Sunday, 26 June 2022

Fire Load Calculation & Estimating Fire Extinguishers Requirement

Hello Everyone, Hope everything is fine...!!

Recently received a request from a professional to write about the fire load calculation, Hence writing this post upto my knowledge. 

Basically, fire-load is the most required parameter while identifying the fire risk relevant to any building / structure.

It will help us to understand the heat liberation while a building / a material is completely burnt.

There shall be certain grading available according to the fire hazards involved based on the fire load. Currently i'm not aware of those gradings, i'll check in my network and would update here.

Fire load can be calculated based on the below formula:

(here i'll define the fire load for a chemical combustion)

Fire-load = Quantity of chemical x Calorific Value / Shop floor area

So, now the next query is What is Calorific value ?

If you can remember your +1 / +2 days, Calorific value shall be defined as Energy contained in a specific material, which can be measured by the heat generated / produced by complete combustion of material.

So, taking this as reference for measuring calorific value of a specific material, please don't burn any pharmaceutical product that you are handling, here we are not taking any responsibility for such events 😄

If you want to estimate the calorific value of a specific product, kindly prefer some well equipped calorimeters with some knowledgeable personnel.

Classification of fire-load:
There shall be three classifications for fire-load and as usually they are low, medium and high.

So, in order to identify the fire load classification, there shall be some classification i.e.,

Low Fire load: < 2, 75, 000 KCal/Sq.mt
Medium Fire load: 2, 75, 000 - 5, 50, 000 KCal/Sq.mt
High Fire load: > 5, 50, 000 KCal/Sq.mt

So, the next thing is taking a practical case and estimating the fire load and required fire extinguishers count.

Just before that lets look into types of fire. Most of you guys know the types of fire and suitable extinguishers, lets have a simple classification below.

Class A fire:
Involves organic combustible materials

Suitable Extinguishers:
Water, Foam, Dry powders & halocarbons

Class B fire:
Involves flammable liquids, liquefiable solids

Suitable Extinguishers:
Foam, Dry powders, Carbon dioxide

Class C fire:
Involves flammable gases, liquified gases

Suitable Extinguishers:
Dry powder, Carbon dioxide

Class D fire:
Involves metal fires (here metal indicates magnesium , potassium, sodium, zinc etc.)

Suitable Extinguishers:
Dry powder (these are metal specific)

Note: Please remember wrong selection of extinguishers will not help you, but they will chase you.

There shall be two types of hazards, which can be classified into two categories i.e., Light & Heavy (Based on IS 2190)

If its Light hazard, mitigation / requirement is one 9 / 10 Liter water expelling/purging extinguisher for every 600 Sq.mt of shop floor  (Class A fire)

For heavy hazard (usually involves burning of solvents / gases), requirement is two 9 / 10 Liter chemical / mechanical foam or 20 Kg dry powder (can be multiple) for every 600 Sq.mt of shop floor (Class B fire)


Lets get into calculation part, in this example i'll consider an intermediate and the quantity available is 600 Kg with a calorific value of 30, 000 KJ/Kg, the area of shop floor is ~170 Sq.mt.

Fire-load = 600 x 30, 000 / 170 = 1, 05, 882.35 KJ/Sq.mt = 25, 306.49 KCal/Sq.mt.

Since the product is an organic intermediate, i'll define it as Class A and its a light hazard

No of extinguishers required = 25306.49 / 600 = 42.17 ~43.

Hope the calculation helps. 

If anything to be added or any queries, please feel free to comment...!!

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

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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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Sunday, 12 June 2022

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.


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

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 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
at 7 comments: