Overall Heat Transfer Co-efficient Calulation based on Trials

Great day Readers.....!!!
This day i wanna demonstrate the way of determining the value of Overall Heat Transfer Co-efficient based on Practical Heating and Cooling trials, Which will return us the perfect value for an heat transfer equipment.

Previously i've demonstrated the Calculation of Overall Heat Transfer Co-efficient based on Properties, which seems to be somewhat difficult when you donno the properties of the Jacket fluids exactly, but this time i'll make it somewhat easy for you as this includes a pure practical approach.

For many Engineers the Overall Heat Transfer Co- efficient calculation remains as a nightmare, because the calculation will involves many terms and conditions and also many formulas, but here i'd like to use minimum formulae, which were well known by all engineer's.

Those well known formulae were,

1. M x Cp x dT - which is used to calculate the heat transfer to a fluid or a matter from source or heat transfer from fluid or a matter to sink.

2. U x A x LMTD - Which defines the overall heat transfer in between to matter of state separated by a material boundary.

May be right now, you came to know how i can derive a relation for calculating the Overall Heat Transfer Co-efficient here, deriving a relation is much more simpler, but incorporating the values i the formulae for calculation is some what sensitive stage in this.

First of all you need to know, how to take a Perfect trial .....!!

Taking trials is so easy, as we simply use water as mass in a batch heat exchanger, i'll demonstrate a simple case study now,
Consider i need to calculate the overall heat transfer coefficient for water as vessel mass and steam/hot water as jacket fluid, let the reactor volume be 5KL and then charge 3KL water into it, and start heating it to 70°C from RT [30°C], then just note down the time taken for heating the 3KL mass, use hot water at 90°C for heating as utility, and it has taken 2 hrs[say].

then try to incorporate the figures in the following relation,

U = M x Cp x dT / A x LMTD,

Here you need to know what is the heat transfer area of a 5KL reactor, then it is too simple to calculate it, follow this post

How To Find Reactor Heat Transfer Area ? 

so, now Heat Transfer Area of the 5KL reactor is 12.55Sq.m [L/D ration is considered as 1.15],

IF JACKET UTILITY IS HOT WATER

LMTD = (88-30)-(90-75)/Ln(88-30)/(90-75) = 31.85°C
 
U = 3000*1*1*(75-30)/(12.55* 2*  31.85) = 168.86KCal/hr.Sq.m.°C.

So, this is for water Vs Hot water at 90°C,

IF JACKET UTILITY IS STEAM

Now i'll Calculate for water vs Steam [ 110°C],

For the both cases only LMTD will vary, everything remains the same.

LMTD = (110-30) - (110-75)/Ln(110-30)/(110-75) =  54.87°C

U = 3000*1*1*(75-30)/(12.55*2*54.87) = 98.02KCal/Sq.m.hr.°C.

That's it, Here you need to know that U value vary depending on the Jacket utility temperature [T] , and the Temperature differences [dT], Time taken for the Heat Transfer [ t ], Heat Transfer Area [ A ],

So, my request to all the readers is that don't blindly go with thumb, if you do so the consequences will be severe to face.

 Some Typical U values, which can meet practical cases by some deviation, shown below,

Thanks for spending your time reading all these, Say cheers if you understood the above,
Any queries feel free to ask us, We will respond you,

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

 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

Outlines for Designing a Chiller Plant

Hello Viewers.....!! Am Back after many days, and today i've selected a topic of Designing a Chiller Plant, This Post may include many sub topics, which i think have covered so far, 

A chiller unit is nothing but just a cooling unit, which can be used to maintain some high temperature variation from room temperatures, Basically a Chiller Plant works on the Principle of Carnot Cycle, and Heat Transfer. Many of you think that the chiller plant as a big mystery to solve, but trust me by the end of this post, i'll make you feel better when you again think about Chiller Plant.

