17 replies to this topic

[US_ChemE]

Hello,

This refers to the ANSI/ API Standard 2000 (Sixth Edition, November 2009). Request to have some clarification for the methodology adopted and formulae derived for estimating out-breathing and in-breathing requirements due to liquid filling and liquid discharge. This will help to correctly apply the principles which API recommends.

Refer Section 4.3.2.2.1. The out-breathing flow volumetric flow rate, Vop, is expressed in SI unit of cubic meter per hour of air, for products stored below 40 °C or with a vapour pressure less than 5.0 kPa, shall be given as

Vop= Vpf, where Vpf is the maximum volumetric filling rate, expressed in cubic meters per hour.

As per above clause is it correct to consider that if the tank is operating at close to atmospheric pressure [Pop] (say, 10 kPa g) and ambient temperature [T] (say, 30 °C), the out-breathing rate shall be Vop (m3/hr) of air at actual operating pressure (10 kPa g) and operating temperature (30 °C)? This flow will be converted into equivalent air at normal conditions (1.013 bara and 0 °C) as per the conversion co-relations given in the standard. If the tank is under nitrogen blanket at same operating pressure and temperature as above, the out-breathing rate Vop will be the nitrogen flow rate at actual operating conditions (10 kPa g and 30 °C) and needs to be converted into equivalent air rate at normal conditions (1.013 bara and 0 °C).

Above query is also applicable to clause 4.3.2.2.2 for in-breathing flow rate estimation.

Since the clauses mention flow rates in terms of air only, the case of blanketed tank is not very clear.

Would be grateful if someone shades light on above queries.


Thanks in advance for your valuable support.

[proinwv]

I can't at this time, look into the details of your question, and I do not have access to the methodolody used. Must of this work was developed in Europe and adopted by API. So, let me say two things.

First, I have found that it is rare for the writing agency to comment at all on their methodology.

Second, I believe that Protego were very much involved in the development of this standard and they may have even published a preliminary paper on it. I recommend your contacting them and see if you can obtain any of the rationale from them. (If you can, please consider sharing your results with this forum.)

[ankur2061]

US_ChemE,

If you are intent in doing precise calculations in terms of nitrogen (remember the molecular weight of N2 is 28 kg / kg-mol and that of air is 28.96 kg / kg-mol and hence there should be hardly any difference in the volume flow rate) then the procedure is simple and as follows:

1. Find the volume flow (Sm³/ h) of air for inbreathing and outbreathing as per API STD 2000.

2. Convert this to mass flow by multiplying with the density of air @standard conditions (remember that mass of the gas remains unchanged)

3. Convert this mass flow rate to nitrogen actual volumetric flow rate by dividing with the density of nitrogen at the actual pressure and temperature or nitrogen volumetric flow rate at standard conditions by dividing with the density of nitrogen @standard conditions.

If you simplify the above statements int terms of an appropriate equation then:

Vol. Flow rate of N2 @std conditions = Air flow rate @std conditions * (Mol Wt of Air / Mol wt of N2)

Hope this helps

Regards,
Ankur.

Edited by ankur2061, 31 January 2011 - 11:10 AM.

[US_ChemE]

Thanks Paul and Ankur for your help. I did see Protego website and could download the program for venting calculation. I have asked them few doubts about the units of measurement in their program and will share once I get response from them.

The doubt which came to my mind was because of a bit confusing language in the main body of API-2000 (Sixth Edition, Nov. 2009). In section 4.3.2.1 of the standard, it is mentioned that the inbreathing and outbreathing flow rates mentioned in this standard are for air at normal or standard conditions.



Sections 4.3.2.3.2 and 4.3.2.3.3 clearly mention the flow rate unit as ‘SI unit of normal cubic meters per hour of air’ making it equivalent quantity. Sections 4.3.2.2.1 and 4.3.2.2.2, however, mention only ‘SI units of cubic meters per hour of air’ for Vop (outbreathing) and Vip (inbreathing) rate. It is bit unclear whether this is actual volume or equivalent quantity.



