Hello Everyone...........!!

Today here i gonna explain you about vent sizing of pressure vessels and process equipments based on the way of usage. Simply take the case of a Batch reactor, what will happen if there is no vent or pressure relief valve during a reaction........??

The result may vary with the size of the reactor, Simply the presence of a Relief valve will relief the over pressure that was developed inside and protects the reactor, Usually the reactor will be having a pressure resistance which is known as Design pressure, if the reactor is felt with more than the design pressure, then there may be a chance for explosion, and i said there may be a chance of explosion because after manufacturing of reactor there will be some recommended tests that were to be done like pressure test, spark test[if its a GLR], etc. While performing a pressure test, the vendor will cross the limit of design pressure and gives a specification called Test Pressure, which means the pressure upto which the test was carried out.

So, now i'll tell you when there is a need for Vent/Relief valve:

Vapour / Gas removal rate < = Vapour / Gas generation rate,

If the above condition is satisfied, then there requires a Relief valve / Vent for the system.

The major factors that will decide the Vent size were:

- Maximum Vapour/Gas Generation rate,
- Type of fluid inside the container, whether its a gas or liquid.

In case of Maximum generation rate, If we need to vent out Vapour / Gas that was generated, then simply calculate the size from below equation:

V = A x S = 0.785 x D x D x S,

here, V = Volumetric flowrate in Cu.m / hr,

D is Diameter of Vent,

A is Cross sectional area of the vent neck,

S is the velocity with which the vapour / gas will escape,

And coming to second case, if the Container / Vessel is containing any gas, then there wont be any problem because if its a solvent then there is a chance that solvent may be transform to vapours and create some vapour pressure, but as it is a Gas then whatever the pressure that is accumulated, that will be kept off like that.

One more important term, named as WCM [ Worst Credible Maloperation ], Which means the highest pressure developed due to the maloperation responsible for the generation of vapour / Gas.

The calculation should obey the criteria, i.e., pressure that was developed during the WCM should safely pass through the vent provided, or the Relief valve should respond to the WCM pressure with a minimal response time

The approach to Relief sizing depends on 2 cases:

- Fire case,
- Reaction case

For a fire,

Vapour generation rate = Heat from fire / Latent heat of vaporization

For a chemical reaction,

Vapour generation rate = Reaction rate x Reaction Enthalpy / Latent heat of vaporization.

So after getting the vapour generation rate, you can go with above mentioned equation, V = A x S,

if there is any problem while calculation, then better go with a direct equation which was developed & Proposed by Leung's Long form eqn.

A = [ m x q ] / G [ { (VxT/m)+(dP/dT)}^0.5 + {Cp x dT}^0.5 ] ^2

m - Initial mass in vessel (Kg),

q - Heat evolution rate per unit mass in vessel(Watt/Kg),

G - Vent flow capacity per unit area at set pressure(Kg/Sq.m),

V - Vessel volume(Cu.m),

T - Vessel temperature(K),

dP/dT - Rate of change of vapour pressure with temperature,Cp - Specific heat ( J/Kg.K),

dT - difference of temperature between Maximum allowable pressure & Set pressure ( °K),

As per the Equilibrium rate model, the Value of G [ vent flow capacity] can be calculated as below,

G = (dP/dT) x SQRT(T/Cp) = hfg / [ Vfg x SQRT( Cp x T) ]

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

Now, here i'll demonstrate the above mentioned formula with a calculation.

Consider a 10 KL reactor, having MOC SS316 with a design pressure of 3.2 Bar, and maximum allowable pressure as 4.2 Bar, i.e., 30% excess to design pressure

| | | Average |

Pressure Bar | 3.2 | 4.16 | |

Bubble point temperature °C | 110 | 120.5 | 115.3 |

Heat release rate ( watt/Kg) | 1150 | 1660 | 1405 |

Liquid density ( Kg/Cu.m) | 847 | 835 | 841 |

Vapor density (Kg/Cu.m) | 3.75 | 4.62 | 4.19 |

Latent heat (KJ/Kg) | 674.9 | 663.0 | 668.95 |

Liquid specific heat (KJ/Kg.K) | 1.96 | 1.96 | 1.96 |

dP/dT | 8300 | 9500 | |

Vfg (Cu.m/Kg) | 0.2655 | 0.2153 | 0.2404 |

Using equation developed from Equilibrium rate model,

At 3.2 Bar pressure, G = 0.5 x (dP/dT) x SQRT(T/C) = 0.5 * 8300 * SQRT( 383/1960)

= 2385 Kg/Sq.m S.

