21 replies to this topic

[Bahrami]

Your explanations will be greatly appreciated.

 

What is the effect of high pressure inlet (e.g. 25 barg) on external floating roof tank?

 

[The question has been clarified @ post #5]

Edited by Bahrami, 03 March 2013 - 10:11 PM.

[kkala]

What is the liquid pressure downstream of entrance nozzle of an atmospheric tank?  Just hydrostatic. Even for a maximum liquid height (HHLL) of 24 m (over centerline of mentioned nozzle), this pressure would be 2.16 kg/cm2 g (if sg=0.9). Neglecting ΔP through the nozzle, say  1/2*9.82 m liq or 0.005 kg/cm2 g (for velocity 1 m/s), pressure upstream nozzle is about 2.16 kg/cm2 g.
If a feeding pump develops a discharge pressure of 25 barg, total ΔP (frictional + static) from discharge to tank nozzle will be 25.0-2.16=22.84 kg/cm2, and probably a partially closed valve will contribute to this ΔP.
Suppose another case, where a pump (to develop flow rate Q against 25 barg discharge pressure) is connected to the line feeding this tank.  For flow Q total ΔP of this line to tank nozzle be 15 bar. This means pressure at tank nozzle = 25-15=10 barg=10.2 kg/cm2 g (not 2.16 kg/cm2 g) for flow rate Q.  Actual pump flow rate will then become (say) 1.15Q and pressure at nozzle 2.16 kg/cm2 g (pump discharge pressure will decrease, since flow rate increases).
At any case the fluid introduced into the tank shall have a pressure of 2.16 kg/cm2 g under the specific conditions of the example. Higher inlet pressure cannot be obtained, as long as floating roof is free to move up and down (*). If a pressure at a point of feeding line is substantially higher than 2.16 kg/cm2 g, there is excessive ΔP between the point and the tank, caused e.g. by intermediate semi closed valves or higher tank elevation.
 
(*) If the floating roof overruns HHLL, excess liquid is expected to be drained through overflow.

Edited by kkala, 02 March 2013 - 02:05 PM.

[Art Montemayor]

Bahrami:

I will be more direct and to the point than Kkala is:

 

Your query is non-sensical.  You either mean to write something else or you don't know how an atmospheric storage tank works.  It is virtually impossible for you to exert a pressure of 25 barg inside an API storage tank because you will have blown off the roof of the tank long before it would be capable of reaching such a high pressure.   That is why your query doesn't make any rational or common sense.

 

A pump builds up pressure only because it is going against a flow resistance.  In order for any crude oil pump to achieve a discharge pressure of 25 barg while pumping into a tank, it must have the tank (or its system) resisting that same pressure - which is an impossibility (and definitely not needed).  Are you mis-interpreting what a pump's discharge pressure is and what the fluid pressure is as it enters the tank?  As Kkala indicates, the pressure of the fluid as it enters the tank is essentially the hydrostatic pressure of the oil height inside the tank (plus, of course, any gas blanketing you might have in the tank).

 

I get the impression that you don't have a solid, accurate idea of what the fluid flow principles involved are in this application.  Please explain why and what the reason for your query are.  Perhaps I am mistaken, but the nature and manner of your question are important if you are seriously working on applying what you believe to be engineering facts.

[kkala]

I think blanketing is not applicable on floating roof tanks, is it?

Of course in the general case of atmospheric tank, gauge pressure at an internal point is hydrostatic pressure plus any gas blanketing.

I have seen gas blanketing applied on fixed roof tanks.

Gas blanketing application seems to have poor chances on the fixed roof of internal floating roof tanks <http://www.eng-tips.....cfm?qid=249886>.

[Bahrami]

It seems that the above brief wording caused misunderstanding of the question. Let me clarify the ambiguity as following:

 

Assume a main multi-user 18" header which is governed by a centrifugal pump. There is a 12" branch from the header equipped with motorized valve for filling an external floating roof tank. The pressure of header at tie-in point (entrance of branch) is 25 barg! and during filling, motorized valve will be fully open and flow rate Q will be provided.

 

As everybody knows, pressure at filling nozzle will be static head of liquid inside the tank which varies from LLLL to HHLL. 

 

The main question is that what will happen to the pressure profile along the pipeline (from tie-in point with 25barg pressure to entrance of the tank with e.g. 2barg pressure)? According to Bernoulli's equation, when levels of first and second point are same, pressure should be converted to momentum (increase velocity) and reduced by friction. Since Crude is in-compressible and sectional area is uniform, due to conservation of mass, velocity will be uniform. Is the pressure drop of branch 25-2=23barg?