As mentioned this works on Carnot Cycle, lets have a look 

Now compare the 4 major units shown in the above pic with those in the Chiller unit,

The Piston can be compared to a Compressor, Condenser will remain the same as Evaporative Condenser, Pump can be compared to as Expansion Valves, and finally the boiler to be Chiller.

A typical Chiller unit will seems like, 

Credits: Pharmacalc.blogspot.com

So, now i'll Explain you the major units of chiller plants and also how they have impact on the cooling of the utility,

1. Compressor: A Compressor is the major part of the chiller plant as the process will start from here, Compressor will be used to compress the chilling agent, usually a chilling agent will be Ammonia, as ammonia will have somewhat higher heat capacity. Immediately you may have a question that when a compressor was made running for many hours, wont it have any wear-tear problem, and for avoiding this problem an oil is circulated to cool the overheated compressor screw, Usually the Ammonia Pressure that is developed in a compressor will be very high, and of the the orders 12-15 Kg/cm2, And for this we need to select pipes having very high Design Pressure, the Red line represnt the high pressure Ammonia.

2.  Evaporative Condenser: This Evaporative Condenser is usually used to cool the high Pressure Ammonia that is forwarded from the compressor, usually water is used as the cooling in the cooling towers as coolant for cooling ammonia from gaseous state to liquid state, as shown in the figure the Water is sprayed over the tube bundle [Not Shown in pic] in which the Ammonia flows, Usually the water after getting into contact with the tube bundle will get vaporized and these vapors are immediately sucked off by the blower, and the water which is not vaporized get collected in the below bed, the water which is collected get re-circulated by the pump, and based on the water level that is being vaporized, a makeup water quantity is added in to the collection tub. 

Makeup water = Evaporation + Blowdown + Windage + Drift

These Windage and Drift will be negligible in large scale,

Makeup water * Cds1 = ( Evaporation + Blowdown ) * Cds2.

Cds1 = Dissolved solids concentration in Makeupwater,

Cds2 = Dissolved solids concentration in Evaporation + Blowdown. 

Consider the Ammonia quantity to be cooled is 800 L per 4 hrs from 20°C to 0°C, and the Heat removed per hr is 800 W, So now i gonna calculate the makeup water here,

Qamm = 0.800 * 0.73 *  0.52 * (20-0) = 6.0637 KCal

Makeup water = Qamm / ( 1 * 45 ) = 6.0637/45 = 0.1347 L 

[ ** here Water is considered to be vaporized at 45°C due to the blower effect as the blower Creates some suction ]

Cooled Ammonia is collected in the Receiver, the Receiver Capacity should be considered as 40% excess to the volume of ammonia processed. 

Basically the ammonia shouldn't contain any oil, but in practical cases the ammonia will carry oil also with it, so after collecting it in receiver, the oil should be removed/separated, The Liquid ammonia is then taken back to a shell and tube heat exchanger which is used to cool the oil from compressor, thos job is done inside the Oil Cooler chamber as given in the above pic, the cooled oil will be transferred again to the compressor.

So, Now lets consider the compressor is  creating a pressure of 15 Kg/cm2 [ 20°C], and in this process 70 L compressor oil [rho = 1300 Kg/Cu.m, heat capacity = 0.8 KCal/kg.°K] is getting heated upto 70°C, and i'm having ammonia cooled to 0°C, so now i need to cool the oil back to 30°C to reprocess it, then we need to calculate the Heat Transfer area of the Required heat exchanger, shown below,

Qoil = 700*1.3*0.8*(70-30) = 29120 KCal,

A = Qoil/U*LMTD,

LMTD = ((70-0)-(30-20))/Ln((70-0)/(30-20)) = 30.8°C

U = 350 KCal/sq.m. hr. °C,

A = 29120 / (350*30.8) = 3 Sq.m

So my required Heat exchanger capacity is 3 Sq.m,

3. Economizer: An Economizer is something like a vessel, but actually its not a vessel, but which helps to stabilize the flow of ammonia towards the expansion valves, and also reduce the load on the compressors, Cold, low-pressure chiller vapor is used to subcool the saturated liquid refrigerant. This decreases the refrigerant circulation rate, and may reduce compressor power. 