I had initially started considering Vop and Vip as actual flows. However, Protego had developed a program (available to download from their site with login input from them) in which they appear to have considered the Vop and Vip flows to have unit of equivalent quantity of air.

Request to advise whether-

1) Vop and Vip are equivalent Nm3/hr of air or actual air (m3/hr at relieving conditions considering volumetric displacement).
2) Also if the tank is blanketed with a gas (say nitrogen), as per API-2000 clauses mentioned above, the inbreathing and outbreathing are still in terms of equivalent Nm3/hr of air. If one has to calculate the actual flowrates (as free gas for blamketing), then Annexure D of API-2000 Sixth edition is the way out.


Thanks.

US_ChemE

[ankur2061]

US_ChemE,

The term "Normal" is archaic and I am still at a loss why API STD 2000 is persisting with it. The correct term is "Standard" used by a lot of texts and the following link should give you a better understanding"

http://en.wikipedia....re_and_pressure

I believe that the term Nm³ / h is still used for air for purely historical reasons. API STD 2000 clearly mentions that the inbreathing / outbreathing requirements are based on normal or standard conditions.

As per the article in the link different standard conditions are utilized by different organizations and even countries / regions

If you are using the term Nm³ / h then API STD 2000 categorically defines it as the volumetric flow rate of air at a temperature of 0 deg C and a pressure of 101.3 kPa.

The actual volume flow rate is simply a function of the temperature and pressure of the gas and is generally not preferred because every time you have to mention the pressure and temperature when mentioning the actual volume flow rate. Using standard volumetric flow rate precludes the requirement of mentioning the pressure and temperature.

Hope this helps.

Regards,
Ankur.

[US_ChemE]

Thanks Ankur for the response. I understand that the venting flowrates obtained as per sections 4.3.2.2.1 and 4.3.2.2.2 are in Nm3/hr of air and thus will also inform us about the mass flow rate as well. However, for calculating the actual flow (Wfl as mass flow rate) for blanketing gas, equation D.35 (Annex D of API-2000 Sixth Edition) shall be used as the quantity of air obtained will require same effective area of discharge as that required by blanketing gas.

Am I correct in this understandiing?

Thanks.

[ankur2061]

US_ChemE,

I have not studied in details the Annex D equation D.35 and cannot make any comments except that since API STD 2000 is universally accepted for normal venting calculations the equations can be accepted without any reservations.

BTW, I have posted an excel sheet for normal venting calculations as per API STD 2000, 6th edition at the following link:

http://sites.google....redirects=0&d=1

Regards,
Ankur.

[US_ChemE]

Thanks for the excel worksheet. It will be of great help to us. Just one final question- the Nm3/hr of air we get are the equivalent Nm3/hr of air - Is this correct? The word equivalent is somehow not being stressed in the main body of API-2000 Sixth edition as against the Annex A of the same standard.

Thanks.
US_ChemE,

[Skeggia]

Hi friends, I have already posted the following message somewhere in the forum, but I think that this is the right place to post it.

"Dear Friends,
I didn't understand how to carry out the venting calculation for refrigerated tanks; in particular I didn't understand how to calculate "Val" that is the flow rate caused by flashing due to changing in atmospheric pressure. API 2000 says that I should evaluate this flow rate through Xgas parameter as explained in par. 5.2.1.3, but I don't know which is the mass flow rate that I should multiply by Xgas to obtain "Val". Can you clarify this calculation???

Thanks a lot and sorry if I didn't explain clearly my problem!

