Technology I Batch Reactor | Chemical Reactors I Industrial Reactors I Pharmaceutical Industry I The Pharma Master
top of page

The Pharma Master

Batch Reactor

Batch Reactors

Reactor
Agitators

This is very popular type of reactor used in pharmaceutical, Chemical and Agro-chemical industry.

Key use of reactor is for performing unit operations like.

  • Reactions

  • Extractions

  • Distillations

  • Crystallization

  • Leaching etc.

Other unit operations like heating, cooling chilling by steam, hot water, cooling water and chilled water etc.

 

Reactor is used for various operations such as reaction, extraction, distillation etc.

There two types of reactor.

  1. Stainless steel reactor

  2. Glass Lined reactor

 

SS reactor :

This Reactor having MOC SS 316 for Internal parts i.e. product contact parts and SS304 for external parts. This reactor have other accessories like condenser, receiver and dosing tank with same MOC as that of reactor. This reactor is used for handling neutral pH or basic pH reaction mass.

Glass Lined reactor :

This Reactor having MOC Glass for Internal parts i.e. product contact parts and SS304 for external parts. This reactor have other accessories like condenser, receiver and dosing tank with same MOC as that of reactor. This reactor is used for handling acidic pH reaction mass.

 

Major Parts of reactor:

 

  1. Agitator: Required for mixing the reaction mass.

  2. Receiver: Required for collection of distilled solvent.

  3. Dosing tank: Required for dosing of reagent into reactor.

  4. Baffles: Required for effective mixing during agitation.

  5. Electric Motor: Required for rotation of agitator during mixing.

  6. Gear box: Required for rpm reduction of electric motor shaft to agitator shaft.

  7. Charging header: Required for charging of solvent / water.

  8. Safety valve: Required to release pressure if reactor pressurize above set pressure.

Agitators are divided into two classes:

Axial-flow impellers:

Generate currents parallel with the axis of impeller.

Pitched Blade Turbine, Propeller, Turbine.

Effective in case of: Where vigorous mixing is essential. Ex- Reaction, Extractions & slurry etc.

Radial-flow impellers:

Generate currents in a tangential or radial direction.

Example :  Anchor.

Effective in case of: Where moderate mixing is essential. Ex- Crystallization.

Mixed-flow impellers:

Generate currents is mixture of axial and tangential or radial direction.

Example: Hydrofoil, MIG.

Effective in case of: Where vigorous mixing is essential. Ex- Reaction, Extractions & slurry etc.

Tip velocity of agitator means rotation speed of agiatator. It is expressed as m/sec. Mixing intensity depends on rotation of agitator.

It means mixing intensity of different agitator is same if tip velocity is same, Answer is no.


Tip Velocity (V) =  π x D x N

Where,
V = Velocity in m/s
D = Diameter of agitator in m
N= Rotation per second

Example :
Reactor with capacity 5 KL agitator running with 75 rpm. Agitator diameter is 1200 mm.
Calculate : Tip velocity in m/sec.

Sol:
Given : rpm = 75, rps = 1.25
Agitator Dia. = 1200 mm = 1.2 m

Tip velocity = π.D.N = 3.14 x 1.2 x 1.25  ( m/sec)
                                 = 4.71 m/sec

Tip velocity

Agiatator Tip Velocity :

Reynold's Number

Reynolds Number

Reynold's number is dimensionless number which represents level of turbulence in fluid dynamics. If value less than 2100 is laminar ( practically seems stagnant) and value above 10000 is considered as turbulent.

Pharmaceutical applications :

  • Reynold's number should be below 10000 for crystallization.​

  • Reynold's number should be more than 10000 for reaction and extractions.

 

Formula :
Nre = D2 x N x ρ

                   μ

 

Where,

D = Diameter of agitator m

N = Rotation per seconds (RPS)

ρ = Density kg/m3

μ = Viscosity kg/m.sec

Conversion:

(1 kg/m.sec = 1000 Centipoise)


Example :
Reactor agitator running with 75 rpm. Agitator diameter is 1200 mm, liquid viscosity is 1200 centipoise and density 1800 kg/m3.