Credits: Jmcampbell.com

Let’s consider removing 1.0×10⁷kJ/h which is equal to 2778 kW (9.479 MMbtu/hr) from a process gas at -35°C (-31°F) and rejecting it to the environment by the condenser at a condensing  temperature of 35°C (95°F). Assuming 5 kPa (0.7 psi) pressure drop in the chiller and 5 kPa in the suction line pressure drop, the compressor suction pressure is 132.4 kPa (19.1 psi). The condenser pressure drop plus the pressure drop in the line from the compressor discharge to the condenser was assumed to be 50 kPa (7.3 psi); therefore, compressor discharge pressure is 1270 kPaa (184.2 psia). The compressor discharge temperature is 66°C (150.8°F). At these conditions, the condenser duty is 4434 kW (15.13 MMbtu/hr). Pure propane is used as the working fluid.

4.  Expansion Valves : The Expansion Valves plays a major role on the degree of cooling in the chiller as the extent of opening of these expansion valves defines the amount/flowrate of chiling agent into the chiller, so as the flowrate into chiller increases the temperature of the process fluid out will be reduced, Usually these are not regularly operated valves, they were kept consistent.

5.  Chiller : This is just a Shell and Tube Heat Exchanger used to Cool the Process fluids that were incoming with the available chilling agent, usually the process fluid will enters the tube side, and the chilling agent will enters shell side, the required capacity of the chiller can be calculated as follows,

Let the flowrate of the water inside is 50 cu.m /hr, and the overall heat transfer coefficient be 450KCal/hr.Sq.m.°C, LMTD be 50°C, and the incoming water to be cooled from 30°C to 5°C, then

Qwater = 50 * 1000 * 1* 25 = 1250000KCal

Req. Heat Transfer Area, A = Qwater / U * LMTD = 1250000 / (450*50) = 55.6 Sq.m

 6.  Surge Tank : Surge tank is like a storage tank which allows to store the ammonia vapors that were produced after heat transfer in the chiller with the process fluid, the size of the surge tank should be 30% less when compared to that off ammonia collection receiver from the evaporative condenser, the residence time of the ammonia vapors in the surge tank depends on the Compressor suction pressure.

So, Now we came to an end and i'll discuss you the Elements that effect the Cooling Rate of the Process fluids in a chiller,

1. Expansion Valves Opening - if the opening is high, then the flowrate will be high into chiller, ultimately leading for high temperature gain from process fluids,

2. Compressor Suction Pressure - if the suction pressure is low then the chilling agent will have some high retention time in chiller which leads for high heat transfer rate.

So, Now if you understand the principle of working of Chiller Plant just say Cheers....!!

If any queries, then feel free to ask us, Comments are most appreciated.........................!!!

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

Pressure and its Altitude Dependence

Good Day Viewers...!!

Back with a post, Today i gonna give you a brief description on pressure, Everyone knows about pressure, but only an Engineer can know what an Excess Pressure can do, and what can lack of pressure can result in....!!!

And also the beginner knows the types of pressure, but in an industry a pressure can do many unremarkable things, and also there is one hidden logic, which is unknown by many that pressure will vary from location to location, in Technical words, Pressure will vary based on Altitude. And i'll explain you how it will vary and what effect it can have on pumping and some industrial operations, even it can severly influence the feasibility studies of an industrial siting. 

Here are Some Typical Interview Queries for an Engineer which will tend to creating some fun in those who donno this logic,

Query 1: Same pumping setup is being provided at visakhapatnam and hyderabad to pump an identical height of 20 m, where the pumping will be completed first ? 

