General Principles of
Freeze Drying (The Lyophilization Process)
Conclusion:
By controlling chamber pressure, you can reduce
the heat transfer coefficient between the warm shelves and the
product. This simple control will greatly improve the energy
transfer and reduce primary drying times. The curve of shelf
temperature is shown in schematic form in B of Fig. 3, and the
vacuum curve in P.
The control of vacuum in the lyophilization process
can become a useful means of controlling heat transfer, and
the means of getting energy to the product. A laboratory example
illustrates the influence of pressure in heat transfer:
A specially fitted freeze dryer, equipped with
the necessary measurement and control equipment was used to
simulate these sublimation phenomena. Freeze drying of the product
was carried out at –20° C, with a controlled pressure
in the chamber in the range of 0.4 torr, and a shelf temperature
of 30° C.
When the injection of non-condensable gas was
terminated, the product cooled rapidly, and the vapor removal
rate slowed. To regain the sublimation temperature of –20°C,
it was necessary to bring the shelf temperature to 125°
C, a difference of 95° C in the heating source to produce
the same sublimation temperature of the product, and virtually
the same evaporation rate!
The chamber pressure acts as a thermal regulator,
which can, in the space of a few moments can produce the same
effect as raising the temperature 95° C.
Rate of Vapor Transfer
to the Condenser:
Thermodynamic formulae enable the calculation
of values of the volume occupied by vapor as a function of chamber
pressure. For one gram of water, this volume is in the region
of 100m3 at a pressure of 10-2 torr, and 1000m3 at a pressure
of 10-3 torr.
Consider a product with a eutectic of about –20°C,
freeze dried under the following conditions:
- 500 ml bottle filled with 300ml liquid product.
- A water content of approximately 250 grams.
- An evaporation surface of about 0.02 sq. meters.
- A chamber pressure of 2 x 10-2 torr.
The length of primary drying (sublimation of the
ice) around ten hours.
One gram of water occupies, under this pressure, a volume of
50m3, and the amount of water vapor which sublimates from the
product per second can be calculated from the following formula:
- 50m3 x 250 = 0.35m3 per/second
- 10 hrs x 3600
The surface area inside the bottle being around
0.02m2, the vapor leaves the product at a speed of:
- 0.350m3/s = 17.5 m/s
0.02 m2
Inside the neck of the bottle, which has a diameter
of about 3mm, or a section of 7cm2 (7 x 10-4m2), the speed is
calculated at:
- 0.350m3/s = 500 m/s
7 x 10-4 m3
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