculation.

What's your overflow rate? You must know the maximum flowrate that you can fill the first tank. That's it, I suppose.

Start with that flowrate,
get n,

http://www.lmnoeng.com/manningn.htm

download spreadsheet thread770-192109
enter n
enter D
enter pipe slope
get the depth of flow, (given as a ratio of pipe Diameter)

do same with a pipe flowing full formula for whenever d/D = 1

http://virtualpipeline.spaces.msn.com

 

[KernOily]

Thanks. As I said above, these lines have no slope, so, no slope, no Manning formula.

The first horizontal segment leaving the tank is easy. I can arbitrarily set that pipe to run half-full as a design constraint and then calc the head loss for that pipe using Darcy with the hyd. radius. Then water will waterfall out the end of this pipe into the first downcomer and partially fill it until the level matches the level in the next riser downstream. And so on it goes.

The downstream horiz segments, though, are another matter. It is not known a priori how full they will run, therefore I can't calc anything, as I see it.

Here is a sketch of the system. I didn't know I could do this or I would've done this earlier... The sketch at the bottom of the page is the whole system.

Thanks!!!

 

[BigInch]

IF you set the tank overflow to a certain rate, that will be the rate in the system. All through the system, as long as the fluid does not overflow the channel, and the level at any back-up point is sufficient to drive that flow through any pipe with a pressure flow. If not, the inlet head to that pipe will increase and velocity will increase until you get the same flowrate coming off the tank or you overflow at the inlet to the last downcomer. You know the flow, you know all the slopes (see next), so the capacity of the system is whatever flowrate you can get by following the available HGL from tank overflow to the next flow back-up point.

With zero channel slope only the difference in water level between inlet and outlet drives flow, since that is also the rate at which the channel extracts energy (slope of HGL) from the water use an artificial channel slope equal to the water slope. Think of it like there is an artificial channel slope in the water, not at the bottom of the channel).

It doesn't matter how full they run, you know the flowrate coming off the tank, so all channels will fill all the way to the tank overflow, if they can to drive flow forward. So they will fill to whatever the "dam" overflow level is holding them back somewhere downstream (as long as the dam level is lower than the tank overflow). When in a steady state condition, the tank overflow = last tank inflow.

Spilling over the "dam" is going at THE same flowrate coming off the tank. (The last dam is the critical point for the system BTW.) When all the flat section is full, the spill over again will = flow from the tank.

Draw the HGL from there to the inlet to the last tank.
The flowrate into the last tank will be the overflow rate from the first tank, you just don't know the velocity. The velocity could be subcritical or super critical, but do you care? The FLOWRATE into the last tank is the same (=) as the overflow rate (OK minus whatever might have left the system by overflowing any of the upstream "open channels".

http://virtualpipeline.spaces.msn.com

 

[katmar]

The way it is drawn, I do not believe it will flow at all. When the level in T-200-03 reaches the overflow it will run down the pipe and flood to the top point of the 4'11" downcomer. The flow will "waterfall" down the 4'11" section and flood the following horizontal section. However, it will not be able to "climb" up the 5'6" riser because there will still be roughly 3'6" of air in the 4'11" downcomer. This means that all that is driving the water up the 5'6" riser is the 2'0" drop from the tank outlet to the top of the 4'11" downcomer, plus the roughly 1'5" of water built up in that downcomer. With 3'5" of head you will never get the water to the top of the 5'6" riser unless the level in T-200-03 rises considerably.

I suppose that if the level in tank T-200-03 rose very rapidly you might get a situation where the air is flushed out of the 4'11" downcomer, but if the level rises gently and the tank starts overflowing slowly you will very neatly get an air-lock, and once it is established it can only be driven out by a large level increase in T-200-03.

In my opinion there should be an air vent at the top of the 4'11" downcomer. 4 or 5 ft of pipe sticking straight up is probably all you need. This will allow the air to escape, and in fact the water will rise a few inches up this pipe to give the required head to drive the water up the 5'6" riser.

If a similar air vent is not put at the top of the 7'6" riser you will probably generate a syphon from the 4'11" downcomer to the destination tank T-75-01. This syphon could pull air down the vent and break the syphon leading to pulsating flow - with quite a long cycle I would imagine. So I would put an air vent at the top of the 7'6" downcomer as well. This means you have approximately 400 ft of piping from T-200-03 to the top of the 7'6" downcomer, with a net drop in elevation of 1'5". By my estimation you would get close to 1500 USGPM flow if the fluid is water.

The pipe section after the 7'6" downcomer has a large fall and it will likely not run full, but it does not matter because the first section of pipe (i.e. from T-200-03 to the top of the 7'6" downcomer) will be the controlling resistance.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com