Query 2: I've a solvent recovery process plant in Ahmedabad and Bangalore both of them will be like twins, and there wont be any difference in between them, not even a flange also, so where can i get more recovery yield ?

When you hear these queries you will be astonished initially but there lies a serious logic behind those queries, as the pressure will vary as per altitude.

So, Before going into the topic, we should be able to differentiate in between types of pressure, the below figure will give clear some of your confusion right now.

 There will be three types of pressure basically, which are primarily known,

1. Atmospheric Pressure,
2. Absolute Pressure,
3. Guage Pressure,

Also we can classify pressure based on direction,

1. Absolute Pressure - Positive direction pressure,
2. Suction Pressure   -  Negative Direction Pressure.

Simple example of these pressure developing equipments were House hold Ceiling fans which will provide pressure while it is rotated in clockwise and Suction while rotated in anti-clockwise direction.

Absolute Pressure: This is the positive pressure which is developed from 0 mm to positive infinite, will usually hold the units like Bar, Atm, mm, Torr, 

Atmospheric Pressure: This will certainly hold some value 1 atm, which equals 760 torr, at 273.15°K. Atmospheric pressure will vary according to Temperature and altitude above Sea level.

Guage Pressure: This is what we use to denote the readings in an industry, generally will be represented with Bar's and Kg/Cm2 

Guage Pressure can be calculated as, Pg = Pab - Pat = Pab -1.

Here, Pg- Guage Pressure, Pab -Absolute pressure, Pat - Atmospheric Pressure.

So, now you are familiar with the types of pressures, and now i'll reveal the procedure to calculate the pressure dependence over altitude above sea level.

For that we need to know the Altitude above sea level for those locations, which is denoted as T.




The Percentage loss in pressure will be, dP = ( 1 - ( 2.25577 x ( 10^-5 ) x T ) ) ^ 5.25588,

So the available pressure in that location would be, P =  760 x dP, in Torr.

Ahmeadabad is located 53 m above sea level, so the dP will be 

= ( 1-(2.25577 x (10^-5) x 53)) ^ 5.25588
= (1-(0.0011955))^5.25588
= (0.9988)^5.25588
= 0.99373

So, the pressure that is available would be, P = 760 x 0.99373 = 755.23 Torr.

So, Now i think you got the logic how pressure varies based on the altitude, So whatever pressure you will be getting in that location will be your atmospheric pressure, and the negative value of that will be your available vacuum.

And now i strongly believe you may have one more doubt that i'm working in a hyderabad industry and my vaccum gauge is showing me 750 mm Hg of Vacuum, how??

And for that my answer will be some what funny, but you need to accept it, you may have seen Dinosaurs in Jurrasic Park movie, and whenever those animals appears in front of your eyes, immediately one word will hit your brain, and that is GRAPHICS, so for your Question your answer will be CALIBRATION.

The persons who will be doing the calibrations will consider the maximum available pressure in that location as full pressure, i.e., 760 mm Hg, and then they will calibrate.

So, if you understand this logic say Cheers.....!!  Or else ...!!

Any queries feel free to ask...!! Always happie to help....!!!

Whatever, Stay Tuned for more!! Share this on your timelines......!!!

Like Us on FaceBook !!


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

Pumps in Series Vs Pumps in Parallel

Good Day Viewers............!!

Back after a gap, Recently i got a query asking me for selecting a better way to choose a pumping network, Whether to select Pumps in series or Pumps in parallel for the secondary pumps in chiller plants to pump into production blocks.

And First of all i'm very happy to share my knowledge with you, and the second thing is its very great experience to receive such mail asking me, but as per regular explanation i'll go on explaining the basics first, and then i'll go deep explaining all those hidden tricks of selection a pump network.

What is a Pump?

A Pump is something which uses the mechanical energy produced by a motor to convert into kinetic energy, simply a pump is used to create some pressure in direction opposite to gravitational force in a system.

