1. Drying principles: a. Water in solids & gases; b. Heat & mass transfer. 2. Dryers description & operations: a. Vacuum-shelf dryer; b. Rotary dryer; c. Spray dryer; d. Freeze dryer.

Lecture Outline

BTE3481 (Separation Processes 1)

Recall: Overview of bioseparations

What is “drying”?

1. Drying Principles

The process of removing all or most of liquids by supplying latent heat to

cause thermal vaporization, i.e. a liquid is converted into a vapour.

Reasons for drying?

1. To prevent degradation of product due to chemical (e.g. deamidation or

hydrolysis) and/or physical (e.g. aggregation or flocculation) reactions

during storage;

2. Help in the preservation of products by preventing fungal or bacterial

growth (their enzymes may lead to product degradation).usually under

10% wt, microb will not grow;

3. Convenience in the final use of the product - it is often desirable that

pharmaceutical drugs be in tablet form;

4. More economical and convenient to store them in dry form rather than

frozen;

5. To reduce bulk & weight, therefore reducing the cost of transportation;

Drug

synthesis Crystallization

Filtration

Drying

Milling Granulation

Excipients

Drying

Lubrication

Tabletting/

Encapsulation

Packaging/

Storage/

Transport

Typical pharmaceutical manufacturing process:

Suspension

Wet solid

Evaporative

dryer

Spray

dryer

- will not remove all moisture because

the solid equilibrates with the moisture

present in the air.

- removal of bound water.

- Also called free water;

- Mainly held in the voids of the solid;

- Exerts its full vapor pressure;

- Easily be removed by evaporation.

How water is held within biological solids?

Water contained

within solids

Unbound

water Bound

water

- Water that is adsorbed on surfaces

of the solid (to form a mono- or bi-

layer) or trapped in capillaries within

solid structure;

- Cannot exert its full vapor pressure;

- Not easily lost by evaporation.

Magnified cross-section of a wet solid

Moisture content of a wet solid can be expressed as g of water presents in 100

g of water-free or dry solid.

Similarly, moisture content of air can be expressed as g of water presents per

100 g of dry air.

Relative humidity (RH) of air

The percentage RH is defined as:

100% RH means that at a given temperature, air has taken up water vapor until

it is saturated.

Amount of water vapor in

air at a given temperature

Maximum amount of water vapor the

air could hold at the same temperature

x 100

The curves give valuable information about the water capacity of solids:

Water remained in solid at 100% RH = bound water

Unbound water = Water contained in wet crystal – bound water

The equilibrium moisture content of a solid exposed to moist air varies with the

RH, as exemplified by the equilibrium-moisture curves below:

Example 1:

The wet antibiotic cefazolin sodium crystals contain 30 g of water per 100 g of

dry antibiotic. Determine the percentages of bound and unbound water in the

wet crystals.

The equilibrium-moisture curve for the

antibiotic can be extrapolated to give a

water content of the solid at 100% RH:

Solution:

23

Since water remained in solid at

100% RH = bound water;

Bound water for the antibiotic crystals

= 23 g/100 g dry solid

23 g/100g

30 g/100g x 100% % bound water =

= 76.7%

% unbound water = (100.0 – 76.7)%

= 23.3%

The properties of air (dry and moist) are provided by the “humidity” or

“psychrometric” chart:

Moisture content of air

(g/kg dry air)

Relative

humidity (%)

Dry bulb

temperature

Example 2:

Air at 1 atm and 25°C with a relative humidity of 50% is to be heated to 50°C

and then to be used in drying wet crystals of the antibiotic cefazolin sodium.

Determine the moisture content of the crystals after drying.

In order to use the equilibrium-moisture

curve for the antibiotic to find the water

content of the crystals after drying (at

50°C), requires the RH value at the

same temperature. This can be found

using the humidity chart:

Solution:

25

Move from a point at a temp of 25°C &

50% RH to a point at a temp of 50°C,

keeping the moisture content constant.

By interpolation, the RH of the air is

read to be 13%.

From the equilibrium-moisture curve for the

antibiotic, the moisture content of the

antibiotic after drying (i.e. at 13% RH) =

Solution (continued):

13

8

8 g water/100g dry solids

Exercise 1:

Wet insulin crystals containing 32 g water per 100 g of dry insulin

need to be dried in air to a moisture level of 5 g water per 100 g of

dry insulin. Determine:

1. The percentage of bound and unbound water in the wet crystals

before drying;

2. The humidity of the air to accomplish the drying;

3. For drying with air at 20°C, what should be the moisture content

of the air (g moisture/g dry air)?

Relative humidity (%)

20 40 60 80 100 0

10

20

30

40

0

10

20

30

40

Wate

r conte

nt

(g/1

00 g

dry

solid

)

29

5

26

20

4