Simone"

[djack77494]

Ankur,
I don't agree with your statement that "Normal" is archaic. In the FPS or Imperial system of units the term "Standard" has a very different meaning than what you would imply would be its use in SI units; that is 0 deg C and a pressure of 101.325 kPa[A]. With people inclined to be less than specific in their use of terminology, I could imagine many conversations making reference to standard conditions. To the SI world, that would mean one thing and to the FPS world it would mean something different. (I think the situation is even more complex in that some industries use even other conditions for their "standard"; as noted in the Wikipedia article you mentioned, there are at least 15 definitions of "standard conditions".) In spite of that, I think that there is basically a single commonly accepted understanding of standard conditions in the chemical engineer's FPS world and a different understanding of normal conditions in the SI world. While it's still confusing, at least our use of the terms "normal" and "standard" tell me whether we're in the SI or FPS world.
Doug

US_ChemE,

The term "Normal" is archaic and I am still at a loss why API STD 2000 is persisting with it. The correct term is "Standard" used by a lot of texts and the following link should give you a better understanding"

http://en.wikipedia....re_and_pressure

I believe that the term Nm³ / h is still used for air for purely historical reasons.

Edited by djack77494, 04 April 2011 - 05:03 AM.

[Skeggia]

Hi friends,
any idea about my problem????

[kkala]

Following views could be useful, subject to comments.
1. Agreeing with Djack77494, term Nm3 is not archaic, at least for SE Europe. It is widely used, referring to 1 Atma and 0 oC. Term std Nm3 (standard Nm3) is not so common here, but at any case it refers to 1 Atma and 60 oF ~ 15.6 oC.
2. For constant weather conditions, tank ingoing flow of A actual m3/h of liquid (of temp<40 oC and vapor pressure < 5 kPa a) will displace A actual m3/h of overhead gas of any nature (Vop=Vpf for air, nitrogen, or other). Actual m3 refers to operating conditions, not differing much to flows reduced to normal or standard gas state for most API 650 tanks (for liquids of temp<40 oC, vap press < 5 kPa a).
3. Thus outbreathing gas rate (= actual m3/h of liquid in) had better be followed with "actual m3/h". If followed with "Nm3/h", or "std m3/h", this will not make an important difference in majority of API 650 tanks.

Note: cryogenic liquids (vap press ~ 1 Atm) or flashing liquids (vap press > 1 Atm) make other cups of tea. Above concerns liquids (e.g. diesel) of stated qualities (temp<40 oC, vap press < 5 kPa a) to API 650 tanks.

Edited by kkala, 04 September 2011 - 11:52 AM.

[chamatkar]

Hi all,
I am touching an old and well discussed topic but I just stumbled into a situation that led me to ask the experts here. According to general understanding, vol displaced from the tank is the vol required to be fed by blanketing gas (irrespective of gas, could be N2, Fuel gas or any) at vessel op conditions. And then one can standardized it using ideal gas law and vessel op P & T. But API2000 6th ed suggests always in m3 air. Now if fuel gas mol wt is 16 and that of air is 28.8, it could make a difference in calculations.

For thermal inbreathing/outbreathing (section 4.3), the equations calculate Nm3 of air irrespective of op conditions of tank/vessel. I am also aware that api2000 is only for atm to low P storage tanks. But considering the scenario when no inflow and liquid out, condensation of vap could occur due to atm conditions change. By standardizing thermal inbreathing same as for liquid displacement considering ideal gas and op P & T of vessel, I can get conservative number for blanket gas instead of taking it Nm3 of air. I want to hear from the experts on this.

Appreciate in advance for sheding some light.

Thanks,

[ankur2061]

chamatkar,

Conversion of Nm3/h of air to Nm3/h of any other gas is easy and is based on Graham's law of effusion. Let us take an example:

Asssumed Blanket Gas Molecualr Weight: = 18 kg/kg-mol
Molecular Weight of Air = 28.96 kg/kg-mol

Now according to Graham's Law:

Vol Flow Rate of Gas 1 / Vol Flow Rate of Gas 2 = square root (Molecular Weight of Gas 2 / Molecular weight of Gas 1)

If Gas 1 is your Blanketing Gas and Gas 2 is Air than

Vol Flow Rate of Blanketing Gas = Vol Flow Rate of Air * square root(Molecular wt of Air / Molecular Weight of Blanketing Gas)

For Graham's law of effusion refer the link below:

http://en.wikipedia....ki/Graham's_law

Hope this helps.

