6 replies to this topic

[Justin Spencer]

I am working on sizing an overflow line for a tank and while I believe I have an acceptable solution, this is the first calculation of this nature that I have performed and I want to ensure that I have not overlooked some critical aspect.

 

The overflow line is inside the tank and has a syphon break, which is exposed to atmospheric pressure (see attached sketch). I have sized the line for the maximum inlet flow rate and an allowable head (h, see sketch) of 0.5 m. I first calculated the required head with the Darcy–Weisbach equation assuming the length of pipe upstream of the syphon break and an equivalent length for the pipe entrance.

 

Following advice given on this and other forums, I have read “Designing Piping for Gravity Flow” by P. D. Hills, Chemical Engineering, Sept. 5, 1983. Equation 3, h=0.811*(Q_L)²/(g*d⁴), is the only equation that might be applicable in this situation, but I remain doubtful as I’m not entirely clear what h denotes.

 

Hills writes that “[s]ingle-phase criteria can be applied to designing sections of outlet piping in which flow can be expected to be flooded.” So it seems like d indeed denotes the inside diameter of the pipe upstream of the syphon break. He then goes on to write that “[t]he criteria for flooded outlets are […] Eq. (3) for outlets from the side of vessels [… where] h is the liquid height above the top of the outlet away from the region of the outlet.” So from this explanation alone, it seems plausible that Eq. (3) is applicable.

 

However, upon looking at (Fig. 2b), to which Eq. 2 and 3 apply, the meaning of h is not very clear. First, the water level appears to be drawn at the same elevation of the syphon break and is denoted as “h > h_min”. It seems to me that h should be the height of the liquid level above the syphon break “top of the outlet”. Secondly, it is not clear what the relationship is between the tank and syphon break pressure in Fig. 2b.

 

Despite my doubts, I calculated h with both the Darcy–Weisbach equation and Eq. (3) for three different diameters of pipe: NPS 8, 10 and 12. I found that Eq. 3 predicts a head requirement roughly twice that from the Darcy method.

 

NPS	10	12	14
h [m], Darcy	0.56	0.28	0.19
h [m], Eq. 3	1.36	0.64	0.40

 

The inlet and outlet lines for this tank are both NPS 12 and the results above indicate that I should choose NPS 12 or 14, depending on the method. I have read elsewhere that overflow lines are generally found to be the same size as the inlet, or one pipe size large.

 

Much of “Designing Piping for Gravity Flow” by P. D. Hills may also be found in the Google preview of the Piping Design Handbook by John J. McKetta pp. 413

Attached Files

•  Tank Overflow Line Sketch.jpg   217.91KB   348 downloads

[ankur2061]

Justin,

Equation 3 in the PD Hills article is for flooded outlet. In real life, overflow lines don't operate flooded. Overflow from a tank is an unsteady state operation depending on the level in the tank that is getting overflowed with a peak flow (flooded) at the tank highest level and a low flow (partially filled overflow pipe) at tank low levels. But having said that, when you are designing the overflow line, the design needs to be considered based on a completely flooded line which means that the equation 3 is applicable. You need to convert the head value that you have calculated from eqn 3 to pressure and check whether it is reasonably below the tank design pressure. Based on that, you can decide to use 12" or 14" overflow pipe size.

The rule of thumb of the overflow line size being at least the size of the largest inlet pipe is also mentioned in a respectable international standard, Norsok Std P-001, Edition 5, "Process Design" in Section 6.1 which is available for free download from the following link:

http://www.standard....ue&preview=true

Hope this helps.

Regards,
Ankur.

[katmar]

Is there a reason why your overflow takes liquid from the bottom of the tank? If it is because the substance in the tank is toxic then your syphon break should not go to atmosphere. If it is because you might have a heavy phase settling to the bottom (commonly done to remove water from fuel tanks) then you have to include the density difference in the upleg to your syphon break and how that affects the tank level. The pressure in the tank will of course also have an influence on the allowable h (h as per your sketch).

