 Dehydration is an operation in which nearly all the
water present in a food is removed by evaporation or
sublimation under controlled conditions.
 Preservation of foods by large reductions in water
activity (many reactions retarded, microbial activity
inhibited). Added advantage is the large savings in
packaging, storage and transportation costs.
What is Dehydration?

Moisture Content expresses the amount of water present
in a moist sample.
Two bases are widely used to express moisture content
Moisture content dry basis
MCdb
Moisture content wet basis
MCwb

…………….(1)
………….(2)
(2)

Moisture content dry basis
MCdb
Moisture content wet basis
MCwb
MCdb
MCdb
1 +
MCwb =
MCwb
MCwb
1 -
MCdb =

MCwb
MCwb
1 -
MCdb =
0.851 -
MCdb =
0.85
MCdb = 5.67
= 567 % db
MCwb = 0.85
From equation

Critical moisture content
• The average moisture throughout a solid
material being dried.
• Its value being related to drying rate,
thickness of material, and the factors that
influence the movement of moisture within the
solid.

Sorption isotherms :
 curves relating the m.c. of the material and
the humidity of the atmosphere with which it
is at equilibrium at different temperatures.
Sorption isotherms of foods are also
expressed as moisture content vs. water
activity.
Sorption behaviour of foods is important in:
1. Studying mechanisms of drying and designing
dehydration processes,
2. Predicting storage stability of dried foods.

Heat and mass transfer when drying in a hot air stream

Constant Rate Period
Falling Rate Period
Settling Period
MoistureContent
Drying Time
Critical Moisture Content
Controlled by
external
resistance
Controlled by both
external and
internal resistance

 The drying curve consists of a number of stages:
 a first stage:
 the constant rate period of drying
 This is the moisture content which is retained at
maximum by hygroscopicity
 a second stage:
 first falling rate period of drying
 At the surface the free water is removed, in the
inner layer there is still free water present
 The moisture content at the surface is reduced
from xhygr.max to xequil
 xequil is the water content not removable by
drying

 a third stage:
 This is the second falling rate period of drying
 During this stage more water is removed from the inner
layers
 This stage starts at value of xhygr.max inside the product
and ends at a value of equilibrium throughout the
product

Constant Rate Period
Settling Period
MoistureContent
Drying Time
Critical Moisture Content
1st Falling Rate Period
2nd Falling Rate Period
• 1st Falling Rate is due to both external and internal resistance
• 2nd Falling Rate is caused by internal resistance only. Drying is very
slow. This may be due to solid-water interaction.

Drying methods
There are three basic types of drying process:
1.Sun drying, Solar drying
2.Atmospheric drying (Hot-air drying)
3.Sub atmospheric dehydration (Drying by
contact with a heated surface)

 Hot-air drying
1. Kiln drier
2. Cabinet, tray or
compartment drier
3. Tunnel drier
4. Conveyor drier
5. Bin drier
6. Fluidized bed
drier
7. Pneumatic drier
8. Rotary drier
9. Spray drier
 Drying by contact with
a heated surface
1. Drum drier
2. Vacuum shelf drier
3. Vacuum band drier

Spray Drying

Spray Drying

1) Atomization of a liquid feed into fine droplets.
2) Mixing of these spray droplets with a heated gas stream,
allowing the liquid to evaporate and leave dried solids.
3) Dried powder is separated from the gas stream and
collected.
Principles of Spray Drying
There are three fundamental steps involved in spray drying.

