8 replies to this topic

[ChemEngrr]

I will conceptualise the issue I am experiencing. Imagine a tank of infinite volume open to atmospheric. The tank has a 100m vertical downpipe filled with water with a 2” full-bore isolation valve at the bottom. The outlet of the valve discharges to atmosphere.

Initial condition - valve closed, static pressure at the isolation valve is circa 10barg, at the 50m elevation circa 5barg,
And so on.

The valve is then partially opened to resulting in flow from the tank. Line remains full.

The static pressure immediately above the valve is no longer 10 barg but a pressure lower due to frictional losses and velocity head. Similarly less at the 50m elevation.

When the valve is fully opened there is no longer a specific point in the system (e.g a partially open valve) which is governing the rate of flow, just resistance of the 2” line over a vertical elevation of 100m and available head. I am having difficultly determining the flow and pressure profile as a result.

Problem:
With atm pressure at the tank, and at the outlet of the valve, what is the static pressure at the 50m elevation point in the line when the valve is fully open I.e now just a continuous 2” downpipe.

If a pressure gauge were installed at the hypothetical 50m elevation, what would it now read?

Is flow stable through the line or are vacuums being pulled in various locations causing vapourisation of the water leading to pulsating flow? Has all head been converted to kinetic energy with static pressure atmospheric throughout?

I feel the issue may be the grey matter associated with gravity flow has long since died, rather than a complex hydraulic issue. Perhaps not.

Any help would be appreciated.

[latexman]

A sketch or drawing would help.

[ChemEngrr]

Essentially a vented tank with line descending vertically down with an open end to atmosphere. The purpose of the valve in the explanation was to show how I have been visualising the system in terms of static pressures. 

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[Pilesar]

This sounds like a student assignment to me. The student forum here is a good place to ask student questions. The answers there are tailored to the perceived knowledge level of the questioner and everyone benefits.

Infinite volume tanks would be very useful, but they cost too much so are not commercially economical. As for pressure calculations, the amount of the fluid above the measurement point matters. In a vacuum, there is no fluid above so the absolute pressure is zero. For earth at sea level, it takes about 14.7 lbs per square inch to hold up the column of air directly above. This is also a convenient reference point for a measurement gauge. So the pressure would be 14.7 psi absolute or when read on a gauge would show as 0 psi gauge pressure. To convert from absolute pressure to gauge pressure, the atmospheric reference pressure at the point of measurement is required. When the fluid above the measured point is not air, the pressure is often represented as 'head' which is the equivalent height of the fluid column. So the gauge pressure at a point 5 meters below the surface of water can be called '5 meters of water pressure.' This can be converted to other units of measure by multiplying the head by the density of the fluid.

[ChemEngrr]

@Pilesar

 

Infinite volume was intended to indicate non-transient flow.

Agree the presentation this way appears academic.

 

Not sure you have understood the question. The same atmospheric conditions are shown on the diagram at top and bottom elevations of the system.

If the bottom of the 100m line were to be isolated there would be 100m H2O static head (gauge) at 0m elevation, 50m (gauge) at 50m, 0m (gauge) at 100m elevation.

 

Question is what static pressure would be measures at 50m elevation if this system were then allowed to flow freely? Your logic would suggest 100m static pressure at bottom of the line, which will in fact be at atmospheric pressure.

[Pilesar]

Here is a short video that I think addresses your questions better than I did: https://www.aft.com/...nation-pressure

[ChemEngrr]

@Pilesar,

 

Well presented video however I do not think that the issue lies in a confusion between stagnation and static pressure.

 

Stagnation pressure would only be observed at the PG if the local fluid velocity at the device were 0 ft/s.

 

This device is connected to a connection normal to the direction of flow.

 

I would dispel the notion that this in an academic query if you can. It is a drain system designed to run liquid filled between 2 atmospherically vented locations by gravity, where I am interested in the static pressure at an intermediate point between the tanks.

 

I feel the issue lies in a misconception of mine that the frictional losses in the system should manifest itself in some way as a back pressure, e.g. a pressure at a location along a horizontal line can be calculated by adding the cumulative frictional loss from the point end the pressure at the discharge location. I suspect that the energy conversion here for the vertical gravity flow example is potential head to velocity, where the pressure at 50m is still 0 barg. 
 

[katmar]

Do not feel alone in being confused.  This question has confused many experienced engineers.  See for example this thread, which BTW illustrates that this is not just an academic problem. If you could put a pressure gauge at the 50 m point it would show zero pressure.  All the static head has been used to overcome the friction and to provide the kinetic energy.  If you start with the valve at the bottom closed and gradually open it to the point where it no longer provides any restriction the flow will keep increasing until the static head exactly matches the sum of the friction head in the pipe and the velocity head.  Where else could the static head go?

 

If you measured the pressure at the 50 m point not with a gauge connected perpendicularly to the pipe but instead used a pitot tube facing upwards then you would measure the stagnation pressure as explained in the video.

 

Another way to think of it is to say that the static head always exactly matches the sum of the friction head and the velocity head. If the valve at the bottom were only 10% open then you would still get some positive reading on your gauge at the 50 m point because the fluid has to preserve some of the static head to overcome the high frictional resistance of the valve.  With the valve 10% open the friction in the pipe would be much less than with the valve fully open, but the sum of the friction head in the pipe and valve (plus the velocity head) would still have to match the static head.

[ChemEngrr]

@katmar - thank you for your explanation, it is much appreciated, you are a digital saint.