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  • Friday 29 March 2019

    Vent Sizing's for Reactor's & Storage vessel

    Hello all....!!!

    Have taken a gap.....  because of my profession!!

    After many days gone through the mail and comments, and most of the requests were related to vent sizing, ejectors. And even in my circle many were expressing the same query related to vent line sizing.

    Also recently have seen a disaster in a chineese chemical plant (Xiangshui), nearly 78 persons were dead and many went missing, which is majorly due to lack of proper safety measures and may be avoidable if adequate planning for relief is available. And especially it can be considered as a black day in every process engineer's career. See Below video:

    As per every management, Process safety is everyone's responsibility, but as per me a Process Engineer holds major accountability than others. Hence being an process engineer, its quite necessary to evaluate every hidden risk before going further with a process. As in a manufacturing company process engineer would be like a head light defining right direction to the operators. 

    Today here i gonna add a small notes related to that, 
    Actually its a very simple task if we have all the data which is required for calculation, as based on assumption its quite waste to calculate because the calculation would be something like satisfying ourselves.

    I've performed the calculation for one of my products, so with that experience i'll elaborate the calculation by altering the values little bit.

    Before that lets have some standard definitions.

    Set pressure: Activation pressure of a relief device 

    Maximum allowable working pressure: Maximum gauge pressure at a given/designated temperature.

    Operating pressure: Gauge pressure, usually 10-12% below the MAWP.

    Over pressure: Pressure above the set pressure during relief process.

    Back pressure: Outlet pressure of the relief device during pressure knocking. Usually it will be equal to atmospheric pressure.

    Accumulated pressure: Pressure raise above MAWP during relief process.

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    Now lets go into the calculation part.

    Vent Size calculation for a reactor:

    Usually the vent size shall be calculated from the data obtained from ARC(Accelerated rate calorimetry), RSST(Reactive System Screening Tool), ARSST(Advanced Reactive System Screening Tool).

    Below is the data required for calculation:

    1. Rate of temperature raise (T, C/min),

    2. Rate of pressure raise (P, psi/min).

    Here is the formula,

    A =                            ( 3.5 x 10^-3 )                        x  V ( T + P )

                     p [1 + (1.98x10^3)/(p^1.75)]^0.285                Cd


    p - Set pressure of relief device (psig),

    V - Volume of gas / reactant (m3),

    Cd - discharge coefficient = Actual flowrate / theoretical flowrate,

    Now, i'll demonstrate with a case study,

    Details of the reaction from the ARC study in run away case:

    Rate of temperature raise = 0.2 C/min,

    Rate of pressure raise      =  0.05 bar/min = 0.73 psi/min,

    Other details:

    Set pressure of the relief device = 4.0 Kg/cm2 = 56.89 psig,

    Volume of the gas = 30 m3, (This may seems higher, but it is quite regular as the gas densities were very low),

    Tentative discharge coefficient (Lets say) = 0.7,

    Then using above formula,

    A =  ( 3.5 x 10^-3 ) / [49.78 x (1 + (1.98*10^3 / 49.78^1.75)^0.285]    x [30*(0.2+0.73)/0.7]

       = 0.00203 m2.

    Dia, d = ((4 / 3.141) x A)^0.5 = 0.0507 m = 5.07 cms = 1.999 in ~2 in.

    Vent size calculation for a storage vessel:

    This is quite lengthy and requires most of the data from the vendor side,

    the details include the constants like discharge coefficient, back pressure correction factor.

    And other details include the vapour/gas compressibility factor, heat capacity ratio, Mass flowrate.

    Here is the formula for the vent size of a storage vessel,

    A = [ w / C x Kd x P1 x Kb ] x ( T x Z / M )^0.5.

    w - mass flowrate of the solvent vapour,

    C = [ j x g / R ] x [ (2 / ( j + 1 )) ^ (( j+1)/(j-1))],

    j - Heat capacity ratio of vapour or gas,

    R - universal gas constant,

    P1 (psia) - Pmax (psig) + 14.7,

    Kd - discharge coefficient (~ 0.7),

    Kb - back pressure correction factor (~1.0),

    Now, let me demonstrate with a case study by calculating the required vent size for a storage vessel of 60 KL capacity containing toluene,

    Heat capacity ration of the vapour be 1.2, Solvent be methanol - mol wt.: 92 g/mole,

    Vessel dia be 2.5 m, height of the vessel be 3 m (torispherical vessel),

    Wetted surface area of the vessel (Awet) = ( 𝜋 d l ) + ( 1.084 x D^2 ) 

    = 30.33 m2 = 326.5 ft2.