Types of Spray Drying
Co-Current Flow Counter Current
Mixed Flow

Designing Parameter for Spray Drying
Dombrowski and Munday Presented various expression for
energy requirements of atomizers. The Energy required for
the pressure nozzle is given by the expression:
Where Q is the liquid flow rate gal/min
For Rotary Atomizer
E = 42.5Q (dLN')2
Where Q is the liquid flow lbm/hr, N is rpm, d is diameter, ft

•Drum drying
is a method used for drying out liquids; for example, milk is
applied as a thin film to the surface of a heated drum, and
the dried milk solids are then scraped off with a knife.
• Types of Drum Drying:
 Single
Double

Single Drum Drying

Double Drum Drying

Designing Parameter
The basic Equation which describes the rate of dehydration
on a drum dryer is as follows:
∆Tm = the mean Temperature difference between roller surface and
product
L = Latent heat of Vaporization BTU/lbm

Problem A drum dryer is being designated for drying of a
product from a n initial moisture content 88% to 4%. An
overall heat transfer coefficient (U) of 300 BTU/hr ft2 oF is
being estimated for the product. An average temperature
difference between the roller surface and the product of
150oF will be used for design process. Determine the
surface area of the roller required to provide a production
rate of 50 lb product/hr.

 These types of drying systems utilize trays or similar product holders to
expose the product to heated air in an enclosed space.
The trays holding the product inside a cabinet or similar enclosure are
exposed to heated air so that the dehydration will proceed
1. Cabinet, tray or compartment drier

 The dehydration system illustrated in Figs. are examples of
tunnel drying.
 The heated drying air is introduced at one end of the tunnel
and moves at an established velocity through trays of products
Tunnel drier

Tunnel drier

 In this system, the product pieces are suspended in the
heated air throughout the time required for drying. As
illustrated in Fig.
 The movement of product through the system is enhance
by the change in mass of individual particles as moisture is
evaporated.
The movement of the product created by fluidized
particles results in equal drying from all product surfaces
Fluidized bed drier

Fluidized bed drier

Dehydration system design

Drying time Prediction
To determine the time required to achieve the desired reduction
in product moisture content, the rate of moisture removal from the
product must be predicted.
For the constant-rate drying period, the following general
expression would apply:
Where wc, is critical moisture content (kg water/kg dry solids) and
tc is time for constant-rate drying.
During falling-rate drying, the following analysis would exist:
Assumption :
………….1
………….2

or
………….3
(3)
or the time for falling-rate- drying becomes

The total drying time becomes
t = tc + tF

Drying Time Prediction
Dehydration System Designing Problem

1. Problem Compare the energy requirements of a pressure nozzle and rotary atomizer for skim
milk at a flow rate of 40 lb/min. The pressure nozzle is operating at 100 lb/in2 , while the rotary
atomizer is operating at a rotation speed of 6000 rpm and the diameter is 5 inch. The product
density is 8 lb/gal.
2. A sample of a food material weighing 20 kg is initially at 450% moisture content dry basis. It is
dried to 25% moisture content wet basis. How much water is removed from the sample per kg
of dry solids?
3. Air enters a counter flow drier at 60°C dry bulb temperature and 25°C dew point temperature.
Air leaves the drier at 40°C and 60% relative humidity. The initial moisture content of the
product is 72% (wet basis). The amount of air moving through the drier is 200 kg of dry air/h.
The mass flow rate of the product is 1000 kg dry solid per hour. What is the final moisture
content of the dried product (in wet basis)?
4. A food solid was dried from 40 to 10% moisture content in 2 h in a batch drier with constant air
conditions. The drying rate remained constant down to a moisture content of 15%. If the
equilibrium moisture content is 2%, calculate the total time required to dry from 40 to 4%
moisture content. All moisture contents are given on a dry basis.

5. The initial moisture content of a food product is 77% (wet basis), and the critical moisture
content is 30% (wet basis). If the constant drying rate is 0.1 kg H 2 O/(m2 s), compute the time
required for the product to begin the falling rate drying period. The product has a cube shape
with 5-cm sides, and the initial product density is 950 kg/m3.
6. A cabinet dryer is being used to dry a food product from 68% moisture content (wet basis) to
5.5% moisture content (wet basis). The drying air enters the system at 54°C and 10% RH and
leaves at 30°C and 70% RH. The product temperature is 25°C throughout drying. Compute the
quantity of air required for drying on the basis of 1 kg of product solids.
7. A product enters a tunnel dryer with 56% moisture content (wet basis) at a rate of 10 kg/h. The
tunnel is supplied with 1500 kg dry air/h at 50°C and 10% RH, and the air leaves at 25°C in
equilibrium with the product at 50% RH. Determine the moisture content of product leaving the
dryer.
8. A cabinet dryer is to be used for drying of a new food product. The product has an initial
moisture content of 75% (wet basis) and requires 10 minutes to reduce the moisture content to
a critical level of 30% (wet basis). Determine the final moisture of the product if a total drying
time of 15 minutes is used.