How to Select a Pump Capacity?

For this query Read this post  How to Select a Pump, Motor??

What are Pumps in parallel?

Pumps in parallel is the way of connecting all the discharge ends of pump to one End, for this the coupling of the pumps must be perfect, otherwise this wont work efficiently.

Recommended for you:

Overall Heat Transfer Co-Efficient Calculation 

Agitation Scaleup 

What are Pumps in Series?

Pumps in series is the way of connecting the Discharge end of first pump is connected to the Suction End of second pump.

So, Now you got the basics required and i'll explain you deeply what will happen with these arrangements.

Pumps in Series

The Above Figure shows you how the pumps in series are arranged, and many of the project engineers will surely get confused over the difference between the pumps in series and pumps in parallel, but while designing a chiller plant if such confusion arises then the result will be unpredictable. As the major part of the chilling plant lies in the pumping of chilled water or brine to production blocks.

Also Read:

Industrial Distillation Column design steps  

What is Steam under Vacuum? 

When pumps are connected in series then the Head of those both pumps can be added, that means if you have two pumps P1 and P2 having flowrates as Q1 and Q2, Head as H1 and H2, then the Total possible head from these arrangement can be H = H1 + H2.

In the above Case study, there were two pumps operating at 200 Cu.m/Hr, having different heads as 100 m and 200 m, then the overall head that would be developed by the pumps arrangement is series would be sum of 100m and 200m, i.e., 300m.

And the Case would something interesting if the pumps were of different, let suppose if the pumps were having some

Pumps in Parallel

Pumps while operating in series would develop other properties irrespective of heads, they will develop some excess flowrate when compared to them individually, Each pump develops the same head H at the corresponding capacity. Thus, the first pump at capacity Q₁ develops the same head H as the second pump at capacity Q₂. This commonality of head across parallel pumps is the most important feature of pumps installed in parallel. If the pump heads are not matched, pumps in parallel will not function properly.

From this above figure, it will be clear that if the pumps are arranged in parallel, then the flowrate will be added as resultant, and the points were indicated representing 3 cases.

Point 1 : Pump 1 operating with 100 Cu.m/Hr,
Point 2 : Pump 2 operating combinedly at 200 Cu.m/Hr
Point 3 : Head and Flow for each Pump equals the point 3 

So, Now you can decide what type of selection you need, still any queries, i'm happy to help you,

That's it........!!!  Cheer's !!

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

You May Also Like:

How to Calculate Energy of steam?  

What does Tonnes of Refrigeration means? 

How to Handle an Out Of Specification[OOS] 

Determine the Power Required for an Operation?  

How to Calculate required Vapour Column Size?

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

Overall Heat Transfer Co-Efficient Calculation

Hello guys........Hope all are enjoying the start of these weekend !!

Today i wanna add a full stop for many peoples fantasies of 'U' value calculation, the most common line that i heard from many engineers was Consider U as 150 for Reactors and 450 for Condensers, in return if you ask them what are the units for those 150 and 450, many of them will take some strategic time out as in IPL.

An Engineer is the one who holds the ability to deal any of the situation with a smart work, but not the one who follow the Thumb blindly, i'm not criticizing the Thumb, as many cases i too prefer the Thumb rules in practical cases but my intention is An Engineer should be able to derive the U value if needed.

 And for that we need to know some basics, before going into topic.

What are Heat Transfer Co-Efficient?

Heat transfer coefficient is a quantitative characteristic of convective heat transfer between a fluid medium (a fluid) and the surface (wall) flowed over by the fluid.

Also Read:

How to Design a Condenser?  

How to Select a Pump , Motor, Line Sizings ?

What is Overall Heat Transfer Co-efficient?

Overall Heat Transfer Co-efficient is the combination of the possible Individual Heat Transfer Co-efficient, which are due to the following Heat Transfer modes,
1. Rate of convective heat transfer from Fluid to Medium Wall, and 
2. Conductive Heat Transfer from Medium Walls one end to other end,
3. Convective Heat Transfer from Medium wall to inside Fluid. 