 

Calculate : Reynold’s No (NRe)

 

Sol:

Given : Viscosity = 1200 centipoise = 1.2 kg/m.sec

            RPM =  75 = 1.25 RPS,  Density = 1800 kg/m3

             Diameter= 1200 mm = 1.2 m

 

NRe = D2 x N x ρ   =  1.22 x 1.25 x 1800   = 2700

                  μ                          1.2

Agitator Power

Agitatator power calculation important for ensuring suitability of use as well as mapping equipment during technolgy transfer for same volume or increased volume.


Pharmaceutical applications :

  • Agitator power should be same for identical reactor and same volume of handling.​

  • Reaction volume required power should be maximum 75 % of rated capacity provided by reactor designer.

Formula :

P = n x Np x ρ x  N^3 x D^5

 

Where,

P = Agitator Power watt ( W)

1 KW = 1000 W

1 HP = 746 W

n = No. agitator stages

Np = Power No.

N = Rotation per sec

D = Dia. Of Agitator in m

ρ = Density in kg/m3

Example :

Reactor Anchor running with 75 rpm. Agitator diameter is 1200 mm, liquid viscosity is 1.2 kg/m.sec and density 1000 kg/m3.

Calculate : Agitator Power

Sol:

Given : rpm = 75/60 = 1.25 RPS,     n = 1,    

Np = 0.6, Density = 1000 kg/m3, Dia. = 1200 = 1.2 m

Agitator Power  = n x Np x ρ x  N^3 x D^5 =1 x 0.6 x 1000 x 1.25^3 x 1.2^5  = 2916 W= 2.91 KW= 3.9 HP

Considering Motor efficiency & transmission losses, Agitator Power = 3.9 / 0.8 = 4.87 i.e. 5 HP

Agitator Pumping Velocity

Agitatator pumping velocity calculation important for ensuring suitability of use as well as mapping equipment during technolgy transfer for same volume or increased volume.


Pharmaceutical applications :

  • Agitator power should be same for identical reactor and same volume of handling.​

  • Reaction volume required power should be maximum 75 % of rated capacity provided by reactor designer.

Formula :

Q = NQ x  N x Da3

 

Where,

Q= Pumping Flow m3/sec

NQ = Pumping No.

N = Rotation persec.

Da = Dia. Of Agitator in m

Pumping Velocity

V(m/s)= Agitator Pumping flow (m3/sec) 

           Reactor C/S Area (m2)

 

Example :

Reactor Anchor running with 75 rpm. Agitator diameter is 600 mm. Reactor Dia. 1100 mm.

Calculate : Agitator Power

Sol:     RPM= 75 = 1.25 RPS,   NQ = 0.6,

             Agitator Dia. = 600 mm= 0.6 m    

             Reactor Dia. = 1100 = 1.1 m

Agitator Pumping = NQ x  N x Da3

Rate Q                  = 0.6 x 1.25 x 0.6^3  = 0.16 m3/sec

 

Reactor C/S Area = 0.785 x D2 = 0.785 x1.1 = 0.86 m2

 

Pumping Velociy (V) = Agitator Pumping rate    = 0.16     = 0.18 m/sec

                                         Reactor C/S Area            0.86

                                      

Reactor Vent Sizing

Agitatator pumping velocity calculation important for ensuring suitability of use as well as mapping equipment during technolgy transfer for same volume or increased volume.


Pharmaceutical applications :

  • Agitator power should be same for identical reactor and same volume of handling.​

  • Reaction volume required power should be maximum 75 % of rated capacity provided by reactor designer.

Formula :

Heat Load (BTU/Hr) = 21000*F*A 0.82

 

Where,

F = Environment factor =1 (for bare vessel)

                                       =0.15 (for insulated vessels)

A= Wetted surface = ft2

 

Conversion:

1Btu/hr = 0.252 Kcal/hr

 

Example :

Design Vent size for stainless reactor having capacity 1.6KL.  Environmental factor F = 0.15. Vapour Velocity (v)20 m/sec. Consider Methanol as Solvent.

