Vent Sizing for Pressure Vessels / Equipments

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....!!



About The Author

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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.
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