Credits: TLV.com

What is Thermal Conductivity?

Thermal Conductivity of a material characterizes the Rate of Heat Transfer of a material, whether the heat is flowing in or out.

The thermal conductivity - k - for some typical materials (varies with temperature)

• Polypropylene PP : 0.1 - 0.22 W/mK
• Stainless steel : 16 - 24 W/mK
• Aluminum : 205 - 250 W/mK
• Glass lining :  1.2 W/mK

What are the units of Overall Heat Transfer Co-efficient and How are they derived? 

British Units for U are Btu/(hr-ft²°F) , SI units for U are Watts/Sq.m .°K or KCal/Sq.m .sec.°K

They are basically derived from the formula Q = U x A x LMTD,

where Q holds the units Watts , A - Sq.m, LMTD - °K or °C.

So, that's how U units were derived.

Credits: engineersedge.com

What will be the Trend of Convective Heat Transfer Co- efficients?

The Convective Heat Transfer Trend will be as follows:

Fluid	Convective heat transfer coefficient (h)
Water	about 1000 W/(m2°C)
Hot Water	1000 – 6000 W/(m2)°C
Steam	6000 – 15000 W/(m2°C)

The convection heat transfer coefficient - h - depends on

1. The type of fluid - gas or liquid,
2. The flow properties such as velocity,
3. Other flow and temperature dependent properties.

Convective heat transfer coefficient for some common fluids:

• Air - 10 to 100 W/m²K
• Water - 500 to 10 000 W/m²K

 So, I think right now you acquired some courage to drive the fantasy of  'U' value calculation from your mind. And, i'll start the main show now.

Recommended Read:

How To Select a Condenser? 

Determine the Power Required for an Operation?  

Industrial Distillation Column Design Steps

U value Calculation:

For calculation of U value we need some basic data regarding the Jacket fluid properties, Vessel Mass properties, Jacket properties, Equipment Physical data  etc.

General Data includes: Vessel Dia, Vessel MOC, Vessel wall thickness.

For Calculating Overall Heat Transfer Coefficient we need Individual Heat Transfer Co-efficients used in the following co-relation.

1 / U = ( 1 / hj ) + ( Fj ) + ( X/K ) + Fv + ( 1/hi )

The hj, hi need to be calculated primarily,

Jacket Side Heat Transfer Co-efficient Calculation:

For Calculation of  hj , we need to know Pipe coil ID, Flowrate in pipe, Fouling factor Fj, Fluid properties[Density, Viscosity, Specific heat, Conductivity],

With an example i'll illustrate here,

Let pipe ID is 2", Flowrate(Q) is 10 Cu.m/hr, Fouling factor(Fj) = 0.0002 h.m².°C/Kcal,

Density (rho) 1000 Kg/Cu.m, Viscosity(μ) 0.01 cP, Specific heat (Cp) 0.9 KCal/Kg.°C, Conductivity(k) 0.2 KCal/m.h.°C,
Viscosity @ wall is (μ_w) = 0.01 cP

Velocity, V = Q/A = 2.7410 m/sec
Reynold's number, Nre = D x V x rho / μ =17638677,
Prandtl's number, Npr = Cp x μ _/ k = 0.2093

using the below co-relation ,

hj x D / K = 0.023 x ( Nre ^ 0.8 ) x ( Npr ^ 0.33 ) x (( μ_/μ_w)^0.14)

hj = 27202.81 Kcal/h.m².°C

Also Read:

Agitation Scale up  

How to find Reactor Heat Transfer Area? 

Vessel Side Heat Transfer Co-efficient Calculation:

For this calculation we need to know the inside fluid properties, which includes Density, Viscosity, Conductivity, Specific Heat, Viscosity @ wall.