 

Also other questions are:

1. Whether pressure regulating of filling stream by PCV is mandatory or not?  As I searched I couldn't find standard or guideline regarding restriction of upstream pressure for filling of API-650 storage tanks.
2. Somebody mentioned that momentum criteria for two phase inflow of separators is applicable for storage tanks filling velocity criteria for selection of inlet device. Is that true?
3. Cause and effects of splashing in Crude floating roof storage tanks with low level side inlet?
4. Effect of pumping pressure on shearing of the oil-water emulsion?

 

Regards

[Bahrami]

Dear kkala,

 

Thank you for your explanation. But there are some ambiguities in the case of post #5 and it will be greatly appreciated if you could help me to understand pressure profile of branch when there is not a specific pump for filling the tank and pressure of the header (tie-in point) is almost  25barg.

Could we consider this case same as "pipeline discharging to atmosphere" and solve as following:

Tie-in pressure: 25barg

Friction loss of branch: 3 bar  ->  Pressure @ filling nozzle of tank inside the pipe = 25-3=22barg  ->  pressure of filling nozzle inside tank=static head=2barg  ->

assume same level for points 1 &2 and 800 kg/m3 for density and 3m/s velocity in pipeline: P₁+¹/₂ Ro.V₁²=P₂+¹/₂ Ro.V₂²   ->  2228600+0.5*800*9=202600+0.5*800*V₂²   ->  V₂=71.2 m/s (neglecting loss of sudden expansion)

 

* Red: Correction due to comment of kkala @ post #7.

Edited by Art Montemayor, 04 March 2013 - 09:37 AM.

[kkala]

I suggest, Bahrami, you restore original post in the place of post no 1; otherwise readers do not know what subject posts no 2 and 3 respond to. Nothing is lost by restoring post no 1, since at present it is same as post no 5 (except 1st line).

Concerning queries of post no 6, let us see two cases, even though only case 2 would be realistic here

Case 1: Pressure at tie-in remains at 25 barg after opening the motorized valve on the branch. That is valves on other branches close (more or less simultaneously) so that pump flow rate and head change, but eventually pressure at tie-in is again 25 barg. Pressure of liquid in tank (inlet nozzle level) is 2 barg.

Frictional pressure drop along the horizontal branch line is 25-2=23 bar. Inlet nozzle ΔP = 3²/19.62=0.46 m liq = 0.036 bar, branch pipe ΔP=22.964 bar, if 3 m/s velocity is assumed. Actually flow rate Q will become such that ΔP=23 bar, e.g. velocity will be roughly 6 m/s in a 12" branch pipe of 2000 m total length (physical+equivalent, viscosity=40 cP). 

Bernoulli expression P+1/2Rv²=constant (R=density, horizontal pipe) supposes no friction in flow, which is not the case.

For incompressible fluids and no change of pipe section, v1=v2 (continuity equation); change in P is just the frictional pressure drop, thus this Bernoulli expression does not seem to offer useful information.

Care of the units is worthwhile, e.g.  just downstream the tie-in P+1/2Rv²=25E5 N/m2+0.5x800 kg/m3 x 3²  m2/s2 = 2500000 N/m2+3600 kg/m2/s2 = 2503600 N/m2=25.036 bar.

Note: Pressure of 0 barg can be also used as zero base (in lieu of 0 bara), since pressure energy is a sort of potential energy.

Case 2: Motorized valve on the branch gets opened, valves on other branches remain as before. Pump total flow rate is expected to increase, thus its discharge pressure will decrease. Pressure at tie-in is expected to get a value Po, lower than 25 barg. Flow Q can be estimated as per case 1, but with ΔP=Po-2 bar along the branch line.

Note: TL net or other network software could define flow rates and pressures in all branches, taking details into account.

 

Other questions, post no 5.

1. I have not seen control valves on the lines to/from offsite tanks in local refineries, only manual valves.

2. See <http://www.cheresour...let-nozzle-size>

3. To my understanding floating roof tanks have no splashing under normal conditions. I do not know the procedure, when filling starts with tank completely empty (after maintenance); of course inlet velocity would be less than 1 m/s, apart from other measures.

4. Not having any idea, it would be interesting to know.

Edited by kkala, 03 March 2013 - 05:21 PM.

[S.AHMAD]

 Bahrami

 

1. You have answered to your own question about the pressure profile: Namely,reassure at the branch is 25 barg and at the tank entrance is 2 barg. This is the pressure profile is it not?

2. I am very sure that you are aware of the pressure drop equation:

DP = f (L_e/D) v² /2g

In your case DP = 25 - 2 = 23 bar (correction) (to use the above equation, need to convert this into m of crude height)

L_e is the equivalent length which the actual pipe length plus equivalent length of fittings and valves between branch and tank entrance

This L_e is known is it not? D the diameter is also known is it not? assuming turbulent then f is constant is it not?

3. Then, from the pressure drop of 23 bar, you can determine what the velocity, can you?

4. Knowing V and D can you determine the total flow Q?

5. If you don't like the figure Q then you need to install a flow control valve between the branch and the tank entrance. So that you have the mechanism to control the flowrate entering the crude tank.

