# Pharma Engineering

For Engineer By Engineer

• ## Wednesday, 8 May 2019

Back with a post related to scale-up, particularly for crystallization operations.

Many a time, during crystallization operations PSD remains the main target and as per me outta all it is the toughest job to do, because we cannot assure the particle size and sometimes we need to meet three tier PSD specifications, and it remains on top.

Have been working a lot on PSD and found some information which is quite valuable i.e., Zweittering equation, which is helpful in scaling up by maintaining similar PSD to that of small scale / to attain a required PSD (mostly single tier specification).

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Also there are some limitations for this too.

Previously for a crystallization operation i use to move ahead by equating power per unit volume but the one which i'm discussing here is an advanced and can replace the power per unit volume.

Before going into this, i'll give some basic things here.

Zweittering equation describes about the suspended speed to particles.

What is the significance and need to develop Zweittering equation ?

PSD is the most important characteristic of an API. As everyone is well known that PSD depends on crystallization pattern and everywhere uniform PSD is desirable. The subject Zweittering equation helps in maintaining the uniform PSD throughout the mixing vessel i.e., till the bottom space between the agitator and bottom dish surface.

What is the suspended speed ?

The minimum stirring speed at which no solid particle remains in its stationary position on the vessel base for more than 2 seconds.

What are the applicable conditions for the Zweittering equation ?

It is applicable for a wide range of reaction mass having solids concentrations between 0.5 - 20 %, with liquid densities varying between 750 - 1600 Kg/m3 and a viscosity ranges between 0.31 x 10^-3 Pas.Sec - 9.3 x 10^-3 Pas.Sec.

What is the purpose of Zweittering equation ?

There are three main objectives:
a. the speed obtain using this equation can help to avoid solids accumulation in a stirred tank,
b. to maximize the contact area between solids and liquid,
c. to ensure solid particles are uniformly distributed through the vessel.

What is the major limitation of this equation ?

Particle / Solids density should be higher than the liquid density. As in this only case the particles would be able to settle on the bottom surface.

Now, we'll go into the topic. As the subject theme is only a calculation, i'll demonstrate with a simple calculation.

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Let the case study be, a particle size of 300µ is required, in our terms d(0.9) need to be less than 300µ. Agitator is a 3 bladed Retreat Curve Impeller and impeller diameter be 0.7 m and the vessel dia be 1.0 m. and the clearance between the agitator center line and bottom base be 0.1 m. Power number is ~ 0.66. Now we need to find the just suspension number to achieve particle size of 250µ as my requirement would be less than 300µ. Solvent is Methanol. During isolation, 300 Kgs will be precipitated in 3000 L of methanol.

Lets start the show,

Just suspension speed Njs = S x (𝝂^0.1) x (g x Δρ/ρL)^0.45 x (dp^0.2) x (X^0.13) / (D^0.85).

𝝂 - Kinematic Viscosity of reaction mass,
Δρ - Density difference between solid and liquid,
ρL - Liquid density,
dp - Particle size,
X - Solids concentration in liquid,
D - agitator diameter.

Kinematic Viscosity = Dynamic viscosity / density = 0.6488 x 10^-6 m2/Sec,
Density difference = 1000 - 792 = 208 Kg/m3  [Bulk density of the material that i've considered is 1 Kg/L],
Required particle Dia = 300µ,
Solids concentration in liquid X = 300 x 100 / (300 + 3000 x 0.792) = 11.21 %,
Zweittering correlation factor = 6.0, [Reference tables were presented at bottom of post]

Lets begin the calculations in-line to above mentioned formula,

Njs = S x (𝝂^0.1) x (g x Δρ/ρL)^0.45 x (dp^0.2) x (X^0.13) / (D^0.85),

= 6.0 x (0.6488 x 10^-6)^0.1 x (9.81 x 208 / 792)^0.45 x (300 x 10^-6)^0.2 x (0.1121)^0.13
(0.7^0.85)

= 0.4431 rps = 0.4431 x 60 = 26.6 RPM ~27 RPM.

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So, based on the above calculation, for attaining a particle size of 300µ, the agitator RPM need to be maintained at 27.

Now, i want to deliver something new which would be a proactive step from my side, as after publishing posts sometimes i'm receiving comments like units are not matching, pl check. As an author while publishing the post, i'll check two to three times. So to avoid those thing, below i'm proving it the dimensional analysis.

Njs will have an units like / Sec.

As per the formula,

kinematic viscosity holds the units m2/sec, g will havee m/sec2, Δρ/ρL wil have Kg/m3,

dp will have m, X will have no units as its a ration of weights, D will have m, S will be unitless.

Substituting the same in the above formula,

Njs = (m2/sec)^0.1 x ( m/sec2 x (Kg/m3 / Kg/m3))^0.45 x ( m )^0.2 x ( m )^-0.85

= m^(2x0.1) x m^0.45 x m^0.2 x m^-0.85
Sec^0.1 x Sec^(2*0.45)

=  m^0.2 x m^0.45 x m^0.2 x m^-0.85
Sec^0.1 x Sec^0.9

=  m^(0.2+0.45+0.2-0.85)
Sec^(0.1+0.9)

=   m^0         =     1  .
Sec^1             Sec

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So, now i think it would be clear for you all.

As mentioned above, i'll give you the Zweitering correlation constants here for reference. For PBT(Pitched Blade Turbine) agitators

That's It........!!!!

What i want to deliver, i've done.

If you got any query or doubts, please free to comment or message me.

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[How to] Perform bond energy calculation ? 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.

1. Dear sir how can maintain free flow of the api

2. if we required d90 , less than 50 micron. then how it will calculate?
=6*(0.6488*10^-6)^0.1*(9.81*208/792)^0.45*(50*10^-6)^0.2*(0.1121)^0.13/0.7^0.85
=~19 RPM.
As i understood , if u required less particle size we have to increase agitator RPM.
But here 300 micron we required-27 RPM & 50 micron require ~19 RPM.

1. Hii nitin,

The equation is applicable for particles having sizes greater than 150 micron.

Regards,
AJAY K

3. Dear Ajay, I have faced a problem in the plant. I have tried equating a pilot plant (0.1 KL) to large scale reactor(2 KL) reactor for crystallization. Particle size of 40-50 micron. RPM of pilot scale is 15 and plant scale is 8. But there is situation where the material is getting settled at the bottom of the reactor due to low rpm. Please help me on this

1. Hiii Sai,

Material Bulk density, Solids concentration in reaction mass, Diameter of the vessel & impeller blades, clearance between impeller blade and bottom dish(both plant case & lab case), type of agitator, density of the reaction mass(approx.), What basis is selected for scaling up from lab to plant ?.

Pl reach me at pharmacalc823@gmail.com with above mentioned data.

Best Regards,
AJAY K

## This Blog is protected by DMCA.com Hi! I am Ajay Kumar Kalva, owner 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.