Within an example i'll illustrate here,

Agitator RPM 36, Agitator Dia 36", Density (rho) 900 Kg/Cu.m, Viscosity(μ) 2cP, Specific Heat(Cp) 1KCal/Kg.°C, Conductivity (k) 0.5 KCal/m.h.°C, Viscosity @wall (μ_w) 1 cP, Fouling factor(Fv) 0.0002 h.m².°C/Kcal,

Reynold's number, Nre = N x D^2 x rho / μ = 313548,
Prandtl's number, Npr = Cp x μ _/ k = 14.4

Now, Individual Heat Transfer Co-efficient hi is Calculated by following co-relation,

hi x D / K = 0.53 x (Nre ^0.66) x ( Npr ^ 0.33) x (( μ_/μ_w)^0.24)

hi = 1441.16 Kcal/h.m².°C

So, Now our task came to an end,

U value Calculation,

1 / U = 1/hj + Fj + X/K + Fv + 1/hi


1 / U = 1/27202.8 + 0.0002 + 0.0007 + 0.0002 + 1/1441.2

U = 537.019 Kcal/h.m².°C


That's it............Task Completed Cheers....!!

Actually one more thing i should have been included in the basic "What are the factor do the  U value depend?", anyway by the end of this tutorial you should have learned this,

Getting mails asking me to demonstrate with some example data, and that's why today i've illustrated the formulae with data,

If you understood clearly then say cheers, or any queries, feel free to contact me, M happy to help,

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

You May Also Like:

How to Calculate Energy of steam?  

What does Tonnes of Refrigeration means? 

How to Handle an Out Of Specification[OOS] 

Determine the Power Required for an Operation?  

How to Calculate required Vapour Column Size?

  


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

Agitation Scale up

Credits: Pharmacalc.blogspot.com

Hello readers, wishing you a very happie day for all of you,

Today i gonna demonstrate a way to Scale up Agitator, which is the main Hero in operations like Reaction, Extractions etc.

A wrong selection of agitator will leads to unimaginable situations, which cannot be tolerated in Pharma industries. Without a proper agitator there lacks efficiency in some operations which will impact the yield of product manufacturing directly/indirectly.

Before going into this topic directly i would like you to have some basic knowledge related to the terms that were used in this topic.




Difference Between Agitation and Mixing?

 Agitation is the process of keeping a mixture that has been mixed in the proper mixed state required for the 'end' product.

Also Read:

How to Select a Condenser?

What Does a TR exactly means? 

How to Calculate the Energy of Steam?

Mixing refers to the actual stirring of different liquids and/or materials to blend them together into an end product or mixture. Once this mixture is 'mixed' it may require agitation to keep the mixture in the proper 'mixed' state.

What is an Agitator?

An agitator is an piece of equipment which is used for mixing operations, it can be used inside reactors, drying equipment's, extraction equipment's etc.

What are the types of Agitator and What is its significance?

For this, read here: Types of Agitators and their Significance

What is the Basic principle involved in selecting an Agitator?

The basic principle involved in agitator selection is the power number resemblance as like in small scale. 

What is Scale of Agitation?

Generally Agitation Scale is nothing but intensity of agitation, this Scale of agitation can be classified based on range of values 

Scale of Agitation Range Values and their Significance over Agitation :

If Scale of Agitation is 1, then the effect of agitation is Mild, and this could be affordable when the operation is a crystallisation, that too if its a supersaturated Crystallisation. 

If Scale of Agitation is 3, then the effect of agitation is Normal, this is most probably used for the Dryer's.

If Scale of Agitation is 6, then the effect of agitation is Vigorous, this means the mass that need to be stirred is somewhat denser but the bulk density will be not much.


If Scale of Agitation is 10, then the effect of agitation is Violent, this is applicable for the mass where there is need for extraction of product in between layers, and also the settling time during this extraction should be higher than regular cases.