6. The above is only in theory, but in real case, the branch pressure is not constant, it varies with flowrate and the tank level

Note: The red words are corrected one after reminder by breizh.

Edited by S.AHMAD, 03 March 2013 - 10:52 PM.

[breizh]

Good day S.Ahmad ,

DP is 23 bar .

Breizh

[S.AHMAD]

Breizh

 

Thank you for the correction. I have made the correction shown in red post#8

Edited by S.AHMAD, 03 March 2013 - 10:53 PM.

[fallah]

Dears,

 

Appears the main question of the OP, as following, is remain unanswered:

 

[quote] What is the effect of high pressure inlet (e.g. 25 barg) on external floating roof tank?[unquote]

 

IMO, being such high pressure in upstream of incoming fluid without any high resistance inline, e.g. control valve,...,will result in very high incoming flowrate to the tank. The cosequence could be an unbalancy in rising the roof. High filling rate will affect bleeder vent sizing and due to high filling speed the rim vent size is going to be high...  

[S.AHMAD]

Normally we limit the inlet velocity low enough ( 1 m/s?) to avoid any turbulent in the crude tank and install inlet distributor if necessary

Edited by S.AHMAD, 04 March 2013 - 04:04 AM.

[kkala]

1. Restoring original post in the place of post no 1 is acknowledged.
2. Used Bernoulli equation in post no 6 may need further explanation. Post no 5 says "when levels of first and second point are same, pressure should be converted to momentum (increase velocity) and reduced by friction", which I think is right; but "reduced by friction" seems not to have been applied in post no 6. Considering the 12" branch pipe, velocity at any cross section along it  has to be same, according to continuity equation for incompressible fluids. In this specific case, there cannot be fluid "acceleration" (that is velocity increase from 3 m/s to 71.2 m/s along the branch pipe).
To interpret last lines of post no 6, assume beginning and end of the 12" horizontal branch line, with velocity v1=3 m/s.
P1=25 barg, P2=2 barg, frictional ΔP=23 bar, v2=v1=3 m/s, Ro=800  kg/m3
Bernoulli equation per post no 6 should be:  (P1+0.5Ro*v1²) - (P2+0.5Ro*v2²) = ΔP, that is (2 500 000+0.5*800*9)-(200 000+0.5*800*9)= 2 300 000 Nt/m2  (*).
3. Branch 12" line has (at least) two valves, one motor operated at the beginning and another isolation valve at tank inlet nozzle (manual or motor operated). Local refinery practice would control flow through these two valves, even if they were gate (post no 7). Partially opening the valve near tie-in can be enough, without installing control valve. Filling of tank seems  intermittent, without need of  precise flow control.
4. It is also noted that inlet / outlet  velocity of flammable liquids is limited here to 1 m/s (maximum),  if moving liquid close to tank nozzle can contact atmospheric air during low liquid levels. This is a legal requirement due to risk from static electricity. For floating roof tanks this happens only during complete emptying for maintenance, or following filling. 

(*) <http://www.cheresour...perature-of-it/> can be useful (mainly posts no 3 and 10 for the Bernoulli equation). Examples consider ΔP~0, which is not the case of present thread.

Edited by kkala, 04 March 2013 - 07:58 AM.

[S.AHMAD]

1. Bernoulli equation can be used if the frictional drop is small/negligible. However in this case, I strongly feel that frictional pressure drop is significant.

2. The modified Bernoulli equation in this case would be:

P₁ + 0.5R_hoV₁² = P₂ + 0.5R_hoV₂² + (0.5f(Le/D) + 0.75)R_hoV₂²

where (0.5f(Le/D) + 0.75) includes entrance and exit losses.

[Bahrami]

Dear Sirs,

 

Thank you for beneficial points.

 

As conclusion, my understanding drafted as following:

 

"

Assume a pipeline branched from a fixed pressure header (25barg) for filling an atmospheric storage tank. Pressure at feed nozzle of tank will be static head of liquid inside the tank e.g. 2barg. 

 

Since velocity is same at inlet and outlet of the branch due to continuity for in-compressible fluids, modified Bernoulli's equation ( P₁ + 0.5R_hoV₁² = P₂ + 0.5R_hoV₂² + DPf ) indicated that pressure difference between inlet and outlet of pipeline (with same elevation) is due to friction loss (DPf) as 25-2=23bar. 