The arrangement you have shown is different from anything Hills discusses, except for the sizing of the downleg after the syphon break which could be self venting. Hills' Equ 3 is for a side outlet, which he unfortunately does not illustrate. The h in Equ 3 is the height above the top of the outlet, but away from the outlet where the level is essentially flat. Near the nozzle the top surface of the liquid will be sloped down to the pipe.

If, for whatever reason, you do have to have the entrance to the overflow pipe low down in the tank then your approach of using Darcy-Weisbach is correct and you don't need to worry about any of Hills equations. You have not explained how the pressure inside the tank is controlled - this is relevant to how the level in the tank affects the internal pressure (as warned by Ankur) and must be considered from a safety point of view.

[Justin Spencer]

Equation 3 in the PD Hills article is for flooded outlet. In real life, overflow lines don't operate flooded.

When I calculated h I assumed that d is the internal diameter of the section upstream of the syphon break, which is flooded as shown in the attached “Fig. 2b Sketch”. Is this correct?

The arrangement you have shown is different from anything Hills discusses, except for the sizing of the downleg after the syphon break which could be self venting. Hills' Equ 3 is for a side outlet, which he unfortunately does not illustrate. The h in Equ 3 is the height above the top of the outlet, but away from the outlet where the level is essentially flat. Near the nozzle the top surface of the liquid will be sloped down to the pipe.

One of the main reasons for posting here was to get clarification on what Hills equations are calculating. I had a couple of thoughts on this running through my mind before posting here, but after reading Hills paper a number of times I was unable to settle on a satisfactory explanation for Eq. 2 and 3.
The size of dip or slight vortex that forms near the nozzle would depend on the velocity within the nozzle and this might explain the relation between h_min and internal diameter if it’s d₁ and not d₂ that is used in Eq. 3 as shown in “Fig. 2d Sketch”. Another possibility was that the upward air velocity in the pipe with a diameter of d₂ as shown in “Fig. 2d Sketch” creates a pressure greater than P­_ATM, thus a static head greater than h_min is required to keep the upleg flooded. Both explanations seem to fit the general trend of the equation that as d decreases, h increases. However, Hills states that the equations are based on the Foude number and from my limited understanding of the Foude number only the second explanation seems to apply.

Is there a reason why your overflow takes liquid from the bottom of the tank? If it is because the substance in the tank is toxic then your syphon break should not go to atmosphere. If it is because you might have a heavy phase settling to the bottom (commonly done to remove water from fuel tanks) then you have to include the density difference in the upleg to your syphon break and how that affects the tank level.

This particular tank is a skim tank for primary oil separation from water and I assumed that the density difference could be neglected as the oil layer is thin. I would like to know more about what typical overflow line configurations are as I have seen P&IDs of storage tanks containing a single, non-toxic liquid phase where the overflow line extends close to the bottom of the tank—similar to what I have shown in “Tank Overflow Line Sketch”. The P&IDs don’t contain any specific elevations and so significance of the line extending “close to the bottom of the tank” is purely pictorial and may not reflect reality.

You have not explained how the pressure inside the tank is controlled - this is relevant to how the level in the tank affects the internal pressure (as warned by Ankur) and must be considered from a safety point of view.

The tank is blanketed with natural gas fed through a control valve set to maintain an atmospheric pressure.

Does the attached “Fig. 2d Sketch” properly illustrate Eq. 3?

Attached Files

•  Fig. 2b Sketch.jpg   52.97KB   144 downloads
•  Fig. 2d Sketch.jpg   98.98KB   113 downloads

[ankur2061]

Justin,

An oil-water skim tank is generally never open to atmosphere and blanketing gas (generally natural gas) maintains a slightly positive pressure of about 50-100 mmWCg depending on the design pressure of the tank. Also no overflow is generally provided to such tanks. Overflow protection is through redundant high level trips using emergency shutdown valves in the inlet lines to prevent high level in the tank. Pressure is controlled in the skim tank using split control of blanketing gas inlet control valve and vent control valve to flare.