At 4.16 Bar Pressure, G = 0.5 * 9500 * SQRT(393.15/1960) = 2128.73 Kg/Sq.m S

The average value of 2385 & 2128.73 for G gives 2256.5 Kg/Sq.m S.

Now, we need to calculate the venting area from Leung's Long form eqn.

dP/dT = ( 4.16 - 3.2 ) x 10^5 / ( 120.5 - 110 ) = 9143 N/Sq.m K

A = [ 1500 * 1405 ] / 2256.5 [ { 10 * 288.5 * 9143/1500 }^0.5 +{1960 * 10.5 }^0.5] ^2

= 0.0122 Sq.m

D = SQRT(0.0122/0.785) = 0.125 m = 5" (inch)

I think, now you got an exact idea of solving the data for getting the venting area for Pressure vessel,

Any queries please feel free to ping me...........................!!

Comments are Most appreciated....!!

**A**bout The Author

Good information

ReplyDeleteGood information

ReplyDeleteThank you Naveen, visit again for more

ReplyDeleteNICE..

ReplyDeleteSir, nice information but we should need how to calculate the vapour nozzel of the reactor.

ReplyDeleteSir,good information sharing but i need how to calculate the reactor vapour nozzel with calculation

ReplyDeletehere is what you need, follow the link

Deletehttp://pharmacalc.blogspot.com/2016/04/calculate-required-vapour-column-size.html

Nice info Sir

ReplyDeleteThanks mate

DeleteRegards,

AJAY K

Sir can you please suggest me how to calculate specific heat capacity of a solvent in a mixture.

ReplyDeleteSpecific heat of binary mixture = ( Specific heat of component 1 x Mole fraction of component 1 ) + ( Specific heat of component 2 x Mole fraction of component 2 ).

DeleteCp(mixture) = ∑[Cp(i) x Mi]

Regards,

AJAY K

Where from the data of 2nd is coming?

ReplyDeleteAnd how we would get the data of heat release rate?

Where from the data of 2nd column is coming?

ReplyDeleteThose are theoretical results derived from spherical chamber explosion test, used in determination of Pmax, Kst and ST class.

DeleteRegards,

AJAY K

Dear Pratim,

ReplyDeleteWe need to get it from RC1 data.

Regards,

AJAY K

Pmax is the max pressure developed...But sir what is Kst and ST class??

ReplyDeleteKst value determines the behavior of the dust cloud which is formed in a confined space.

Deleteplease follow the link for a brief explanation about Kst: http://www.adinex.be/en/dust-explosions/explosion-indices.htm

The test stratergy of determining the Pmax & Kst can be found here: https://www.fauske.com/blog/bid/316347/kst-and-pmax-tests-for-combustible-dust-who-or-what-are-they

Regards,

AJAY K

Dear Sir...

ReplyDeleteIn my plant in certain reaction ,reaction mass to be centrifuged get chocked in the reactor so ro remove rm pocking is done by ss stick,if we develop nitrogen point from the bottom of that reactor ..Would it be an alternative solution for that purging nitrogen from the bottom valve of the reactor as the reactor two valve so in between two valve can nitrogen pressure be given remaining another valve keep closed?????Please reply

If possible add baby blades/teflon extension blades to the existing agitator which will reduce the minimum stirring volume, so that under stirring the reaction mass wont enter the bottom.

DeleteHope this works.

Hai ajay

ReplyDeleteduring the pH adjustment for product isolation , foam formation which is further difficult for filteration in centrifuge. what ate the ways to control the foam formation . i reduce the RPM of agitator during the pH adjustment. but not controlled

If its an anchor agitator, then get foam cutters installed, while feeding maintain pressure above 0.5 Kg/cm2,

DeleteNext time kindly comment with your good name plz,

Regards,

AJAY K