 

To reach this pressure drop, there are 2 alternatives according to Pressure drop equation ( DPf = f (L_e/D) v² /2g ) with specific  f and D and constant g:

1. Higher velocity (v), higher pressure drop. But line sizing criteria restrict velocity in pipeline upto a specific value e.g. 3m/s and also pressure drop through pipeline e.g. 4.5 bar/km to prevent erosion, vibration,etc and also API recommend velocity for initial filling upto 1m/s . So in this case for 23bar pressure drop, increasing of velocity is not practicable.
2. Higher equivalent length of pipe and fitting (Le): An appropriate valve, restricted orifice or combination of them will increase pressure drop and satisfy the mentioned Equations.

"

 

Your confirmation or comments will be greatly appreciated.

Edited by Bahrami, 05 March 2013 - 05:46 AM.

[S.AHMAD]

1. You are almost there. Let me summarize

2. You need to determine the maximum flow (Q_max) using the pressure drop equation

3. If this Q_max is too high (i.e. high velocity) that beyond allowable limits then you need to add additional pressure drop such as a flow control valve or use gate valve or globe valve with orifice meter so that you can control the flow within the allowable limit or you size the pipeline such that Q_max is within the allowable limits.

4. I hope by now you have good idea about the situation and precisely know what to do.

5. Good luck

[fallah]

AHMAD,

 

[quote]...you need to add additional pressure drop such as a flow control valve or use gate valve or globe valve with orifice meter so that you can control the flow within the allowable limit...[unquote]

 

It would normally be done by motor operated valves which basically are trunnion mounted and reduced bore ball valves produce additional pressure drop even in full open position...

[Sherif Morsi]

Hi,

 

What if the atmospheric tank is with fixed roof and opened to the atmosphere (air breather connection)? Would it be possible to send a fluid at 25 barg for example to the atmospheric tank and rely on the fact that the pressure will drop across the line up until it reaches the tank?

 

Your answer is appreciated

 

Sherif

[Art Montemayor]

Sherif:

 

You don't send a fluid at 25 barg into a storage tank without controlling it.  You are missing the whole point and you are getting all confused - as the O.P. is.

 

You have to supply a throttle (usually a globe valve or similar device) to ensure that you are in control of any upstream high pressure prior to the tank.  It is that simple.

[andrithzabalam86]

Hi, 

 

I have an additional question about this topic. 

 

I have a recent case in a crude oil dehydration system using gun barrels. They are installing a screw pump with a capacity of 10000 barrels per day and the motor power is 200 hp. Plant administrators found this screw pump unused in other crude oil processing plant so they borrowed it. 

 

Nevertheless, according to my simulation a 30 hp power pump is enough to increase pressure from 2 kg/cm2 g to 7 kg/cm2 g (requiered for the piping system considering a 16 API crude oil and viscosities of 1300 - 1500 cP). 

 

If I connect this pump with a 200 hp motor to the system the pressure will increase over 60 kg/cm2 g, and the system does not have regulating valves in the pump discharge.

 

As you explained in the previous posts, for a bigger dP in the piping system the flow will increase to match this dP, but the pump vendor indicated that the maximum flow rate that could be given by this pump is 12 000 bpd, so my doubt is: 

 

What will happend in the gun barrel tank when this pump is turned on with this 60 kg/cm2 discharge pressure and a maximum flow rate of 12000 bpd?

 

Your explanations will be greatly appreciated.

[Art Montemayor]

Andrith:

 

You, also, are missing a very important point regarding pumps:  Pumps do not create any pressure.  Pumps create flow of liquids.  The resultant pressure increase seen in the discharge port of a pump is the result of any resistance imposed on the discharged liquid by the system.   For example, you can have a screw pump pumping crude oil out of an open pit in the ground and into another open pit next to it.  The discharge pressure of the screw pump in this case can be practically zero if you make the discharge piping large enough and short enough, without any throttling valves.  I am sure you can visualize this scenario as being true.   I am sure that you can also visualize that if you start to throttle the pump's discharge, the pressure on the upstream side of the valve will increase while that downstream of the valve will remain the same atmospheric pressure.  Bear in mind that you are dealing with a positive displacement pump and not with a centrifugal type - therefore, you can't continue the throttling for very long without experiencing a continuing pressure increase.  Positive displacement pumps will do exactly that; they will not allow accumulation of flow without a pressure increase.

 

Now to your specific topic.  I don't understand how you can expect to hook up a screw pump (a positive displacement type) without knowing its characteristic performance curve - or in other words, what speed does it have to have to pump the flow rate you want?  This is not a centrifugal pump that you can throttle and adjust to the flow rate you desire.  The horsepower of the required motor is easy to estimate, but you must identify if your screw pump will pump the flow rate required - and at what speed.  Knowing the speed, you then find a motor that matches that speed or employ a reduction gear box to achieve that same speed.   I think you need to do some homework on your knowledge of pumps and how they operate.

[Waqasuet]

If tank inlet pressure is high due to an external source, a redundant PCV is a must in tank inlet herader. In addition to that The surge relive protection system is also compulsory for protection of the system. This analysis is based on my experience.