If you have understood the above things clearly you are ready for next step, if not any queries feel free to ask.




Agitation ScaleUp:

For this We need some basic Data of Small Scale process like Vessel dia ( T1), Liquid level ( Z1 ), Liquid Density ( rho ), Liquid Viscosity ( myu ), Shaft Speed ( N1 ), Type of Agitator, D/T ration.

If you can gather all this data then our mission can be completed in minutes. And one major input we need to know is How Many times we need to Scale-Up.

Scale-Up ratio Calculation:

The Cubth Root value ratio of  Large scale batch volume to small scale batch volume is Scale-Up ratio.

S = ( V2 / V1 ) ^ (1/3).

Shaft Speed Scaling have to be done based on the Scaling exponent, which depends upon the way of Scale up.

Scale-up to be done based on equaling atleast one of the following parameters,
1. Reynolds number
2. Power number.
3. Froude's number,
4. Shaft Speed,
5. Tip Speed.

The Scaling exponent will differs for all these 5, and they were tabulated below.

PARAMETER	SCALING EXPONENT
Reynolds Number	2
Tip Speed	1
Power per Volume	2/3
Froude Number	1/2
Shaft Speed	0

And now there holds a relation between the scaling exponent and the shaft speed.

                                    n/3                             n
N2 = N1 x ( V1 / V2 )         =  N1 x ( 1 / R )


So, Based on the Value obtained from the above relation we can calculate the following, if needed.




Reynolds number, Froude Number, Power per Volume, Pumping Rate, Pumping Velocity, Scale of Agitation.

Reynolds Number: Nre = D x D x N x rho / myu.

Tip Speed: Ts = 3.141 x D x N

Froude Number: Nfr = Nq x N x ( D ^ 3 )

Power Per Volume : P / V = Np x rho x  ( N^3 ) x ( D^5 )  

Pumping Rate: Q = Ts x ( 0.785 x ( T ^ 2 ) )

Pumping Velocity: Vs = Q / A.

Scale of Agitation : R = Pumping Velocity(Ft/min) / 6

Recommended Posts For You:

How to Scale Up?  

Industrial Column design steps.........!! 

How to calculate time cycle of Batch reactor distillation? 

The Values for Flow Number Nq and Power Number Np can be found from the below tables.

For Example, i'll give you a sample case, with inputs as T1 = 250 mm, Z1 = 300 mm, N1 = 600, D / T = 0.4, water as solvent. and i wish to do scale up 100 times to the small scale.

Now, Lets start the show.

Scale Up ration,                            R = ( V2/ V1) ^(1/3) = ( 100 )^(1/3) = 4.64,
This scale up, i wish to do with equaling power / volume, so my scaling exponent is 2/3.
So RPM calculation,                    N2 = N1 x ( 1 / R )^n = 600 x ( 1 / 4.64 )^(2/3) = 216,
Reynold's number calculation,    Nre = D^2  x N x rho / myu = 774264,
Tip Speed,                                    G= pi x D x N = 5.24,
Flow number,                              Nq = 0.68 [For PBT, refer to the agitator curves above],
Pumping Rate,                             Q = Nq x N x D^3 = 0.24 m/sec,
Power Number,                          Np = 1.37 [for PBT],
Power required,                          P = Np x rho x N^3 x D^5 = 1.37 KW,
Power per Volume,                  P / V = 0.93 KW / Cu.m,
Bulk velocity,                           Vc = 4 x Q / ( 3.141 x Teq^2) = 0.2 m /sec,
Scale of Agitation,                     S = Vc / 0.03048 = 6.56

So, our Scale of Agitation need to be Vigorous.  

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

Any queries happy to help,

Comments are Most appreciated....!!

Also Read:

How to Design a Condenser?  

How to Select a Pump , Motor, Line Sizings ? 

How To Select a Condenser? 

Determine the Power Required for an Operation?  

Industrial Distillation Column Design Steps

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