Hope someone can help here. We recently did a flow test on a drain line by connecting it to a tanker filled with water. The tanker had 2 compartments each open to atmosphere and each with its own control valve. A hose was connected to each of the compartment and rolled out down a hill to a 'Y' piece which was then connected to the single drain line. What happened next was not expected. Both valves at the tanker were opened and water allowed to flow, but rather than the tanks compartments emptying together, one emptied fully before the other started. Its like it was give preference. Now I do remember reading something in the past, but its gone. A pointer would be great.
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Assuming both tanks had the same original surface leve, you'd have to get the valves fully open at the exact same second in time, otherwise the one with the least pressure drop across its valve would establish flow first. Even if you did, probably one tank would manage to establish a precedence anyway via turbulence from its stream exiting at the Y. Once arriving at the Y, the path of least resistance for that established flow would be out the bottom of the Y. If that original tank was able to maintain flow, which it probably would be able to do when open to atmosphere, the flow through the Y would be the maximum possible via gravity alone from that original tank, leaving no further "room" in the Y for the other path's flow to enter, especially since the first path had already established an entry momentum into the Y, ie. it had already a greater total head, not from fluid level pressure alone, but from that plus its velocity head as well.
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
BrianCNTFM
Mechanical
Just tried to duplicate your scenario with twin kitchen sinks without success!One sink drained ahead of other. As Big Inch suggested for levels to drop simultaneously everything would need to mirrored/duplicated. I suspect there were other factors debris/equipment acting as non return valve in this instance. Would be interested to learn if anyone else has seen this occurence before?
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The Scenario envisaged if does not have anything to do with plugging/blockages internally in the not draing tank(compartment)'s flow path
then other idea coming to mind is that one flow stream in a way serves as higher velocity carrier stream due to geometrical trajectory and
sucks-in other flow stream thus iducing faster rate draining of second stream taking place and
utill and unless this drain-out completes first other draining does not take any real effect!
Just hypothetically envisaged case might prove helpful!
Best Regards
Qalander(Chem)
then other idea coming to mind is that one flow stream in a way serves as higher velocity carrier stream due to geometrical trajectory and
sucks-in other flow stream thus iducing faster rate draining of second stream taking place and
utill and unless this drain-out completes first other draining does not take any real effect!
Just hypothetically envisaged case might prove helpful!
Best Regards
Qalander(Chem)
Just out of curiosity, what are the dimensions of the "Y" inlet and outlet diameters? Are the inlet diameters smaller than the outlet?
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
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It may be that you don't have enough pressure drop to effect a complete acceleration at the Y. Just looking at a Bernoulli analysis it would appear that you'd need around a 20 ft drop to the Y to double your flow, ignoring pressure loss due to flow. Less than that and velocity from one inlet may be impacting on flow from the other and results could be unpredictable, similar to outflow into the leg of a T, where opposing velocity heads would tend to cancel, although to perhaps a lesser extent in Y. An increased outlet diameter would help too.
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
I find it hard to imagine that two water sources with roughly equal static heads above them would not flow in approximately equal quantities through the Y-piece.
The only way that I can see this happening is if there is a vapor lock. If tank A is opened first and its water flows down towards the Y-piece, the water will try to flow up the line to B as well as down the drain pipe. The degree to which the water from A will flow up the line to B will of course depend on the pressure drop in the line beyond the Y-piece.
Depending on the geometry of the line, the vapor trapped in the B line can stay there, even after the B valve is opened. We've all seen this happen. The water from B has to flow through the trapped air, which effectively functions as a restriction in line B because the water cannot fill the line at some point. Once the water from A stops flowing there will be no pressure drop after the Y-piece and the pressure available to flush out the trapped air suddenly increases and tank B starts to drain.
I suppose the way to test this is to open valve A and establish flow. Then close the valve at the end of the drain line and close valve A, leaving the A line full of water (no trapped air). Now open valve B and then the end of line valve to establish flow from B. Finally open A again and see if both tanks drain equally.
Katmar Software
Engineering & Risk Analysis Software
The only way that I can see this happening is if there is a vapor lock. If tank A is opened first and its water flows down towards the Y-piece, the water will try to flow up the line to B as well as down the drain pipe. The degree to which the water from A will flow up the line to B will of course depend on the pressure drop in the line beyond the Y-piece.
Depending on the geometry of the line, the vapor trapped in the B line can stay there, even after the B valve is opened. We've all seen this happen. The water from B has to flow through the trapped air, which effectively functions as a restriction in line B because the water cannot fill the line at some point. Once the water from A stops flowing there will be no pressure drop after the Y-piece and the pressure available to flush out the trapped air suddenly increases and tank B starts to drain.
I suppose the way to test this is to open valve A and establish flow. Then close the valve at the end of the drain line and close valve A, leaving the A line full of water (no trapped air). Now open valve B and then the end of line valve to establish flow from B. Finally open A again and see if both tanks drain equally.
Katmar Software
Engineering & Risk Analysis Software
Compositepro
Chemical
One possible explanation would be a vortex in the main flow, which would result in a back pressure on any branch line coming into it.
In a T flowing in from the top left, given you've had just a little flow from one side first, you'd have static head + velocity head. In the right, only static head. Total head would be higher on the left than on the right and there would be little reason for that flow to enter, in fact some of the flow from the left might actually enter the right side and recirculate until it happened to near the downward outlet. That effect would probably be reduced in a Y, but by how much. Probably by the cosine of 1/2 the crotch angle. I think it could be one of those things where Mother Nature doesn't get the theory exactly right and the first bit of turbulence sets up flow on one side and it takes over, especially since the Y outlet isn't twice the area of the inlets. Perhaps somebody should do some more experiments. We already have one kitchen sink experiment above confirming the effect is possible.
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
Bernoulli tells me the total head has to be the same in both legs. In the left leg the static head has been converted to velocity head to satisfy Bernoulli's requirement that the total remain constant. The static head on the left should therefore be less than on the right. This is how a venturi ejector works. If anything, the flow from the right should be sucked into the Y-piece.
Katmar Software
Engineering & Risk Analysis Software
I'd tend to agree it would be sucked in, if it had someplace to be sucked into.
How would you explain the "experimental" results?
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
How would you explain the "experimental" results?
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
With my vapor lock theory.
I tried to do the twin sink experiment in my kitchen, and I got an intermediate result. One sink did drain slightly faster than the other, but my drain piping is not symmetrical and the sink with the more direct piping drained faster.
@makeup - is there any chance of redoing the experiment the way I described in my post of April 21?
Katmar Software
Engineering & Risk Analysis Software
Isn't your post allowing for both scenarios. No stars for sitting on the fence. 
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
Qalander
a star for you ... mostly because I think the "little pink (or purple) stars are funny.
Feel better?
Patricia Lougheed
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a star for you ... mostly because I think the "little pink (or purple) stars are funny.
Feel better?
Patricia Lougheed
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Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.
There's another star just for effort.
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
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"The problem isn't finding the solution, its trying to get to the real question." BigInch
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