    Total heat absorption, Q = 34500 x (Awet)^0.82 = 3973775.31 BTU / hr.

    Let the rate of vaporization be 20 Kgs/sec [the value is quite lower than the actual, usually for hydrocarbons it will vary between 30-50 Kg/sec as they possess lower latent heats] ,

    P1 = Pmax  + 14.7 = (Pmawp x 1.21) + 14.7 = (4 x 1.21) + 1.013 

    = 5.853 bar abs = 5.853 x 10^5 N/m2.

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    C = [ (1.3 x 1 / 8314) x (2/(1.3+1))^(1.3/0.3)]^0.5 = 7.32*10^-3,

    A = ( 20 / (7.32*10^-3 x 0.7 x 5.853 x 10^5 x 1)) x ((273.15+110.65) x 1 / 92)^0.5

        =  0.0067 x 2.042 = 0.0136 m2.

    Dia, d = ( 4 x 0.0136 / 3.141 )^0.5 = 0.132 m = 13.2 cms = 5.185 in ~5.5 / 6 in.

    The above demonstrated calculation would work great if relevant authenticated safety reports are available. If not, then as an engineer we should be in a position to provide a temporary way forward without giving a major break in production.

    Vent Size Calculation without Safety Data:

    In such cases we should depend on the theoretical material balance,

    I'll give you a case study with a random chemical reaction,

    Yeah, i got a Balz Schiemann reaction, where a Aniline undergoes diazotization with Nitrous acid followed by reaction with fluoroboric acid giving aryl diazonium salts and thermal decomposition of dry aryl diazonium salts gives aryl fluoride and bi-product of the reaction is Nitrogen gas and boron trifluoride.

    Below is the scheme of the reaction:

    C6H5NH2 + HNO2 + HBF4  ----- > C6H5F + N2 + BF3

    Note: Property of Pharma Engineering
    Now lets suppose, we are manufacturing fluorobenzene with a batch size of 100 Kgs.

    Below are the raw materials quantities used in the batch manufacturing:

    Aniline                       -   100 Kgs,
    HNO2                        -   50 Kgs,
    Fluoroboric acid      -    150   Kgs.

    Below is the material balance for the above scheme with the raw material Qty.'s:

    So based on the material balance, it can be said that

    29.80 Kgs of Nitrogen and 72.13 Kgs of Boron trifluoride is liberating.

    As mostly the Diazonium salts are highly unstable at lower temperatures, the preferable temperature of addition would be below 0° C. Hence the first reaction would be formation of Diazo salts, and nitrogen would be liberating in this point and it depends on the addition rate of HNO2, as the Nitrogen formation is attributed by Nitrous acid addition.
    Reaction - 1: Formation of Diazonium salts

    Hence in the first step, Nitrogen would be liberating and the Qty.'s is 29.80 Kgs.

    Converting into volumes, it would be 29.80 / 1.25 = 23.84 m3, which is quite higher.

    and let the addition of nitrous acid took 2 hours, 

    In 1st case, lets consider Nitrogen is liberating uniformly throughout the addition,

    i.e., 23.84 / (2 x 60) = 0.199 m3/min = 199 L /min.

    But while performing the calculations through material balance, we need to consider the worst scenario's, as liberation would go in 10 minutes with a peak liberation of 90%.

    The worst scenario would be 23.84 x 0.9 / 10 = 2.15 m3/min.

    Now to position ourselves in a safe state during addition, we shall make the system ready to knock out 2.15 m3 of nitrogen per minute,

    So, now one more assumption we need to consider is velocity of nitrogen, i.e., based on my experience i'll consider as 10 m/sec.

    So, the relief device(vent) area shall be (A) = (2.15/60) / 10 = 0.0036 m2.

    Dia, d = ( 0.0036 x 4 / 3.141 )^0.5 = 0.0675 m = 6.75 cms = 2.66 in ~3 in.

    Hence for the first reaction we should provide a vent size of 3 inch.

    Reaction - 2: Formation of aryl fluoride

    Now enter the dragon, lets move to next reaction,

    i.e., formation of fluorobenzene, where boron trifluoride is a bi-product and the gas is quite toxic and the same can't be left into atmosphere, it need to be scrubbed.

    If you need a scrubber design, pl refer below link, in this subject post i'll describe only about vent size.