Regards,
Ankur

[kkala]

Just a note on the original topic: readers may find http://www.cheresour..._hl__ api 2000 '> http://www.cheresour..._hl__ api 2000 useful.

[kkala]

Concerning Graham's law of effusion, mentioned in post No 14, some additional notes with examples can be found in http://www.ausetute.com.au/grahaml.html.
Graham's law of diffusion looks different. It was applied in the enrichment of gaseous UF6 to isotope U²³⁵ in 1944/45, to make the atomic bomb having (alas!) destroyed Hiroshima (also reported in wikipedia reference), http://web.ead.anl.gov/uranium/guide/depletedu/enrich/index.cfm.
Graham's law of effusion seems to consider frictional head through a vent as proportional to the square of volumetric (or molar) flow rate of gases, assuming them incompressible, close to atmospheric pressure, and at same conditions.
Consider constant Pi and P (Pi=pressure inside tank, P=atmospheric pressure). And head = k*(Pi-P)/(ρg), ρ=gas density, k=constant, g=gravitational constant. Two different gases 1, 2, would give V1^2 / V2^2 = [(Pi-P)/(ρ1g)]/[(Pi-P)/(ρ2g)] = ρ2/ρ1, where V=volumetric flow rate of gas through the vent referred. Hence V1/V2 = SQRT(ρ2/ρ1) = SQRT (MW2/MW1).
Above is about flow of different gases through an orifice (vent, valve, etc).
My (elementary) understanding suggests that atmospheric tank blanketing should cover (α) inbreathing (β) liquid pumped out (γ) tank outbreathing (δ) liquid pumped in, or combination of these cases. In e.g. (β) and (δ) displaced volume rate equals blanketing gas rate in or out (volumewise), without "disturbing" tank min / max gauge design pressure.
So Grahams law of effusion seems to concern blanketing gas flow to atmosphere under constant Pi (vent), or to tank through a valve under constant upstream / downstream pressure. Comments would be welcomed.

Edited by kkala, 10 July 2012 - 12:26 PM.

[Mark-TR]

Hi all,

I have also been interested on this subject.

I actually calculated the relieving flow of Hidrocarbon Nm3/h due to the heat input during fire scenario as per API 2000 equation. Then I compared this value to the discharge flow in Nm3/h of air as per the same code.

In fact there is a temperature correction in the analysis, grahams law assumes constant temperature and pressure, as stated above.

Details of effusion equation are explained at http://en.wikipedia.org/wiki/Effusion

As in fire scenario is not in normal temperature conditions , the change in temperature shall be considered in the analysis. Pressure is very close to atmosphere, therefore it is credible to assume constant pressure.

Consequently, the relation between hydrocarbon flow and air flow follows the next equation:

Nm3/h of air= SQRT((Mw*T)/(28.96*273.15))*Nm3/h of Hydrocarbon

Where Mw is the molecular weight and T is temperature in Kelvin degrees.

I´m not an expert in the physics behind this subject, but I´ve notice that the ratio between Nm3/h of air and Nm3 of hydrocarbon follows the above equation. It is my understanding that the MW is introduced to follow graham´s law, and temperature is used to compensate and to put the two fluids at the same energy level.

Just like in flow control:

http://profmaster.bl...orrections.html

Kind regards

Edited by Mark-TR, 03 September 2012 - 05:46 AM.

[Raj Mehta]

Sorry for the late participation in this discussion.

 

I have read all the posts in this discussion. I was convinced with post # 14 initially but later found in the subsequent posts that grahams law of effusion is applicable only at constant pressure and temperature conditions. 

 

In case of venting the pressure or temperature inside tank changes (basically they are the prime reasons why we need venting) and as post # 16 indicates the pressure changes need to be taken into account (Pi-P). But, won't this term be same again for both the gases and hence cancel out, giving the same expression finally as highlighted by Mr. Ankur in post #14.  

 

Whats the final call ? Experts kindly pitch in and clear this pending discussion with final comments. 

 

Thanks.