 

Sol:

Reactor Surface Area (m2) = 4.8 x (Reactor Capacity in KL)0.68 = 4.7 x 1.60.68   = 6.43m2 = 69.5 ft2

Heat Load (BTU/Hr) =21000*F*A 0.82 = 21000x0.15x69.50.68 = 56345 Btu/hr= 14198kcal /hr

 

Vapour generation rate (m)  =  14198/ 263 = 54 kg/hr = 0.015 kg/sec

 

Vapour density (ρ) = Pressure x Mol. Wt = 1 (atm) x 32 (kmol) =1.18 kg/m3

                                           R X T                  0.08206  x 328 (K)

 

V(Flow)=M/ ρ =0.015/1.18 =0.013 m3/sec

 

 

Flow  = Cross Sectional area x Velocity = 0.7854 d2 x V

d = (Flow/0.7854 x V)0.5= (0.013 /0.7854 x 20) 0.5 = 0.029 m=29 mm= 1.16 inch i.e. 1.5 inch.

 

Cooling water flow required (assuming 5°C temp. drop i.e. dt)

 = heat to be removed / Cpwater dt =74762/ 1 x 5 = 14952 kg/hr = 15m3/hr

Utility line size required (assuming 1.4 m/s velocity in pipe)

 = 0.62 (Flow  Rate)0.5 = 0.62(15)0.5 = 2.4  equv. 2.5 inch.

Condensor Sizing

Agitatator pumping velocity calculation important for ensuring suitability of use as well as mapping equipment during technolgy transfer for same volume or increased volume.


Pharmaceutical applications :

  • Agitator power should be same for identical reactor and same volume of handling.​

  • Reaction volume required power should be maximum 75 % of rated capacity provided by reactor designer.

Formula :

Heat Transfer (KCAL /HR) = U x A x LMTD

 

 

Where,

U =Overall heat transfer coefficient =  Kcal/hr.m2.°C

 

A=  Surface area = m2

 

LMTD = Log mean Temp. difference

             = (T2-T1)/ ln (T2/T1) = °C

 

Example :

A stainless reactor having capacity 4 KL. Reactor is subjected to maximum +25°C temperature (LMTD =20°C) jacket side and 25°C at condenser side. Consider methanol as distillation solvent.

Calculate :

1. Condenser size required in m2.        2.Utility flow rate for condenser m3/hr  3.Utility line size inch.

 

 

Sol:  SS reactor U = 300kcal/hr.m2.°c

         SS Condenser U = 400kcal/hr.m2.°c

         Methanol Hvap = 263 kcal/kg, Methanol Cp = 0.69 kcal/kg

          Reactor KMTD = 20°C, Condenser LMTD = 25°C

 

Reactor surface area = 4.7 X 40.68 = 12.06 m2

 

Reactor Heat Transfer (kcal/hr) = U A LMTD =300X12.06X20 = 72360kcal/hr

Methanol vapours (kg/hr)  = 72360/263 =277 kg/hr

 

Condenser has to condens vapours and assume to subcool by about 10°C (dt).

 

Condenser heat transfer (kcal/kg) = condensation + Sub-cooling

                                                                   = m Hvap  + mcpdt = (277x 263 ) + (277x0.69 x 10)

                                                                    =74762kcal/hr

 

Condenser size (m2) = Condensor heat duty / U LMTD

                                       = 74762/ (400X 25)  = 7.5 M2 = 8.25 equv. 9 m2 (10% extra area)

 

 

Cooling water flow required (assuming 5°C temp. drop i.e. dt)

 = heat to be removed / Cpwater dt =74762/ 1 x 5 = 14952 kg/hr = 15m3/hr

 

 

Utility line size required (assuming 1.4 m/s velocity in pipe)

 = 0.62 (Flow  Rate)0.5 = 0.62(15)0.5 = 2.4  equv. 2.5 inch.

bottom of page