Estimating Booster Pump Requirement for Vacuum System

Img Credits: A&J Vacuum Services

Good day everyone....!!

Hope everyone is doing good and safe.

Due to my schedules and ongoing trainings, not able to find time for blogging. Hence after a long time writing a post on selection of booster pump capacity for a vacuum system.

Basically vacuum shall be considered as heart of pharmaceutical manufacturing, the reason to say this is most of the pharmaceutical products are thermally sensitive, hence to knock out the volatile matter at reduced temperatures vacuum is required. Some times if we are going to handle products which are of high sensitive then the requirement of vacuum will increase and there is a necessity for us to maintain high vacuum consistently for longer period of times. In such cases we need some booster to maintain high vacuum for prolonged periods.

In my career i've worked with a booster of capacity 5000 m3/hr to attain fine vacuum. The requirement will differ from operation to operation i.e., if its distillation the requirement can be low and if its drying the requirement may go high. So based on the scenario, we need to install a booster to vacuum pump. 

Basically, the vacuum pump to which we need to install booster shall be referred as backup pump. The capacity of the booster depends on the non-condensing vapour load and the air seepage that might result through the vessel MOC. The efficiency of the booster majorly depends on the backup pump working. Boosters can be aligned even with oil ring vacuum pumps and some times with ejectors.

Today, lets start our calculation on estimating the booster capacity for a specific operation.

Before going into the topic, lets check out some basic things which shall make things easier.

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NC Load: Non - condensable load

Carry over load: The load attributed by the gases on the vacuum system which are not condensable at process condenser.

Ultimate vacuum: The maximum attainable vacuum on the system, most of the times these are going to be specified in absolute vacuum.

Pump down time: The time taken by a pump to reach the desired vacuum level.

For performing the calculation, we need some data,

So here i'm assuming some data.

Inlet vapour temperature (Tv) = 50℃ = 323°K

Vacuum required on system (V) = 2 torr

Non - condensable vapour load from process (Nc) = 10 Kg/hr

Molecular weight of the vapour (Mc): 50 g/mole

Carry over air load (Ca) = 4 Kg/hr

Molecular weight of air (Ma) = 28.84 g/mole

Capacity of booster required = ((R x Tv) / (1.33 x V)) x ((Nc / Mc) + (Ca / Ma))

= ((83.14 x  323) / (1.33 x 2)) x ((10 / 50) + (4/28.84)) = 3419 m3/hr.

Now, lets see the basis for equation and how it is derived,

To build some complex equation, most of the times I'll prefer the basic equations, in this case I've preferred universal gas law PV = nRT

V is the volumetric flowrate that we need to estimate,

So, V = n R T / P,

V = (R x T / P) x n

V = (R x T / P) x (Moles of non - condensable gases + Moles of air seepage through MOC)

Upon simplifying the above equation we got the below,

V = (83.14 x T / 1.33 x P) x ((Wt. of gases/Mol. Wt. of gases) + (Wt. of air / Mol. Wt. of air))

Units:

V in m3/hr

T in degree kelvin,

P or V in torr (Absolute vacuum),

R = 83.14 cm3-bar/gmole. K,

Wt. in Kgs; Mol. Wt. in g/mole.

Usually i like leaving some puzzles in my post (for example the constant 1.33 in this post) to which most of the readers were found commenting, please don't mind about that. To solve the riddle, perform dimensionless analysis 😀

Note: You can further customize the above equation by adding other volatile content in terms of moles (if required).

Now, we need to extend our calculation for estimating the pump down time (t),

t = k x 2.3 x (V / S) x log(Pi/Pf)

k - Empirical factor which depends on the pressure range

t - Effective time in Seconds

V - Volume of system in liters

Pi - Initial pressure

Pf - Final pressure

S - Pumping speed in LPS

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k values for different pressure ranges are as follows,

k                 Pressure Range

1.1                    760 - 1

1.5                  1   -  0.01

4                    0.1 - 0.001

System volume is 15 KL, 

t = 1.1 x 2.3 x (15 x 1000 / 3419 x 0.2778) x log (760/2) = 103.07 seconds = 1.72 minutes.

That's it ....!!!

If any queries, feel free to write a mail or comment

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

QuizMaker

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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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