Regards,
Ankur.

[katmar]

As explained by Ankur, your outlet is not really an overflow. It is more of a self regulating level controller on the outlet and in the light of what Ankur has said it is clear why this outlet needs to remain flooded.

The size of dip or slight vortex that forms near the nozzle would depend on the velocity within the nozzle and this might explain the relation between h_min and internal diameter if it’s d₁ and not d₂ that is used in Eq. 3 as shown in “Fig. 2d Sketch”.

Yes, it is d₁ that is used in Equation 3. d₂ (as per your sketch 2d) is the diameter required to ensure self venting flow after the syphon break. d₂ has no influence on keeping the outlet from the tank flooded. It is just to ensure that air is not sucked down the syphon break, which in any case might not be a problem - it just depends on whether entrained air would be a problem downstream.

Another possibility was that the upward air velocity in the pipe with a diameter of d₂ as shown in “Fig. 2d Sketch” creates a pressure greater than P­_ATM, thus a static head greater than h_min is required to keep the upleg flooded.

I have never come across this as a problem with pipes designed for self venting flow..

Both explanations seem to fit the general trend of the equation that as d decreases, h increases. However, Hills states that the equations are based on the Foude number and from my limited understanding of the Foude number only the second explanation seems to apply.

For an understanding of the Froude number think first of another dimensionless number we all know well - Reynolds Number. We know from empirical evidence that if Re > 4000 we have turbulent flow. Similarly, from empirical evidence we know that for Re < 2000 we have laminar flow.

We know (again, empirically) that for Fr < 0.31 flow is self venting an no syphon forms - but for Fr > 1.0 syphons do form readily. Hills has defined his J*_L in Equation 1 as a variation of the Froude Number, but it is still a dimensionless ratio. Hills (or his sources) have found empirically that for J*_L < (2h/d)^0.5 a side outlet will be flooded, and this leads to Equation 3. Solving Eq 3 gives h_min and then you can design your P-trap (inverted U) so that h > h_min.

Remember that the trend of h increasing as d decreases is for a constant volumetric outlet flow Q_L. What this really means is that for a higher velocity in the outlet pipe you need a higher level to keep it flooded.

Some general comments
For a thin layer of oil I agree that there will be no significant effect on the height of the P-trap.

It is better from a maintenance point of view to install your piping externally as you have shown in Sketch 2d than to have it internal to the tank as shown originally.

[Justin Spencer]

Thank you, Ankur and katmar, for your helpful replies in explaining Hills’ article.

I just have a couple of comments regarding the tank itself, as I may not have provided quite enough information in my previous posts. I’ve uploaded another sketch that I hope provides all the necessary information.

Overflow protection is through redundant high level trips using emergency shutdown valves in the inlet lines to prevent high level in the tank.

The majority of the inlet flow is not controlled and there are no ESD valves on the inlet. Instead, the water and oil levels are each controlled downstream of the tank. I’m assuming that the overflow is a safeguard employed in lieu of an inlet ESD.

As explained by Ankur, your outlet is not really an overflow. It is more of a self regulating level controller on the outlet and in the light of what Ankur has said it is clear why this outlet needs to remain flooded.

The tank in the new sketch I’ve uploaded has three outlets. Where you thinking of a scenario where a tank with only skim oil and overflow outlets would operate with the level maintained near the top of the overflow line? To be honest, your comment confused me a little bit, which led me to believe that I might not have provided all the proper information

It is better from a maintenance point of view to install your piping externally as you have shown in Sketch 2d than to have it internal to the tank as shown originally.

This seems like a good idea, regardless of the tank configuration. Any thoughts why this wouldn't be the case for any tank I've come across? This particular tank is site fabricated, but for shop fabricated tanks Is it easier to have the vendor pre-fabricate it internally?

Attached Files

•  Tank Overflow Line Sketch 2.jpg   101.91KB   139 downloads