    72.13 Kgs of Boron trifluoride is being liberated,

    Lets follow the same procedure like above,

    Converting it to volume,  it will be 72.13 / 2.76 = 26.13 m3.

    Same 90% of peak load in 10 minutes,

    it will be   26.13 x 0.9 / 10 = 2.35 m3 /min.

    And as this time, i'll be supported by scrubber, hence i'll consider the velocity as 10 m/sec,

    Here you might get a doubt that, "why am i considering velocity as 10 m/sec even having a scrubber advantage?".

    Answering your query, "dear engineers, if you can see the density above, it is 2.76 Kg/m3 which is more than double of nitrogen(1.25 Kg/m3)". Hope you got it.

    Let's get back into calculation, 2.35 m3 of BF3 to be knocked out in one minute.

    Relief device(vent) area shall be (A) = (2.35/60) / 10 = 0.0039 m2.

    Dia, d = ( 4 x 0.0039 / 3.141 ) ^ 0.5 = 0.071 m = 7.065 cms = 2.78 in ~ 3 inches.

    Hence, from the reaction 2 it is observed that a minimum size of 3 inches is required for both reactions.

    And here the lucky fact is both are liberating in different reactions, or else it would be a complicate case to solve.

    Reaction - 3: Knocking out the Unreacted compounds

    And also we need to check the vent sizing during quenching operations, as in the above case, there is an excess reagent used in process i.e., HBF4,

    If water is used for quenching the unreacted HBF4, then the reaction would be as below:

    So here, in this case, lets suppose 100 lts of water used for quenching the unreacted Fluoroboric acid. products will be boric acid (Solid) and hydrogen fluoride (gas).

    And the sane calculation shall be repeated. [I'll leave it and let it be an exercise to the readers].

    That's it......!!!

    Hope everything is understandable to all.......!!!

    Any queries feel free to comment / message....!!!

    Comments are most appreciated.....!!!

    This post is dedicated to my Dearest brother Govindarajan V.

    Finally, I'm expressing my view here, "If you are unable to Knockout a process risk, you'll be knocked out by it".

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    About The Author

    Hi! I am Ajay Kumar Kalva, Currently serving as the CEO of this site, a tech geek by passion, and a chemical process engineer by profession, i'm interested in writing articles regarding technology, hacking and pharma technology.
    Follow Me on Twitter AjaySpectator & Computer Innovations


    1. Thank you so much sir, you are really doing a great work.sir if possible then can you write something on ejector system both water and steam and calculation involved in it's design and selection

    2. A = ( 3.5 x 10^-3 ) / [3.5 x (1 + (1.98*10^3 / 49.78^1.75)^0.285] x [30*(0.2+0.73)/0.7]

      = 0.00185 m2.
      in above calculation, u r taken pressure in PSI & kg/cm2 in some case, i am not understand how it will come.

      1. Dear Nitin,

        Thanks for correcting, modified the calculation, the vent size increased from 1.911 inch to 1.999 inch.

        Best Regards,
        AJAY K

    3. How to interpret heat of reaction

      1. Hii prasad,

        Once go through these posts:

        If any queries feel free to connect with me at

        AJAY K

    4. Thanks for the information. Just wanted to ask where can I find the derivation or explanation of the formula you used above.

    5. Thanks for the info. Can you give me the link for the explanation or derivation of the formula you used above?

    6. Ajay this post is very nice, now i am working on a project size selection of PSV& rupture disc.this post is very helpful for me.

      I have a one quary regarding W you put w value 20 kg/sec.pls elabrate.

      In A = ( 20 / (7.32*10^-3 x 0.7 x 5.853 x 10^5 x 1)) x ((273.15+110.65) x 1 / 92)^0.5

      When Q=3973775.31 BTU / hr.

      1. Dear ,

        Q is the vaporation losses, in the above line i've mentioned it, thats a considered value only.

        Next time pl comment with your good name.

    7. Dear ajay,
      Sorry for that i didn't mention my name next time it will not done.
      I m reffering Crosby engineering handbook
      Where in Pg. No 7.23

      W = Q / Hvap

      As per this book we have to calculate Q also. We can't consider .
      Please brief me.


      1. Are you referring to the vaporation losses ?

    8. If you have any psv /rupture disc sizing
      Calculation sheet please share with me.


      1. Hii Raj,

        Pl share your id at


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    Hi! I am Ajay Kumar Kalva, owner of this site, a tech geek by passion, and a chemical process engineer by profession, i'm interested in writing articles regarding technology, hacking and pharma technology.

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