Freeze-Drying

Principles of Freeze drying:
Freeze drying is a drying process where water is removed
from a product below the freezing point.
The principle of freeze/sublimation-drying is based on this
physical fact. The ice in the product is directly converted into
water vapor (without passing through the “fluid state”) if
the ambient partial water vapor pressure is lower than the
partial pressure of the ice at its relevant temperature.
The operation is usually carried out in three steps:
• freezing,
• primary drying by sublimation in vacuum,
• secondary drying at elevated temperature to remove
traces of water.

A. Pre-freezing:
 Freezing should be rapid and produce small ice crystals
resulting in a fine porous structure through which water vapour
can diffuse quickly.
 It is the free water in the food which is frozen most easily and
therefore free water which is sublimated in the primary drying
stage.
 A plate freezer can be used for regular solids or alternatively
immersion freezing in liquid nitrogen can be employed.
 Liquid foods may be frozen on the surface of a drum freezer or
plate freezer.

B. Primary drying:
 If heat is supplied under vacuum then the ice crystals which are formed in the
freezing stage will sublime.
 A low pressure (i.e. a high vacuum) is required to increase the sublimation rate
and absolute pressures of 130–260 Pa are commonly used.
 The heat of sublimation of water, equal to approximately 2840 kj / kg , must be
supplied either by conduction through the plate supporting the food (which may
take the form of hollow shelves through which a hot fluid circulates) or from
radiative elements above the food.
 The water vapour which is generated at the ice front then diffuses out through
the already dried layer.
 Water vapour is condensed in a vapour trap placed before the vacuum pump in
order to ensure that water can not be added back to the dried product.

C. Secondary drying:
 The residual moisture is now desorbed at ambient
temperature, but still under vacuum, using the same
heating mechanisms as in the primary drying stage.
 Bound water which is held by hydrogen bonding can
be removed in this step but electrostatically bound
water will remain in the ‘dry’ product.
 When drying is complete the vacuum is broken with a
dry inert gas such as nitrogen followed by packing
and storage.

The advantages of freeze-dried food
 The process at low temperature and low pressure makes
freeze drying an effective way to keep the color, smell, flavor
and heat-sensitive nutrients of food and also eliminates the
surface hardening of food.
 Freeze-dried food is porous and easy to be rehydrated and
instantly dissolved. It can be consumed directly or after
rehydration.
 Since freeze-dried food contains very low moisture content,
it has relatively small density and is easy to be transported.
The freeze-dried food can be preserved at room
temperature for a long time, while the cost of transportation
is much lower than that of frozen food.
 No additives are added into the food during the freeze
drying process.

The disadvantages of freeze-dried food
 If exposed directly to air, freeze-dried food will be rehydrated
quickly and a series of chemical reactions will happen,
resulting in the deterioration of food.
 The freeze-dried products have to be vacuum-packaging or
vacuum-nitrogen charged packaging. The packaging
materials should be waterproof.
 During transportation and sale process, freeze-dried food is
easy to be powdered or cracked for its loose porous
structure.
 Freeze-drying is a time-consuming and energy-consuming
process, which lead to higher product costs of freeze-dried
food.

Various factor that affect the rate of dehydration of
fruits and vegetables
1.Composition
2.Size, shape and arrangement of stacking
3.Temperature as well as humidity and
velocity of air
4.Pressure
5.Heat transfer to surface (conductive,
convective or radiative)

Complexities in specific applications
of food dehydration:
1. Food constituents; proteins, carbohydrates,
fats, vitamins, enzymes, inorganic salts, have
different hydration properties.
2. During drying solubles move with water.
3. The food shrinks.
4. Case hardening.